nt face="Times New Roman">Introduction

4.1 Probability Density Functions

4.2 Cumulative Distribution Functions and Expected Values

4.3 The Normal Distribution

4.4 The Exponential and Gamma Distributions

4.5 Other Continuous Distributions

4.6 Probability Plots

Supplementary Exercises

Bibliography

5 Joint Probability Distributions and Random Samples

Introduction

5.1 Jointly Distributed Random Variables

5.2 Expected Values, Covariance, and Correlation

5.3 Statistics and Their Distributions

5.4 The Distribution of the Sample Mean

5.5 The Distribution of a Linear Combination

Supplementary Exercises

Bibliography

6 Point Estimation

Introduction

Bibliography

12 Simple Linear Regression and Correlation

Introduction

12.1 The Simple Linear Regression Model

12.2 Estimating Model Parameters

12.3 Inferences About the Slope Parameter β_1

12.4 Inferences Concerning μY • x* and the Prediction of Future Y Values

12.5 Correlation

Supplementary Exercises

Bibliography

13 Nonlinear and Multiple Regression

Introduction

13.1 Aptness of the Model and Model Checking

13.2 Regression with Transformed Variables

13.3 Polynomial Regression

13.4 Multiple Regression Analysis

13.5 Other Issues in Multiple Regression

Supplementary Exercises

Bibliography

14 Goodness-of-Fit Tests and Categorical Data Analysis

Bibliography

Appendix Tables

A.1 Cumulative Binomial Probabilities

A.2 Cumulative Poisson Probabilities

A.3 Standard Normal Curve Areas

A.4 The Incomplete Gamma Function

A.5 Critical Values for t Distributions

A.6 Tolerance Critical Values for Normal Population Distributions

A.7 Critical Values for Chi-Squared Distributions

A.8 t Curve Tail Areas

A.9 Critical Values for F Distributions

A.10 Critical Values for Studentized Range Distributions

A.11 Chi-Squared Curve Tail Areas

A.12 Critical Values for the Ryan–Joiner Test of Normality

A.13 Critical Values for the Wilcoxon Signed-Rank Test

A.14 Critical Values for the Wilcoxon Rank-Sum Test

A.15 Critical Values for the Wilcoxon Signed-Rank Interval

A.16 Critical Values for the Wilcoxon Rank-Sum Interval

A.17 β Curves for t Tests

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Preface

Purpose

The use of probability models and statistical methods for analyzing data has become common practice in virtually all scientific disciplines. This book attempts to provide a comprehensive introduction to those models and methods most likely to be encountered and used by students in their careers in engineering and the natural sciences. Although the examples and exercises have been designed with scientists and engineers in mind, most of the methods covered are basic to statistical analyses in many other disciplines, so that students of business and the social sciences will also profit from reading the book.

Approach

Students in a statistics course designed to serve other majors may be initially skeptical of the value and relevance of the subject matter, but my experience is that students can be turned on to statistics by the use of good examples and exercises that blend their everyday experiences with their scientific interests. Consequently, I have worked hard to find examples of real, rather than artificial, data—data that someone thought was worth collecting and analyzing. Many of the methods presented, especially in the later chapters on statistical inference, are illustrated by analyzing data taken from a published source, and many of the exercises also involve working with such data. Sometimes the reader may be unfamiliar with the context of a particular problem (as indeed I often was), but I have found that students are more attracted by real problems with a somewhat strange context than by patently artificial problems in a familiar setting.

Mathematical Level

The exposition is relatively modest in terms of mathematical development. Substantial use of the calculus is made only in Chapter 4 and parts of Chapters 5 and 6. In particular, with the exception of an occasional remark or aside, calculus appears in the inference part of the book only in the second section of Chapter 6. Matrix algebra is not used at all. Thus almost all the exposition should be accessible to those whose mathematical background includes one semester or two quarters of differential and integral calculus.

Content

Chapter 1 begins with some basic concepts and terminology—population, sample, descriptive and inferential statistics, enumerative versus analytic studies, and so on—and continues with a survey of important graphical and numerical descriptive methods. A rather traditional development of probability is given in Chapter 2, followed by probability distributions of discrete and continuous random variables in Chapters he presence of uncertainty and variation. Without uncertainty or variation, there would be little need for statistical methods or statisticians. If every component of a particular type had exactly the same lifetime, if all resistors produced by a certain manufacturer had the same resistance value, if pH determinations for soil specimens from a particular locale gave identical results, and so on, then a single observation would reveal all desired information.

An interesting manifestation of variation arises in the course of performing emissions testing on motor vehicles. The expense and time requirements of the Federal Test Procedure (FTP) preclude its widespread use in vehicle inspection programs. As a result, many agencies have developed less costly and quicker tests, which it is hoped replicate FTP results. According to the journal article “Motor Vehicle Emissions Variability” (J. of the Air and Waste Mgmt. Assoc., 1996: 667–675), the acceptance of the FTP as a gold standard has led to the widespread belief that repeated measurements on the same vehicle would yield identical (or nearly identical) results. The authors of the article applied the FTP to seven vehicles characterized as “high emitters.” Here are the results for one such vehicle:

The substantial variation in both the HC and CO measurements casts considerable doubt on conventional wisdom and makes it much more difficult to make precise assessments about emissions levels.

How can statistical techniques be used to gather information and draw conclusions? Suppose, for example, that a materials engineer has developed a coating for retarding corrosion in metal pipe under specified circumstances. If this coating is applied to different segments of pipe, variation in environmental conditions and in the segments themselves will result in more substantial corrosion on some segments than on others. Methods of statistical analysis could be used on data from such an experiment to decide whether the average amount of corrosion exceeds an upper specification limit of some sort or to predict how much corrosion will occur on a single piece of pipe.

Alternatively, suppose the engineer has developed the coating in the belief that it will be superior to the currently used coating. A comparative experiment could be carried out to investigate this issue by applying the current coating to some segments of pipe and the new coating to other segments. This must be done with care lest the wrong conclusion emerge. For example, perhaps the average amount of corrosion is identical for the two coatings. However, the new coating may be applied to segments that have superior ability to resist corrosion and under less stressful environmental conditions compared to the segments and conditions for the current coating. The investigator would then likely observe a difference between the two coatings attributable not to the coatings themselves, but just to extraneous variation. Statistics offers not only methods for analyzing the results of experiments once they have been carried out but also suggestions for how experiments can be performed in an efficient manner to mitigate the effects of variation and have a better chance of producing correct conclusions.

Figure 1.1 A MINITAB stem-and-leaf display and histogram of the O-ring temperature data

The lowest temperature is 31 degrees, much lower than the next-lowest temperature, and this is the observation for the Challenger disaster. The presidential investigation discovered that warm temperatures were needed for successful operation of the O-rings, and that 31 degrees was much too cold. In Chapter 13 we will develop a relationship between temperature and the likelihood of a successful launch.

Having obtained a sample from a population, an investigator would frequently like to use sample information to draw some type of conclusion (make an inference of some sort) about the population. That is, the sample is a means to an end rather than an end in itself. Techniques for generalizing from a sample to a population are gathered within the branch of our discipline called inferential statistics.

Example 1.2

Material strength investigations provide a rich area of application for statistical methods. The article “Effects of Aggregates and Microfillers on the Flexural Properties of Concrete” (Magazine of Concrete Research, 1997: 81–98) reported on a study of strength properties of high-performance concrete obtained by using superplasticizers and certain binders. The compressive strength of such concrete had previously been investigated, but not much was known about flexural strength (a measure of ability to resist failure in bending). The accompanying data on flexural strength (in MegaPascal, MPa, where 1 Pa (Pascal) = 1.45 × 10”4 psi) appeared in the article cited:

Suppose we want an estimate of the average value of flexural strength for all beams that could be made in this way (if we conceptualize a population of all such beams, we are trying to estimate the population mean). It can be shown that, with a high degree of confidence, the population mean strength is between 7.48 MPa and 8.80 MPa; we call this a confidence interval or interval estimate. Alternatively, this data could be used to predict the flexural strength of a single beam of this type. With a high degree of confidence, the strength of a single such beam will exceed 7.35 MPa; the number 7.35 is called a lower prediction bound.

The main focus of this book is on presenting and illustrating methods of inferential statistics that are useful in scientific work. The most important ich the sample was selected.

In the lap belt example, the population is well defined and concrete: all drivers of cars equipped in a certain way in a particular metropolitan area. In Example 1.1, however, a sample of O-ring temperatures is available, but it is from a population that does not actually exist. Instead, it is convenient to think of the population as consisting of all possible temperature measurements that might be made under similar experimental conditions. Such a population is referred to as a conceptual or hypothetical population. There are a number of problem situations in which we fit questions into the framework of inferential statistics by conceptualizing a population.

Enumerative Versus Analytic Studies

W. E. Deming, a very influential American statistician who was a moving force in Japan’s quality revolution during the 1950s and 1960s, introduced the distinction between enumerative studies and analytic studies. In the former, interest is focused on a finite, identifiable, unchanging collection of individuals or objects that make up a population. A sampling frame—that is, a listing of the individuals or objects to be sampled—is either available to an investigator or else can be constructed. For example, the frame might consist of all signatures on a petition to qualify a certain initiative for the ballot in an upcoming election; a sample is usually selected to ascertain whether the number of valid signatures exceeds a specified value. As another example, the frame may contain serial numbers of all furnaces manufactured by a particular company during a certain time period; a sample may be selected to infer something about the average lifetime of these units. The use of inferential methods to be developed in this book is reasonably noncontroversial in such settings (though statisticians may still argue over which particular methods should be used).

An analytic study is broadly defined as one that is not enumerative in nature. Such studies are often carried out with the objective of improving a future product by taking action on a process of some sort (e.g., recalibrating equipment or adjusting the level of some input such as the amount of a catalyst). Data can often be obtained only on an existing process, one that may differ in important respects from the future process. There is thus no sampling frame listing the individuals or objects of interest. For example, a sample of five turbines with a new design may be experimentally manufactured and tested to investigate efficiency. These five could be viewed as a sample from the conceptual population of all prototypes that could be manufactured under similar conditions, but not necessarily as representative of the population of units manufactured once regular production gets underway. Methods for using sample information to draw conclusions about future production units may be problematic. Someone with expertise in the area of turbine design and engineering (or whatever other subject area is relevant) should be called upon to judge whether such extrapolation is sensible. A good exposition of these issues is contained in the article “Assumptions for Statistical Inference” by Gerald Hahn and William Meeker (The American Statistician, 1993: 1–11).

When data collection entails selecting individuals or objects from a frame, the simplest method for ensuring a representative selection is to take a simple random sample. This is one for which any particular subset of the specified size (e.g., a sample of size 100) has the same chance of being selected. For example, if the frame consists of 1,000,000 serial numbers, the numbers 1, 2, …, up to 1,000,000 could be placed on identical slips of paper. After placing these slips in a box and thoroughly mixing, slips could be drawn one by one until the requisite sample size has been obtained. Alternatively (and much to be preferred), a table of random numbers or a computer’s random number generator could be employed.

Sometimes alternative sampling methods can be used to make the selection process easier, to obtain extra information, or to increase the degree of confidence in conclusions. One such method, stratified sampling, entails separating the population units into nonoverlapping groups and taking a sample from each one. For example, a manufacturer of DVD players might want information about customer satisfaction for units produced during the previous year. If three different models were manufactured and sold, a separate sample could be selected from each of the three corresponding strata. This would result in information on all three models and ensure that no one model was over- or underrepresented in the entire sample.

Frequently a “convenience” sample is obtained by selecting individuals or objects without systematic randomization. As an example, a collection of bricks may be stacked in such a way that it is extremely difficult for those in the center to be selected. If the bricks on the top and sides of the stack were somehow different from the others, resulting sample data would not be representative of the population. Often an investigator will assume that such a convenience sample approximates a random sample, in which case a statistician’s repertoire of inferential methods can be used; however, this is a judgment call. Most of the methods discussed herein are based on a variation of simple random sampling described in Chapter 5.

Engineers and scientists often collect data by carrying out some sort of designed experiment. This may involve deciding how to allocate several different treatments (such as fertilizers or coatings for corrosion protection) to the various experimental units (plots of land or pieces of pipe). Alternatively, an investigator may h4>

1. Give one possible sample of size 4 from each of the following populations:

a. All daily newspapers published in the United States

b. All companies listed on the New York Stock Exchange

c. All students at your college or university

d. All grade point averages of students at your college or university

2. For each of the following hypothetical populations, give a plausible sample of size 4:

a. All distances that might result when you throw a football

b. Page lengths of books published 5 years from now

c. All possible earthquake-strength measurements (Richter scale) that might be recorded in California during the next year

d. All possible yields (in grams) from a certain chemical reaction carried out in a laboratory

3. Consider the population consisting of all computers of a certain brand and model, and focus on whether a computer needs service while under warranty.

a. Pose several probability questions based on selecting a sample of 100 such computers.

b. What inferential statistics question might be answered by determining the number of such computers in a sample of size 100 that need warranty service?

4. a. Give three different examples of concrete populations and three different examples of hypothetical populations.

b. For one each of your concrete and your hypothetical populations, give an example of a probability question and an example of an inferential statistics question.

5. Many universities and colleges have instituted supplemental instruction (SI) programs, in which a student facilitator meets regularly with a small group of students enrolled in the course to promote discussion of course material and enhance subject mastery. Suppose that students in a large statistics course (what else?) are randomly divided into a control group

a. Are the scores from the SI group a sample from an existing population? If so, what is it? If not, what is the relevant conceptual population?

b. What do you think is the advantage of randomly dividing the students into the two groups rather than letting each student choose which group to join?

c. Why didn’t the investigators put all students in the treatment group? Note: The article “Supplemental Instruction: An Effective Component of Student Affairs Programming” (J. of College Student Devel., 1997: 577–586) discusses the analysis of data from several SI programs.

6. The California State University (CSU) system consists of 23 campuses, from San Diego State in the south to Humboldt State near the Oregon border. A CSU administrator wishes to make an inference about the average distance between the hometowns of students and their campuses. Describe and discuss several different sampling methods that might be employed. Would this be an enumerative or an analytic study? Explain your reasoning.

7. A certain city divides naturally into ten district neighborhoods. How might a real estate appraiser select a sample of single-family homes that could be used as a basis for developing an equation to predict appraised value from characteristics such as age, size, number of bathrooms, distance to the nearest school, and so on? Is the study enumerative or analytic?

8. The amount of flow through a solenoid valve in an automobile’s pollution-control system is an important characteristic. An experiment was carried out to study how flow rate depended on three factors: armature length, spring load, and bobbin depth. Two different levels (low and high) of each factor were chosen, and a single observation on flow was made for each combination of levels.

a. The resulting data set consisted of how many observations?

b. Is this an enumerative or analytic study? Explain your reasoning.

9. In a famous experiment carried out in 1882, Michelson and Newcomb obtained 66 observations on the time it took for light to travel between two locations in Washington, D.C. A few of the measurements (coded in a certain manner) were 31, 23, 32, 36, -2, 26, 27, and 31.

a. Why are these measurements not identical?

b. Is this an enumerative study? Why or why not?

1.2 Pictorial and Tabular Methods in Descbe divided into two general subject areas. In this section, we consider representing a data set using visual techniques. In Sections 1.3 and 1.4, we will develop some numerical summary measures for data sets. Many visual techniques may already be familiar to you: frequency tables, tally sheets, histograms, pie charts, bar graphs, scatter diagrams, and the like. Here we focus on a selected few of these techniques that are most useful and relevant to probability and inferential statistics.

Notation

Some general notation will make it easier to apply our methods and formulas to a wide variety of practical problems. The number of observations in a single sample, that is, the sample size, will often be denoted by n, so that n = 4 for the sample of universities {Stanford, Iowa State, Wyoming, Rochester} and also for the sample of pH measurements {6.3, 6.2, 5.9, 6.5}. If two samples are simultaneously under consideration, either m and n or n_1 and n_2 can be used to denote the numbers of observations. Thus if {29.7, 31.6, 30.9} and {28.7, 29.5, 29.4, 30.3} are thermal-efficiency measurements for two different types of diesel engines, then m = 3 and n = 4.

Given a data set consisting of n observations on some variable x, the individual observations will be denoted by x_1, x_2, x_3, …, xn . The subscript bears no relation to the magnitude of a particular observation. Thus x_1 will not in general be the smallest observation in the set, nor will xn typically be the largest. In many applications, x_1 will be the first observation gathered by the experimenter, x_2 the second, and so on. The ith observation in the data set will be denoted by xi .

Stem-and-Leaf Displays

Consider a numerical data set x_1, x_2, …, xn for which each xi consists of at least two digits. A quick way to obtain an informative visual representation of the data set is to construct a stem-and-leaf display.

Steps for Constructing a Stem-and-Leaf Display

1. Select one or more leading digits for the stem values. The trailing digits become the leaves.

2. List possible stem values in a vertical column.

3. Record the leaf for every observation beside the corresponding stem value.

4. Indicate the units for stems and leaves someplace in the display.

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A stem-and-leaf display conveys information about the following aspects of the data:

- identification of a typical or representative value

- extent of spread about the typical value

- presence of any gaps in the data

- extent of symmetry in the distribution of values

- number and location of peaks

- presence of any outlying values

Example 1.6

Figure 1.5 presents stem-and-leaf displays for a random sample of lengths of golf courses (yards) that have been designated by Golf Magazine as among the most challenging in the United States. Among the sample of 40 courses, the shortest is 6433 yards long, and the longest is 7280 yards. The lengths appear to be distributed in a roughly uniform fashion over the range of values in the sample. Notice that a stem choice here of either a single digit (6 or 7) or three digits (643, …, 728) would yield an uninformative display, the first because of too few stems and the latter because of too many.

Statistical software packages do not generally produce displays with multiple-digit stems. The MINITAB display in Figure 1.5(b) results from truncating each observation by deleting the ones digit.

Figure 1.5 Stem-and-leaf displays of golf course yardages: (a) two-digit leaves; (b) display from MINITAB with truncated one-digit leaves

Dotplots

A dotplot is an attractive summary of numerical data when the data set is reasonably small or there are relatively few distinct data values. Each observation is represented by a dot above the corresponding location on a horizontal measurement scale. When a value occurs more than once, there is a dot for each occurrence, and these dots are stacked vertically. As with a stem-and-leaf display, a dotplot gives information about location, spread, extremes, and gaps.

Example 1.7

Figure 1.6 shows a dotplot for the O-ring temperature data introduced in Example 1.1 in the previous section. A representative temperature value is one in the mid-60s (°F), and there is quite a bit of spread about the center. The data stretches out more at the lower end than at the upper end, and the smallest observation, 31, can fairly be described as an outlier.

Figure 1.6 A dotplot of the O-ring temperature data (°F)

The resulting histogram appears in Figure 1.10. The right or upper tail stretches out much farther than does the left or lower tail—a substantial departure from symmetry.

Figure 1.10 A MINITAB density histogram for the bond strength data of Example 1.10

When class widths are unequal, not using a density scale will give a picture with distorted areas. For equal-class widths, the divisor is the same in each density calculation, and the extra arithmetic simply results in a rescaling of the vertical axis (i.e., the histogram using relative frequency and the one using density will have exactly the same appearance). A density histogram does have one interesting property. Multiplying both sides of the formula for density by the class width gives

relative frequency = (class width)(density) = (rectangle width)(rectangle height)

= rectangle area

That is, the area of each rectangle is the relative frequency of the corresponding class. Furthermore, since the sum of relative frequencies should be 1, the total area of all rectangles in a density histogram is l. It is always possible to draw a histogram so that the area equals the relative frequency (this is true also for a histogram of discrete data)—just use the density scale. This property will play an important role in creating models for distributions in Chapter 4.

Histogram Shapes

Histograms come in a variety of shapes. A unimodal histogram is one that rises to a single peak and then declines. A bimodal histogram has two different peaks. Bimodality can occur when the data set consists of observations on two quite different kinds of individuals or objects. For example, consider a large data set consisting of driving times for automobiles traveling between San Luis Obispo, California and Monterey, California (exclusive of stopping time for sightseeing, eating, etc.). This histogram would show two peaks, one for those cars that took the inland route (roughly 2.5 hours) and another for those cars traveling up the coast (3.5–4 hours). However, bimodality does not automatically follow in such situations. Only if the two separate histograms are “far apart” relative to their spreads will bimodality occur in the histogram of combined data. Thus a large data set consisting of heights of college students should not result in a bimodal histogram because the typical male height of about 69 inches is not far enough above the typical female height of about 64–65 inches. A histogram with more than two peaks is said to be ight="1em" width="-1em">
a. Construct a stem-and-leaf display of the data by first deleting (truncating) the tenths digit and then repeating each stem value five times (once for leaves 1 and 2, a second time for leaves 3 and 4, etc.). Why is it relatively easy to identify a representative strength value?

b. Construct a histogram using equal-width classes with the first class having a lower limit of 122 and an upper limit of 124. Then comment on any interesting features of the histogram.

14. The accompanying data set consists of observations on shower-flow rate (L/min) for a sample of n = 129 houses in Perth, Australia (“An Application of Bayes Methodology to the Analysis of Diary Records in a Water Use Study,” J. Amer. Stat. Assoc., 1987: 705–711):

a. Construct a stem-and-leaf display of the data.

b. What is a typical, or representative, flow rate?

c. Does the display appear to be highly concentrated or spread out?

d. Does the distribution of values appear to be reasonably symmetric? If not, how would you describe the departure from symmetry?

e. Would you describe any observation as being far from the rest of the data (an outlier)?

15. A Consumer Reports article on peanut butter (Sept. 1990) reported the following scores for various brands:

Construct a comparative stem-and-leaf display by listing stems in the middle of your page and then displaying the creamy leaves out to the right and the crunchy leaves out to the left. Describe similarities and differences for the two types.

16. The article cited in Example 1.2 also gave the accompanying strength observations for cylinders:

a. Construct a comparative stem-and-leaf display (see the previous exercise) of the beam and cylinder data, and then answer the questions in parts (b)ew Roman">

23. In a study of warp breakage during the weaving of fabric (Technometrics, 1982: 63), 100 specimens of yarn were tested. The number of cycles of strain to breakage was determined for each yarn specimen, resulting in the following data:

a. Construct a relative frequency histogram based on the class intervals 0–<100, 100–<200, …, and comment on features of the histogram.

b. Construct a histogram based on the following class intervals: 0–<50, 50–<100, 100–<150, 150–<200, 200–<300, 300–<400, 400–<500, 500–<600, and 600–<900.

c. If weaving specifications require a breaking strength of at least 100 cycles, what proportion of the yarn specimens in this sample would be considered satisfactory?

24. The accompanying data set consists of observations on shear strength (lb) of ultrasonic spot welds made on a certain type of alclad sheet. Construct a relative frequency histogram based on ten equal-width classes with boundaries 4000, 4200, …. [The histogram will agree with the one in “Comparison of Properties of Joints Prepared by Ultrasonic Welding and Other Means” (J. of Aircraft, 1983: 552–556).] Comment on its features.

25. A transformation of data values by means of some mathematical function, such as or 1/x, can often yield a set of numbers that has “nicer” statistical properties than the original data. In particular, it may be possible to find a function for which the histogram of transformed values is more symmetric (or, even better, more like a bell-shaped curve) than the original data. As an example, the article “Time Lapse Cinematographic Analysis of Beryllium-Lung Fibroblast Interactions” (Environ. Research, 1983: 34–43) reported the results of experiments designed to study the behavior of certain individual cells that had been exposed to beryllium. An important characteristic of such an individual cell is its interdivision time (IDT). IDTs were determined for a large number of cells both in exposed (treatment) and unexposed (control) conditions. The authors of the article used a logarithmic transformation, that is, transformed value = log(original value). Consider the following representative IDT data:

Use class intervals 10–<20, 20–<30, … to coneight="0em">

26. Automated electron backscattered diffraction is now being used in the study of fracture phenomena. The following information on misorientation angle (degrees) was extracted from the article “Observations on the Faceted Initiation Site in the Dwell-Fatigue Tested Ti-6242 Alloy: Crystallographic Orientation and Size Effects (Metallurgical and Materials Trans., 2006: 1507–1518).

a. Is it true that more than 50% of the sampled angles are smaller than 15°, as asserted in the paper?

b. What proportion of the sampled angles are at least 30°?

c. Roughly what proportion of angles are between 10° and 25°?

d. Construct a histogram and comment on any interesting features.

27. The paper “Study on the Life Distribution of Microdrills” (J. of Engr. Manufacture, 2002: 301–305) reported the following observations, listed in increasing order, on drill lifetime (number of holes that a drill machines before it breaks) when holes were drilled in a certain brass alloy.

a. Why can a frequency distribution not be based on the class intervals 0–50, 50–100, 100–150, and so on?

b. Construct a frequency distribution and histogram of the data using class boundaries 0, 50, 100, …, and then comment on interesting characteristics.

c. Construct a frequency distribution and histogram of the natural logarithms of the lifetime observations, and comment on interesting characteristics.

d. What proportion of the lifetime observations in this sample are less than 100? What proportion of the observations are at least 200?

28. Human measurements provide a rich area of application for statistical methods. The article “A Longitudinal Study of the Development of Elementary School Children’s Private Speech” (Merrill-Palmer Q. , 1990: 443–463) reported on a study of children talking to themselves (private speech). It was thought that private speech would be related to IQ, because IQ is supposed to measure mental maturity, and it was known that private speech decreases as students progress through the primary grades. The study included 33ew Roman">Figure 1.13 shows a stem-and-leaf display of the data; a crack length in the low 20 s appears to be “typical.”

Figure 1.13 A stem-and-leaf display of the crack-length data

With Σxi = 444.8, the sample mean is

a value consistent with information conveyed by the stem-and-leaf display.

A physical interpretation of demonstrates how it measures the location (center) of a sample. Think of drawing and scaling a horizontal measurement axis, and then represent each sample observation by a 1-lb weight placed at the corresponding point on the axis. The only point at which a fulcrum can be placed to balance the system of weights is the point corresponding to the value of (see Figure 1.14).

Figure 1.14 The mean as the balance point for a system of weights

Just as represents the average value of the observations in a sample, the average of all values in the population can be calculated. This average is called the population mean and is denoted by the Greek letter μ. When there are N values in the population (a finite population), then μ = (sum of the N population values)/N . In Chapters 3 and 4, we will give a more general definition for μ that applies to both finite and (conceptually) infinite populations. Just as is an interesting and important measure of sample location, μ is an interesting and important (often the most important) characteristic of a population. In the chapters on statistical inference, we will present methods based on the sample mean for drawing conclusions about a population mean. For example, we might use the sample mean = 21.18 computed in Example 1.12 as a point estimate (a single number that is our “best” guess) of μ = the true average crack length for all specimens treated as described.

The mean suffers from one deficiency that makes it an inappropriate measure of center under some circumstances: Its value can be greatly affected by the presence of even a single outlier (unusually large or small observation)he outliers 15.2, 16.2, and 20.4.

The data in Example 1.13 illustrates an important property of in contrast to : The sample median is very insensitive to outliers. If, for example, we increased the two largest xi s from 16.2 and 20.4 to 26.2 and 30.4, respectively, would be unaffected. Thus, in the treatment of outlying data values, and are at opposite ends of a spectrum.

Because the large values in the sample of Example 1.13 affect more than , < for that data. Although and both provide a measure for the center of a data set, they will not in general be equal because they focus on different aspects of the sample.

Analogous to as the middle value in the sample is a middle value in the population, the population median, denoted by . As with and μ, we can think of using the sample median to make an inference about . In Example 1.13, we might use = 10.05 as an estimate of the median concentration in the entire population from which the sample was selected. A median is often used to describe income or salary data (because it is not greatly influenced by a few large salaries). If the median salary for a sample of engineers were = $66,416, we might use this as a basis for concluding that the median salary for all engineers exceeds $60,000.

Figure 1.15 Three different shapes for a population distribution

The population mean μ and median will not generally be identical. If the population distribution is positively or negatively skewed, as pictured in Figure 1.15, then μ ≠ . When this is the case, in making inferences we must first decide which of the two population characteristics is of greater interest and then proceed accordingly.

Other Measures of Location: Quartiles, Percentiles, and Trimmed Means

The median (population or sample) divides the data set into two parts of equal size. To obtain finer measures of location, we could divide the data into more than two such parts. Roughly sen 5% and 25%—will yield a measure of center that is neither as sensitive to outliers as is the mean nor as insensitive as the median. If the desired trimming percentage is 100a% and nΝ a is not an integer, the trimmed mean must be calculated by interpolation. For example, consider a = .10 for a 10% trimming percentage and n = 26 as in Example 1.14. Then tr (10) would be the appropriate weighted average of the 7.7% trimmed mean calculated there and the 11.5% trimmed mean resulting from trimming three observations from each end.

Categorical Data and Sample Proportions

When the data is categorical, a frequency distribution or relative frequency distribution provides an effective tabular summary of the data. The natural numerical summary quantities in this situation are the individual frequencies and the relative frequencies. For example, if a survey of individuals who own digital cameras is undertaken to study brand preference, then each individual in the sample would identify the brand of camera that he or she owned, from which we could count the number owning Canon, Sony, Kodak, and so on. Consider sampling a dichotomous population—one that consists of only two categories (such as voted or did not vote in the last election, does or does not own a digital camera, etc.). If we let x denote the number in the sample falling in category 1, then the number in category 2 is n − x. The relative frequency or sample proportion in category 1 is x/n and the sample proportion in category 2 is 1 − x/n. Let’s denote a response that falls in category 1 by a 1 and a response that falls in category 2 by a 0. A sample size of n = 10 might then yield the responses, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1. The sample mean for this numerical sample is (since number of 1s = x = 7)

More generally, focus attention on a particular category and code the sample results so that a 1 is recorded for an observation in the category and a 0 for an observation not in the category. Then the sample proportion of observations in the category is the sample mean of the sequence of 1s and 0s. Thus a sample mean can be used to summarize the results of a categorical sample. These remarks also apply to situations in which categories are defined by grouping values in a numerical sample or population (e.g., we might be interested in knowing whether individuals have owned their present automobile for at least 5 years, rather than studying the exact length of ownership).

Analogous to the sample proportion x/n of individuals or objects falling in a particular category, let p represent the proportion of those in the entire population falling in the category. As with x/n, p is a quantity between 0 and 1, and while x/n is a sample characteristic, p is a characteristic of the population. The relationship between the two parallels the relationship between and and between and μ. In particular, we will subsequently use x/n to make inferences about p. If, for example, a sample of 100 car os a point estimate of the proportion of all owners who have owned their car at least 5 years. We will study the properties of x/n as an estimator of p and see how x/n can be used to answer other inferential questions. With k categories (k > 2), we can use the k sample proportions to answer questions about the population proportions p_1, …, pk .

EXERCISES Section 1.3 (33–43)

33. The article “The Pedaling Technique of Elite Endurance Cyclists” (Int. J. of Sport Biomechanics, 1991: 29–53) reported the accompanying data on single-leg power at a high workload:

a. Calculate and interpret the sample mean and median.

b. Suppose that the first observation had been 204 rather than 244. How would the mean and median change?

c. Calculate a trimmed mean by eliminating the smallest and largest sample observations. What is the corresponding trimming percentage?

d. The article also reported values of single-leg power for a low workload. The sample mean for n = 13 observations was = 119.8 (actually 119.7692), and the 14th observation, somewhat of an outlier, was 159. What is the value of for the entire sample?

34. Exposure to microbial products, especially endotoxin, may have an impact on vulnerability to allergic diseases. The article “Dust Sampling Methods for Endotoxin—An Essential, But Underestimated Issue” (Indoor Air, 2006: 20–27) considered various issues associated with determining endotoxin concentration. The following data on concentration (EU/mg) in settled dust for one sample of urban homes and another of farm homes was kindly supplied by the authors of the cited article.

a. Determine the sample mean for each sample. How do they compare?

b. Determine the sample median for each sample. How do they compare? Why is the median for the urban sample so different from the mean for that sample?

c. Calculate the trimmed mean for each sample by deleting the smallest and largest observation. What are the corresponding trimming percentages? How do the values of these trimmed means compare to the corresvations, obtained using spectrophotogrammetry, on leaf reflectance under specified experimental conditions.

The computational formula now gives

from which s 2 = Sxx /(n—1) = 54.85/14 = 3.92 and s = 1.98.

Both the defining formula and the computational formula for s 2 can be sensitive to rounding, so as much decimal accuracy as possible should be used in intermediate calculations.

Several other properties of s 2 can enhance understanding and facilitate computation.

PROPOSITION

Let x_1, x_2, …, xn be a sample and c be any nonzero constant.

1. If y_1 = x_1 + c,y_2 = x_2 + c, …, yn = xn + c, then = , and

2. If y_1 = cx_1, …, yn = cxn , then = c 2, sy = | c | sx ,

where is the sample variance of the x’s and is the sample variance of the y’ s.

In words, Result 1 says that if a constant c is added to (or subtracted from) each data value, the variance is unchanged. This is intuitive, since adding or subtracting c shifts the location of the data set but leaves distances between data values unchanged. According to Result 2, multiplication of each xi by c results in s 2 being multiplied by a factor of c^2 . These properties can be proved by noting in Result 1 that = + c and in Result 2 that = c .

Boxplotlign="center">

The values of both the mean and median suggest that the cancer sample is centered somewhat to the right of the no-cancer sample on the measurement scale. The mean, however, exaggerates the magnitude of this shift, largely because of the observation 210 in the cancer sample. The values of s suggest more variability in the cancer sample than in the no-cancer sample, but this impression is contradicted by the fourth spreads. Again, the observation 210, an extreme outlier, is the culprit. Figure 1.22 shows a comparative boxplot from the S-Plus computer package. The no-cancer box is stretched out compared with the cancer box (fs = 18 vs. fs = 11), and the positions of the median lines in the two boxes show much more skewness in the middle half of the no-cancer sample than the cancer sample. Outliers are represented by horizontal line segments, and there is no distinction between mild and extreme outliers.

EXERCISES Section 1.4 (44–61)

44. The article “Oxygen Consumption During Fire Suppression: Error of Heart Rate Estimation” (Ergonomics, 1991: 1469–1474) reported the following data on oxygen consumption (mL/kg/min) for a sample of ten firefighters performing a fire-suppression simulation:

29.5 49.3 30.6 28.2 28.0 26.3 33.9 29.4 23.5 31.6

Compute the following:

a. The sample range

b. The sample variance s 2 from the definition (i.e., by first computing deviations, then squaring them, etc.)

c. The sample standard deviation

d. s 2 using the shortcut method

45. The value of Young’s modulus (GPa) was determined for cast plates consisting of certain intermetallic substrates, resulting in the following sample observations (“Strength and Modulus of a Molybdenum-Coated Ti-25Al-10Nb-3U-1Mo Intermetallic,” J. of Materials Engr. and Performance, 1997: 46–50):

116.4 115.9 114.6 115.2 115.8

a. Calculate
49. A study of the relationship between age and various visual functions (such as acuity and depth perception) reported the following observations on area of scleral lamina (mm^2) from human optic nerve heads (“Morphometry of Nerve Fiber Bundle Pores in the Optic Nerve Head of the Human,” Experimental Eye Research, 1988: 559–568):

a. Calculate Σxi and .

b. Use the values calculated in part (a) to compute the sample variance s 2 and then the sample standard deviation s.

50. In 1997 a woman sued a computer keyboard manufacturer, charging that her repetitive stress injuries were caused by the keyboard (Genessy v. Digital Equipment Corp .). The injury awarded about $3.5 million for pain and suffering, but the court then set aside that award as being unreasonable compensation. In making this determination, the court identified a “normative” group of 27 similar cases and specified a reasonable award as one within two standard deviations of the mean of the awards in the 27 cases. The 27 awards were (in $1000s) 37, 60, 75, 115, 135, 140, 149, 150, 238, 290, 340, 410, 600, 750, 750, 750, 1050, 1100, 1139, 1150, 1200, 1200, 1250, 1576, 1700, 1825, and 2000, from which Σxi = 20,179, = 24,657,511. What is the maximum possible amount that could be awarded under the two-standard-deviation rule?

51. The article “A Thin-Film Oxygen Uptake Test for the Evaluation of Automotive Crankcase Lubricants” (Lubric. Engr., 1984: 75–83) reported the following data on oxidation-induction time (min) for various commercial oils:

a. Calculate the sample variance and standard deviation.

b. If the observations were reexpressed in hours, what would be the resulting values of the sample variance and sample standard deviation? Answer without actually performing the reexpression.

52. The first four deviations from the mean in a sample of n = 5 reaction times were .3, .9, 1.0, and 1.3. What is the fifth deviation from the mean? Give a sample for which these are the five deviations from the mean.

53. Reconsider the data on area of scleral lamina g>b. Calculate the value of the fourth spread.

c. If the two largest sample values, 4.33 and 4.52, had instead been 5.33 and 5.52, how would this affect fs ? Explain.

d. By how much could the observation 2.34 be increased without affecting fs ? Explain.

e. If an 18th observation, x_18 = 4.60, is added to the sample, what is fs ?

54. Consider the following observations on shear strength (MPa) of a joint bonded in a particular manner (from a graph in the article “Diffusion of Silicon Nitride to Austenitic Stainless Steel without Interlayers,” Metallurgical Trans., 1993:

a. What are the values of the fourths, and what is the value of fs ?

b. Construct a boxplot based on the five-number summary, and comment on its features.

c. How large or small does an observation have to be to qualify as an outlier? As an extreme outlier?

d. By how much could the largest observation be decreased without affecting fs ?

55. Here is a stem-and-leaf display of the escape time data introduced in Exercise 36 of this chapter.

a. Determine the value of the fourth spread.

b. Are there any outliers in the sample? Any extreme outliers?

c. Construct a boxplot and comment on its features.

d. By how much could the largest observation, currently 424, be decreased without affecting the value of the fourth spread?

56. The amount of aluminum contamination (ppm) in plastic of a certain type was determined for a sample of 26 plastic specimens, resulting in the following data (“The Lognormal Distribution for Modeling Quality Data when the Mean Is Near Zero,” J. of Quality Technology, 1990: 105–110):

±€

Construct a boxplot that shows outliers, and comment on its features.

57. A sample of 20 glass bottles of a particular type was selected, and the internal pressure strength of each bottle was determined. Consider the following partial sample information:

a. Are there any outliers in the sample? Any extreme outliers?

b. Construct a boxplot that shows outliers, and comment on any interesting features.

58. A company utilizes two different machines to manufacture parts of a certain type. During a single shift, a sample of n = 20 parts produced by each machine is obtained, and the value of a particular critical dimension for each part is determined. The comparative boxplot at the top of page 41 is constructed from the resulting data. Compare and contrast the two samples.

59. Blood cocaine concentration (mg/L) was determined both for a sample of individuals who had died from cocaine-induced excited delirium (ED) and for a sample of those who had died from a cocaine overdose without excited delirium; survival time for people in both groups was at most 6 hours.

The accompanying data was read from a comparative box-plot in the article “Fatal Excited Delirium Following Cocaine Use” (J. of Forensic Sciences, 1997: 25–31).

a. Determine the medians, fourths, and fourth spreads for the two samples.

b. Are there any outliers in either sample? Any extreme outliers?

c. Construct a comparative boxplot, and use it as a basis for comparing and contrasting the ED and non-ED samples.

60. Observations on burst strength (lb/in^2) were obtained both for test nozzle closure welds and for production cannister nozzle welds (“Proper Procedures Are the Key to Welding Radioactive Waste Cannisters,” Welding J., Aug. 1997: 61–67).

aluating Tunnel Kiln Performance” (Amer. Ceramic Soc. Bull., Aug. 1997: 59–63).

a. Construct a histogram based on relative frequencies, and comment on any interesting features.

b. What proportion of the strength observations are at least 85? Less than 95?

c. Roughly what proportion of the observations are less than 90?

66. A deficiency of the trace element selenium in the diet can negatively impact growth, immunity, muscle and neuromuscular function, and fertility. The introduction of selenium supplements to dairy cows is justified when pastures have low selenium levels. Authors of the paper “Effects of Short-Term Supplementation with Selenised Yeast on Milk Production and Composition of Lactating Cows” (Australian J. of Dairy Tech ., 2004: 199–203) supplied the following data on milk selenium concentration (mg/L) for a sample of cows given a selenium supplement and a control sample given no supplement, both initially and after a 9-day period.

a. Do the initial Se concentrations for the supplement and control samples appear to be similar? Use various techniques from this chapter to summarize the data and answer the question posed.

b. Again use methods from this chapter to summarize the data and then describe how the final Se concentration values in the treatment group differ from those in the control group.

67. Aortic stenosis refers to a narrowing of the aortic valve in the heart. The paper “Correlation Analysis of Stenotic Aortic Valve Flow Patterns Using Phase Contrast MRI” (Annals of Biomed. Engr, 2005: 878–887) gave the following data on aortic root diameter (cm) and gender for a sample of patients having various degrees of aortic stenosis:

a. Compare and contrast the diameter observations for the two genders.

b. Calculate a 10% trimmed mean for each of the two samples, and compare to other measures of center (for the male sample, the interpolation method mentioned in Section 1.3 must be used).

68. a. For what value of c is the quantil to 0, and solve.]

b. Using the result of part (a), which of the two quantities Σ(xi - )2 and Σ(xi - μ)2 will be smaller than the other (assuming that ≠ μ)?

69. a. Let a and b be constants and let yi = axi + b for i = 1, 2, …, n. What are the relationships between and and between . and ?

b. A sample of temperatures for initiating a certain chemical reaction yielded a sample average (°C) of 87.3 and a sample standard deviation of 1.04. What are the sample average and standard deviation measured in °F? [Hint: F = C + 32.]

70. Elevated energy consumption during exercise continues after the workout ends. Because calories burned after exercise contribute to weight loss and have other consequences, it is important to understand this process. The paper “Effect of Weight Training Exercise and Treadmill Exercise on Post-Exercise Oxygen Consumption” (Medicine and Science in Sports and Exercise, 1998: 518–522) reported the accompanying data from a study in which oxygen consumption (liters) was measured continuously for 30 minutes for each of 15 subjects both after a weight training exercise and after a treadmill exercise.

a. Construct a comparative boxplot of the weight and treadmill observations, and comment on what you see.

b. Because the data is in the form of (x, y) pairs, with x and y measurements on the same variable under two different conditions, it is natural to focus on the differences within pairs: d_1 = x_1 − y_1, …, dn = xn − yn . Construct a boxplot of the sample differences. What does it suggest?

71. Here is a description from MINITAB of the strength data given in Exercise 13.

a. Comment on any interesting features (the quartiles and fourths are virtually idr reasoning.

76. The three measures of center introduced in this chapter are the mean, median, and trimmed mean. Two additional measures of center that are occasionally used are the midrange, which is the average of the smallest and largest observations, and the midfourth, which is the average of the two fourths. Which of these five measures of center are resistant to the effects of outliers and which are not? Explain your reasoning.

77. Consider the following data on active repair time (hours) for a sample of n = 46 airborne communications receivers:

Construct the following:

a. A stem-and-leaf display in which the two largest values are displayed separately in a row labeled HI.

b. A histogram based on six class intervals with 0 as the lower limit of the first interval and interval widths of 2, 2, 2, 4, 10, and 10, respectively.

78. Consider a sample x_1, x_2, …, xn and suppose that the values of , s 2, and s have been calculated.

a. Let yi = xi— for i = 1, …, n. How do the values of s 2 and s for the yis compare to the corresponding values for the xi s? Explain.

b. Let zi = (xi − )/s for i = 1, …, n. What are the values of the sample variance and sample standard deviation for the zi s?

79. Let n and denote the sample mean and variance for the sample x_1, …, xn and let n +1 and +1 denote these quantities when an additional observation xn +1 is added to the sample.

a. Show how n +1 can be computed from n and xn +1.

so that +1 can be computed from xn +1, n , and .

c. Suppose that a sample of 15 strands of drapery yarn has resulted in a sample mean thread elongation of 12.58σmand a sample standard deviation of .512 mm. A 16th strand results in an elongation value of 11.8. What are the values of the sample mean and sample standard deviation for all 16 elongation observations?

80. Lengths of bus routes for any particular transit system will typically vary from one route to another. The article “Planning of City Bus Routes” (J. of the Institution of Engineers, 1995: 211–215) gives the following information on lengths (km) for one particular system:

a. Draw a histogram corresponding to these frequencies.

b. What proportion of these route lengths are less than 20? What proportion of these routes have lengths of at least 30?

c. Roughly what is the value of the 90th percentile of the route length distribution?

d. Roughly what is the median route length?

81. A study carried out to investigate the distribution of total braking time (reaction time plus accelerator-to-brake movement time, in ms) during real driving conditions at 60 km/hr gave the following summary information on the distribution of times (“A Field Study on Braking Responses during Driving,” Ergonomics , 1995: 1903–1910):

What can you conclude about the shape of a histogram of this data? Explain your reasoning.

82. The sample data x_1, x_2, …, xn sometimes represents a time series, where xt = the observed value of a response variable x at time t. Often the observed series shows a great deal of random variation, which makes it difficult to study longer-term behavior. In such situations, it is desirable to produce a smoothed version of the series. One technique for doing so involves exponential smoothing. The value of a smoothing constant a is chosen (0 < a < 1). Then with t = smoothed vex="00055" alt="Image">, = αxt + (1 − α )t -1.

a. Consider the following time series in which xt = temperature (°F) of effluent at a sewage treatment plant on day t: 47, 54, 53, 50, 46, 46, 47, 50, 51, 50, 46, 52, 50, 50. Plot each xt against t on a two-dimensional coordinate system (a time-series plot). Does there appear to be any pattern?

b. Calculate the ts using α = .1. Repeat using α = .5. Which value of a gives a smoother t series?

c. Substitute t −1 = αxt −1 + (1 − α )t −2 on the right-hand side of the expression for t , then substitute t −2 in terms of t −2 and t −3, and so on. On how many of the values xt , xt -1, …, x_1 does , depend? What happens to the coefficient on xt-k as k increases?

d. Refer to part (c). If t is large, how sensitive is t to the initialization 1 = x_1? Explain.

[Note: A relevant reference is the article “Simple Statistics for Interpreting Environmental Data,” Water Pollution Control Fed. J., 1981: 167–175.]

83. Consider numerical observations x_1, …, xn . It is frequently of interest to know whether the xi s are (at least approximately) symmetrically distributed about some value. If n is at least moderately large, the extent of symmetry can be assessed from a stem-and-leaf display or histogram. However, if n is not very large, such pictures are not particularly informative. Consider the following alternative. Let y_1 denote the smallest xi, y_2 the second smallest xi , and so on. Then plot the following pairs as points on a two-dimensional coordinate system: (yn − , − y_1), (yn−1 − , − y_2), (yn −2 − , n − 1)/2 when n is odd.

a. What does this plot look like when there is perfect symmetry in the data? What does it look like when observations stretch out more above the median than below it (a long upper tail)?

b. The accompanying data on rainfall (acre-feet) from 26 seeded clouds is taken from the article “A Bayesian Analysis of a Multiplicative Treatment Effect in Weather Modification” (Technometrics, 1975: 161–166). Construct the plot and comment on the extent of symmetry or nature of departure from symmetry.

Bibliography

Chambers, John, William Cleveland, Beat Kleiner, and Paul Tukey, Graphical Methods for Data Analysis, Brooks/Cole, Pacific Grove, CA, 1983. A highly recommended presentation of various graphical and pictorial methodology in statistics.

Cleveland, William, Visualizing Data, Hobart Press, Summit, NJ, 1993. An entertaining tour of pictorial techniques.

Devore, Jay, and Roxy Peck, Statistics: The Exploration and Analysis of Data (5th ed.), Thomson Brooks/Cole, Belmont, CA, 2005. The first few chapters give a very nonmathematical survey of methods for describing and summarizing data.

Freedman, David, Robert Pisani, and Roger Purves, Statistics (3rd ed.), Norton, New York, 1998. An excellent, very nonmathematical survey of basic statistical reasoning and methodology.

Hoaglin, David, Frederick Mosteller, and John Tukey, Understanding Robust and Exploratory Data Analysis, Wiley, New York, 1983. Discusses why, as well as how, exploratory methods should be employed; it is good on details of stem-and-leaf displays and boxplots.

Moore, David, and William Notz, Statistics: Concepts and Controversies (6th ed.), Freeman, San Francisco, 2006. An extremely readable and entertaining paperback that contains an intuitive discussion of problems connected with sampling and designed experiments.

Peck, Roxy, et al. (eds.), Statistics: A Guide to the Unknown (4th ed.), Thomson Brooks/Cole, Belmont, CA, 2006. Contains many short nontechnical articles describing various applications of statistics.

Verzani, John, Using R for Introductory Statistics, Chapman and Hall/CRC, Boca Raton, FL, 2005. A very nice introduction to the R software package.

The term probability refers to the study of randomness and uncertainty. In any situation in which one of a number of possible outcomes may occur, the discipline of probability provides methods for quantifying the chances, or likelihoods, associated with the various outcomes. The language of probability is constantly used in an informal manner in both written and spoken contexts. Examples include such statements as “It is likely that the Dow-Jones average will increase by the end of the year,” “There is a 50–50 chance that the incumbent will seek reelection,” “There will probably be at least one section of that course offered next year,” “The odds favor a quick settlement of the strike,” and “It is expected that at least 20,000 concert tickets will be sold.” In this chapter, we introduce some elementary probability concepts, indicate how probabilities can be interpreted, and show how the rules of probability can be applied to compute the probabilities of many interesting events. The methodology of probability will then permit us to express in precise language such informal statements as those given above.

The study of probability as a branch of mathematics goes back over 300 years, where it had its genesis in connection with questions involving games of chance. Many books are devoted exclusively to probability, but our objective here is to cover only that part of the subject that has the most direct bearing on problems of statistical inference.

2.1 Sample Spaces and Events

An experiment is any action or process whose outcome is subject to uncertainty. Although the word experiment generally suggests a planned or carefully controlled laboratory testing situation, we use it here in a much wider sense. Thus experiments that may be of interest include tossing a coin once or several times, selecting a card or cards from a deck, weighing a loaf of bread, ascertaining the commuting time from home to work on a particular morning, obtaining blood types from a group of individuals, or measuring the compressive strengths of different steel beams.

The Sample Space of an Experiment

DEFINITION

The sample space of an experiment, denoted by S , is the set of all possible outcomes of that experiment.

Example 2.1

The simplest experiment to which probability applies is one with two possible outcomes. One such experiment consists of examining a single fuse to see whether it is defective. The sample space for this experiment can be abbreviated as S = {N , D }, where N represents not defective, D represents defective, and the braces are used to enclose the elements of a set. Another such experiment would involve tossing a thumbtack and noting whether it landed point t="0em">

Example 2.2

If we examine three fuses in sequence and note the result of each examination, then an outcome for the entire experiment is any sequence of N’s and D’s of length 3, so

S = {NNN, NND, NDN, NDD, DNN, DND, DDN, DDD }

If we had tossed a thumbtack three times, the sample space would be obtained by replacing N by U in S above, with a similar notational change yielding the sample space for the experiment in which the genders of three newborn children are observed.

Example 2.3

Two gas stations are located at a certain intersection. Each one has six gas pumps. Consider the experiment in which the number of pumps in use at a particular time of day is determined for each of the stations. An experimental outcome specifies how many pumps are in use at the first station and how many are in use at the second one. One possible outcome is (2, 2), another is (4, 1), and yet another is (1, 4). The 49 outcomesin S are displayed in the accompanying table. The sample space for the experiment in which a six-sided die is thrown twice results from deleting the 0 row and 0 column from the table, giving 36 outcomes.

Example 2.4

If a new type-D flashlight battery has a voltage that is outside certain limits, that battery is characterized as a failure (F ); if the battery has a voltage within the prescribed limits, it is a success (S ). Suppose an experiment consists of testing each battery as it comes off an assembly line until we first observe a success. Although it may not be very likely, a possible outcome of this experiment is that the first 10 (or 100 or 1000 or …) are F ’s and the next one is an S . That is, for any positive integer n , we may have to examine n batteries before seeing the first S . The sample space is S = {S, FS, FFS, FFFS …}, which contains an infinite number of possible outcomes. The same abbreviated form of the sample space is appropriate for an experiment in which, starting at a specified time, the gender of each newborn infant is recorded until the birth of a male is observed.

Events

In our study of probability, we will be interested not only in the individual outcomes of S but alsoew Roman">DEFINITION

An event is any collection (subset) of outcomes contained in the sample space S . An event is simple if it consists of exactly one outcome and compound if it consists of more than one outcome.

When an experiment is performed, a particular event A is said to occur if the resulting experimental outcome is contained in A . In general, exactly one simple event will occur, but many compound events will occur simultaneously.

Example 2.5

Consider an experiment in which each of three vehicles taking a particular freeway exit turns left (L ) or right (R ) at the end of the exit ramp. The eight possible outcomes that comprise the sample space are LLL, RLL, LRL, LLR, LRR, RLR, RRL , and RRR . Thus there are eight simple events, among which are E_1 = {LLL} and E_5 = {LRR} . Some compound events include

A = {RLL, LRL, LLR } = the event that exactly one of the three vehicles turns right

B = {LLL, RLL, LRL, LLR } = the event that at most one of the vehicles turns right

C = {LLL, RRR } = the event that all three vehicles turn in the same direction

Suppose that when the experiment is performed, the outcome is LLL . Then the simple event E_1 has occurred and so also have the events B and C (but not A).

Example 2.6 (Example 2.3 continued)

When the number of pumps in use at each of two six-pump gas stations is observed, (Example 2.3 there are 49 possible outcomes, so there are 49 simple events: E_1 = {(0, 0)}, continued) E_2 = {(0, 1)}, …, E_49 = {(6, 6)}. Examples of compound events are

A = {(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} = the event that the number of pumps in use is the same for both stations

B = {(0, 4), (1, 3), (2, 2), (3, 1), (4, 0)} = the event that the total number of pumps in use is four

C = {(0, 0), (0, 1), (1, 0), (1, 1)} = the event that at most one pump is in use at each station

Example 2.7 (Example 2.4 continued)

The sample space for the}5 v>

A ′= {FFFS, FFFFS, FFFFFS , …}, C ′={S, FFS, FFFFS , …}

A ∪ B = {S, FS, FFS, FFFFS }, A ∩ B = {S, FFS}

Sometimes A and B have no outcomes in common, so that the intersection of A and B contains no outcomes.

DEFINITION

Let denote the null event (the event consisting of no outcomes whatsoever). When A ∩ B = , A and B are said to be mutually exclusive or disjoint events.

Example 2.10

A small city has three automobile dealerships: a GM dealer selling Chevrolets, Pontiacs, and Buicks; a Ford dealer selling Fords and Mercurys; and a Chrysler dealer selling Plymouths and Chryslers. If an experiment consists of observing the brand of the next car sold, then the events A = {Chevrolet, Pontiac, Buick} and B = {Ford, Mercury} are mutually exclusive because the next car sold cannot be both a GM product and a Ford product.

The operations of union and intersection can be extended to more than two events. For any three events A, B , and C , the event A ∪ B ∪ C is the set of outcomes contained in at least one of the three events, whereas A ∩ B ∩ C is the set of outcomes contained in all three events. Given events A_1, A_2, A_3, …, these events are said to be mutually exclusive (or pairwise disjoint) if no two events have any outcomes in common.

A pictorial representation of events and manipulations with events is obtained by using Venn diagrams. To construct a Venn diagram, draw a rectangle whose interior will represent the sample space S . Then any event A is represented as the interior of a closed curve (often a circle) contained in S . Figure 2.1 shows examples of Venn diagrams.

Figure 2.1 Venn diagrams

EXERCISES Section 2.1 (1–10)

1. Four universities—1, 2, 3en the two winners will play for the championship, and the two losers will also play. One possible outcome can be denoted by 1324 (1 beats 2 and 3 beats 4 in first-round games, and then 1 beats 3 and 2 beats 4).

a. List all outcomes in S .

b. Let A denote the event that 1 wins the tournament. List outcomes in A .

c. Let B denote the event that 2 gets into the championship game. List outcomes in B .

d. What are the outcomes in A ∪ B and in A ∩ B ? What are the outcomes in A ′?

2. Suppose that vehicles taking a particular freeway exit can turn right (R) , turn left (L) , or go straight (S) . Consider observing the direction for each of three successive vehicles.

a. List all outcomes in the event A that all three vehicles go in the same direction.

b. List all outcomes in the event B that all three vehicles take different directions.

c. List all outcomes in the event C that exactly two of the three vehicles turn right.

d. List all outcomes in the event D that exactly two vehicles go in the same direction.

e. List outcomes in D′, C ∪ D , and C ∩ D .

3. Three components are connected to form a system as shown in the accompanying diagram. Because the components in the 2–3 subsystem are connected in parallel, that subsystem will function if at least one of the two individual components functions. For the entire system to function, component 1 must function and so must the 2–3 subsystem.

The experiment consists of determining the condition of each component [S (success) for a functioning component and F (failure) for a nonfunctioning component].

a. What outcomes are contained in the event A that exactly two out of the three components function?

b. What outcomes are contained in the event B that at least two of the components function?

c. unctions?

d. List outcomes in C ′, A ∪ C , A ∩ C, B ∪ C , and B ∩ C .

4. Each of a sample of four home mortgages is classified as fixed rate (F ) or variable rate (V ).

a. What are the 16 outcomes in S ?

b. Which outcomes are in the event that exactly three of the selected mortgages are fixed rate?

c. Which outcomes are in the event that all four mortgages are of the same type?

d. Which outcomes are in the event that at most one of the four is a variable-rate mortgage?

e. What is the union of the events in parts (c) and (d), and what is the intersection of these two events?

f. What are the union and intersection of the two events in parts (b) and (c)?

5. A family consisting of three persons—A, B , and C—belongs to a medical clinic that always has a doctor at each of stations 1, 2, and 3. During a certain week, each member of the family visits the clinic once and is assigned at random to a station. The experiment consists of recording the station number for each member. One outcome is (1, 2, 1) for A to station 1, B to station 2, and C to station 1.

a. List the 27 outcomes in the sample space.

b. List all outcomes in the event that all three members go to the same station.

c. List all outcomes in the event that all members go to different stations.

d. List all outcomes in the event that no one goes to station 2.

6. A college library has five copies of a certain text on reserve. Two copies (1 and 2) are first printings, and the other three (3, 4, and 5) are second printings. A student exami

b. Suppose there is no outcome common to all three of the events A, B , and C . Are these three events necessarily mutually exclusive? If your answer is yes, explain why; if your answer is no, give a counterexample using the experiment of Example 2.10.

2.2 Axioms, Interpretations, and Properties of Probability

Given an experiment and a sample space S, the objective of probability is to assign to each event A a number P(A), called the probability of the event A, which will give a precise measure of the chance that A will occur . To ensure that the probability assignments will be consistent with our intuitive notions of probability, all assignments should satisfy the following axioms (basic properties) of probability.

AXIOM 1

For any event A, P(A) ≥ 0.

AXIOM 2

P(S) = 1.

AXIOM 3

If A_1, A_2, A_2, … is an infinite colletion of disjoint events, then

You might wonder why the third axiom contains no reference to a finite collection of disjoint events. It is because the corresponding property for a finite collection can be derived from our three axioms. We want our axiom list to be as short as possible and not contain any property that can be derived from others on the list. Axiom 1 reflects the intuitive notion that the chance of A occurring should be nonnegative. The sample space is by definition the event that must occur when the experiment is performed (S contains all possible outcomes), so Axiom 2 says that the maximum possible probability of 1 is assigned to S . The third axiom formalizes the idea that if we wish the probability that at least one of a number of events will occur and no two of the events can occur simultaneously, then the chance of at least one occurring is the sum of the chances of the individual events.

PROPOSITION

P () = 0 where is the null event (the event containing no outcomes whatsoever). This in turn implies that the property contained in Axiom 3 is valid for a finite collection of events., …. Since ∩ = , the events in this collection are disjoint and ∪ Ai = . The third axiom then gives P () = Σ P ()

This can happen only if P () = 0.

Now suppose that A_1, A_2, …, Ak are disjoint events, and append to these the infinite collection Ak +1 = , Ak +2 = , Ak +3 = , …. Again invoking the third axiom,

as desired.

Example 2.11

Consider tossing a thumbtack in the air. When it comes to rest on the ground, either its point will be up (the outcome U ) or down (the outcome D ). The sample space for this event is therefore S = {U, D }. The axioms specify P (S ) = 1, so the probability assignment will be completed by determining P (U ) and P (D ). Since U and D are disjoint and their union is S , the foregoing proposition implies that

1 = P (S ) = P (U ) + P (D )

It follows that P (D ) = 1 − P (U ). One possible assignment of probabilities is P (U ) = .5, P (D ) = .5, whereas another possible assignment is P (U ) = .75, P (D ) = .25. In fact, letting p represent any fixed number between 0 and 1, P (U ) = p, P(D) = 1 − p is an assignment consistent with the axioms.

Example 2.12

Let’s return to the experiment in Example 2.4, in which batteries coming off an assembly line are tested one by one until one having a voltage within prescribed limits is found. The simple events are E_1 = {S }, E_2 = {FS }, E_3 = {FFS }, E_4 = { FFFS }, …. Suppose the probability of any particular battery being satisfactory is .99. Then it can be shown that P (E_1) = .99, P (E_2) = (.01)(.99), P (E_3) = (.01)2such as “The probability of that coin landing with the head facing up when it is tossed is .5” to mean that in a large number of such tosses, a head will appear on approximately half the tosses and a tail on the other half.

This relative frequency interpretation of probability is said to be objective because it rests on a property of the experiment rather than on any particular individual concerned with the experiment. For example, two different observers of a sequence of coin tosses should both use the same probability assignments since the observers have nothing to do with limiting relative frequency. In practice, this interpretation is not as objective as it might seem, since the limiting relative frequency of an event will not be known. Thus we will have to assign probabilities based on our beliefs about the limiting relative frequency of events under study. Fortunately, there are many experiments for which there will be a consensus with respect to probability assignments. When we speak of a fair coin, we shall mean P (H ) = P (T ) = .5, and a fair die is one for which limiting relative frequencies of the six outcomes are all , suggesting probability assignments P ({1}) = … = P ({6}) = .

Because the objective interpretation of probability is based on the notion of limiting frequency, its applicability is limited to experimental situations that are repeatable. Yet the language of probability is often used in connection with situations that are inherently unrepeatable. Examples include: “The chances are good for a peace agreement”; “It is likely that our company will be awarded the contract”; and “Because their best quarterback is injured, I expect them to score no more than 10 points against us.” In such situations we would like, as before, to assign numerical probabilities to various outcomes and events (e.g., the probability is .9 that we will get the contract). We must therefore adopt an alternative interpretation of these probabilities. Because different observers may have different prior information and opinions concerning such experimental situations, probability assignments may now differ from individual to individual. Interpretations in such situations are thus referred to as subjective . The book by Robert Winkler listed in the chapter references gives a very readable survey of several subjective interpretations.

More Probability Properties

PROPOSITION

For any event A , P (A ) + P (A ′) = 1, from which P (A ) = 1 − P (A ′).

Proof

In Axiom 3, let k = 2, A_1 = A , and A_2 = A ′. Since by definition of A ′, A ∪ A ′= S while A and A ′ are disjoint, 1 = P (S ) = P (A ∪ A ′) = P (A ) + P (A ′).

This proposition is surprisingly useful because there are many situations in which P(A′ ) is more easily obtained by direct methods than is P(A).

Example 2.13

Consider a system of five identical components connected in series, as illustrated in Figure 2.3.

Figure 2.3 A system of five components connected in series

Denote a component that fails by F and one that doesn’t fail by S (for success). Let A be the event that the system fails. For A to occur, at least one of the individual components must fail. Outcomes in A include SSFSS (1, 2, 4, and 5 all work, but 3 does not), FFSSS , and so on. There are in fact 31 different outcomes in A . However, A′, the event that the system works, consists of the single outcome SSSSS . We will see in Section 2.5 that if 90% of all these components do not fail and different components fail independently of one another, then P (A ′) = P (SSSSS ) = .9^5 = .59. Thus P (A ) = 1 − .59 = .41; so among a large number of such systems, roughly 41% will fail.

In general, the foregoing proposition is useful when the event of interest can be expressed as “at least …,” since then the complement “less than …” may be easier to work with (in some problems, “more than …” is easier to deal with than “at most …”). When you are having difficulty calculating P(A) directly, think of determining P (A ′).

PROPOSITION

For any event A , P (A ) ≤ 1.

This is because 1 = P (A ) + P (A ′) ≥ P (A ) since P (A ′) ≥ 0.

When events A and B are mutually exclusive, P (A ∪ B ) = P (A ) + P (B ). For events that are not mutually exclusive, adding P (A ) and P (B ) results in “double-counting” outcomes in the intersection. The next result shows how to correct for this.

PROPOSITION

For any two>P (A ) + P (B ) − P (A ∩ B )

Proof

Note first that A ∪ B can be decomposed into two disjoint events, A and B ∩ A ′; the latter is the part of B that lies outside A. Furthermore, B itself is the union of the two disjoint events A ∩ B and A ′ ∩ B , so P (B ) = P (A ∩ B ) + P (A ′ ∩ B ). Thus

Figure 2.4 Representing A ∪ B as a union of disjoint events

Example 2.14

In a certain residential suburb, 60% of all households subscribe to the metropolitan newspaper published in a nearby city, 80% subscribe to the local paper, and 50% of all households subscribe to both papers. If a household is selected at random, what is the probability that it subscribes to (1) at least one of the two newspapers and (2) exactly one of the two newspapers?

With A = {subscribes to the metropolitan paper} and B = {subscribes to the local paper}, the given information implies that P (A ) = .6, P (B ) = .8, and P (A ∩ B ) = .5. The foregoing proposition now yields

P (subscribes to at least one of the two newspapers)

= P (A ∪ B) = P (A ) + P (B ) − P (A ∩ B ) = .6 +.8 -.5 = .9

The event that a household subscribes only to the local paper can be written as A ′ ∩ B [(not metropolitan) and local]. Now Figure 2.4 implies that

.9 = P(A ∪ B) = P (A ) + P (A ′ ∩ B ) = .6 + P (A ′ ∩ B )

from which P (A ′ ∩ B ) = .3. Similarly, P (A ∩ B ′) = P (A ∪ B ) − P (B ) = .1. This is all illustrated in Figure 2.5, from which we see that1 + 2p_1 + 4p_1 + 2p_1 + p_1 = 10p_1

implying p_1 = p_5 = .1, p_2 = p_4 = .2, p_3 = .4. The probability that one of the three middle cars is selected (a compound event) is then p_2 + p_3 + p_4 = .8.

Equally Likely Outcomes

In many experiments consisting of N outcomes, it is reasonable to assign equal probabilities to all N simple events. These include such obvious examples as tossing a fair coin or fair die once or twice (or any fixed number of times), or selecting one or several cards from a well-shuffled deck of 52. With p = P (Ei) for every i ,

That is, if there are N equally likely outcomes, the probability for each is 1/N .

Now consider an event A , with N (A ) denoting the number of outcomes contained in A . Then

Thus when outcomes are equally likely, computing probabilities reduces to counting: determine both the number of outcomes N(A ) in A and the number of outcomes N in S , and form their ratio.

Example 2.16

When two dice are rolled separately, there are N = 36 outcomes (delete the first row and column from the table in Example 2.3). If both the dice are fair, all 36 outcomes are equally likely, so P (Ei) = . Then the event A = {sum of two numbers = 7} consists of the six outcomes (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1), so

(EXERCISES Section 2.2 (11–28)

11. A mutual fund company offers its customers several different funds: a money-market fund, three different bond funds (short, intermediate, and long-term), two stock funds (moderate and high-risk), and a balanced fund. Among customers who own shares in just one fund, the percentages of customers in the different funds are as follows:

A customer who owns shares in just one fund is randomly selected.

a. What is the probability that the selected individual owns shares in the balanced fund?

b. What is the probability that the individual owns shares in a bond fund?

c. What is the probability that the selected individual does not own shares in a stock fund?

12. Consider randomly selecting a student at a certain university, and let A denote the event that the selected individual has a Visa credit card and B be the analogous event for a MasterCard. Suppose that P (A ) = .5, P (B ) = .4, and P (A ∩ B ) = .25.

a. Compute the probability that the selected individual has at least one of the two types of cards (i.e., the probability of the event A ∪ B ).

b. What is the probability that the selected individual has neither type of card?

c. Describe, in terms of A and B , the event that the selected student has a Visa card but not a MasterCard, and then calculate the probability of this event.

13. A computer consulting firm presently has bids out on three projects. Let Ai = {awarded project i} , for i = 1, 2, 3, and suppose that P (A_1) = .22, P (A_2) = .25, P (A_3) = .28, P (A_1 ∩ A_2) = .11, P (A_1 ∩ A_3) = .05, P (A_2 ∩ A_3) = .07, P (A_1 ∩ A_2 ∩ A_3) = .01. Express in words each of the following events, and compute the probability of each event:

a. A_1 ∪ A_2

b. [Hint: (A_1 ∪ A_2)′= ]

c. A_1 ∪ A_2 ∪ A_3

Suppose an individual having both types of policies is randomly selected.

a. What is the probability that the individual has a medium auto deductible and a high homeowner’s deductible?

b. What is the probability that the individual has a low auto deductible? A low homeowner’s deductible?

c. What is the probability that the individual is in the same category for both auto and homeowner’s deductibles?

d. Based on your answer in part (c), what is the probability that the two categories are different?

e. What is the probability that the individual has at least one low deductible level?

f. Using the answer in part (e), what is the probability that neither deductible level is low?

22. The route used by a certain motorist in commuting to work contains two intersections with traffic signals. The probability that he must stop at the first signal is .4, the analogous probability for the second signal is .5, and the probability that he must stop at at least one of the two signals is .6. What is the probability that he must stop

a. At both signals?

b. At the first signal but not at the second one?

c. At exactly one signal?

23. The computers of six faculty members in a certain department are to be replaced. Two of the faculty members have selected laptop machines and the other four have chosen desktop machines. Suppose that only two of the setups can be done on a particular day, and the two computers to be set up are randomly selected from the six (implying 15 equally likely outcomes; if the computers are numbered 1, 2,…, 6, then one outcome consists of computers 1 and 2, another consists of computers 1 and 3, and so on).

a. What is the probability that both selected setups are for laptop computers?

d. What is the probability that at least one computer of each type is chosen for setup?

24. Show that if one event A is contained in another event B (i.e., A is a subset of B ), then P (A ) ≤ P (B ). [Hint: For such A and B, A and B ∩ A′ are disjoint and B = A ∪ (B ∩ A′) , as can be seen from a Venn diagram.] For general A and B , what does this imply about the relationship among P (A ∩ B) , P (A ), and P (A ∪ B) ?

25. The three major options on a certain type of new car are an automatic transmission (A) , a sunroof (B) , and a stereo with compact disc player (C ). If 70% of all purchasers request A , 80% request B , 75% request C , 85% request A or B , 90% request A or C , 95% request B or C , and 98% request A or B or C , compute the probabilities of the following events. [Hint: “A or B “ is the event that at least one of the two options is requested; try drawing a Venn diagram and labeling all regions.]

a. The next purchaser will request at least one of the three options.

b. The next purchaser will select none of the three options.

c. The next purchaser will request only an automatic transmission and not either of the other two options.

d. The next purchaser will select exactly one of these three options.

26. A certain system can experience three different types of defects. Let Ai (i = 1, 2, 3) denote the event that the system has a defect of type i . Suppose that

P (A_1) = .12 P (A_2) = .07 P (A_3) = .05

P (A_1 ∪ A_2) = .13 P (A_1 ∪ A_3) = .14

P (A_2 ∪ A_3) = .10 P (A_1 ∩ A_2 ∩ A_3) = ./sub>, Pj ) for which Oi and Pj are associated with the same clinic. Because there are four obstetricians, n_1 = 4, and for each there are three choices of pediatrician, so n_2 = 3. Applying the product rule gives N = n_1n_2 = 12 possible choices.

In many counting and probability problems, a configuration called a tree diagram can be used to represent pictorially all the possibilities. The tree diagram associated with Example 2.18 appears in Figure 2.7. Starting from a point on the left side of the diagram, for each possible first element of a pair a straight-line segment emanates rightward. Each of these lines is referred to as a first-generation branch. Now for any given first-generation branch we construct another line segment emanating from the tip of the branch for each possible choice of a second element of the pair. Each such line segment is a second-generation branch. Because there are four obstetricians, there are four first-generation branches, and three pediatricians for each obstetrician yields three second-generation branches emanating from each first-generation branch.

Figure 2.7 Tree diagram for Example 2.18

Generalizing, suppose there are n_1 first-generation branches, and for each first-generation branch there are n_2 second-generation branches. The total number of second-generation branches is then n_1n_2. Since the end of each second-generation branch corresponds to exactly one possible pair (choosing a first element and then a second puts us at the end of exactly one second-generation branch), there are n_1n_2 pairs, verifying the product rule.

The construction of a tree diagram does not depend on having the same number of second-generation branches emanating from each first-generation branch. If the second clinic had four pediatricians, then there would be only three branches emanating from two of the first-generation branches and four emanating from each of the other two first-generation branches. A tree diagram can thus be used to represent pictorially experiments other than those to which the product rule applies.

A More General Product Rule

If a six-sided die is tossed five times in succession rather than just twice, then each possible outcome is an ordered collection of five numbers such as (1, 3, 1, 2, 4) or (6, 5, 2, 2, 2). We will call an ordered collection of k objects a k-tuple (so a pair is a 2-tuple and a triple is a 3-tuple). Each outcome of the die-tossing experiment is then a 5-tuple.1 possible choices for the first element; for each choice of the first element, there are n_2 possible choices of the second element; …; for each possible choice of the first k − 1 elements, there are nk choices of the kth element. Then there are n_1n_2….nk possible k-tuples.

This more general rule can also be illustrated by a tree diagram; simply construct a more elaborate diagram by adding third-generation branches emanating from the tip of each second generation, then fourth-generation branches, and so on, until finally k th-generation branches are added.

Example 2.19 (Example 2.17 continued)

Suppose the home remodeling job involves first purchasing several kitchen appliances. They will all be purchased from the same dealer, and there are five dealers in the area. With the dealers denoted by D_1, …, D_5, there are N = n_1n_2n_3 = (5)(12)(9) = 540 3-tuples of the form (Di, Pj, Qk), so there are 540 ways to choose first an appliance dealer, then a plumbing contractor, and finally an electrical contractor.

Example 2.20 (Example 2.18 continued)

If each clinic has both three specialists in internal medicine and two general surgeons, there are n_1n_2n_3n_4 = (4)(3)(3)(2) = 72 ways to select one doctor of each type such that all doctors practice at the same clinic.

Permutations and Combinations

Consider a group of n distinct individuals or objects (“distinct” means that there is some characteristic that differentiates any particular individual or object from any other). How many ways are there to select a subset of size k from the group? For example, if a Little League team has 15 players on its roster, how many ways are there to select 9 players to form a starting lineup? Or if you have 10 unread mysteries on your bookshelf and want to select 3 to take on a short vacation, how many ways are there to do this?

An answer to the general question just posed requires that we distinguish between two cases. In some situations, such as the baseball scenario, the order of selection is important. For example, Angela being the pitcher and Ben the catcher gives a different lineup from the one in which Angela is catcher and Ben is pitcher. Often, though, order is not important and one is interested only in which individuals or objects are selected, as wouldd/or spades (no red cards) is = 26!/13!13! = 10,400,600. One of these hands consists entirely of spades, and one consists entirely of clubs, so there are hands that consist entirely of clubs and spades with both suits represented in the hand. Suppose a bridge hand is dealt from a well-shuffled deck (i.e., 13 cards are randomly selected from among the 52 possibilities) and let

A = {the hand consists entirely of spades and clubs with both ">

EXERCISES Section 2.3 (29–44)

29. As of April 2006, roughly 50 million .com web domain names were registered (e.g., yahoo.com).

a. How many domain names consisting of just two letters in sequence can be formed? How many domain names of length two are there if digits as well as letters are permitted as characters? [Note: A character length of three or more is now mandated.]

b. How many domain names are there consisting of three letters in sequence? How many of this length are there if either letters or digits are permitted? [Note: All are currently taken.]

c. Answer the questions posed in (b) for four-character sequences.

d. As of April 2006, 97,786 of the four-character sequences using either letters or digits had not yet been claimed. If a four-character name is randomly selected, what is the probability that it is already owned?

30. A friend of mine is giving a dinner party. His current wine supply includes 8 bottles of zinfandel, 10 of merlot, and 12 of cabernet (he only drinks red wine), all from different wineries.

a. If he wants to serve 3 bottles of zinfandel and serving order is important, how many ways are there to do this?

b. If 6 bottles of wine are to be randomly selected from the 30 for serving, how many ways are there to do this?

c. If 6 bottles are randomly selected, how many ways are there to obtain two bottles of each variety?

d. If 6 bottles are randomly selected, what is the probability that this results in two bottles of each variety being chosen?

e. If 6 bottles are randomly selected, what is the probability that all of them are the same variety.

31. a. Beethoven wrote 9 symphonies and Mozart wrote 27 piano concertos. If a university radio station announcer wishes to play first a Beethoven symphony and then a Mozart concerto, in how many ways can this be done?

32. A stereo store is offering a special price on a complete set of components (receiver, compact disc player, speakers, cassette deck). A purchaser is offered a choice of manufacturer for each component:

Receiver: Kenwood, Onkyo, Pioneer, Sony, Sherwood Compact disc player: Onkyo, Pioneer, Sony, Technics Speakers: Boston, Infinity, Polk Cassette deck: Onkyo, Sony, Teac, Technics

A switchboard display in the store allows a customer to hook together any selection of components (consisting of one of each type). Use the product rules to answer the following questions:

a. In how many ways can one component of each type be selected?

b. In how many ways can components be selected if both the receiver and the compact disc player are to be Sony?

c. In how many ways can components be selected if none is to be Sony?

d. In how many ways can a selection be made if at least one Sony component is to be included?

e. If someone flips switches on the selection in a completely random fashion, what is the probability that the system selected contains at least one Sony component? Exactly one Sony component?

33. Again consider a Little League team that has 15 players on its roster.

a. How many ways are there to select 9 players for the starting lineup?

b. How many ways are there to select 9 players for the starting lineup and a batting order for the 9 starters?

c. Suppose 5 of the 15 players are left-handed. How many ways are there to select 3 left-handed outfielders and have all 6 other positions occupied by right-handed players?

34. Shortly after being put into service, some buses manufactured by a certain company have developed cracks on the underside of the main frame. Suppose a particular city has 25 of these buses, and cracks have actually appeared in 8 of them.

a. ly 4 with visible cracks?

c. If a sample of 5 buses is chosen at random, what is the probability that exactly 4 of the 5 will have visible cracks?

d. If buses are selected as in part (c), what is the probability that at least 4 of those selected will have visible cracks?

35. A production facility employs 20 workers on the day shift, 15 workers on the swing shift, and 10 workers on the graveyard shift. A quality control consultant is to select 6 of these workers for in-depth interviews. Suppose the selection is made in such a way that any particular group of 6 workers has the same chance of being selected as does any other group (drawing 6 slips without replacement from among 45).

a. How many selections result in all 6 workers coming from the day shift? What is the probability that all 6 selected workers will be from the day shift?

b. What is the probability that all 6 selected workers will be from the same shift?

c. What is the probability that at least two different shifts will be represented among the selected workers?

d. What is the probability that at least one of the shifts will be unrepresented in the sample of workers?

36. An academic department with five faculty members narrowed its choice for department head to either candidate A or candidate B . Each member then voted on a slip of paper for one of the candidates. Suppose there are actually three votes for A and two for B . If the slips are selected for tallying in random order, what is the probability that A remains ahead of B throughout the vote count (e.g., this event occurs if the selected ordering is AABAB , but not for ABBAA) ?

37. An experimenter is studying the effects of temperature, pressure, and type of catalyst on yield from a certain chemical reaction. Three different temperatures, four different pressures, and five different catalysts are under consideration.

a. If any particular experimental run involves the use of a single temperature, pressure, and catalyst, how many experimental runs are possible?

b. How many experimental runs are there that involve use of the lowest temperature and two lowest pressures?

Example 2.24

Complex components are assembled in a plant that uses two different assembly lines, A and A′ . Line A uses older equipment than A′ , so it is somewhat slower and less reliable. Suppose on a given day line A has assembled 8 components, of which 2 have been identified as defective (B ) and 6 as nondefective (B′ ), whereas A′ has produced 1 defective and 9 nondefective components. This information is summarized in the accompanying table.

Unaware of this information, the sales manager randomly selects 1 of these 18 components for a demonstration. Prior to the demonstration

However, if the chosen component turns out to be defective, then the event B has occurred, so the component must have been 1 of the 3 in the B column of the table. Since these 3 components are equally likely among themselves after B has occurred,

In Equation (2.2), the conditional probability is expressed as a ratio of unconditional probabilities: The numerator is the probability of the intersection of the two events, whereas the denominator is the probability of the conditioning event B . A Venn diagram illuminates this relationship (Figure 2.8).

Figure 2.8 Motivating the definition of conditional probability

Given that B has occurred, the relevant sample space is no longer S but consists of outcomes in B; A has occurred if and only if one of the outcomes in the intersection occurred, so the conditional probability of A given B is proportional to P (A ∩ B ). The proportionality constant 1/P (B ) is used to ensure that the probability P (B | B ) of the new sample space B equals 1.

The Definition of Conditional Probability

Example 2.24 demonstrates that when outcomes are equally likely, computation of conditional probabilities can be based on intuition. When experiments are more complicated, though, i face="Times New Roman">

When the experiment of interest consists of a sequence of several stages, it is convenient to represent these with a tree diagram. Once we have an appropriate tree diagram, probabilities and conditional probabilities can be entered on the various branches; this will make repeated use of the multiplication rule quite straightforward.

Example 2.29

A chain of video stores sells three different brands of DVD players. Of its DVD player sales, 50% are brand 1 (the least expensive), 30% are brand 2, and 20% are brand 3. Each manufacturer offers a 1-year warranty on parts and labor. It is known that 25% of brand 1’s DVD players require warranty repair work, whereas the corresponding percentages for brands 2 and 3 are 20% and 10%, respectively.

1. What is the probability that a randomly selected purchaser has bought a brand 1 DVD player that will need repair while under warranty?

2. What is the probability that a randomly selected purchaser has a DVD player that will need repair while under warranty?

3. If a customer returns to the store with a DVD player that needs warranty repair work, what is the probability that it is a brand 1 DVD player A brand 2 DVD player? A brand 3 DVD player?

The first stage of the problem involves a customer selecting one of the three brands of DVD player. Let Ai = {brand i is purchased}, for i = 1, 2, and 3. Then P (A_1) = .50, P (A_2) = .30, and P (A_3) = .20. Once a brand of DVD player is selected, the second stage involves observing whether the selected DVD player needs warranty repair. With B = {needs repair} and B ′ = {doesn’t need repair}, the given information implies that P (B | A_1) = .25, P (B | A_2) = .20, and P (B | A_3) = .10.

The tree diagram representing this experimental situation is shown in Figure 2.10. The initial branches correspond to different brands of DVD players; there are two second-generation branches emanating from the tip of each initial branch, one for “needs repair” and the other for “doesn’t need repair.” The probability P (Ai ) appears on the ith initial branch, whereas the conditional probabilities P (B | Ai ) and P (B ′ | Ai ) appear on the second-generation branches. To the right of each second-generation branch corresponding to the occurrence of B , we display the product of probabilities on the branches leading out to that point. This is simply the multiplication rule in action. The answer to the question posed in 1 is thus P (A_1 ∩ B ) = P (B
P(B ) = P [(brand 1 and repair) or (brand 2 and repair) or (brand 3 and repair)]

= P (A_1 ∩ B ) + P (A_2 ∩ B ) + P (A_3 ∩ B )

= .125 +.060 +.020 = .205

Figure 2.10 Tree diagram for Example

Finally,

and

P (A_3|B ) = 1 − P (A_1|B ) − P (A_2 | B ) = .10

The initial or prior probability of brand 1 is.50. Once it is known that the selected DVD player needed repair, the posterior probability of brand 1 increases to.61. This is because brand 1 DVD players are more likely to need warranty repair than are the other brands. The posterior probability of brand 3 is P (A_3 |B ) = .10, which is much less than the prior probability P (A_3) = .20.

Bayes’ Theorem

The computation of a posterior probability P (Aj | B ) from given prior probabilities P (Ai ) and conditional probabilities P (B | Ai ) occupies a central position in elementary probability. The general rule for such computations, which is really just a simple application of the multiplication rule, goes back to Reverend Thomas Bayes, who lived in the eighteenth century. To state it we first need another result. Recall that events A_1, …, Ak are mutually exclusive if no two have any common outcomes. The events are exhaustive if one Ai must occur, so that A_1 ∪… ∪ Ak = S .

The Law of Total Probability

Let A_1,…, Ak be mutually exclusive and exhaustive events. Then for any other event B ,

Proof

a. What is the probability that the next shirt sold is a medium, long-sleeved, print shirt?

b. What is the probability that the next shirt sold is a medium print shirt?

c. What is the probability that the next shirt sold is a short-sleeved shirt? A long-sleeved shirt?

d. What is the probability that the size of the next shirt sold is medium? That the pattern of the next shirt sold is a print?

e. Given that the shirt just sold was a short-sleeved plaid, what is the probability that its size was medium?

f. Given that the shirt just sold was a medium plaid, what is the probability that it was short-sleeved? Long-sleeved?

51. One box contains six red balls and four green balls, and a second box contains seven red balls and three green balls. A ball is randomly chosen from the first box and placed in the second box. Then a ball is randomly selected from the second box and placed in the first box.

a. What is the probability that a red ball is selected from the first box and a red ball is selected from the second box?

b. At the conclusion of the selection process, what is the probability that the numbers of red and green balls in the first box are identical to the numbers at the beginning?

52. A system consists of two identical pumps, #1 and #2. If one pump fails, the system will still operate. However, because of the added strain, the extra remaining pump is now more likely to fail than was originally the case. That is, r = P (#2 fails | #1 fails) > P (#2 fails) = q . If at least one pump fails by the end of the pump design life in 7% of all systems and both pumps fail during that period in only 1%, what is the probability that pump #1 will fail during the pump design life?

53. A certain shop repairs both audio and video components. Let A denote the event that the next component brought in for repair is an audio component, and let B be the event that the next component is a compact disc player (so the event B is contained in A ). Suppose that P (A ) = .6 and P (B ) = .05. What is P (B | A )?

54. In Exercise 13, Ai = {awarded project i }, for i = 1, 2, 3. Use the probabilities given there to compute the following probabilities, and explain in words the meaning of em>1)

b. P (A_2 ∩ A_3 | A_1)

c. P (A_2 ∪ A_3 | A_1)

d. P (A_1 ∩ A_2 ∩ A_3 | A_1 ∪ A_2 ∪ A_3).

55. Deer ticks can be carriers of either Lyme disease or human granulocytic ehrlichiosis (HGE). Based on a recent study, suppose that 16% of all ticks in a certain location carry Lyme disease, 10% carry HGE, and 10% of the ticks that carry at least one of these diseases in fact carry both of them. If a randomly selected tick is found to have carried HGE, what is the probability that the selected tick is also a carrier of Lyme disease?

56. For any events A and B with P (B ) > 0, show that P (A | B ) + P (A ′ |B ) = 1.

57. If P (B | A ) > P (B ), show that P (B ′ |A ) < P (B ′). [Hint: Add P (B ′ |A ) to both sides of the given inequality and then use the result of Exercise 56.]

58. Show that for any three events A, B , and C with P (C ) > 0, P (A ∪ B | C ) = P (A | C ) + P (B | C ) − P (A ∩ B | C ).

59. At a certain gas station, 40% of the customers use regular gas (A_1), 35% use plus gas (A_2), and 25% use premium (A_3). Of those customers using regular gas, only 30% fill their tanks (event B ). Of those customers using plus, 60% fill their tanks, whereas of those using premium, 50% fill their tanks.

a. What is the probability that the next customer will request plus gas and fill the tank (A_2 ∩ B) ?

b. What is the probability that the next customer fills the tank?

c. If the next customer fills the tank, what is the probability that regular gas is requested? Plus? Premium?

60. Seventy percent of the light aircra ) originally assigned to A when we are subsequently informed that another event B has occurred; the new probability of A is P (A | B ). In our examples, it was frequently the case that P (A | B ) differed from the unconditional probability P (A ), indicating that the information “B has occurred” resulted in a change in the chance of A occurring. Often the chance that A will occur or has occurred is not affected by knowledge that B has occurred, so that P (A | B ) = P (A ). It is then natural to regard A and B as independent events, meaning that the occurrence or nonoccurrence of one event has no bearing on the chance that the other will occur.

DEFINITION

Two events A and B are independent if P (A | B ) = P (A ) and are dependent otherwise.

The definition of independence might seem “unsymmetric” because we do not also demand that P (B | A ) = P (B ). However, using the definition of conditional probability and the multiplication rule,

The right-hand side of Equation (2.7) is P (B ) if and only if P (A | B ) = P (A ) (independence), so the equality in the definition implies the other equality (and vice versa). It is also straightforward to show that if A and B are independent, then so are the following pairs of events: (1) A ′ and B , (2) A and B ′, and (3) A ′ and B ′.

Example 2.31

Consider a gas station with six pumps numbered 1, 2, …, 6 and let Ei denote the simple event that a randomly selected customer uses pump i (i = 1, …, 6). Suppose that P (E_1) = P (E_6) = .10, P (E_2) = P (E_5) = .15, and P (E_3) = P (E_4) = .25. Define events A, B, C by A = {2, 4, 6}, B = {1, 2, 3}, and C = {2, 3, 4, 5}. We then have P (A ) = .50, P (A | B ) = .30, and P (A | C ) = .50. That is, events A and B are dependent, whereas events A and C are independent. Intuitively, A and C are independent because the relative division of probability among even- and odd-numbered pumps is the same among pumps 2, 3, 4, 5 as it is among all six pumps.

Example 2.32

Let A and B be any two mutually exclusive events with P (A ) t="0em" align="right">

It is straightforward to show that A and B are independent iff A′ and B are independent, A and B′ are independent, and A′ and B′ are independent. Thus in Example 2.33, the probability that neither machine needs service is

P (A ′ ∩ B ′) = P (A ′) · P (B ′) = (.70)(.90) = .63

Example 2.34

Each day, Monday through Friday, a batch of components sent by a first supplier arrives at a certain inspection facility. Two days a week, a batch also arrives from a second supplier. Eighty percent of all supplier 1’s batches pass inspection, and 90% of supplier 2’s do likewise. What is the probability that, on a randomly selected day, two batches pass inspection? We will answer this assuming that on days when two batches are tested, whether the first batch passes is independent of whether the second batch does so. Figure 2.13 displays the relevant information.

Figure 2.13 Tree diagram for Example 2.34

P (two pass) = P (two received ∩ both pass)

= P (both pass| two received) · P (two received)

= [(.8)(.9)](.4) = .288

* iff is an abbreviation for “if and only if.”

Independence of More Than Two Events

The notion of independence of two events can be extended to collections of more than two events. Although it is possible to extend the definition for two independent events by working in terms of conditional and unconditional probabilities, it is more direct and less cumbersome to proceed along the lines of the last proposition.

DEFINITION

Events A_1, …, An are mutually independent if for every k (k = 2, 3, …, n ) and every subset of indices i_1, i_2, …, ik ,

To paraphrase the definition, the events are mutually independent if the f any subset of the n events is equal to the product of the individual probabilities. In using the multiplication property for more than two independent events, it is legitimate to replace one or more of the s by their complements (e.g., if A_1, A_2, and A_3 are independent events, so are , , and ). As was the case with two events, we frequently specify at the outset of a problem the independence of certain events. The probability of an intersection can then be calculated via multiplication.

Example 2.35

The article “Reliability Evaluation of Solar Photovoltaic Arrays” (Solar Energy , 2002: 129-141) presents various configurations of solar photovoltaic arrays consisting of crystalline silicon solar cells. Consider first the system illustrated in Figure 2.14(a).

Figure 2.14 System configurations for Example 2.35: (a) series-parallel; (b) total-cross-tied

There are two subsystems connected in parallel, each one containing three cells. In order for the system to function, at least one of the two parallel subsystems must work. Within each subsystem, the three cells are connected in series, so a subsystem will work only if all cells in the subsystem work. Consider a particular lifetime value t_0, and supose we want to determine the probability that the system lifetime exceeds t_0. Let Ai denote the event that the lifetime of cell i exceeds t_0 (i = 1, 2,…, 6). We assume that the Ai s are independent events (whether any particular cell lasts more than t_0 hours has no bearing on whether or not any other cell does) and that P (Ai ) = .9 for every i since the cells are identical. Then

P (system lifetime exceeds t_0) = P [(A_1 ∩ A_2 ∩ A_3) ∪ (A_4 ∩ A_5 ∩ A_6)]

= P (A_1 ∩ A_2 ∩ A_3) + P (A_4 ∩ A_5 ∩ A_6)

- P [(A_1 ∩ A_2 ∩ A_3) ∩ (A_4 ∩ A_5 ∩ A_6)]

= (.9)(.9)(.9) + (.9)(.9)(.9) − (.9)(.9)(.9)(.9)(.9)(.9) = .927

Alternatively,

P

73. If A and B are independent events, show that A ′ and B are also independent. [Hint: First establish a relationship between P (A ′ ∩ B ), P (B ), and P (A ∩ B ).]

74. Suppose that the proportions of blood phenotypes in a particular population are as follows:

Assuming that the phenotypes of two randomly selected individuals are independent of one another, what is the probability that both phenotypes are O? What is the probability that the phenotypes of two randomly selected individuals match?

75. One of the assumptions underlying the theory of control charting (see Chapter 16) is that successive plotted points are independent of one another. Each plotted point can signal either that a manufacturing process is operating correctly or that there is some sort of malfunction. Even when a process is running correctly, there is a small probability that a particular point will signal a problem with the process. Suppose that this probability is .05. What is the probability that at least one of 10 successive points indicates a problem when in fact the process is operating correctly? Answer this question for 25 successive points.

76. The probability that a grader will make a marking error on any particular question of a multiple-choice exam is .1. If there are ten questions and questions are marked independently, what is the probability that no errors are made? That at least one error is made? If there are n questions and the probability of a marking error is p rather than .1, give expressions for these two probabilities.

77. An aircraft seam requires 25 rivets. The seam will have to be reworked if any of these rivets is defective. Suppose rivets are defective independently of one another, each with the same probability.

a. If 20% of all seams need reworking, what is the probability that a rivet is defective?

b. How small should the probability of a defective rivet be to ensure that only 10% of all seams need reworking?

78. A boiler has five identical relief valves. The probability that any particular valve will open on demand is .95. Assuming independent operation of the valves, calculate P (at least one valve opens) and P (at least one valve fails to open).

79. Two pumps connected in parallel fail independeree inspected fails)

c. P (exactly one of the next three inspected passes)

d. P (at most one of the next three vehicles inspected passes)

e. Given that at least one of the next three vehicles passes inspection, what is the probability that all three pass (a conditional probability)?

85. A quality control inspector is inspecting newly produced items for faults. The inspector searches an item for faults in a series of independent fixations, each of a fixed duration. Given that a flaw is actually present, let p denote the probability that the flaw is detected during any one fixation (this model is discussed in “Human Performance in Sampling Inspection,” Human Factors , 1979: 99-105).

a. Assuming that an item has a flaw, what is the probability that it is detected by the end of the second fixation (once a flaw has been detected, the sequence of fixations terminates)?

b. Give an expression for the probability that a flaw will be detected by the end of the nth fixation.

c. If when a flaw has not been detected in three fixations, the item is passed, what is the probability that a flawed item will pass inspection?

d. Suppose 10% of all items contain a flaw [P (randomly chosen item is flawed) = .1]. With the assumption of part (c), what is the probability that a randomly chosen item will pass inspection (it will automatically pass if it is not flawed, but could also pass if it is flawed)?

e. Given that an item has passed inspection (no flaws in three fixations), what is the probability that it is actually flawed? Calculate for p = .5.

86. a. A lumber company has just taken delivery on a lot of 10,000 2 × 4 boards. Suppose that 20% of these boards (2,000) are actually too green to be used in first-quality construction. Two boards are selected at random, one after the other. Let A = {the first board is green} and B = {the second board is green}. Compute P(A ), P (B ), and P (A ∩ B ) (a tree diagram might help). Are A and B independent?

b. With A and B independent and P (A ) = P(B ) = .2, what is P (A ∩ B )? How much difference is there between this answer and P (A ∩ B ) in part (a)? For purposes of calculating P (A ∩ B) , can we assume that A and B of part (a) are independent to obtain essentially the correct probability>

c. Suppose the lot consists of ten boards, of which two are green. Does the assumption of independence now yield approximately the correct answer for P (A ∩ B )? What is the critical difference between the situation here and that of part (a)? When do you think that an independence assumption would be valid in obtaining an approximately correct answer to P (A ∩ B )?

87. Refer to the assumptions stated in Exercise 80 and answer the question posed there for the system in the accompanying picture. How would the probability change if this were a subsystem connected in parallel to the subsystem pictured in Figure 2.14(a)?

88. Professor Stan der Deviation can take one of two routes on his way home from work. On the first route, there are four railroad crossings. The probability that he will be stopped by a train at any particular one of the crossings is .1, and trains operate independently at the four crossings. The other route is longer but there are only two crossings, independent of one another, with the same stoppage probability for each as on the first route. On a particular day, Professor Deviation has a meeting scheduled at home for a certain time. Whichever route he takes, he calculates that he will be late if he is stopped by trains at at least half the crossings encountered.

a. Which route should he take to minimize the probability of being late to the meeting?

b. If he tosses a fair coin to decide on a route and he is late, what is the probability that he took the four-crossing route?

89. Suppose identical tags are placed on both the left ear and the right ear of a fox. The fox is then let loose for a period of time. Consider the two events C_1 = {left ear tag is lost} and C_2 = {right ear tag is lost}. Let π = P (C_1) = P (C_2), and assume C_1 and C_2 are independent events. Derive an expression (involving π) for the probability that exactly one tag is lost given that at most one is lost (“Ear Tag Loss in Red Foxes,” J. Wildlife Mgmt. , 1976: 164-167). [Hint: Draw a tree diagram in which the two initial branches refer to whether the left ear tag was lost.]

SUPPLEMENTARY EXERCISES (90–114)

90. A small manufacturing company will start operating a night shift. There are 20 machinisy one of six possible categories from which questions will be asked. Suppose P (contestant requests category i ) = 767 and successive contestants choose their categories independently of one another. If there are three contestants on each show and all three contestants on a particular show select different categories, what is the probability that exactly one has selected category 1?

99. Fasteners used in aircraft manufacturing are slightly crimped so that they lock enough to avoid loosening during vibration. Suppose that 95% of all fasteners pass an initial inspection. Of the 5% that fail, 20% are so seriously defective that they must be scrapped. The remaining fasteners are sent to a recrimping operation, where 40% cannot be salvaged and are discarded. The other 60% of these fasteners are corrected by the recrimping process and subsequently pass inspection.

a. What is the probability that a randomly selected incoming fastener will pass inspection either initially or after recrimping?

b. Given that a fastener passed inspection, what is the probability that it passed the initial inspection and did not need recrimping?

100. One percent of all individuals in a certain population are carriers of a particular disease. A diagnostic test for this disease has a 90% detection rate for carriers and a 5% detection rate for noncarriers. Suppose the test is applied independently to two different blood samples from the same randomly selected individual.

a. What is the probability that both tests yield the same result?

b. If both tests are positive, what is the probability that the selected individual is a carrier?

101. A system consists of two components. The probability that the second component functions in a satisfactory manner during its design life is .9, the probability that at least one of the two components does so is .96, and the probability that both components do so is .75. Given that the first component functions in a satisfactory manner throughout its design life, what is the probability that the second one does also?

102. A certain company sends 40% of its overnight mail parcels via express mail service E_1 Of these parcels, 2% arrive after the guaranteed delivery time (denote the event “late delivery” by L ). If a record of an overnight mailing is randomly selected from the company’s file, what is the probability that the parcel went via E_1 and was late?

103. Refer to Exercise 102. Suppose that 50% of the overnight parcels are sent via express mail service E_2 and the remaining 10% are sent via E_3 Of those sent via E_2, only 1% arrive late, whereas 5% of the parcels handled by E_3 m">
a. Show that P (granite| R_1 < R_2 < R_3) > P (basalt| R_1 < R_2 < R_3). If measurements yielded R_1 < R_2 < R_3, would you classify the rock as granite or basalt?

b. If measurements yielded R_1 < R_3 < R_2, how would you classify the rock? Answer the same question for R_3 < R_1 < R_2.

c. Using the classification rules indicated in parts (a) and (b), when selecting a rock from this region, what is the probability of an erroneous classification? [Hint: Either G could be classified as B or B as G , and P (B ) and P (G ) are known.]

d. If P(granite) = p rather than .25, are there values of p (other than 1) for which one would always classify a rock as granite?

107. A subject is allowed a sequence of glimpses to detect a target. Let Gi = {the target is detected on the ith glimpse}, with pi = P (G ). Suppose the Gi s are independent events and write an expression for the probability that the target has been detected by the end of the nth glimpse. [Note: This model is discussed in “Predicting Aircraft Detectability,” Human Factors , 1979: 277-291.]

108. In a Little League baseball game, team A’s pitcher throws a strike 50% of the time and a ball 50% of the time, successive pitches are independent of one another, and the pitcher never hits a batter. Knowing this, team B’s manager has instructed the first batter not to swing at anything. Calculate the probability that

a. The batter walks on the fourth pitch

b. The batter walks on the sixth pitch (so two of the first five must be strikes), using a counting argument or constructing a tree diagram

c. The batter walks

d. The first batter up scores while no one is out (assuming that each batter pursues a no-swing strategy)

109. Four engineers, A, B, C, and D, have been scheduled for job interviews at 10 A.M . on Friday, January 13, at Random Sampling, Inc. The personnel manager has scheduled the four for interview rooms 1, 2, 3, and 4, respectively. However, the manager/em>3 are independent, and that A_2 and A_3 are also independent (this is pairwise independence). However, show that P (A_1 ∩ A_2 ∩ A_3) ≠ P (A_1) · P (A_2) · P (A_3), so the three events are not mutually independent.

114. Show that if A_1, A_2, and A_3 are independent events, then P (A_1 | A_2 ∩ A_3) = P (A_1).

Bibliography

Durrett, Richard, The Essentials of Probability , Duxbury Press, Belmont, CA, 1993. A concise presentation at a slightly higher level than this text.

Mosteller, Frederick, Robert Rourke, and George Thomas, Probability with Statistical Applications (2nd ed.), Addison-Wesley, Reading, MA, 1970. A very good precalculus introduction to probability, with many entertaining examples; especially good on counting rules and their application.

Olkin, Ingram, Cyrus Derman, and Leon Gleser, Probability Models and Application (2nd ed.), Macmillan, New York, 1994. A comprehensive introduction to probability, written at a slightly higher mathematical level than this text but containing many good examples.

Ross, Sheldon, A First Course in Probability (6th ed.), Macmillan, New York, 2002. Rather tightly written and more mathematically sophisticated than this text but contains a wealth of interesting examples and exercises.

Winkler, Robert, Introduction to Bayesian Inference and Decision , Holt, Rinehart & Winston, New York, 1972. A very good introduction to subjective probability.

3 Discrete Random Variables and Probability Distributions

INTRODUCTION

Whether an experiment yields qualitative or quantitative outcomes, methods of statistical analysis require that we focus on certain numerical aspects of the data (such as a sample proportion x/n, mean , or standard deviation s). The concept of a random variable allows us to pass from the experimental outcomes themselves to a numerical function of the outcomes. There are two fundamentally different types of random variables—discrete random variables and continuous random variables. In this chapter, we examine the basic properties and discuss the most important examples of discrete variables.

2. No possible value of the variable has positive probability, that is, P (X = c ) = 0 for any possible value c.

Although any interval on the number line contains an infinite number of numbers, it can be shown that there is no way to create an infinite listing of all these values—there are just too many of them. The second condition describing a continuous random variable is perhaps counterintuitive, since it would seem to imply a total probability of zero for all possible values. But we shall see in Chapter 4 that intervals of values have positive probability; the probability of an interval will decrease to zero as the width of the interval shrinks to zero.

Example 3.6

All random variables in Examples 3.1–3.4 are discrete. As another example, suppose we select married couples at random and do a blood test on each person until we find a husband and wife who both have the same Rh factor. With X = the number of blood tests to be performed, possible values of X are D = {2, 4, 6, 8, …}. Since the possible values have been listed in sequence, X is a discrete rv.

To study basic properties of discrete rv’s, only the tools of discrete mathematics—summation and differences—are required. The study of continuous variables requires the continuous mathematics of the calculus—integrals and derivatives.

EXERCISES Section 3.1 (1 –10)

1. A concrete beam may fail either by shear (S ) or flexure (F ). Suppose that three failed beams are randomly selected and the type of failure is determined for each one. Let X = the number of beams among the three selected that failed by shear. List each outcome in the sample space along with the associated value of X.

2. Give three examples of Bernoulli rv’s (other than those in the text).

3. Using the experiment in Example 3.3, define two more random variables and list the possible values of each.

4. Let X = the number of nonzero digits in a randomly selected zip code. What are the possible values of X ? Give three possible outcomes and their associated X values.

5. If the sample space S is an infinite set, does this necessarily imply that any rv X defined from S will have an infinite set of possible values? If yes, say why. If no,g>

We can now use elementary probability properties to calculate other probabilities of interest. For example, the probability that at most 2 pumps are in use is

P (X ≤ 2) = P (X = 0 or 1 or 2) = p (0) + p (1) + p (2) = .05 + .10 + .15 = .30

Since the event at least 3 pumps are in use is complementary to at most 2 pumps are in use,

P (X ≥ 3) = 1 –P (X ≤ 2) = 1 –.30 = .70

which can, of course, also be obtained by adding together probabilities for the values, 3, 4, 5, and 6. The probability that between 2 and 5 pumps inclusive are in use is

P (2 ≤ X ≤ 5) = P (X = 2, 3, 4, or 5) = .15 + .25 + .20 +.15 = .75

whereas the probability that the number of pumps in use is strictly between 2 and 5 is

P (2 < X < 5) = P (X = 3 or 4) = .25 + .20 = .45

DEFINITION

The probability distribution or probability mass function (pmf) of a discrete rv is defined for every number x by p (x ) = P (X = x ) = P (all s ∈ S: X(s) = x).

In words, for every possible value x of the random variable, the pmf specifies the probability of observing that value when the experiment is performed. The conditions p (x ) ≥ 0 and ∑all possible xp (x) = 1 are required of any pmf.

The pmf of X in the previous example was simply given in the problem description. We now consider several examples in which various probability properties are exploited to obtain the desired distribution.

Example 3.8

Six lots of components are ready to be shipped by a certain supplier. The number of defective components in each lot is as follows:

One of these lots is to be randomly selected for shipment to a particular customer. Let X be the number of defectives in the selected lot. The three possible X values are 0, 1, and 2. Of the six equally likely simple events, three result in X = 0, one in X = 1, and the other two in X

That is, a probability of .500 is distributed to the X value 0, a probability of .167 is placed on the X value 1, and the remaining probability, .333, is associated with the X value 2. The values of X along with their probabilities collectively specify the pmf. If this experiment were repeated over and over again, in the long run X = 0 would occur one-half of the time, X = 1 one-sixth of the time, and X = 2 one-third of the time.

Example 3.9

Consider whether the next person buying a computer at a university book store buys a laptop or a desktop model. Let

If 20% of all purchasers during that week select a laptop, the pmf for X is

p (0) = P (X = 0) = P (next customer purchases a desktop model) = .8

p (1) = P (X = 1) = P (next customer purchases a laptop model) = .2

p (x) = P (X = x ) = 0 for x ≠ 0 or 1

An equivalent description is

Figure 3.2 is a picture of this pmf, called a line graph. X is, of course, a Bernoulli rv and p (x ) is a Bernoulli pmf.

Figure 3.2 The line graph for the pmf in Example 3.9

Example 3.10

Consider a group of five potential blood donors—a, b, c, d, and e—of whom only a and b have type O+ blood. Five blood samples, one from each individual, will be typed in random order until an O+ individual is identified. Let the rv Y = the number of typings necessary to identify an O+ individual. Then the pmf of Y is

p (1) = P (Y = 1) = P (a or b typed firs height="0em">

Example 3.11

Consider selecting at random a student who is among the 15,000 registered for the current term at Mega University. Let X = the number of courses for which the selected student is registered, and suppose that X has the following pmf:

One way to view this situation is to think of the population as consisting of 15,000 individuals, each having his or her own X value; the proportion with each X value is given by p (x ). An alternative viewpoint is to forget about the students and think of the population itself as consisting of the X values: There are some 1s in the population, some 2s, …, and finally some 7s. The population then consists of the numbers 1, 2, …, 7 (so is discrete), and p (x ) gives a model for the distribution of population values.

Once we have such a population model, we will use it to compute values of population characteristics (e.g., the mean μ) and make inferences about such characteristics.

A Parameter of a Probability Distribution

In Example 3.9, we had p (0) = .8 and p (1) = .2 because 20% of all purchasers selected a laptop computer. At another bookstore, it may be the case that p (0) = .9 and p (1) = .1. More generally, the pmf of any Bernoulli rv can be expressed in the form p (1) = α and p (0) = 1 –α, where 0 < α < 1. Because the pmf depends on the particular value of α, we often write p (x ; α) rather than just p (x ):

Then each choice of α in Expression (3.1) yields a different pmf.

DEFINITION

Suppose p (x ) depends on a quantity that can be assigned any one of a number of possible values, with each different value determining a different probability distribution. Such a quantity is called a parameter of the distribution. The collection of all probability distributions for different values of the parameter is called a family of probability distributions.

The quantity α in Expression (3.1) is a parameter. Each different number α between 0 and 1 determines a different meign="center">

For X a discrete rv, the graph of F (x ) will have a jump at every possible value of X and will be flat between possible values. Such a graph is called a step function.

Example 3.14

In Example 3.12, any positive integer was a possible X value, and the pmf was

For any positive integer x,

To evaluate this sum, we use the fact that the partial sum of a geometric series is

Using this in Equation (3.4), with a = 1 –p and k = x –1, gives

Since F is constant in between positive integers,

where [x ] is the largest integer ≤ x (e.g., [2.7] = 2). Thus if p = .51 as in the birth example, then the probability of having to examine at most five births to see the first boy is F (5) = 1 –(.49)5 = 1 –.0282 = .9718, whereas F (10) ≈ 1.0000. This cdf is graphed in Figure 3.6.

Figure 3.6 A graph of F (x ) for Example 3.14

In examples thus far, the cdf has been derived from the pmf. This process can be reversed to obtain the pmf from the cdf whenever the latter function is available. For example, consider again the rv of Example 3.7 (the number of pumps in use at a gas station); possible X vonsumer organization that evaluates new automobiles customarily reports the number of major defects in each car examined. Let X denote the number of major defects in a randomly selected car of a certain type. The cdf of X is as follows:

Calculate the following probabilities directly from the cdf:

a. p (2), that is, P (X = 2)

b. P (X > 3)

c. P (2 ≤ X ≤ 5)

d. P (2 < X < 5)

24. An insurance company offers its policyholders a number of different premium payment options. For a randomly selected policyholder, let X = the number of months between successive payments. The cdf of X is as follows:

a. What is the pmf of X ?

b. Using just the cdf, compute P (3 ≤ X ≤ 6) and P (4 ≤ X ).

25. In Example 3.12, let Y = the number of girls born before the experiment terminates. With p = P (B ) and 1 –p = P (G ), what is the pmf of Y ? [Hint: First list the possible values of Y, starting with the smallest, and proceed until you see a general formula.]

26. Alvie Singer lives at 0 in the accompanying diagram and has four friends who live at A, B, C, and D. One day Alvie decides to go visiting, so he tosses a fair coin twice to decide which of the four to visit. Once at a friend’s house, he will either return home or else proceed to one of the two adjacent houses (such as 0, A, or C when at B ), with each of the three possibilities having probability . In this way, Alvie continues to visit friends until he returns home.

a. Let X = the number of times that Alvie visits a friend. Derive the pmf of X.

Y?

c. Suppose that female friends live at A and C and male friends at B and D. If Z = the number of visits to female friends, what is the pmf of Z ?

27. After all students have left the classroom, a statistics professor notices that four copies of the text were left under desks. At the beginning of the next lecture, the professor distributes the four books in a completely random fashion to each of the four students (1, 2, 3, and 4) who claim to have left books. One possible outcome is that 1 receives 2’s book, 2 receives 4’s book, 3 receives his or her own book, and 4 receives 1’s book. This outcome can be abbreviated as (2, 4, 3, 1).

a. List the other 23 possible outcomes.

b. Let X denote the number of students who receive their own book. Determine the pmf of X.

28. Show that the cdf F (x ) is a nondecreasing function; that is, x_1 < x_2 implies that F (x_1) ≤ F (x_2). Under what condition will F (x_1) = F (x_2)?

3.3 Expected Values

Consider a university having 15,000 students and let X = the number of courses for which a randomly selected student is registered. The pmf of X follows. Since p (1) = .01, we know that (.01) • (15,000) = 150 of the students are registered for one course, and similarly for the other x values.

The average number of courses per student, or the average value of X in the population, results from computing the total number of courses taken by all students and dividing by the total number of students. Since each of 150 students is taking one course, these 150 contribute 150 courses to the total. Similarly, 450 students contribute 2(450) courses, and so on. The population average value of X is then

Since 150/15,000 = .01 = p (1), 450/15,000 = .03 = p (2), and so on, an alternative expression for (3.7) is

Expression (3.8) shows that to compute the population average value of X, we need only the possible values of X alongs 1, …, 7, where the weights are the probabilities of those values.

The Expected Value of X

DEFINITION

Let X be a discrete rv with set of possible values D and pmf p (x ). The expected value or mean value of X, denoted by E(X) or is

When it is clear to which X the expected value refers, μ rather than μx is often used.

Example 3.16

For the pmf in (3.6),

μ = 1 · p (1) + 2 · p (2) + … + 7 · p (7)

= (1)(.01) + 2(.03) + … + (7)(.02)

= .01 + .06 + .39 + 1.00 + 1.95 + 1.02 + .14 = 4.57

If we think of the population as consisting of the X values 1, 2, …, 7, then μ = 4.57 is the population mean. In the sequel, we will often refer to μ as the population mean rather than the mean of X in the population.

In Example 3.16, the expected value μ was 4.57, which is not a possible value of X. The word expected should be interpreted with caution because one would not expect to see an X value of 4.57 when a single student is selected.

Example 3.17

Just after birth, each newborn child is rated on a scale called the Apgar scale. The possible ratings are 0, 1, …, 10, with the child’s rating determined by color, muscle tone, respiratory effort, heartbeat, and reflex irritability (the best possible score is 10). Let X be the Apgar score of a randomly selected child born at a certain hospital during the next year, and suppose that the pmf of X is

Then the mean value of X is

E (X ) = μ = 0(.002) + 1(.001) + 2(.002) + … + 8(.25) + 9(.12) + 10(.01)

= 7.15

(Or, using alternative notation, μαX+b = a · μx + b .)

To paraphrase, the expected value of a linear function equals the linear function evaluated at the expected value E(X). Since h (X ) in Example 3.23 is linear and E (X ) = 2, E [h (X )] = 800(2) –900 = $700, as before.

Proof

Two special cases of the proposition yield two important rules of expected value.

Multiplication of X by a constant a changes the unit of measurement (from dollars to cents, where a = 100, inches to cm, where a = 2.54, etc.). Rule 1 says that the expected value in the new units equals the expected value in the old units multiplied by the conversion factor a. Similarly, if a constant b is added to each possible value of X, then the expected value will be shifted by that same constant amount.

The Variance of X

The expected value of X describes where the probability distribution is centered. Using the physical analogy of placing point mass p (x ) at the value x on a one-dimensional axis, if the axis were then supported by a fulcrum placed at μ, there would be no tendency for the axis to tilt. This is illustrated for two different distributions in Figure 3.7.

Figure 3.7 Two different probability distributions with μ = 4

Although both distributions pictured in Figure 3.7 have the same center μ , the distribution of Figure 3.7(b) has greater spread or variability or dispersion than does that of Figure 3.7(a). We will use the variance of X to assess the amount of variability in (the distribution of) X, just as s 2 was used in Chapter 1 to measure variability in a sample.

DEFINITIhe number of cylinders X on the next car to be tuned at a certain facility was given in Example 3.24 as p (4) = .5, p (6) = .3, and p (8) = .2, from which μ = 5.4 and

E (X 2) = (4^2)(.5) + (6^2)(.3) + (8^2)(.2) = 31.6

Thus σ 2 = 31.6 –(5.4)2 = 2.44 as in Example 3.24.

Proof of the Shortcut Formula

Expand (x –μ )2 in the definition of σ 2 to obtain x 2 –2μx + μ 2, and then carry ∑ through to each of the three terms:

Rules of Variance

The variance of h (X ) is the expected value of the squared difference between h (X ) and its expected value:

When h (X ) = aX + b, a linear function,

h (x ) –E [h (X )] = ax + b –(aμ + b ) = a (x –μ )

Substituting this into (3.13) gives a simple relationship between V [h (X )] and V (X ):

PROPOSITION

V (aX + b ) = = a 2 • and σaX+b = |a | · σx

In particular,

The absolute value is necessary because a might be negative, yet a standard deviation cannot be. Usually multiplication by a corresponds to a change in the unit of measurement (e.g., kg to lb or dollars to euros). According to the first relation in (3.14), the sd in the new unit is the original sd multiplied by the conversion factor. The second relation says that adding or subtracting a constant does not impact variability; it j Roman">Example 3.26

In the computer sales problem of Example 3.23, E (X ) = 2 and

E (X 2) = (0)2(.1) + (1)2(.2) + (2)2(.3) + (3)2(.4) = 5

so V (X ) = 5 –(2)2 = 1. The profit function h (X ) = 800X –900 then has variance (800)2 · V (X ) = (640,000)(1) = 640,000 and standard deviation 800.

EXERCISES Section 3.3 (29–45)

29. The pmf for X = the number of major defects on a randomly selected appliance of a certain type is

Compute the following:

a. E (X )

b. V (X ) directly from the definition

c. The standard deviation of X

d. V (X ) using the shortcut formula

30. An individual who has automobile insurance from a certain company is randomly selected. Let Y be the number of moving violations for which the individual was cited during the last 3 years. The pmf of Y is

a. Compute E (Y ).

b. Suppose an individual with Y violations incurs a surcharge of $100 Y 2. Calculate the expected amount of the surcharge.

31. Refer to Exercise 12 and calculate V (Y ) and σγ. Then determine the probability that Y is within 1 standard deviation of its mean value.

32. An appliance dealer sells three different models of upright freezem> has pmf

a. Compute E (X ), E (X 2), and V (X ).

b. If the price of a freezer having capacity X cubic feet is 25X – 8.5, what is the expected price paid by the next customer to buy a freezer?

c. What is the variance of the price 25X – 8.5 paid by the next customer?

d. Suppose that although the rated capacity of a freezer is X, the actual capacity is h (X ) = X –.01X 2. What is the expected actual capacity of the freezer purchased by the next customer?

33. Let X be a Bernoulli rv with pmf as in Example 3.18.

a. Compute E (X 2).

b. Show that V (X ) = p (1 –p ).

c. Compute E (X 79).

34. Suppose that the number of plants of a particular type found in a rectangular region (called a quadrat by ecologists) in a certain geographic area is an rv X with pmf

Is E (X ) finite? Justify your answer (this is another distribution that statisticians would call heavy-tailed).

35. A small market orders copies of a certain magazine for its magazine rack each week. Let X = demand for the magazine, with pmf

Suppose the store owner actually pays $1.00 for each copy of the magazine and the price to customers is $2.00. If magazines left at the end of the week have no salvage value, is it better to order three or four copies of the magazine? [Hint: For both three and four copies ordered, express net revenue as a function of demand X, and then compute the expected revenue.]

36. Let X be the damage incurred (in $) in a certain type of accident during a given year. Possible X values are 0, 100000, with probabilities .8, .1, .08, and .02, respectively. A particular company offers a $500 deductible policy. If the company wishes its expected profit to be $100, what premium amount should it charge?

37. The n candidates for a job have been ranked 1, 2, 3, …, n. Let X = the rank of a randomly selected candidate, so that X has pmf

(this is called the discrete uniform distribution ). Compute E (X ) and V (X ) using the shortcut formula. [Hint: The sum of the first n positive integers is n (n + 1)/2, whereas the sum of their squares is n (n + 1)(2n + 1)/6.]

38. Let X = the outcome when a fair die is rolled once. If before the die is rolled you are offered either (1/3.5) dollars or h (X ) = 1/X dollars, would you accept the guaranteed amount or would you gamble? [Note: It is not generally true that 1/E (X ) = E (1/X ).]

39. A chemical supply company currently has in stock 100 lb of a certain chemical, which it sells to customers in 5-lb lots. Let X = the number of lots ordered by a randomly chosen customer, and suppose that X has pmf

Compute E (X ) and V (X ). Then compute the expected number of pounds left after the next customer’s order is shipped and the variance of the number of pounds left. [Hint: The number of pounds left is a linear function of X .]

40. a. Draw a line graph of the pmf of X in Exercise 35. Then determine the pmf of –X and draw its line graph. From these two pictures, what can you say about V (X ) and V (–X )?

b. Use the proposition involving V (aX + b ) to establish a general relationship between V (X ) and V (–X ).

41. Use the definition in Expression (3.13) to prove that V (aX + b ) = a 2 • [Hint: With h (X ) = aX + b , E [h (X )] = aμ + b where μ = E (X ).]

42. Suppose E (X ) = 5 and E [X (X –1)] = 27.5. What is

up>2)? [Hint: E [X (X –1)] = E [X 2 –X ] = E (X 2) –E (X )]?

b. V (X )?

c. The general relationship among the quantities E (X ), E [X (X –1)], and V (X )?

43. Write a general rule for E (X –c ) where c is a constant. What happens when you let c = μ, the expected value of X?

44. A result called Chebyshev’s inequality states that for any probability distribution of an rv X and any number k that is at least 1, P (|X –μ | ≥ kσ ) ≤ 1/k 2. In words, the probability that the value of X lies at least k standard deviations from its mean is at most 1/k 2.

a. What is the value of the upper bound for k = 2? k = 3? k = 4? k = 5? k = 10?

b. Compute μ and σ for the distribution of Exercise 13. Then evaluate P (|X –μ | ≥ kσ ) for the values of k given in part (a). What does this suggest about the upper bound relative to the corresponding probability?

c. Let X have possible values –1, 0, and 1, with probabilities , and respectively. What is P (|X –μ | ≥ 3σ ), and how does it compare to the corresponding bound?

d. Give a distribution for which P (|X –μ | ≥ 5σ ) = .04.

45. If a ≤ X ≤ b , show that a ≥ E (X ) ≤ b .

3.4 The Binomial Probability Distribution

There are many experiments that conform either exactly or approximately to the following list of requirements:

1. The experiment consists of a sequence of n smaller experiments called trials, where n is fixed in advance of the experiment.

2. Each trial can result in one of the same two possibleoutcomes (dichotomous trials), which we denote by success (S ) and failure (F ) .

3. The trials are independent, so that the outcome on any particular trial does not influence the outcome on any other trial.

4. The probability of success is constant from trial to trial; we denote this probability by p.

DEFINITION

An experiment for which Conditions 1–4 are satisfied is called a binomial experiment.

Example 3.27

The same coin is tossed successively and independently n times. We arbitrarily use S to denote the outcome H (heads) and F to denote the outcome T (tails). Then this experiment satisfies Conditions 1–4. Tossing a thumbtack n times, with S = point up and F = point down, also results in a binomial experiment.

Many experiments involve a sequence of independent trials for which there are more than two possible outcomes on any one trial. A binomial experiment can then be created by dividing the possible outcomes into two groups.

Example 3.28

The color of pea seeds is determined by a single genetic locus. If the two alleles at this locus are AA or Aa (the genotype), then the pea will be yellow (the phenotype), and if the allele is aa, the pea will be green. Suppose we pair off 20 Aa seeds and cross the two seeds in each of the ten pairs to obtain ten new genotypes. Call each new genotype a success S if it is aa and a failure otherwise. Then with this identification of S and F, the experiment is binomial with n = 10 and p = P (aa genotype). If each member of the pair is equally likely to contribute a or A, then p = P (a) · P (a) =

Example 3.29

Suppose a certain city has 50 licensed restaurants, of which 15 currently have at least one serious health code violation and the other 35 have no serious violations. There are five inspectors, each of whom will inspect one restaurant during the coming week. The name of each restaurant is written on a different slip of paper, and after the slips are thoroughly mixed, each inspector in turn draws one of the slips without replacement. Label the ith trial as a success if the ith restaurant selected (i = 1, …, 5) has no serious violations. Then

and

P (S on second trial) = P (SS ) + P (FS )

= P (second S | first S ) P (first S ) + P (second S | first F ) P (first F )

Similarly, it can be shown that P (S on ith trial) = .70 for i = 3, 4, 5. However,

whereas

The experiment is not binomial because the trials are not independent. In general, if sampling is without replacement, the experiment will not yield independent trials. If each slip had been replaced after being drawn, then trials would have been independent, but this might have resulted in the same restaurant being inspected by more than one inspector.

Example 3.30

A certain state has 500,000 licensed drivers, of whom 400,000 are insured. A sample of 10 drivers is chosen without replacement. The ith trial is labeled S if the ith driver chosen is insured. Although this situation would seem identical to that of Example 3.29, the important difference is that the size of the population being sampled is very large relative to the sample size. In this case

and

These calculations suggest that although the trials are not exactly independent, the conditional probabilities differ so slightly from one another that for practical purposes the trials can be regarded as independent with constant P(S) = .8. Thus, to a very good approximation, the experiment is binomial with n = 10 and p = .8.

We will use the following rule of thumb
Consider sampling without replacement from a dichotomous population of size N. If the sample size (number of trials) n is at most 5% of the population size, the experiment can be analyzed as though it were exactly a binomial experiment.

By “analyzed,” we mean that probabilities based on the binomial experiment assumptions will be quite close to the actual “without-replacement” probabilities, which are typically more difficult to calculate. In Example 3.29, n /N = 5/50 = .1 > .05, so the binomial experiment is not a good approximation, but in Example 3.30, n /N = 10/500,000 < .05.

The Binomial Random Variable and Distribution

In most binomial experiments, it is the total number of S ’s, rather than knowledge of exactly which trials yielded S ’s, that is of interest.

DEFINITION

The binomial random variable X associated with a binomial experiment consisting of n trials is defined as

X = the number of S ’s among the n trials

Suppose, for example, that n = 3. Then there are eight possible outcomes for the experiment:

SSS SSF SFS SFF FSS FSF FFS FFF

From the definition of X, X (SSF ) = 2, X (SFF ) = 1, and so on. Possible values for X in an n -trial experiment are x = 0, 1, 2, …, n. We will often write X ~ Bin(n, p ) to indicate that X is a binomial rv based on n trials with success probability p.

NOTATION

Because the pmf of a binomial rv X depends on the two parameters n and p, we denote the pmf by b (x; n, p ).

Consider first the case n = 4 for which each outcome, its probability, and corresponding x value are listed in Table 3.1. For example,

P (SSFS ) = P (S ) · P (S ) · P (F ) · P (S ) (independent trials)

Table 3.1 Outcomes and Probabilities for a Binomial Experiment with four Trials

In this special case, we wish b (x; 4, p ) for x = 0, 1, 2, 3, and 4. For b (3; 4, p ), let’s identify which of the 16 outcomes yield an x value of 3 and sum the probabilities associated with each such outcome:

b (3; 4, p ) = P (FSSS ) + P (SFSS ) + P (SSFS ) + P (SSSF ) = 4p 3(1 –p )

There are four outcomes with x = 3 and each has probability p 3(1 –p ) (the order of S ’s and F ’s is not important, but only the number of S ’s), so

Similarly, b (2; 4, p ) = 6p 2(1 –p )2, which is also the product of the number of outcomes with X = 2 and the probability of any such outcome.

In general,

Since the ordering of S ’s and F ’s is not important, the second factor in the previous equation is px (1 –p )n–x (e.g., the first x trials resulting in S and the last n –x resulting in F ). The first factor is the number of ways of choosing x of the n trials to be S ’s—that is, the number of combinations of size x that can be constructed from n distinct objects (trials here).

THEOREM

Example 3.31

Each of six randomly selected cola drinkers is given a glass containing cola S and one containing cola F. The glasses are identical in appearance except for a code on the bottom to identify the cola. Suppose there is actually no tendency among cola drinkers to prefer one cola to the other. Then p = P (a selected individual prefers S ) = .5, so with X = the number among the six who prefer S, X ~ Bin(6, .5).

Thus

The prob9 –.996 = .003.

3. The probability that at least 8 fail is

P (X ≥ 8) = 1 –P (X ≤ 7) = 1 –B (7; 15, .2)

= 1 –.996 = .004

4. Finally, the probability that between 4 and 7, inclusive, fail is

P (4 ≤ X ≤ 7) = P (X = 4, 5, 6, or 7) = P (X ≤ 7) –P (X ≤ 3)

= B (7; 15, .2) –B (3; 15, .2) = .996 –.648 = .348

Notice that this latter probability is the difference between entries in the x = 7 and x = 3 rows, not the x = 7 and x = 4 rows.

Example 3.33

An electronics manufacturer claims that at most 10% of its power supply units need service during the warranty period. To investigate this claim, technicians at a testing laboratory purchase 20 units and subject each one to accelerated testing to simulate use during the warranty period. Let p denote the probability that a power supply unit needs repair during the period (the proportion of all such units that need repair). The laboratory technicians must decide whether the data resulting from the experiment supports the claim that p ≤ .10. Let X denote the number among the 20 sampled that need repair, so X ~ Bin(20, p ). Consider the decision rule

Reject the claim that p < .10 in favor of the conclusion that p > .10 if x > 5 (where x is the observed value of X ), and consider the claim plausible if x –4.

The probability that the claim is rejected when p = .10 (an incorrect conclusion) is

P (X ≥ 5 when p = .10) = 1 –B (4; 20, .1) = 1 − .957 = .043

The probability that the claim is not rejected when p = .20 (a different type of incorrect conclusion) is

P (X ≤ 4 when p = .2) = B (4; 20, .2) = .630

The first probability is rather small, but the second is intolerably large. When p<#8217;s claim being judged plausible!

One might think that the probability of this second type of erroneous conclusion could be made smaller by changing the cutoff value 5 in the decision rule to something else. However, although replacing 5 by a smaller number would yield a probability smaller than .630, the other probability would then increase. The only way to make both “error probabilities” small is to base the decision rule on an experiment involving many more units.

The Mean and Variance of X

For n = 1, the binomial distribution becomes the Bernoulli distribution. From Example 3.18, the mean value of a Bernoulli variable is μ = p, so the expected number of S ’s on any single trial is p. Since a binomial experiment consists of n trials, intuition suggests that for X ~ Bin(n, p ), E (X ) = np, the product of the number of trials and the probability of success on a single trial. The expression for V (X ) is not so intuitive.

PROPOSITION

If X ~ Bin(n, p ), then E (X ) = np , V (X ) = np (1 –p ) = npq, and σx = (where q = 1 –p ).

Thus, calculating the mean and variance of a binomial rv does not necessitate evaluating summations. The proof of the result for E (X ) is sketched in Exercise 64.

Example 3.34

If 75% of all purchases at a certain store are made with a credit card and X is the number among ten randomly selected purchases made with a credit ch (X ) = the revenue when X of the 25 purchasers want new copies. Express this as a linear function.]

53. Exercise 30 (Section 3.3) gave the pmf of Y, the number of traffic citations for a randomly selected individual insured by a particular company. What is the probability that among 15 randomly chosen such individuals

a. At least 10 have no citations?

b. Fewer than half have at least one citation?

c. The number that have at least one citation is between 5 and 10, inclusive?*

* “Between a and b, inclusive” is equivalent to (a ≤ X ≤ b ).

54. A particular type of tennis racket comes in a midsize version and an oversize version. Sixty percent of all customers at a certain store want the oversize version.

a. Among ten randomly selected customers who want this type of racket, what is the probability that at least six want the oversize version?

b. Among ten randomly selected customers, what is the probability that the number who want the oversize version is within 1 standard deviation of the mean value?

c. The store currently has seven rackets of each version. What is the probability that all of the next ten customers who want this racket can get the version they want from current stock?

55. Twenty percent of all telephones of a certain type are submitted for service while under warranty. Of these, 60% can be repaired, whereas the other 40% must be replaced with new units. If a company purchases ten of these telephones, what is the probability that exactly two will end up being replaced under warranty?

56. The College Board reports that 2% of the 2 million high school students who take the SAT each year receive special accommodations because of documented disabilities (Los Angeles Times, July 16, 2002). Consider a random sample of 25 students who have recently taken the test.

d. What is the probability that the number among the 25 who received a special accommodation is within 2 standard deviations of the number you would expect to be accommodated?

e. Suppose that a student who does not receive a special accommodation is allowed 3 hours for the exam, whereas an accommodated student is allowed 4.5 hours. What would you expect the average time allowed the 25 selected students to be?

57. Suppose that 90% of all batteries from a certain supplier have acceptable voltages. A certain type of flashlight requires two type-D batteries, and the flashlight will work only if both its batteries have acceptable voltages. Among ten randomly selected flashlights, what is the probability that at least nine will work? What assumptions did you make in the course of answering the question posed?

58. A very large batch of components has arrived at a distributor. The batch can be characterized as acceptable only if the proportion of defective components is at most .10. The distributor decides to randomly select 10 components and to accept the batch only if the number of defective components in the sample is at most 2.

a. What is the probability that the batch will be accepted when the actual proportion of defectives is .01? .05? .10? .20? .25?

b. Let p denote the actual proportion of defectives in the batch. A graph of P (batch is accepted) as a function of p, with p on the horizontal axis and P (batch is accepted) on the vertical axis, is called the operating characteristic curve for the acceptance sampling plan. Use the results of part (a) to sketch this curve for 0 ≤ p ≤ 1.

c. Repeat parts (a) and (b) with “1” replacing “2” in the acceptance sampling plan.

d. Repeat parts (a) and (b) with “15” replacing “10” in the acceptance sampling plan.

e. Which of the three sampling plans, that of part (a), (c), or (d), appears most satisfactory, and why?

59. An ordinance requiring that a smoke detector be installed in all previously constructed houses has been in effect in a particular city for 1 year. The fire department is concerned that many houses remain without detectors. Let p = the true proportion of such houses having detectors, and suppose that a random sample of 25 homes is inspected. If the sample strongly indicates that fewer than 80% of all houses have a detector, the fire department will campem> ) as a sum with lower limit x = 1. Then factor out np, let y = x –1 so that the sum is from y = 0 to y = n –1, and show that the sum equals 1.]

65. Customers at a gas station pay with a credit card (A ), debit card (B ), or cash (C ). Assume that successive customers make independent choices, with P (A ) = .5, P (B ) = .2, and P (C ) = .3.

a. Among the next 100 customers, what are the mean and variance of the number who pay with a debit card? Explain your reasoning.

b. Answer part (a) for the number among the 100 who don’t pay with cash.

66. An airport limousine can accommodate up to four passengers on any one trip. The company will accept a maximum of six reservations for a trip, and a passenger must have a reservation. From previous records, 20% of all those making reservations do not appear for the trip. Answer the following questions, assuming independence wherever appropriate.

a. If six reservations are made, what is the probability that at least one individual with a reservation cannot be accommodated on the trip?

b. If six reservations are made, what is the expected number of available places when the limousine departs?

c. Suppose the probability distribution of the number of reservations made is given in the accompanying table.

Let X denote the number of passengers on a randomly selected trip. Obtain the probability mass function of X.

67. Refer to Chebyshev’s inequality given in Exercise 44. Calculate P (|X –μ | ≥ ka ) for k = 2 and k = 3 when X ~ Bin(20, .5), and compare to the corresponding upper bound. Repeat for X ~ Bin(20, .75).

3.5 Hypergeometric and Negative Binomial Distributions

The hypergeometric and negative binomial distributions are both related to the binomial distribution. Whereas the binomial distribution is the approximate probability model for sampling without replacement from a finite dichotomous (S –F ) population, the hypergeometric distribution is the exact probability model for the number of S ’s in the sample. The binomial rv X is the number of S ’s when the number n of trials is fixed, whereas the negative binomial distribution arises from fixing t">

Comprehensive tables of the hypergeometric distribution are available, but because the distribution has three parameters, these tables require much more space than tables for the binomial distribution. MINITAB and other statistical software packages will easily generate hypergeometric probabilities.

As in the binomial case, there are simple expressions for E (X ) and V (X ) for hypergeometric rv’s.

PROPOSITION

The mean and variance of the hypergeometric rv X having pmf h (x ; n, M, N ) are

The ratio M/N is the proportion of S ’s in the population. If we replace M /N by p in E (X ) and V (X ), we get

E (X ) = np

Expression (3.16) shows that the means of the binomial and hypergeometric rv’s are equal, whereas the variances of the two rv’s differ by the factor (N –n )/(N –1), often called the finite population correction factor. This factor is less than 1, so the hypergeometric variable has smaller variance than does the binomial rv. The correction factor can be written (1 –n /N ) /(1 –1/N ), which is approximately 1 when n is small relative to N.

Example 3.37 (Example 3.36 continued)

In the animal-tagging example, n = 10, M = 5, and N = 25, so p = = .2 and

E (X ) = 10(.2) = 2

If the sampling was carried out with replacement, V (X ) = 1.6.

Suppose the population size N is not actually known, so the value x is observed and we wish to estimate N. It is reasonable to equate the observed sample proportion of S ’s, x /n, with the population proportion, M /N, giving the estimate

Our general rule of thumb in Section 3.4 stated that if sampling was without replacement but n /N was at most .05, then the binomial distribution could be used to compute approximate probabilities involving the number of S ’s in the sample. A more precise statement is as follows: Let the population size, N, and number of population S ’s, M, get large with the ratio M /N approaching p. Then h (x; n, M, N ) approaches b (x; n, p ); so for n /N small, the two are approximately equal provided that p is not too near either 0 or 1. This is the rationale for our rule of thumb.

The Negative Binomial Distribution

The negative binomial rv and distribution are based on an experiment satisfying the following conditions:

1. The experiment consists of a sequence of independent trials.

2. Each trial can result in either a success (S ) or a failure (F ).

3. The probability of success is constant from trial to trial, so P (S on trial i ) = p for i = 1, 2, 3 ….

4. The experiment continues (trials are performed) until a total of r successes have been observed, where r is a specified positive integer.

The random variable of interest is X = the number of failures that precede the rth success; X is called a negative binomial random variable because, in contrast to the binomial rv, the number of successes is fixed and the number of trials is random.

Possible values of X are 0, 1, 2, …. Let nb (x; r, p ) denote the pmf of X. The event {X = x } is equivalent to {r –1 S ’s in the first (x + r –1) trials and an S on the (x + r )th trial} (e.g., if r = 5 and x = 10, then there must be four S ’s in the first 14 trials and trial 15 must be an S ). Since trials are independent,

The first probability on the far right of Expression (3.17) is the binomial probability

where P (S ) = p

PROPOSITION

The S) is

Example 3.38

A pediatrician wishes to recruit 5 couples, each of whom is expecting their first child, to participate in a new natural childbirth regimen. Let p = P (a randomly selected couple agrees to participate). If p = .2, what is the probability that 15 couples must be asked before 5 are found who agree to participate? That is, with S = {agrees to participate}, what is the probability that 10 F ’s occur before the fifth S ? Substituting r = 5, p = .2, and x = 10 into nb (x; r, p ) gives

The probability that at most 10 F ’s are observed (at most 15 couples are asked) is

In some sources, the negative binomial rv is taken to be the number of trials X + r rather than the number of failures.

In the special case r = 1, the pmf is

In Example 3.12, we derived the pmf for the number of trials necessary to obtain the first S, and the pmf there is similar to Expression (3.18). Both X = number of F ’s and Y = number of trials (= 1 + X ) are referred to in the literature as geometric random variables, and the pmf in Expression (3.18) is called the geometric distribution.

In Example 3.19, the expected number of trials until the first S was shown to be 1/p, so that the expected number of F ’s until the first S is (1/p ) –1 = (1 –p )/p . Intuitively, we would expect to see r • (1 –p )/p F ’s before the rth S, and this is indeed E (X ). There is also a simple formula for V (X ).

PROPOSITION

If X is a negative binomial rv with pmf nb (x; r, p ), then

Finally, by expanding the binomial coefficient in front of p r(1 –p )x and doing some cancellation, it can be seen that nb (x; r, p ined even when r is not an integer. This generalized negative binomial distribution has been found to fit observed data quite well in a wide variety of applications.

EXERCISES Section 3.5 (68–78)

68. A certain type of digital camera comes in either a 3-megapixel version or a 4-megapixel version. A camera store has received a shipment of 15 of these cameras, of which 6 have 3-megapixel resolution. Suppose that 5 of these cameras are randomly selected to be stored behind the counter; the other 10 are placed in a storeroom. Let X = the number of 3-megapixel cameras among the 5 selected for behind-the-counter storage.

a. What kind of a distribution does X have (name and values of all parameters)?

b. Compute P (X = 2), P (X ≤ 2), and P (X ≥ 2).

c. Calculate the mean value and standard deviation of X .

69. Each of 12 refrigerators of a certain type has been returned to a distributor because of an audible, high-pitched, oscillating noise when the refrigerator is running. Suppose that 7 of these refrigerators have a defective compressor and the other 5 have less serious problems. If the refrigerators are examined in random order, let X be the number among the first 6 examined that have a defective compressor. Compute the following:

a. P (X = 5)

b. P (X ≤ 4)

c. The probability that X exceeds its mean value by more than 1 standard deviation.

d. Consider a large shipment of 400 refrigerators, of which 40 have defective compressors. If X is the number among 15 randomly selected refrigerators that have defective compressors, describe a less tedious way to calculate (at least approximately) P (X ≤ 5) than to use the hypergeo-metric pmf.

70. An instructor who taught two sections of engineering statistics last term, the first with 20 students and the second with 30, decided to assign a term project. After all projects had been turned in, the instructor randomly ordered them before grading. Consider the first 15 graded projects.

a. What is the probability that exactly 10 of these are from the second section? height="1em" width="-1em">
c. What is the probability that at least 10 of these are from the same section?

d. What are the mean value and standard deviation of the number among these 15 that are from the second section?

e. What are the mean value and standard deviation of the number of projects not among these first 15 that are from the second section?

71. A geologist has collected 10 specimens of basaltic rock and 10 specimens of granite. The geologist instructs a laboratory assistant to randomly select 15 of the specimens for analysis.

a. What is the pmf of the number of granite specimens selected for analysis?

b. What is the probability that all specimens of one of the two types of rock are selected for analysis?

c. What is the probability that the number of granite specimens selected for analysis is within 1 standard deviation of its mean value?

72. A personnel director interviewing 11 senior engineers for four job openings has scheduled six interviews for the first day and five for the second day of interviewing. Assume that the candidates are interviewed in random order.

a. What is the probability that x of the top four candidates are interviewed on the first day?

b. How many of the top four candidates can be expected to be interviewed on the first day?

73. Twenty pairs of individuals playing in a bridge tournament have been seeded 1, …, 20. In the first part of the tournament, the 20 are randomly divided into 10 east–west pairs and 10 north–south pairs.

a. What is the probability that x of the top 10 pairs end up playing east–west?

b. What is the probability that all of the top five pairs end up playing the same direction?

c. If there are 2n pairs, what is the pmf of X = the number among the top n pairs who end up playing east–west? What are E (X ) and V (X )?

74. A second-stage smog alert has been called in a certain area of Los Angng>

A random variable X is said to have a Poisson distribution with parameter λ (λ > 0) if the pmf of X is

The value of λ is frequently a rate per unit time or per unit area. The letter e in P (X ; λ) represents the base of the natural logarithm system; its numerical value is approximately 2.71828. Because λ must be positive, P (X ; λ) > 0 for all possible x values. The fact that is a consequence of the Maclaurin infinite series expansion of e λ, which appears in most calculus texts:

If the two extreme terms in Expression (3.19) are multiplied by e –λ and then e –λ is placed inside the summation, the result is

which shows that p (X ; λ) fulfills the second condition necessary for specifying a pmf.

Example 3.39

Let X denote the number of creatures of a particular type captured in a trap during a given time period. Suppose that X has a Poisson distribution with λ = 4.5, so on average traps will contain 4.5 creatures. [The article “Dispersal Dynamics of the Bivalve Gemma Gemma in a Patchy Environment (Ecological Monographs, 1995: 1–20) suggests this model; the bivalve Gemma gemma is a small clam.] The probability that a trap contains exactly five creatures is

The probability that a trap has at most five creatures is

The Poisson Distribution as a Limit

The rationale for using the Poisson distribution in many situations is provided by the following proposition.

PROPOSITION

Suppose that in the binomial pmf b (x; n, p ), we let n → ∞ and p → 0 in such a way that np approaches a value λ > 0. Then b (x; n, p ) → P (X ; &#s New Roman">According to this proposition, in any binomial experiment in which n is large and p is small, b (x ; n, p ) ≈ p (X ; λ), where λ = np. As a rule of thumb, this approximation can safely be applied if n > 50 and np < 5.

Example 3.40

If a publisher of nontechnical books takes great pains to ensure that its books are free of typographical errors, so that the probability of any given page containing at least one such error is .005 and errors are independent from page to page, what is the probability that one of its 400-page novels will contain exactly one page with errors? At most three pages with errors?

With S denoting a page containing at least one error and F an error-free page, the number X of pages containing at least one error is a binomial rv with n = 400 and p = .005, so np = 2. We wish

The binomial value is b (1; 400, .005) = .270669, so the approximation is very good.

Similarly,

= .135335 + .270671 + .270671 + .180447

= .8571

and this again is quite close to the binomial value P (X ≤ 3) = .8576.

Table 3.2 shows the Poisson distribution for λ= 3 along with three binomial distributions with np = 3, and Figure 3.8 (from S-Plus) plots the Poisson along with the first two binomial distributions. The approximation is of limited use for n = 30, but of course the accuracy is better for n = 100 and much better for n = 300.

Table 3.2 Comparing the Poisson and Three Binomial Distributions

Figure 3.8 Comparing a Poisson and two binomial distributions

Appendix Table A.2 exhibits the cdf F (x ; λ) for λ = .1, .2, …, le 3.40, whereas P (X = 3) = F (3; 2) –F (2; 2) = .180. Alternatively, many statistical computer packages will generate p (x ; λ) and F (x ; λ) upon request.

The Mean and Variance of X

Since b (x; n, p ) → P (X ; λ) as n → ∞, p → 0, np → λ, the mean and variance of a binomial variable should approach those of a Poisson variable. These limits are np → λ and np (1 –p ) → λ.

PROPOSITION

If X has a Poisson distribution with parameter λ, then E (X ) = V (X ) = λ.

These results can also be derived directly from the definitions of mean and variance.

Example 3.41 (Example 3.39 continued)

Both the expected number of creatures trapped and the variance of the number trapped equal 4.5, and σX = = 2.12.

The Poisson Process

A very important application of the Poisson distribution arises in connection with the occurrence of events of some type over time. Events of interest might be visits to a particular website, pulses of some sort recorded by a counter, email messages sent to a particular address, accidents in an industrial facility, or cosmic ray showers observed by astronomers at a particular observatory. We make the following assumptions about the way in which the events of interest occur:

1. There exists a parameter α > 0 such that for any short time interval of length t, the probability that exactly one event is received is α • t + o (t ).*

* A quantity is o (t ) (read “little o of delta t ”) if, as t approaches 0, so does o (t )/t . That is, o (t ) is even more negligible (approaches 0 faster) than t itself. The quantity (t ) does not.

2. The probability of more than one event being received during t is o (t ) [which, along with Assumption 1, implies that the probability of no events during t is 1 –α • t –o (t )].

3. The number of events received during the time interval t is independent of the number received prior to this time interval.

Informally, Assumption 1 says that for a short interval of time, the probability of receiving a single event is approximately proportional to the length of the time interval, where α is the constant of proportionality. Now let Pk(t ) denote the probability that k events will be observed during any particular time interval of length t.

PROPOSITION

Pk(t ) = e–αt • (αt )k/k !, so that the number of events during a time interval of length t is a Poisson rv with parameter λ = αt. The expected number of events during any such time interval is then αt , so the expected number during a unit interval of time is α.

The occurrence of events over time as described is called a Poisson process; the parameter α specifies the rate for the process.

Example 3.42

Suppose pulses arrive at a counter at an average rate of six per minute, so that α = 6. To find the probability that in a .5-min interval at least one pulse is received, note that the number of pulses in such an interval has a Poisson distribution with parameter αt = 6(.5) = 3 (.5 min is used because α is expressed as a rate per minute). Then with X = the number of pulses received in the 30-sec interval,

Instead of observing events over time, consider observing events of some type that occur in a two- or three-dimensional region. For example, we might select on a map a certain region R of a forest, go to that region, and count the number of trees. Each tree would represent an event occurring at a particular point in space. Under assumptions similar to 1–3, it can be shown that the number of events occurring in a region R has a Poisson distribution with parameter α height="0em">

EXERCISE. Section 3.6 (79–93)

79. Let X , the number of flaws on the surface of a randomly selected boiler of a certain type, have a Poisson distribution with parameter λ = 5. Use Appendix Table A.2 to compute the following probabilities:

a. P (X ≤ 8)

b. P (X = 8)

c. P (9 ≤ X )

d. P (5 ≤ X ≤ 8)

e. P (5 < X < 8)

80. Suppose the number X of tornadoes observed in a particular region during a 1-year period has a Poisson distribution with λ = 8.

a. Compute P (X ≤ 5).

b. Compute P (6 ≤ X ≤ 9).

c. Compute P (10 ≤ X ).

d. What is the probability that the observed number of tornadoes exceeds the expected number by more than 1 standard deviation?

81. Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a Poisson distribution with parameter λ = 20 (suggested in the article “Dynamic Ride Sharing: Theory and Practice,” J. of Transp. Engr., 1997: 308–312). What is the probability that the number of drivers will

a. Be at most 10?

b. Exceed 20?

c. Be between 10 and 20, inclusive? Be strictly between 10 and 20?

d. Be within 2 standard deviations of the mean value?

82. Consider writing onto a computer disk and then sending it through a certifier that counts the number of /em> = .2. (Suggested in “Average Sample Number for Semi-Curtailed Sampling Using the Poisson Distribution,” J. Quality Technology, 1983: 126–129.)

a. What is the probability that a disk has exactly one missing pulse?

b. What is the probability that a disk has at least two missing pulses?

c. If two disks are independently selected, what is the probability that neither contains a missing pulse?

83. An article in the Los Angeles Times (Dec. 3, 1993) reports that 1 in 200 people carry the defective gene that causes inherited colon cancer. In a sample of 1000 individuals, what is the approximate distribution of the number who carry this gene? Use this distribution to calculate the approximate probability that

a. Between 5 and 8 (inclusive) carry the gene.

b. At least 8 carry the gene.

84. Suppose that only .10% of all computers of a certain type experience CPU failure during the warranty period. Consider a sample of 10,000 computers.

a. What are the expected value and standard deviation of the number of computers in the sample that have the defect?

b. What is the (approximate) probability that more than 10 sampled computers have the defect?

c. What is the (approximate) probability that no sampled computers have the defect?

85. Suppose small aircraft arrive at a certain airport according to a Poisson process with rate α = 8 per hour, so that the number of arrivals during a time period of t hours is a Poisson rv with parameter λ = 8t.

a. What is the probability that exactly 6 small aircraft arrive during a 1-hour period? At least 6? At least 10?

b. What are the expected value and standard deviation of the number of small aircraft that arrive during a 90-min period?

c. What is the probability that at least 20 small aircraft arrive during a hour period? That at most 10 arrive during this period?

86. The number of people arriving for tr>

d. It can be shown in a manner similar to parts (a) and (b) that the Pk(t )s must satisfy the system of differential equations

k = 1, 2, 3, …

Verify that Pk(t ) = e–αt (αt )k/k! satisfies the system. (This is actually the only solution.)

SUPPLEMENTARY EXERCISES (94–122)

94. Consider a deck consisting of seven cards, marked 1, 2, …, 7. Three of these cards are selected at random. Define an rv W by W = the sum of the resulting numbers, and compute the pmf of W. Then compute μ and σ 2. [Hint: Consider outcomes as unordered, so that (1, 3, 7) and (3, 1, 7) are not different outcomes. Then there are 35 outcomes, and they can be listed. (This type of rv actually arises in connection with a hypothesis test called Wilcoxon’s rank-sum test, in which there is an x sample and a y sample and W is the sum of the ranks of the x ’s in the combined sample.)]

95. After shuffling a deck of 52 cards, a dealer deals out 5. Let X = the number of suits represented in the five-card hand.

a. Show that the pmf of X is

[Hint: p (1) = 4P (all are spades), p (2) = 6P (only spades and hearts with at least one of each suit), and p (4) = 4P (2 spades ∩ one of each other suit).]

b. Compute μ, σ^2, and σ.

96. The negative binomial rv X was defined as the number of F ’s preceding the rth S. Let Y = the number of trials necessary to obtain the rth S. In the same manner in which the pmf of X was derived, derive the pmf of Y.

97. Of all customers purchasing automatic garage-door openers, 75% purchase a chain-driven model. Let X = the number among the next 15 purchasers who select the chain-driven model.

a. What ir acquittal. If the jurors leave the jury room in random order and each of the first four leaving the room is accosted by a reporter in quest of an interview, what is the pmf of X = the number of jurors favoring acquittal among those interviewed? How many of those favoring acquittal do you expect to be interviewed?

109. A reservation service employs five information operators who receive requests for information independently of one another, each according to a Poisson process with rate α = 2 per minute.

a. What is the probability that during a given 1-min period, the first operator receives no requests?

b. What is the probability that during a given 1-min period, exactly four of the five operators receive no requests?

c. Write an expression for the probability that during a given 1-min period, all of the operators receive exactly the same number of requests.

110. Grasshoppers are distributed at random in a large field according to a Poisson distribution with parameter α = 2 per square yard. How large should the radius R of a circular sampling region be taken so that the probability of finding at least one in the region equals .99?

111. A newsstand has ordered five copies of a certain issue of a photography magazine. Let X = the number of individuals who come in to purchase this magazine. If X has a Poisson distribution with parameter λ = 4, what is the expected number of copies that are sold?

112. Individuals A and B begin to play a sequence of chess games. Let S = {A wins a game}, and suppose that outcomes of successive games are independent with P (S ) = p and P (F ) = 1 –p (they never draw). They will play until one of them wins ten games. Let X = the number of games played (with possible values 10, 11, …, 19).

a. For x = 10, 11, …, 19, obtain an expression for P (X ) = P (X = x ).

b. If a draw is possible, with p = P (S ), q = P (F ), 1 –p –q = P (draw), what are the possible values of X ? What is P (20 ≤ X )? [Hint: P (20 ≤ X ) = 1 –P (X < 20).]

113. A test for the presence of a certain disease has probability .20 of giving a false-positive reading (indicating that an individual has the disease when this is not the case) and probability .10 of giving a false-negative result. Suppose that ten individuals are tested, five of whom have the disease and five of whom donumber of positive readings that result.

a. Does X have a binomial distribution? Explain your reasoning.

b. What is the probability that exactly three of the ten test results are positive?

114. The generalized negative binomial pmf is given by

nb (x; r, p ) = k (r, x ) · p r(1 –p )x

x = 0, 1, 2, …

Let X , the number of plants of a certain species found in a particular region, have this distribution with p = .3 and r = 2.5. What is P (X = 4)? What is the probability that at least one plant is found?

115. Define a function p (x; λ, μ ) by

a. Show that p (x; λ, μ ) satisfies the two conditions necessary for specifying a pmf, [Note: If a firm employs two typists, one of whom makes typographical errors at the rate of λ per page and the other at rate μ per page and they each do half the firm’s typing, then p (x; λ, μ ) is the pmf of X = the number of errors on a randomly chosen page.]

b. If the first typist (rate λ ) types 60% of all pages, what is the pmf of X of part (a)?

c. What is E (X ) for p (x; λ, μ ) given by the displayed expression?

d. What is σ 2 for p (x; λ, μ ) given by that expression?

116. The mode of a discrete random variable X with pmf p (x ) is that value x * for which p (x ) is largest (the most probable x value).

a. Let X ~ Bin(n, p ). By considering the ratio b (x + 1; n, p )/b (x; n, p ), show that b (x; n, p ) increases with x as long as x < np –(how that if X has a Poisson distribution with parameter λ, the mode is the largest integer less than λ. If λ is an integer, show that both λ –1 and λ are modes.

117. A computer disk storage device has ten concentric tracks, numbered 1, 2, …, 10 from outermost to innermost, and a single access arm. Let pi = the probability that any particular request for data will take the arm to track i (i = 1, …, 10). Assume that the tracks accessed in successive seeks are independent. Let X = the number of tracks over which the access arm passes during two successive requests (excluding the track that the arm has just left, so possible X values are x = 0, 1,…,9). Compute the pmf of X . [Hint: P (the arm is now on track i and X = j ) = P (X = j | arm now on i ) · pi. After the conditional probability is written in terms of p_1, …, p_10, by the law of total probability, the desired probability is obtained by summing over i .]

118. If X is a hypergeometric rv, show directly from the definition that E (X ) = nM /N (consider only the case n < M ). [Hint: Factor nM /N out of the sum for E (X ), and show that the terms inside the sum are of the form h (y; n –1, M –1, N –1), where y = x –1.]

119. Use the fact that

to prove Chebyshev’s inequality given in Exercise 44.

120. The simple Poisson process of Section 3.6 is characterized by a constant rate α at which events occur per unit time. A generalization of this is to suppose that the probability of exactly one event occurring in the interval [t , t + t] is α(t ) • t + o (t ). It can then be shown that the number of events occurring during an interval [t_1, t_2] has a Poisson distribution with parameter

The occurrence of events over time in this situation is called a nonhomogeneous Poisson process. The article “Inference Based on Retrospective Ascertainment,” J. Amer. Stat. Assoc., 1989: 360–372, considers the intensity function

α (t ) = ea+bt

x ≤6.5, but X would still be continuous.

Example 4.3 Let X represent the amount of time a randomly selected customer spends waiting for a haircut before his/her haircut commences. Your first thought might be that X is a continuous random variable, since a measurement is required to determine its value. However, there are customers lucky enough to have no wait whatsoever before climbing into the barber’s chair. So it must be the case that P(X = 0) >0. Conditional on no chairs being empty, though, the waiting time will be continuous since X could then assume any value between some minimum possible time A and a maximum possible time B. This random variable is neither purely discrete nor purely continuous but instead is a mixture of the two types.

One might argue that although in principle variables such as height, weight, and temperature are continuous, in practice the limitations of our measuring instruments restrict us to a discrete (though sometimes very finely subdivided) world. However, continuous models often approximate real-world situations very well, and continuous mathematics (the calculus) is frequently easier to work with than mathematics of discrete variables and distributions.

Probability Distributions for Continuous Variables

Suppose the variable X of interest is the depth of a lake at a randomly chosen point on the surface. Let M = the maximum depth (in meters), so that any number in the interval [0, M] is a possible value of X. If we “discretize” X by measuring depth to the nearest meter, then possible values are nonnegative integers less than or equal to M. The resulting discrete distribution of depth can be pictured using a probability histogram. If we draw the histogram so that the area of the rectangle above any possible integer k is the proportion of the lake whose depth is (to the nearest meter) k, then the total area of all rectangles is 1. A possible histogram appears in Figure 4.1(a).

If depth is measured much more accurately and the same measurement axis as in Figure 4.1(a) is used, each rectangle in the resulting probability histogram is much narrower, though the total area of all rectangles is still 1. A possible histogram is pictured in Figure 4.1(b); it has a much smoother appearance than the histogram in Figure 4.1(a). If we continue in this way to measure depth more and more finely, the resulting sequence of histograms approaches a smooth curve, such as is pictured in Figure 4.1(c). Because for each histogram the total area of all rectangles equals 1, the total area under the smooth curve is also 1. The probability that the depth at a randomly chosen point is belign="center">P (0 ≤ X ≤90) +P (270 ≤ X <360) = .25 +.25 = .50

Figure 4.3 The pdf and probability from Example 4.4

Because whenever 0 ≤ α ≤ b ≤360 in Example 4.4, P (α ≤ X ≤b ) depends only on the width b − α of the interval, X is said to have a uniform distribution.

DEFINITION

A continuous rv X is said to have a uniform distribution on the interval [A, B] if the pdf of X is

The graph of any uniform pdf looks like the graph in Figure 4.3 except that the interval of positive density is [A , B] rather than [0, 360].

In the discrete case, a probability mass function (pmf) tells us how little “blobs” of probability mass of various magnitudes are distributed along the measurement axis. In the continuous case, probability density is “smeared” in a continuous fashion along the interval of possible values. When density is smeared uniformly over the interval, a uniform pdf, as in Figure 4.3, results.

When X is a discrete random variable, each possible value is assigned positive probability. This is not true of a continuous random variable (that is, the second condition of the definition is satisfied) because the area under a density curve that lies above any single value is zero:

The fact that P(X = c) = 0 when X is continuous has an important practical consequence: The probability that X lies in some interval between α and b does not depend on whether the lower limit α or the upper limit b is included in the probability calculation:

If X is discrete and both α and b are possible values (e.g., X is binomial with n = 20 and α = 5, b = 10), then all four of these probabilities are different.

The zero probability condition has a physical analog. Consider a solid circular rod with cross-sectional area = 1 in^2. Place the rod alongside a measurement axis and suppose that the density of the rod at any pm>b and this segment is removed, the amount of mass removed is dx; if the rod is sliced just at the point c , no mass is removed. Mass is assigned to interval segments of the rod but not to individual points.

Example 4.5

“Time headway” in traffic flow is the elapsed time between the time that one car finishes passing a fixed point and the instant that the next car begins to pass that point. Let X = the time headway for two randomly chosen consecutive cars on a freeway during a period of heavy flow. The following pdf of X is essentially the one suggested in “The Statistical Properties of Freeway Traffic” (Transp. Res., vol. 11: 221-228):

The graph of f (x ) is given in Figure 4.4; there is no density associated with headway times less than .5, and headway density decreases rapidly (exponentially fast) as x increases from .5. Clearly, f (x ) ≥ 0; to show that dx = 1, we use the calculus result dx = (1/k)e−k•;a. Then

Figure 4.4 The density curve for headway time in Example 4.5

The probability that headway time is at most 5 sec is

Unlike discrete distributions such as the binomial, hypergeometric, and negative binomial, the distribution of any given continuous rv cannot usually be derived using simple probabilistic arguments. Instead, one must make a judicious choice of pdf based on prior knowledge and available data. Fortunately, there are some general families of pdf s that have been found to fit well in a wide variety of experimental situations; several of these are discussed later in the chapter.

Just as in the discrete case, it is often helpful to think of the population of interest as consisting of X values rather than individuals or objects. The pdf is then a model for the distribution of values in this numerical population, and from this model various population characteristics (such as the mean) can be calculated.

EXERCISES Section 4.1 (1–10)

1. Let X denote the amount of time for which a book on 2-hour reserve at a college library is checked out by a randomly selckquote>b. Suppose θ = 100 (a value suggested by a graph in the article). What is the probability that X is at most 200? Less than 200? At least 200?

c. What is the probability that X is between 100 and 200 (again assuming θ = 100)?

d. Give an expression for P(X ≤x).

5. A college professor never finishes his lecture before the end of the hour and always finishes his lectures within 2 min after the hour. Let X = the time that elapses between the end of the hour and the end of the lecture and suppose the pdf of X is

f(x )

a. Find the value of k and draw the corresponding density curve. [Hint: Total area under the graph of f(x ) is 1.]

b. What is the probability that the lecture ends within 1 min of the end of the hour?

c. What is the probability that the lecture continues beyond the hour for between 60 and 90 sec?

d. What is the probability that the lecture continues for at least 90 sec beyond the end of the hour?

6. The actual tracking weight of a stereo cartridge that is set to track at 3 g on a particular changer can be regarded as a continuous rv X with pdf

a. Sketch the graph of f (x ).

b. Find the value of k.

c. What is the probability that the actual tracking weight is greater than the prescribed weight?

d. What is the probability that the actual weight is within .25 g of the prescribed weight?

e. What is the probability that the actual weight differs from the prescribed weight by more than .5 g?

7. The time X (min) for a lab assistant to prepare the equipment for a certain experiment is believed to have a uniform distrong> Determine the pdf of X and sketch the corresponding density curve.

b. What is the probability that preparation time exceeds 33 min?

c. What is the probability that preparation time is within 2 min of the mean time? [Hint: Identify μ from the graph of f(x).]

d. For any α such that 25<a<a + 2 <35, what is the probability that preparation time is between α and α +2 min?

8. In commuting to work, I must first get on a bus near my house and then transfer to a second bus. If the waiting time (in minutes) at each stop has a uniform distribution with A = 0 and B = 5, then it can be shown that my total waiting time Y has the pdf

a. Sketch a graph of the pdf of Y .

b. Verify that f(y) dy = 1.

c. What is the probability that total waiting time is at most 3 min?

d. What is the probability that total waiting time is at most 8 min?

e. What is the probability that total waiting time is between 3 and 8 min?

f. What is the probability that total waiting time is either less than 2 min or more than 6 min?

9. Consider again the pdf of X = time headway given in Example 4.5. What is the probability that time headway is

a. At most 6 sec?

b. More than 6 sec? At least 6 sec?

c. Between 5 and 6 sec?

10. A family of pdf’s that has been used to approximate the distribution of income, city population size, and size of firms is the Pareto family. The family has two parameters, k and θ , both ≥ 0, and the pdf is

b. Verify that the total area under the graph equals 1.

c. If the rv X has pdf f(x; k, θ), for any fixed b > θ, obtain an expression for P(X ≤b).

d. For θ < α < b, obtain an expression for the probability P(a ≤ X ≤b).

4 2 Cumulative Distribution Functions and Expected Values

Several of the most important concepts introduced in the study of discrete distributions also play an important role for continuous distributions. Definitions analogous to those in Chapter 3 involve replacing summation by integration.

The Cumulative Distribution Function

The cumulative distribution function (cdf) F(x ) for a discrete rv X gives, for any specified number x, the probability P(X ≤ x). It is obtained by summing the pmf p(y ) over all possible values y satisfying y ≤ x. The cdf of a continuous rv gives the same probabilities P(X ≤ x ) and is obtained by integrating the pdf f(y ) between the limits -∞ and x.

DEFINITION

The cumulative distribution function F(x ) for a continuous rv X is defined for every number x by

For each x, F(x ) is the area under the density curve to the left of x. This is illustrated in Figure 4.5, where F(x ) increases smoothly as x increases.

Figure 4.5 A pdf and associated cdf

Example 4.6

Let X, the thickness of a certain metal sheet, have a uniform distribution on [A , B]. The density function is shown in Figure 4.6. For x < A, F(x ) = 0, since there is no area under the graph of the density function to the left of such an x. For x ≥ B, F(x ) = 1, since all the area is accumulated to the left of such an x. Finally, for A ≤

Thus

The graphs of f (x ) and F(x ) are shown in Figure 4.9. The probability that the load is between 1 and 1.5 is

The probability that the load exceeds 1 is

Figure 4.9 The pdf and cdf for Example 4.7

Once the cdf has been obtained, any probability involving X can easily be calculated without any further integration.

Obtaining f(x) from F(x )

For X discrete, the pmf is obtained from the cdf by taking the difference between two F(x ) values. The continuous analog of a difference is a derivative. The following result is a consequence of the Fundamental Theorem of Calculus.

PROPOSITION

If X is a continuous rv with pdf f(x ) and cdf F(x), then at every x at which the derivative F′(x ) exists, F′(x ) = f(x).

Example 4.8 (Example 4.6 continued)

When X has a uniform distribution, F(x ) is differentiable except at x = A and x = B, where the graph of F(x ) has sharp corners. Since F(x ) = 0 for x < A and F(x ) = 1 for x > B, F′(x ) = 0 = f(x ) for such x. For A < x <B,

Percentiles of a Continuous Distribution

When we say that an individual’s test score was at the 85th percentile of the population, we mean that 85% of all population scores were below that score and 15% were above. Similarly, the 40th percentile is the score that exceeds 40% of all scores and is exceeded by 60% of all scores.

median of a continuous distribution, denoted by , is the 50th percentile, so satisfies .5 = F (). That is, half the area under the density curve is to the left of and half is to the right of .

A continuous distribution whose pdf is symmetric—the graph of the pdf to the left of some point is a mirror image of the graph to the right of that point—has median β equal to the point of symmetry, since half the area under the curve lies to either side of this point. Figure 4.12 gives several examples. The error in a measurement of a physical quantity is often assumed to have a symmetric distribution.

Figure 4.12 Medians of symmetric distributions

Expected Values

For a discrete random variable X, E(X ) was obtained by summing x · p(x ) over possible X values. Here we replace summation by integration and the pmf by the pdf to get a continuous weighted average.

DEFINITION

The expected or mean value of a continuous rv X with pdf f(x ) is

Example 4.10 (Example 4.9 continued)

The pdf of weekly gravel sales X was

so

When the pdf f(x ) specifies a model for the distribution of values in a numerical population, then μ is the population mean, which is the most frequently used measure of population location or center.

Often we wish to compute the expected value of some function h(X ) of the rv X. If we think of h(X ) as a new rv Y, techniques from mathematical statistics can be used to derive the pdf of Y, and E(Y ) can be computed from the definition. Fortunately, as in the discrete case, there is 415" alt="Image">

When h(X ) = aX + b, the expected value and variance of hX ) satisfy the same properties as in the discrete case: E[h(X)] = α μ + b and V [h (X )] = a^2 · σ^2 .

EXERCISES Section 4.2 (11–27)

11. The cdf of checkout duration X as described in Exercise 1 is

Use this to compute the following:

a. P(X < 1)

b.P (.5 < X < 1)

c. P(X > .5)

d. The median checkout duration [solve .5 = F (]

e. F′(x ) to obtain the density function f(x)

f.E(X )

g. V(X ) and σX

h. If the borrower is charged an amount h(X) = X^2 when checkout duration is X, compute the expected charge E[h(X)].

12. The cdf for X (=measurement error) of Exercise 3 is

a. Compute P(X < 0).

b. Compute P (-1 ≤ X ≤1).

cfont>

b. What can you say about E(X ) if k = 1?

c. If k >2, show that V(X ) = kQ^2(k − 1)−2(k −>2)−1.

d. If k = 2, what can you say about V(X)?

e. What conditions on k are necessary to ensure that E(X ) is finite?

25. Let X be the temperature in °C at which a certain chemical reaction takes place, and let Y be the temperature in °F (so Y = 1.8 X +32).

a. If the median of the X distribution is μ, show that 1.8/1 +32 is the median of the Y distribution.

b. How is the 90th percentile of the Y distribution related to the 90th percentile of the X distribution? Verify your conjecture.

c. More generally, if Y = aX + b, how is any particular percentile of the Y distribution related to the corresponding percentile of the X distribution?

26. Let X be the total medical expenses (in 1000s of dollars) incurred by a particular individual during a given year. Although X is a discrete random variable, suppose its distribution is quite well approximated by a continuous distribution with pdf f (x ) =k (1 +x/2.5 )−7 for x ≥ 0.

a. What is the value of k?

b. Graph the pdf of X.

c. What are the expected value and standard deviation of total medical expenses?

d. This individual is covered by an insurance plan that entails a $500 deductible provision (so the first $500 worth of expenses are paid by the individual). Then the plan will pay 80% of any additional expenses exceeding $500, and the maximum payment by the individual (including the deductible amount) is $2500. Let Y denote the amount of this individual’s medical expenses paid by the insurance company. What is the expected value of Y ?

[Hint: First figure out what value of X corresponds to the maximum out-of-pocket expense of $2500. Then write an expression for Y as a function of X (which involves several different pieces) the curve (the points at which the curve changes from turning downward to turning upward). Large values of σ yield graphs that are quite spread out about μ, whereas small values of σ yield graphs with a high peak above μ and most of the area under the graph quite close to μ. Thus a large σ implies that a value of X far from μ may well be observed, whereas such a value is quite unlikely when σ is small.

Figure 4.13 Normal density curves

The Standard Normal Distribution

To compute P(a ≤ X ≤ b ) when X is a normal rv with parameters μ and σ, we must determine

None of the standard integration techniques can be used to evaluate Expression (4.4). Instead, for μ = 0 and σ = 1, Expression (4.4) has been calculated using numerical techniques and tabulated for certain values of α and b. This table can also be used to compute probabilities for any other values of μ and σ under consideration.

DEFINITION

The normal distribution with parameter values μ =0 and σ =1 is called the standard normal distribution. A random variable having a standard normal distribution is called a standard normal random variable and will be denoted by Z . The pdf of Z is

The graph of f(z ; 0, 1) is called the standard normal (or z ) curve. The cdf of Z is P(Z ≤z ) f (y ; 0, 1) dy , which we will denote by Φ(z ).

The standard normal distribution does not frequently serve as a model for a naturally arising population. Instead, it is a reference distribution from which information about other normal distributions can be obtained. Appendix Table A.3 gives Φ(z ) = P(Z ≤z ), the area under the standard normal density curve to the left of z, for z = −3.49, −3.48, …, 3.48, 3.49. Figure 4.14 illustrates the type of cumulative area (probability) tabulated in Table A.3. From this table, various other probabilities involving Z can be calculated.

Figure 4.14 Standard normal cumulative areas tabulated in Appendix Table A.3

Ex0em">
Let’s determine the following standard normal probabilities: (a) P (Z ≤ 1.25), (b) P(Z < 1.25), (c) P (Z ≤ −1.25), and (d) P (− .38 ≤Z ≤1.25).

a. P (Z < 1.25) = Φ(1.25), a probability that is tabulated in Appendix Table A.3 at the intersection of the row marked 1.2 and the column marked .05. The number there is .8944, so P(Z ≤1.25) = .8944. Figure 4.15(a) illustrates this probability.

Figure 4.15 Normal curve areas (probabilities) for Example 4.13

b. P(Z >1.25) = 1 − P(Z ≤1.25) = 1 −Φ(1.25), the area under the z curve to the right of 1.25 (an upper-tail area). Then Φ(1.25) = .8944 implies that P(Z >1.25) = .1056. Since Z is a continuous rv, P(Z ≥ 1.25) = .1056. See Figure 4.15(b).

c.P (Z < −1.25) = Φ(−1.25), a lower-tail area. Directly from Appendix Table A.3, Φ(−1.25) = .1056. By symmetry of the z curve, this is the same answer as in part (b).

d. P (−.38 ≤ Z ≤1.25) is the area under the standard normal curve above the interval whose left endpoint is −.38 and whose right endpoint is 1.25. From Section 4.2, if X is a continuous rv with cdf F(x), then P(a ≤ X ≤b ) =F (b ) −F (α). Thus P (−.38 ≤Z ≤1.25) = Φ(1.25) − Φ(−.38) = .8944 −.3520 = .5424. (See Figure 4.16.)

Figure 4.16 P (-.38 ≤ Z ≤1.25) as the difference between two cumulative areas

Percentiles of the Standard Normal Distribution

For any p between 0 and 1, Appendix Table A.3 can be used to obtain the (100p )th percentile of the standard normal distribution.

Example 4.14 The 99th percentile of the standard normal distribution is that value on the horizontal axis such that the area under the z curve to the left of the value is .9900. Appendix Table A.3 gives for fixed z the area under the standard normal curve to the left of z, wh5; is also α. The zαs are usually referred to as z critical values. Table 4.1 lists the most useful z percentiles and zα values.

Table 4.1 Standard Normal Percentiles and Critical Values

Example 4.15

z.05 is the 100(1 −.05)th = 95th percentile of the standard normal distribution, so z.05 = 1.645. The area under the standard normal curve to the left of −z.05 is also .05. (See Figure 4.20.)

Figure 4.20 Finding z.05

Nonstandard Normal Distributions

When X ˜ N (μ, σ^2), probabilities involving X are computed by “standardizing.” The standardized variable is (X − μ)/σ. Subtracting μ shifts the mean from μ to zero, and then dividing by σ scales the variable so that the standard deviation is 1 rather than σ.

PROPOSITION

has a standard normal distribution. Thus

The key idea of the proposition is that by standardizing, any probability involving X can be expressed as a probability involving a standard normal rv Z , so that Appendix Table A.3 can be used. This is illustrated in Figure 4.21. The proposition can be proved by writing the cdf of Z = (X − μ )/ σ as

Using a result from calculus, this integral can be differentiated with respect to z to yield the desired pdf f (z ; 0, 1).

Figure 4.21 Equality of nonstandard and standard normal curve areas

Figure 4.23.

Figure 4.23 Distribution of amount dispensed for Example 4.18

The Normal Distribution and Discrete Populations

The normal distribution is often used as an approximation to the distribution of values in a discrete population. In such situations, extra care should be taken to ensure that probabilities are computed in an accurate manner.

Example 4.19

IQ in a particular population (as measured by a standard test) is known to be approximately normally distributed with μ =100 and σ =15. What is the probability that a randomly selected individual has an IQ of at least 125? Letting X = the IQ of a randomly chosen person, we wish P(X ≥ 125). The temptation here is to standardize X ≥ 125 as in previous examples. However, the IQ population distribution is actually discrete, since IQs are integer-valued. So the normal curve is an approximation to a discrete probability histogram, as pictured in Figure 4.24.

The rectangles of the histogram are centered at integers, so IQs of at least 125 correspond to rectangles beginning at 124.5, as shaded in Figure 4.24. Thus we really want the area under the approximating normal curve to the right of 124.5. Standardizing this value gives P(Z ≥ 1.63) = .0516, whereas standardizing 125 results in P(Z ≥ 1.67) = .0475. The difference is not great, but the answer .0516 is more accurate. Similarly, P(X = 125) would be approximated by the area between 124.5 and 125.5, since the area under the normal curve above the single value 125 is zero.

Figure 4.24 A normal approximation to a discrete distribution

The correction for discreteness of the underlying distribution in Example 4.19 is often called a continuity correction. It is useful in the following application of the normal distribution to the computation of binomial probabilities.

Approximating the Binomial Distribution

Recall that the mean value and standard deviation of a binomial random variable X are μX = np and σX = respectively. Figure 4.25 displays a binomial probability histogram for the binomial distribution with n = 20, p = .6, p>

The exact probabilities are .2622 and .8348, respectively, so the approximations are quite good. In the last calculation, the probability P (5 ≤ X ≤15) is being approximated by the area under the normal curve between 4.5 and 15.5—the continuity correction is used for both the upper and lower limits.

When the objective of our investigation is to make an inference about a population proportion p, interest will focus on the sample proportion of successes X/n rather than on X itself. Because this proportion is just X multiplied by the constant 1/n , it will also have approximately a normal distribution (with mean μ = p and standard deviation σ = provided that both np ≥ 10 and nq ≥ 10. This normal approximation is the basis for several inferential procedures to be discussed in later chapters.

EXERCISES Section 4.3 (28–58)

28. Let Z be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate.

a. P(0 ≤Z ≤2.17)

b. P(0 ≤ Z ≤1)

c.P (-2.50 ≤ Z ≤0)

d.P (-2.50 ≤ Z ≤2.50)

e.P (Z ≤ 1.37)

f.P (-1.75 ≤ Z )

g. P(-1.50 ≤ Z ≤2.00)

h.P (1.37 ≤Z ≤2.50)

i.P (1.50 ≤Z )

j. P (|Z | ≤ 2.50)

29. In each case, determine the value of the constant c that mrong>a.Φ(c ) = .9838

b. P(0 ≤ Z ≤ c ) = .291

c. P(c ≤Z ) = .121

d.P (-c ≤ Z ≤ c ) = .668

e. P(c ≤ |Z|) = .016

30. Find the following percentiles for the standard normal distribution. Interpolate where appropriate.

a.91st

b.9th

c.75th

d.25th

e.6th

31. Determine zα for the following:

a. α = .0055

b. α = .09

c. α = .663

32. If X is a normal rv with mean 80 and standard deviation 10, compute the following probabilities by standardizing:

a.P(X ≤100)

b. P(X ≤80)

c.P (65 ≤ X ≤100)

d.P (70 ≤ X )

e.P (85 ≤ X ≤95)

f.P (|X − 80 | ≤ 10)

33. Suppose the force acting on a column that helps to support a building is normally distributed with mean 15.0 kips and standard deviation 1.25 kips. What is the probability that the force
">b. If the acceptable range of hardness is (70 −c , 70 +c ), for what value of c would 95% of all specimens have acceptable hardness?

c. If the acceptable range is as in part (a) and the hardness of each of ten randomly selected specimens is independently determined, what is the expected number of acceptable specimens among the ten?

d. What is the probability that at most eight of ten independently selected specimens have a hardness of less than 73.84? [Hint: Y = the number among the ten specimens with hardness less than 73.84 is a binomial variable; what is p ?]

47. The weight distribution of parcels sent in a certain manner is normal with mean value 12 lb and standard deviation 3.5 lb. The parcel service wishes to establish a weight value c beyond which there will be a surcharge. What value of c is such that 99% of all parcels are at least 1 lb under the surcharge weight?

48. Suppose Appendix Table A.3 contained Φ(z ) only for z ≥ 0. Explain how you could still compute

a.P (-1.72 ≤ Z ≤−.55)

b.P (-1.72 ≤ Z ≤.55)

Is it necessary to table Φ(z ) for z negative? What property of the standard normal curve justifies your answer?

49. Consider babies born in the “normal” range of 37–43 weeks gestational age. Extensive data supports the assumption that for such babies born in the United States, birth weight is normally distributed with mean 3432 g and standard deviation 482 g. [The article “Are Babies Normal?” (The American Statistician (1999): 298-302) analyzed data from a particular year; for a sensible choice of class intervals, a histogram did not look at all normal but after further investigations it was determined that this was due to some hospitals measuring weight in grams and others measuring to the nearest ounce and then converting to grams. A modified choice of class intervals that allowed for this gave a histogram that was well described by a normal distribution.]

a. What is the probability that the birth weight of a randomly selected baby of this type exceeds 4000 grams? Is between 3000 and 4000 grams?

b. What is the probability that the birth weight of a randomly selected baby of this type is either less than 2000 grams or greater than 5000 grams?

c. What is the probability that the birth weight of a randomly selected baby of this type exceeds 7 lb?

d. How would youweights?

e. If X is a random variable with a normal distribution and a is a numerical constant (a Φ 0), then Y = aX also has a normal distribution. Use this to determine the distribution of birth weight expressed in pounds (shape, mean, and standard deviation), and then recalculate the probability from part (c). How does this compare to your previous answer?

50. In response to concerns about nutritional contents of fast foods, McDonald’s has announced that it will use a new cooking oil for its french fries that will decrease substantially trans fatty acid levels and increase the amount of more beneficial polyunsaturated fat. The company claims that 97 out of 100 people cannot detect a difference in taste between the new and old oils. Assuming that this figure is correct (as a long-run proportion), what is the approximate probability that in a random sample of 1000 individuals who have purchased fries at McDonald’s,

a. At least 40 can taste the difference between the two oils?

b. At most 5% can taste the difference between the two oils?

51. Chebyshev’s inequality, (see Exercise 44 Chapter 3), is valid for continuous as well as discrete distributions. It states that for any number k satisfying k ≥ 1, P(|X − μ | ≥ kσ ) ≤1/k 2 (see Exercise 44 in Chapter 3 for an interpretation). Obtain this probability in the case of a normal distribution for k = 1, 2, and 3, and compare to the upper bound.

52. Let X denote the number of flaws along a 100-m reel of magnetic tape (an integer-valued variable). Suppose X has approximately a normal distribution with μ =25 and σ =5. Use the continuity correction to calculate the probability that the number of flaws is

a. Between 20 and 30, inclusive.

b. At most 30. Less than 30.

53. Let X have a binomial distribution with parameters n = 25 and p. Calculate each of the following probabilities using the normal approximation (with the continuity correction) for the cases p = .5, .6, and .8 and compare to the exact probabilities calculated from Appendix Table A.1>c.P (20 ≤X )

54. Suppose that 10% of all steel shafts produced by a certain process are nonconforming but can be reworked (rather than having to be scrapped). Consider a random sample of 200 shafts, and let X denote the number among these that are nonconforming and can be reworked. What is the (approximate) probability that X is

a. At most 30?

b. Less than 30?

c. Between 15 and 25 (inclusive)?

55. Suppose only 75% of all drivers in a certain state regularly wear a seat belt. A random sample of 500 drivers is selected. What is the probability that

a. Between 360 and 400 (inclusive) of the drivers in the sample regularly wear a seat belt?

b. Fewer than 400 of those in the sample regularly wear a seat belt?

56. Show that the relationship between a general normal per-centile and the corresponding z percentile is as stated in this section.

57. a. Show that if X has a normal distribution with parameters μ and σ, then Y = aX + b (a linear function of X ) also has a normal distribution. What are the parameters of the distribution of Y [i.e., E(Y ) and V(Y)]? [Hint: Write the cdf of Y, P(Y ≤ y), as an integral involving the pdf of X, and then differentiate with respect to y to get the pdf of Y.]

b. If when measured in °C, temperature is normally distributed with mean 1w Roman">The expected time between successive calls is 1/.5 = 2 days.

Another important application of the exponential distribution is to model the distribution of component lifetime. A partial reason for the popularity of such applications is the “memoryless” property of the exponential distribution. Suppose component lifetime is exponentially distributed with parameter λ. After putting the component into service, we leave for a period of t_0 hours and then return to find the component still working; what now is the probability that it lasts at least an additional t hours? In symbols, we wish P(X ≥ t + t_0 | X ≥ t_0). By the definition of conditional probability,

But the event X ≥ t_0 in the numerator is redundant, since both events can occur if and only if X ≥ t + t_0. Therefore,

This conditional probability is identical to the original probability P(X ≥ t ) that the component lasted t hours. Thus the distribution of additional lifetime is exactly the same as the original distribution of lifetime, so at each point in time the component shows no effect of wear. In other words, the distribution of remaining lifetime is independent of current age.

Although the memoryless property can be justified at least approximately in many applied problems, in other situations components deteriorate with age or occasionally improve with age (at least up to a certain point). More general lifetime models are then furnished by the gamma, Weibull, and lognormal distributions (the latter two are discussed in the next section).

The Gamma Function

To define the family of gamma distributions, we first need to introduce a function that plays an important role in many branches of mathematics.

DEFINITION

For α >0, the gamma function Γ (α) is defined by

The most important properties of the gamma function are the following:

1. For any α > 1, Γ (α) =(α −1) • Γ(α − 1) [via integration by parts]

2ecindex="00469" alt="Image">

By Expression (4.6), if we let

then f (x ; α) ≤; 0 and f (x ; α) dx = Γ(α)/Γ(α) = 1, so f (x ; α) satisfies the two basic properties of a pdf.

The Gamma Distribution

DEFINITION

A continuous random variable X is said to have a gamma distribution if the pdf of X is

where the parameters α and β satisfy α > 0, β > 0. The standard gamma distribution has β = 1, so the pdf of a standard gamma rv is given by (4.7).

The exponential distribution results from taking α = 1 and β = 1/λ .

Figure 4.27(a) illustrates the graphs of the gamma pdf f (x ; α, β )(4.8) for several (α, β ) pairs, whereas Figure 4.27(b) presents graphs of the standard gamma pdf. For the standard pdf, when α ≤ l, f (x ; α ) is strictly decreasing as x increases from 0; when α > 1, f (x ; α ) rises from 0 at x = 0 to a maximum and then decreases. The parameter β in (4.8) is called the scale parameter because values other than 1 either stretch or compress the pdf in the x direction.

Figure 4.27 (a) Gamma density curves; (b) standard gamma density curves

The mean and variance of a random variable X having the gamma distribution f (x ; α, β ) are

E (X ) = μ = αβ V (X ) = σ 2 = αβ 2

When X is a standard gamma rv, the cdf of X,

c. What is the 99th percentile of the lifetime distribution?

d. Suppose the test will actually be terminated after t weeks. What value of t is such that only .5% of all transistors would still be operating at termination?

68. The special case of the gamma distribution in which σ is a positive integer n is called an Erlang distribution. If we replace β by 1/λ in Expression (4.8), the Erlang pdf is

It can be shown that if the times between successive events are independent, each with an exponential distribution with parameter λ, then the total time X that elapses before all of the next n events occur has pdf f (x ; λ, n ).

a. What is the expected value of X ? If the time (in minutes) between arrivals of successive customers is exponentially distributed with λ =.5, how much time can be expected to elapse before the tenth customer arrives?

b. If customer interarrival time is exponentially distributed with λ =.5, what is the probability that the tenth customer (after the one who has just arrived) will arrive within the next 30 min?

c. The event {X ≤ t } occurs iff at least n events occur in the next t units of time. Use the fact that the number of events occurring in an interval of length t has a Poisson distribution with parameter X t to write an expression (involving Poisson probabilities) for the Erlang cdf F (t; X, n ) = P (X ≤ t ).

69. A system consists of five identical components connected in series as shown:

As soon as one component fails, the entire system will fail. Suppose each component has a lifetime that is exponentially distributed with λ = .01 and that components fail independently of one another. Define events Ai = {ith component lasts at least t hours}, i = 1, …, 5, so that the Ais are independent events. Let X = the time at which the system fails−that is, the shortest (minimum) lifetime among the five components.

Figure 4.28. β is a scale parameter, so different values stretch or compress the graph in the x direction.

Integrating to obtain E (X ) and E (X 2) yields

The computation of μ and σ^2 thus necessitates using the gamma function.

The integration dy is easily carried out to obtain the cdf of X

The cdf of a Weibull rv having parameters α and β is

Example 4.25

In recent years the Weibull distribution has been used to model engine emissions of various pollutants. Let X denote the amount of NOx emission (g/gal) from a randomly selected four-stroke engine of a certain type, and suppose that X has a Weibull distribution with α = 2 and β = 10 (suggested by information in the article “Quantification of Variability and Uncertainty in Lawn and Garden Equipment NOx and Total Hydrocarbon Emission Factors,” J. of the Air and Waste Management Assoc, 2002: 435–448). The corresponding density curve looks exactly like the one in Figure 4.28 for α = 2, β = 1 except that now the values 50 and 100 replace 5 and 10 on the horizontal axis (because β is a “scale parameter”). Then

P (X ≤ 10) = F (10; 2, 10) = 1 − e −(10/10)2 = 1 − e −1 = .632

Similarly, P (X ≤ 25) = .998, so the distribution is almost entirely concentrated on values between 0 and 25. The value c , which separates the 5% of all engines having the largest amounts of NOx emissions from the remaining 95%, satisfies

.95 = 1 − e −(c /10)2

Isolating the exponential term on one side, taking logarithms, and soleight="0em" align="right">

In practical situations, a Weibull model may be reasonable except that the smallest possible X value may be some value γ not assumed to be zero (this would also apply to a gamma model). The quantity γ can then be regarded as a third parameter of the distribution, which is what Weibull did in his original work. For, say, γ = 3, all curves in Figure 4.28 would be shifted 3 units to the right. This is equivalent to saying that X − γ has the pdf (4.11), so that the cdf of X is obtained by replacing x in (4.12) by x − γ .

Figure 4.28 Weibull density curves

Example 4.26

Let X = the corrosion weight loss for a small square magnesium alloy plate immersed for 7 days in an inhibited aqueous 20% solution of MgBr_2. Suppose the minimum possible weight loss is y = 3 and that the excess X −3 over this minimum has a Weibull distribution with α = 2 and β = 4. (This example was considered in “Practical Applications of the Weibull Distribution,” Industrial Quality Control, Aug. 1964: 71-78; values for α and β were taken to be 1.8 and 3.67, respectively, though a slightly different choice of parameters was used in the article.) The cdf of X is then

Therefore,

P (X > 3.5) = 1 –F (3.5; 2, 4, 3) = e -.0156 = .985

and

P (7 ≤ X ≤ 9) = 1 –e -2.25 − (1 − e -1) = .895 –.632 = .263

The Lognormal Distribution

DEFINITION

A nonnegative rv X is said to have a lognormal distribution if the rv Y = ln(X ) has a normal distribution. The resulting pdf of a lognormal rv when ln(X ) is normally distributed with parameters μ and σ is

X but of ln(X ). The mean and variance of X can be shown to be

E (X ) = e μ+σ^2/2 V (X ) = e 2μ+σ^2 • (e σ^2 −1)

In Chapter 5, we will present a theoretical justification for this distribution in connection with the Central Limit Theorem, but as with other distributions, the log-normal can be used as a model even in the absence of such justification. Figure 4.29 illustrates graphs of the lognormal pdf; although a normal curve is symmetric, a log-normal curve has a positive skew.

Because ln(X ) has a normal distribution, the cdf of X can be expressed in terms of the cdf Φ(z ) of a standard normal rv Z .

Figure 4.29 Lognormal density curves

Example 4.27

The lognormal distribution is frequently used as a model for various material properties. The article “Reliability of Wood Joist Floor Systems with Creep” (J. of Structural Engr , 1995: 946-954) suggests that the lognormal distribution with μ = .375 and σ =.25 is a plausible model for X = the modulus of elasticity (MOE, in 10^6 psi) of wood joist floor systems constructed from #2 grade hem-fir. The mean value and variance of MOE are

E (X ) = e .375 + (.25)2/2 = e .40625 = 150

V (X ) = e .8125(e .0625 − 1) = .1453

The probability that MOE is between 1 and 2 is

P (1 ≤ X ≤ 2) =P (ln(1) ≤ ln(X ) ≤ ln(2))

=P (0 ≤ ln(X ) ≤ .693)

= Φ(1.27) –Φ (-1.50) = .8312

What value c is such that only 1% of all days to lay the foundation is

The standard beta distribution is commonly used to model variation in the proportion or percentage of a quantity occurring in different samples, such as the proportion of a 24-hour day that an individual is asleep or the proportion of a certain element in a chemical compound.

EXERCISES Section 4.5 (72–86)

72. The lifetime X (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters α = 2 and β = 3. Compute the following:

a. E (X ) and V (X )

b. P (X ≤ 6)

c. P (1.5 ≤ X ≤ 6)

(This Weibull distribution is suggested as a model for time in service in “On the Assessment of Equipment Reliability: Trading Data Collection Costs for Precision,” J. Engr. Manuf ., 1991: 105-109.)

73. The authors of the article “A Probabilistic Insulation Life Model for Combined Thermal-Electrical Stresses” (IEEE Trans. on Elect. Insulation , 1985: 519-522) state that “the Weibull distribution is widely used in statistical problems relating to aging of solid insulating materials subjected to aging and stress.” They propose the use of the distribution as a model for time (in hours) to failure of solid insulating specimens subjected to AC voltage. The values of the parameters depend on the voltage and temperature; suppose α = 2.5 and β = 200 (values suggested by data in the article).

a. What is the probability that a specimen’s lifetime is at most 250? Less than 250? More than 300?

b. What is the probability that a specimen’s lifetime is between 100 and 250?

c. What value is such that exactly 50% of all specimens have lifetimes exceeding that value?

74. Let X = the time (in 10-1 weeks) from shipment of a defective product until the customer returns the product. Suppose that the minimum return time is γ = 3.5 and that the excess X −3.5 ovity Control article referenced in Example 4.26).

a. What is the cdf of X ?

b. What are the expected return time and variance of return time? [Hint: First obtain E (X − 3.5) and V (X − 3.5).]

c. Compute P (X > 5).

d. Compute P (5 ≤ X ≤ 8).

75. Let X have a Weibull distribution with the pdf from Expression (4.11). Verify that μ = β Γ(1 + 1/α). [Hint: In the integral for E (X ), make the change of variable y = (x /β )α, so that x = βy 1/α.]

76. a. In Exercise 72, what is the median lifetime of such tubes? [Hint: Use Expression (4.12).]

b. In Exercise 74, what is the median return time?

c. If X has a Weibull distribution with the cdf from Expression (4.12), obtain a general expression for the (100p )th per-centile of the distribution.

d. In Exercise 74, the company wants to refuse to accept returns after t weeks. For what value of t will only 10% of all returns be refused?

77. The authors of the paper from which the data in Exercise 1.27 was extracted suggested that a reasonable probability model for drill lifetime was a lognormal distribution with μ = 4.5 and σ = .8.

a. What are the mean value and standard deviation of lifetime?

b. What is the probability that lifetime is at most 100?

c. What is the probability that lifetime is at least 200? Greater than 200?

78. The article “On Assessing the Accuracy of Offshore Wind Turbine Reliability-Based Design Loads from the Environmental Contour Method” (Intl. J. of Offshore and Polar Engr , 2005: 132-140) proposes the Weibull distribution with α = 1.8>

b. What is the probability that wave height exceeds its mean value by more than one standard deviation?

c. What is the median of the wave-height distribution?

d. For 0 < p < 1, give a general expression for the 100pth percentile of the wave-height distribution.

79. Let X = the hourly median power (in decibels) of received radio signals transmitted between two cities. The authors of the article “Families of Distributions for Hourly Median Power and Instantaneous Power of Received Radio Signals” (J. Research National Bureau of Standards , vol. 67D, 1963: 753-762) argue that the lognormal distribution provides a reasonable probability model for X . If the parameter values are μ = 3.5 and σ = 1.2, calculate the following:

a. The mean value and standard deviation of received power.

b. The probability that received power is between 50 and 250 dB.

c. The probability that X is less than its mean value. Why is this probability not .5?

80. a. Use Equation (4.13) to write a formula for the median of the lognormal distribution. What is the median for the power distribution of Exercise 79?

b. Recalling that zα is our notation for the 100(1 –α) percentile of the standard normal distribution, write an expression for the 100(1 –α) percentile of the lognormal distribution. In Exercise 79, what value will received power exceed only 5% of the time?

81. A theoretical justification based on a certain material failure mechanism underlies the assumption that ductile strength X of a material has a lognormal distribution. Suppose the parameters are μ = 5 and σ = .1.

a. Compute E (X ) and V (X ).

b. Compute P (X > 125).

c. Compute P (110 ≤ X ≤ 125).

b. Compute P (8 ≤ Y ≤ 12).

c. Compute the probability that the bar snaps more than 2 in. from where you expect it to.

4.6 Probability Plots

An investigator will often have obtained a numerical sample x_1, x_2,…, xn and wish to know whether it is plausible that it came from a population distribution of some particular type (e.g., from a normal distribution). For one thing, many formal procedures from statistical inference are based on the assumption that the population distribution is of a specified type. The use of such a procedure is inappropriate if the actual underlying probability distribution differs greatly from the assumed type. Additionally, understanding the underlying distribution can sometimes give insight into the physical mechanisms involved in generating the data. An effective way to check a distributional assumption is to construct what is called a probability plot. The essence of such a plot is that if the distribution on which the plot is based is correct, the points in the plot will fall close to a straight line. If the actual distribution is quite different from the one used to construct the plot, the points should depart substantially from a linear pattern.

Sample Percentiles

The details involved in constructing probability plots differ a bit from source to source. The basis for our construction is a comparison between percentiles of the sample data and the corresponding percentiles of the distribution under consideration. Recall that the (100p )th percentile of a continuous distribution with cdf F (•) is the number η(p ) that satisfies F (η(p )) = p . That is, η(p ) is the number on the measurement scale such that the area under the density curve to the left of η(p ) is p . Thus the 50th percentile η(.5) satisfies F (η(.5)) = .5, and the 90th percentile satisfies F (η(.9)) = .9. Consider as an example the standard normal distribution, for which we have denoted the cdf by Φ(•). From Appendix Table A.3, we find the 20th percentile by locating the row and column in which .2000 (or a number as close to it as possible) appears inside the table. Since .2005 appears at the intersection of the -.8 row and the .04 column, the 20th percentile is approximately -.84. Similarly, the 25th percentile of the standard normal distribution is (using linear interpolation) approximately -.675.

Roughly speaking, sample percentiles are defined in the same way that percentiles of a population distribution are defined. The 50th-sample percentile should separate the smallest 50% of the sample from the largest 50%, the 90th percentile should be such that 90% of the sample lies below that value and 10% lies above, and so on. Unfortunately, we run into problems when we actually try to compute the sample percentiles for a particular sample of n observations. If, for example, n = 10, we can split offy 23% of these ten observations. To proceed further, we need an operational definition of sample percentiles (this is one place where different people do slightly different things). Recall that when n is odd, the sample median or 50th-sample percentile is the middle value in the ordered list, for example, the sixth largest value when n =11. This amounts to regarding the middle observation as being half in the lower half of the data and half in the upper half. Similarly, suppose n = 10. Then if we call the third smallest value the 25th percentile, we are regarding that value as being half in the lower group (consisting of the two smallest observations) and half in the upper group (the seven largest observations). This leads to the following general definition of sample percentiles.

DEFINITION

Order the n sample observations from smallest to largest. Then the ith smallest observation in the list is taken to be the [100(i –.5)/n ]th sample percentile.

Once the percentage values 100(i –.5)/n (i = 1, 2,…, n ) have been calculated, sample percentiles corresponding to intermediate percentages can be obtained by linear interpolation. For example, if n = 10, the percentages corresponding to the ordered sample observations are 100(1 –.5)/10 = 5%, 100(2 –.5)/10 =1 5%, 25%, …, and 100(10 –.5)/10 =95%. The 10th percentile is then halfway between the 5th percentile (smallest sample observation) and the 15th percentile (second smallest observation). For our purposes, such interpolation is not necessary because a probability plot will be based only on the percentages 100(i –.5)/n corresponding to the n sample observations.

A Probability Plot

Suppose now that for percentages 100(i –.5)/n (i = 1, …, n ) the percentiles are determined for a specified population distribution whose plausibility is being investigated. If the sample was actually selected from the specified distribution, the sample percentiles (ordered sample observations) should be reasonably close to the corresponding population distribution percentiles. That is, for i = 1, 2, … , n there should be reasonable agreement between the ith smallest sample observation and the [100(i –.5)/n ]th percentile for the specified distribution. Consider the (population percentile, sample percentile) pairs–that is, the pairs

for i = 1, …, n . Each such pair can be plotted as a point on a two-dimensional coordinate system. If the sample percentiles are close to the corresponding population distribution percentiles, the first number in each pair will be roughly equal to the second number. The plotted points will then fall close to a 45° line. Substantial deviations of the plotted points from a 45° line cast doubt on the assumption that the distribution under consideration is the correct one.

Example 4.29 The value of a certain physical constant is known to an experimensessing the plausibility of an entire family. If the plot deviates substantially from a straight line, no member of the family is plausible. When the plot is quite straight, further work is necessary to estimate values of the parameters that yield the most reasonable distribution of the specified type.

Let’s focus on a plot for checking normality. Such a plot is useful in applied work because many formal statistical procedures give accurate inferences only when the population distribution is at least approximately normal. These procedures should generally not be used if the normal probability plot shows a very pronounced departure from linearity. The key to constructing an omnibus normal probability plot is the relationship between standard normal (z ) percentiles and those for any other normal distribution:

Consider first the case, μ = 0. If each observation is exactly equal to the corresponding normal percentile for some value of σ, the pairs (σ • [z percentile], observation) fall on a 45° line, which has slope 1. This implies that the (z percentile, observation) pairs fall on a line passing through (0, 0) (i.e., one with y -intercept 0) but having slope σ rather than 1. The effect of a nonzero value of μ is simply to change the y -intercept from 0 to μ.

A plot of the n pairs

([100(i –.5)/n ]th z percentile, ith smallest observation)

on a two-dimensional coordinate system is called a normal probability plot. If the sample observations are in fact drawn from a normal distribution with mean value μ and standard deviation σ, the points should fall close to a straight line with slope σ and intercept μ. Thus a plot for which the points fall close to some straight line suggests that the assumption of a normal population distribution is plausible.

Example 4.30

The accompanying sample consisting of n = 20 observations on dielectric breakdown voltage of a piece of epoxy resin appeared in the article “Maximum Likelihood Estimation in the 3-Parameter Weibull Distribution (IEEE Trans. on Dielectrics and Elec. Insul ., 1996: 43-55). The values of (i –.5)/n for which z percentiles are needed are (1 –.5)/20 = .025, (2 –.5)/20 = .075, …, and .975.

Figure 4.33963; of the normal family are location and scale parameters, respectively. Changing μ shifts the location of the bell-shaped density curve to the right or left, and changing σ amounts to stretching or compressing the measurement scale (the scale on the horizontal axis when the density function is graphed). Another example is given by the cdf

F (x ; θ_1, θ_2) = 1 − e -e(x-θ_1)/θ_2 -∞ < x < ∞

A random variable with this cdf is said to have an extreme value distribution . It is used in applications involving component lifetime and material strength.

Although the form of the extreme value cdf might at first glance suggest that θ_1 is the point of symmetry for the density function, and therefore the mean and median, this is not the case. Instead, P (X ≤ θ_1) = F (θ_1; θ_1, θ_2) = 1 –e -1 = .632, and the density function f (x ; θ_1, θ_2) = F ’(x ; θ_1, θ_2) is negatively skewed (a long lower tail). Similarly, the scale parameter θ_2 is not the standard deviation (μ = θ_1 −.5772θ_2 and σ = 1.283θ_2). However, changing the value of θ_1 does change the location of the density curve, whereas a change in θ_2 rescales the measurement axis.

The parameter β of the Weibull distribution is a scale parameter, but α is not a location parameter. The parameter a is usually referred to as a shape parameter. A similar comment applies to the parameters a and β of the gamma distribution. In the usual form, the density function for any member of either the gamma or Weibull distribution is positive for x > 0 and zero otherwise. A location parameter can be introduced as a third parameter γ (we did this for the Weibull distribution) to shift the density function so that it is positive if x > γ and zero otherwise.

When the family under consideration has only location and scale parameters, the issue of whether any member of the family is a plausible population distribution can be addressed via a single, easily constructed probability plot. One first obtains the percentiles of the standard distribution , the one with θ_1 = 0 and θ_2 =1, for percentages 100(i –.5)/n (i = 1, …, n ). The n (standardized percentile, observation) pairs give the points in the plot. This is exarong>a. Construct a normal probability plot. Is normality plausible?

b. Construct a Weibull probability plot. Is the Weibull distribution family plausible?

93. Construct a probability plot that will allow you to assess the plausibility of the lognormal distribution as a model for the rainfall data of Exercise 83 (Chapter 1).

94. The accompanying observations are precipitation values during March over a 30-year period in Minneapolis-St. Paul.

a. Construct and interpret a normal probability plot for this data set.

b. Calculate the square root of each value and then construct a normal probability plot based on this transformed data. Does it seem plausible that the square root of precipitation is normally distributed?

c. Repeat part (b) after transforming by cube roots.

95. Use a statistical software package to construct a normal probability plot of the tensile ultimate strength data given in Exercise 13 of Chapter 1, and comment.

96. Let the ordered sample observations be denoted by y_1, y_2,…, yn (y_1 being the smallest and yn the largest). Our suggested check for normality is to plot the (Φ-1((i –.5)/n ), yi ) pairs. Suppose we believe that the observations come from a distribution with mean 0, and let w_1,…, wn be the ordered absolute values of the xis . A half-normal plot is a probability plot of the wi s. More specifically, since P (|Z | ≤ w ) = P (–w ≤ Z ≤ w ) = 2Φ(w ) –1, a half-normal plot is a plot of the (Φ-1{[(i − .5)/n + 1]/2}, wi ) pairs. The virtue of this plot is that small or large outliers in the original sample will now appear only at the upper end of the plot rather than at both ends. Construct a half-normal plot for the following sample of measurement errors, and comment: -3.78, -1.27, 1.44, -.39, 12.38, -43.40, 1.15, -3.96, -2.34, 30.84.

97. The following failure time observations (1000s of hours) resulted from accelerated life testing of 16 integrated circuit chips of a certain type:

Use the corresponding percentiles of the exponential distribution with λ =1 to construct a probability plot. Then explain why the plot assesses the plausibility of the sample having been generated from any exponential distribution.

SUPPLEMENTARY EXERCISES (98–128)

98. Let X = the time it takes a read/write head to locate a desired record on a computer disk memory device once the head has been positioned over the correct track. If the disks rotate once every 25 millisec, a reasonable assumption is that X is uniformly distributed on the interval [0, 25].

a. Compute P (10 ≤ X ≤ 20).

b. Compute P (X ≥ 10).

c. Obtain the cdf F (X ).

d. Compute E (X ) and σX.

99. A 12-in. bar that is clamped at both ends is to be subjected to an increasing amount of stress until it snaps. Let Y = the distance from the left end at which the break occurs. Suppose Y has pdf

Compute the following:

a. The cdf of Y , and graph it.

b. P (Y ≤ 4), P (Y > 6), and P (4 ≤ Y ≤ 6)

c. E (Y ), E (Y 2), and V (Y )

d. The probability that the break point occurs more than 2 in. from the expected break point

e. The expected length of the shorter segment when the break occurs

100. Let X denote the time to failure (in years) of a certain hydraulic component. Suppose the pdf of X is f (x ) = 32/(x + 4)3 for x > 0.

a. Verify that f (x ) is a legitimate pdf.

b

c. Use the result of part (b) to calculate the probability that time to failure is between 2 and 5 years.

d. What is the expected time to failure?

e. If the component has a salvage value equal to 100/(4 + x ) when its time to failure is x , what is the expected salvage value?

101. The completion time X for a certain task has cdf F (x ) given by

a. Obtain the pdf f (x ) and sketch its graph.

b. Compute P (.5 ≤ X ≤ 2).

c. Compute E (X ).

102. The breakdown voltage of a randomly chosen diode of a certain type is known to be normally distributed with mean value 40 V and standard deviation 1.5 V.

a. What is the probability that the voltage of a single diode is between 39 and 42?

b. What value is such that only 15% of all diodes have voltages exceeding that value?

c. If four diodes are independently selected, what is the probability that at least one has a voltage exceeding 42?

103. The article “Computer Assisted Net Weight Control” (Quality Progress , 1983: 22-25) suggests a normal distribution with mean 137.2 oz and standard deviation 1.6 oz for the actual contents of jars of a certain type. The stated contents was 135 oz.

a. What is the probability that a single jar contains more than the stated contents?

b. Among ten randomly selected jars, what is the probability that at least eight contain more than the stated contents?

c. Assuming that the mean remains at 137.2, to what value would the standard deviation have to be changed so that 95% of all jars contain more than the stated contents?

104. When cir condition of any particular board is independent of that of any other board.

a. What is the approximate probability that at least 10% of the boards in the batch are defective?

b. What is the approximate probability that there are exactly 10 defectives in the batch?

105. The article “Characterization of Room Temperature Damping in Aluminum-Indium Alloys” (Metallurgical Trans ., 1993: 1611-1619) suggests that Al matrix grain size (μm) for an alloy consisting of 2% indium could be modeled with a normal distribution with a mean value 96 and standard deviation 14.

a. What is the probability that grain size exceeds 100?

b. What is the probability that grain size is between 50 and 80?

c. What interval (a, b ) includes the central 90% of all grain sizes (so that 5% are below a and 5% are above b)?

106. The reaction time (in seconds) to a certain stimulus is a continuous random variable with pdf

a. Obtain the cdf.

b. What is the probability that reaction time is at most 2.5 sec? Between 1.5 and 2.5 sec?

c. Compute the expected reaction time.

d. Compute the standard deviation of reaction time.

e. If an individual takes more than 1.5 sec to react, a light comes on and stays on either until one further second has elapsed or until the person reacts (whichever happens first). Determine the expected amount of time that the light remains lit. [Hint: Let h (X ) = the time that the light is on as a function of reaction time X .]

107. Let X denote the temperature at which a certain chemical reaction takes place. Suppose that X has pdf

c. Is 0 the median temperature at which the reaction takes place? If not, is the median temperature smaller or larger than 0?

d. Suppose this reaction is independently carried out once in each of ten different labs and that the pdf of reaction time in each lab is as given. Let Y = the number among the ten labs at which the temperature exceeds 1. What kind of distribution does Y have? (Give the name and values of any parameters.)

108. The article “Determination of the MTF of Positive Photoresists Using the Monte Carlo Method” (Photographic Sci. and Engr , 1983: 254–260) proposes the exponential distribution with parameter λ = .93 as a model for the distribution of a photon’s free path length (μm) under certain circumstances. Suppose this is the correct model.

a. What is the expected path length, and what is the standard deviation of path length?

b. What is the probability that path length exceeds 3.0? What is the probability that path length is between 1.0 and 3.0?

c. What value is exceeded by only 10% of all path lengths?

109. The article “The Prediction of Corrosion by Statistical Analysis of Corrosion Profiles” (Corrosion Science , 1985: 305-315) suggests the following cdf for the depth X of the deepest pit in an experiment involving the exposure of carbon manganese steel to acidified seawater.

F (x ; α , β) =e -e -(x -α)/β -∞ < x < ∞

The authors propose the values α = 150 and β = 90. Assume this to be the correct model.

a. What is the probability that the depth of the deepest pit is at most 150? At most 300? Between 150 and 300?

b. Below what value will the depth of the maximum pit be observed in 90% of all such experiments?

c. What is the density function of X ?

d. The density function can be shown to be unimodal (a single peak). Above what value on the measurement axis does this peak occur? (This value is the mode.)

e. It can be shown that Em> and sketch it.

c. Compute P (X ≤ 0), P (X ≤ 2), P (-1 ≤ X ≤ 2), and the probability that an error of more than 2 miles is made.

113. In some systems, a customer is allocated to one of two service facilities. If the service time for a customer served by facility i has an exponential distribution with parameter λi (i =1, 2) and p is the proportion of all customers served by facility 1, then the pdf of X = the service time of a randomly selected customer is

This is often called the hyperexponential or mixed exponential distribution. This distribution is also proposed as a model for rainfall amount in “Modeling Monsoon Affected Rainfall of Pakistan by Point Processes” (J. Water Resources Planning and Mgmnt ., 1992: 671-688).

a. Verify that f (x ; λ1, λ_2, p ) is indeed a pdf.

b. What is the cdf F (x ; λ_1, λ_2, p )?

c. If X has f (x ; λ1, λ_2, p ) as its pdf, what is E (X )?

d. Using the fact that E (X 2) = 2/λ^2 when X has an exponential distribution with parameter A, compute E (X 2) when X has pdf f (x ; λ1, λ_2, p ). Then compute V (X ).

e. The coefficient of variation of a random variable (or distribution) is CV = σ/μ. What is CV for an exponential rv? What can you say about the value of CV when X has a hyperexponential distribution?

f. What is CV for an Erlang distribution with parameters λ and n as defined in Exercise 68? [Note: In applied work, the sample CV is used to decide which of the three distributions might be appropriate.]

114. Suppose a particular state allows individuals filing tax returns to itemize deductions only if the total of all itemized deductions is at least $5000. Let X (in 1000s of dollars) be the total of itemized deductions on a randomly chosen form. Assume that X has the pdf

a. Find the value of k . What restriction on a is necessary?

b. What is the cdf of X ?

c. What is the expected total deduction on a randomly chosen form? What restriction on α is necessary for E (X ) to be finite?

d. Show that ln(X /5) has an exponential distribution with parameter α –1.

115. Let Ii be the input current to a transistor and I_0 be the output current. Then the current gain is proportional to ln(I_0/Ii). Suppose the constant of proportionality is 1 (which amounts to choosing a particular unit of measurement), so that current gain = X = ln(I_0/Ii). Assume X is normally distributed with μ = 1 and σ = .05.

a. What type of distribution does the ratio I_0/Ii have?

b. What is the probability that the output current is more than twice the input current?

c. What are the expected value and variance of the ratio of output to input current?

116. The article “Response of SiCf/Si_3N_4 Composites Under Static and Cyclic Loading–An Experimental and Statistical Analysis” (J. of Engr. Materials and Technology , 1997: 186-193) suggests that tensile strength (MPa) of composites under specified conditions can be modeled by a Weibull distribution with α = 9 and β = 180.

a. Sketch a graph of the density function.

b. What is the probability that the strength of a randomly selected specimen will exceed 175? Will be between 150 and 175?

c. If two randomly selected specimens are chosen and their strengths are independent of one another, what is the probability that at least one has a strength between 150 and 175?

d. What strength value separates the weakest 10% of all specimens from the remaining 90%?

117. Let Z have a standard normal distribution and define a new rv Y by Y = σ Z + μ. Show that Y has a normal distribution with parameters μ and σ. [Hint: Y ≤ y id then differentiate it with respect to y ]

118. a. Suppose the lifetime X of a component, when measured in hours, has a gamma distribution with parameters a and β . Let Y = the lifetime measured in minutes. Derive the pdf of Y [Hint: Y ≤ y iff X ≤ y /60. Use this to obtain the cdf of Y and then differentiate to obtain the pdf.]

b. If X has a gamma distribution with parameters α and β , what is the probability distribution of Y = cX?

119. In Exercises 111 and 112, as well as many other situations, one has the pdf f (x ) of X and wishes to know the pdf of Y = h (X ). Assume that h (•) is an invertible function, so that y = h (x ) can be solved for x to yield x = k (y ). Then it can be shown that the pdf of Y is

g (y ) = f [k (y )] • |k’(y )|

a. If X has a uniform distribution with A = 0 and B = 1, derive the pdf of Y = –ln(X ).

b. Work Exercise 117, using this result.

c. Work Exercise 118(b), using this result.

120. Based on data from a dart-thro wing experiment, the article “Shooting Darts” (Chance , Summer 1997, 16-19) proposed that the horizontal and vertical errors from aiming at a point target should be independent of one another, each with a normal distribution having mean 0 and variance σ^2. It can then be shown that the pdf of the distance V from the target to the landing point is

a. This pdf is a member of what family introduced in this chapter?

b. If σ = 20 σm (close to the value suggested in the paper), what is the probability that a dart will land within 25 σm (roughly 1 in.) of the target?

121. The article “Three Sisters Give Birth on the Same Day” (Chance , Spring 2001, 23-25) used the fact that three Utah sisters had all given birth on March 11, 1998 as a basis for posing some interesting questions regarding birth coincidences.

a. Disregarding leap year and assuming that the other 365 days are equally likely, what is the probability that thre="0em"> Pined on the sample space S of an experiment. The joint probability mass function p (x, y ) is defined for each pair of numbers (x, y ) by

p (x, y ) = P (X = x and Y = y )

It must be the case that p (x, y ) ≥ 0 and p (x, y ) = 1.

Now let A be any set consisting of pairs of (x, y ) values (e.g., A = {(x, y ): x + y = 5} or {(x, y ): max(x, y ) ≤ 3}). Then the probability P [(X, Y ) ∈ A] is obtained by summing the joint pmf over pairs in A:

Example 5.1

A large insurance agency services a number of customers who have purchased both a homeowner’s policy and an automobile policy from the agency. For each type of policy, a deductible amount must be specified. For an automobile policy, the choices are $100 and $250, whereas for a homeowner’s policy, the choices are 0, $100, and $200. Suppose an individual with both types of policy is selected at random from the agency’s files. Let X = the deductible amount on the auto policy and Y = the deductible amount on the homeowner’s policy. Possible (X, Y ) pairs are then (100, 0), (100, 100), (100, 200), (250, 0), (250, 100), and (250, 200); the joint pmf specifies the probability associated with each one of these pairs, with any other pair having probability zero. Suppose the joint pmf is given in the accompanying joint probability table:

Then p (100, 100) = P (X = 100 and Y = 100) = P ($100 deductible on both policies) = .10. The probability P (Y ≥ 100) is computed by summing probabilities of all (x, y ) pairs for which y ≥ 100:

P (Y ≥ 100) = p (100, 100) + p (250, 100) + p (100, 200) + p (250, 200)

= .75

The pmf of one of the variables alone is obtained by summing p (x, y ) over values of the other variable. The result is called a marginal pmf because when the p (x, y ) values appear in a rectangular table, the sums are just marginal (row or column) totals.

DEFINITION

Y (y ), respectively, are given by

Thus to obtain the marginal pmf of X evaluated at, say, x = 100, the probabilities p (100, y) are added over all possible y values. Doing this for each possible X value gives the marginal pmf of X alone (without reference to Y ). From the marginal pmf’s, probabilities of events involving only X or only Y can be computed.

Example 5.2 (Example 5.1 continued)

The possible X values are x = 100 and x = 250, so computing row totals in the joint probability table yields

pX(100) = p (100, 0) + p (100, 100) + p (100, 200) = .50

and

pX (250) = p (250, 0) + p (250, 100) + p (250, 200) = .50

The marginal pmf of X is then

Similarly, the marginal pmf of Y is obtained from column totals as

so P (Y ≥ 100) = pY (100) + pY (200) = .75 as before.

Two Continuous Random Variables

The probability that the observed value of a continuous rv X lies in a one-dimensional set A (such as an interval) is obtained by integrating the pdf f (x ) over the set A. Similarly, the probability that the pair (X, Y ) of continuous rv’s falls in a two-dimensional set A (such as a rectangle) is obtained by integrating a function called the joint density function.

DEFINITION

Let X and Y be continuous rv’s. A joint probability density function f (x, y ) for these two variables is a function satisfying f (x, y ) ≥ 0 and dx dy = 1. Then for any two-dimensional set A

In particular, if A is : a ≤ x ≤ b, c ≤ y ≤ d}, then

We can think of f (x, y ) as specifying a surface at height f (x, y ) above the point (x, y ) in a three-dimensional coordinate system. Then P [(X, Y ) ∈ A ] is the volume underneath this surface and above the region A, analogous to the area under a curve in the one-dimensional case. This is illustrated in Figure 5.1.

Figure 5.1 P [(X, Y ) ∈ A = volume under density surface above A

Example 5.3

A bank operates both a drive-up facility and a walk-up window. On a randomly selected day, let X = the proportion of time that the drive-up facility is in use (at least one customer is being served or waiting to be served) and Y = the proportion of time that the walk-up window is in use. Then the set of possible values for (X, Y ) is the rectangle D = {(x, y ): 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}. Suppose the joint pdf of (X, Y ) is given by

To verify that this is a legitimate pdf, note that f (x, y ) ≥ 0 and

The probability that neither facility is busy more than one-quarter of the time is

As with joint pmf’s, from the joint pdf of X and Y , each of the two marginal density functions can be computed.

DEFINITION

The marginal probability density functions of X and Y , denoted by fX (x ) and fY (y ), respectively, are given by

Example 5.4 (Example 5.3 continued)

The marginal pdf of X, which gives the probability distribution of busy time for the dt the system lasts past time t is

Therefore,

P(system lifetime ≤ t) = 1 − e-nλt for t ≥ 0

which shows that system lifetime has an exponential distribution with parameter nλ; the expected value of system lifetime is 1/n λ.

In many experimental situations to be considered in this book, independence is a reasonable assumption, so that specifying the joint distribution reduces to deciding on appropriate marginal distributions.

Conditional Distributions

Suppose X = the number of major defects in a randomly selected new automobile and Y = the number of minor defects in that same auto. If we learn that the selected car has one major defect, what now is the probability that the car has at most three minor defects—that is, what is P (Y ≤ 3 | X = 1)? Similarly, if X and Y denote the lifetimes of the front and rear tires on a motorcycle, and it happens that X = 10,000 miles, what now is the probability that Y is at most 15,000 miles, and what is the expected lifetime of the rear tire “conditional on” this value of X ? Questions of this sort can be answered by studying conditional probability distributions.

DEFINITION

Let X and Y be two continuous rv’s with joint pdf f (x, y ) and marginal X pdf fX(x ). Then for any X value x for which fX(x ) ≥ 0, the conditional probability density function of Y given that X = x is

If X and Y are discrete, replacing pdf’s by pmf’s in this definition gives the conditional probability mass function of Y when X = x.

Notice that the definition of fY |X (y | x ) parallels that of P (B | A ), the conditional probability that B will occur, given that A has occurred. Once the conditional pdf or pmf has been determined, questions of the type posed at the outset of this subsection can be answered by integrating or summing over an appropriate set of Y values.

Example 5.12of time that a bank’s drive-up facility is busy and Y = the analogous proportion for the walk-up window. The conditional pdf of Y given that X = .8 is

The probability that the walk-up facility is busy at most half the time given that X= .8 is then

Using the marginal pdf of Y gives P (Y—.5) = .350. Also E (Y ) = .6, whereas the expected proportion of time that the walk-up facility is busy given that X= .8 (a conditional expectation) is

EXERCISES Section 5.1 (1–21)

Y(3) = .05, pY(4) = .2, display the joint pmf of (X , Y ) in a joint probability table.

b. Compute P (X ≤ 1 and Y ≤ 1) from the joint probability table, and verify that it equals the product P (X ≤ 1) · P (Y ≤ 1).

c. What is P (X + Y = 0) (the probability of no violations)?

d. Compute P (X + Y ≤ 1).

3. A certain market has both an express checkout line and a superexpress checkout line. Let X_1 denote the number of customers in line at the express checkout at a particular time of day, and let X_2 denote the number of customers in line at the superexpress checkout at the same time. Suppose the joint pmf of X_1 and X_2 is as given in the accompanying table.

a. What is P(X_1 = 1, X_2 = 1), that is, the probability that there is exactly one customer in each line?

b. What is P(X_1 = X_2), that is, the probability that the numbers of customers in the two lines are identical?

c. Let A denote the event that there are at least two more customers in one line than in the other line. Express A in terms of X_1 and X_2, and calculate the probability of this event.

d. What is the probability that the total number of customers in the two lines is exactly four? At least four?

4. Return to the situation described in Exercise 3.

a. Determine the marginal pmf of X_1, and then calculate the expected number of customers in line at the express checkout.

b. Determine the marginal pmf of X_2.

c. By inspection of the probabilities P (X_1 = 4), P (X_2 = 0), and P (X_1 = 4, X_2 = 0), are 17;s? Explain.

8. A stockroom currently has 30 components of a certain type, of which 8 were provided by supplier 1, 10 by supplier 2, and 12 by supplier 3. Six of these are to be randomly selected for a particular assembly. Let X = the number of supplier 1s components selected, Y = the number of supplier 2s components selected, and p (x, y ) denote the joint pmf of X and Y.

a. What is p (3, 2)? [Hint: Each sample of size 6 is equally likely to be selected. Therefore, p (3, 2) = (number of outcomes with X = 3 and Y = 2)/(total number of outcomes). Now use the product rule for counting to obtain the numerator and denominator.]

b. Using the logic of part (a), obtain p (x, y). (This can be thought of as a multivariate hypergeometric distribution—sampling without replacement from a finite population consisting of more than two categories.)

9. Each front tire on a particular type of vehicle is supposed to be filled to a pressure of 26 psi. Suppose the actual air pressure in each tire is a random variable —X for the right tire and Y for the left tire, with joint pdf

a. What is the value of K?

b. What is the probability that both tires are underfilled?

c. What is the probability that the difference in air pressure between the two tires is at most 2 psi?

d. Determine the (marginal) distribution of air pressure in the right tire alone.

e. Are X and Y independent rv’s?

10. Annie and Alvie have agreed to meet between 5:00 P.M. and 6:00 P.M. for dinner at a local health-food restaurant. Let X = Annie’s arrival time and Y = Alvie’s arrival time. Suppose X and Y are independent with each uniformly distributed on the interval [5, 6].

a. What is the joint pdf of X and Y?

b. What is the probability that they both arrive between 5:15 and 5:45?

c. If the f/em> The event of interest is A = {(x, y ): | x –y | ≤ }.]

11. Two different professors have just submitted final exams for duplication. Let X denote the number of typographical errors on the first professor’s exam and Y denote the number of such errors on the second exam. Supposet: Draw a picture of the region A = {(x, y ): x ≥ 0, y ≥ 0, x + y –2} before integrating.]

d. What is the probability that the total lifetime is between 1 and 2?

14. Suppose that you have ten lightbulbs, that the lifetime of each is independent of all the other lifetimes, and that each lifetime has an exponential distribution with parameter λ.

a. What is the probability that all ten bulbs fail before time t?

b. What is the probability that exactly k of the ten bulbs fail before time t?

c. Suppose that nine of the bulbs have lifetimes that are exponentially distributed with parameter λ and that the remaining bulb has a lifetime that is exponentially distributed with parameter θ (it is made by another manufacturer). What is the probability that exactly five of the ten bulbs fail before time t?

15. Consider a system consisting of three components as pictured. The system will continue to function as long as the first component functions and either component 2 or component 3 functions. Let X_1, X_2, and X_3 denote the lifetimes of components 1, 2, and 3, respectively. Suppose the Xis are independent of one another and each Xi has an exponential distribution with parameter λ.

a. Let Y denote the system lifetime. Obtain the cumulative distribution function of Y and differentiate to obtain the pdf. [Hint: F (y ) = P (Y ≤ y); express the event {Y ≤ y } in terms of unions and/or intersections of the three events {X_1 ≤ y }, {X_2 ≤ y }, and {X_3 ≤ y }.]

b. Compute the expected system lifetime.

16. a. For f (x_1, x_2, x_3) as given in Example 5.10, compute the joint marginal density function of X_1 and X_3 alone (by integrating over x_2).

b. What is the probability that rocks of types 1 and 3 together make up at most 50% of the sample? [Hint: Use the result of part (a).]

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c. If the pressure in the right tire is found to be 22 psi, what is the expected pressure in the left tire, and what is the standard deviation of pressure in this tire?

20. Let X_1, X_2, X_3, X_4, X_5, and X_6 denote the numbers of blue, brown, green, orange, red, and yellow M&M candies, respectively, in a sample of size n. Then these Xis have a multinomial distribution. According to the M&M web site, the color proportions are p_1 = .24, p_2 = .13, p_3 = .16, p_4 = .20, p_5 = .13, and p_6 = .14.

a. If n = 12, what is the probability that there are exactly two M&Ms of each color?

b. For n = 20, what is the probability that there are at most five orange candies? [Hint: Think of an orange candy as a success and any other color as a failure.]

c. In a sample of 20 M&Ms, what is the probability that the number of candies that are blue, green, or orange is at least 10?

21. Let X_1, X_2, and X_3 be the lifetimes of components 1, 2, and 3 in a three-component system.

a. How would you define the conditional pdf of X_3 given that X_1 = x_1 and X_2 = x_2?

b. How would you define the conditional joint pdf of X_2 and X_3 given that X_1 = x_1?

5.2 Expected Values, Covariance, and Correlation

We previously saw that any function h (X ) of a single rv X is itself a random variable. However, to compute E [h (X )], it was not necessary to obtain the probability distribution of h (X ); instead, E [h (X )] was computed as a weighted average of h (x ) values, where the weight function was the pmf p (x ) or pdf f (x ) of X. A similar result holds for a function h (X, Y ) of two jointly distributed random variables.

PROPOSITION

Let X and Y be jointly distributed rv’s with pmf p (x, y ) or pdf f (x, y ) according to whether the variables are dis )] or μh (X Y ), is given by

Example 5.13

Five friends have purchased tickets to a certain concert. If the tickets are for seats 1-5 in a particular row and the tickets are randomly distributed among the five, what is the expected number of seats separating any particular two of the five? Let X and Y denote the seat numbers of the first and second individuals, respectively. Possible (X , Y ) pairs are {(1, 2), (1, 3),…, (5, 4)}, and the joint pmf of (X, Y ) is

The number of seats separating the two individuals is h (X , Y ) = |X − Y| – 1. The accompanying table gives h (x, y ) for each possible (x, y ) pair.

Thus

Example 5.14

In Example 5.5, the joint pdf of the amount X of almonds and amount Y of cashews in a 1-lb can of nuts was

If 1 lb of almonds costs the company $1.00, 1 lb of cashews costs $1.50, and 1 lb of peanuts costs $.50, then the total cost of the contents of a can is

h (X, Y ) = (1)X + (1.5)Y + (.5)(1 − X − Y ) = .5 + .5X + Y

(since 1 − X − Y of the weight consists of peanuts). The expected total cost is

The method of computing the expected value of a function h(X_1,…, Xn) of n random variables is similar to that for two random variables. If the Xis are discrete, E [h (X_1,…, Xn )] is an n -dimensional sum; is X and Y are not independent, it is frequently of interest to assess how strongly they are related to one another.

DEFINITION

The covariance between two rv’s X and Y is

That is, since X –μX and Y –μY are the deviations of the two variables from their respective mean values, the covariance is the expected product of deviations. Note that Cov(X, X ) = E [(X –μX )2] = V (X ).

The rationale for the definition is as follows. Suppose X and Y have a strong positive relationship to one another, by which we mean that large values of X tend to occur with large values of Y and small values of X with small values of Y. Then most of the probability mass or density will be associated with (x –and μX ) and (y –μY ), either both positive (both X and Y above their respective means) or both negative, so the product (x –μX )(y –μY ) will tend to be positive. Thus for a strong positive relationship, Cov(X, Y ) should be quite positive. For a strong negative relationship, the signs of (x –μX ) and (y –μY ) will tend to be opposite, yielding a negative product. Thus for a strong negative relationship, Cov (X, Y ) should be quite negative. If X and Y are not strongly related, positive and negative products will tend to cancel one another, yielding a covariance near 0. Figure 5.4 illustrates the different possibilities. The covariance depends on both the set of possible pairs and the probabilities. In Figure 5.4, the probabilities could be changed without altering the set of possible pairs, and this could drastically change the value of Cov(X, Y).

Figure 5.4 p (x, y ) = 1/10 for each of ten pairs corresponding to indicated points; (a) positive covariance; (b) negative covariance; (c) covariance near zero

Example 5.15

The joint and marginal pmf’s for X = automobile policy deductible amount and Y = homeowner policy deductible amount in Example 5.1 were

= (100 − 175)(0 − 125)(.20) + …

+ (250 − 175)(200 − 125)(.30)

= 1875

The following shortcut formula for Cov (X, Y ) simplifies the computations.

PROPOSITION

Cov(X , Y ) = E (XY )—μX · μY

According to this formula, no intermediate subtractions are necessary; only at the end of the computation is μX · μY , subtracted from E (XY ). The proof involves expanding (X—μX ) (Y—μY ) and then taking the expected value of each term separately.

Example 5.16 (Example 5.5 continued)

The joint and marginal pdf’s of X = amount of almonds and Y = amount of cashews were

with fY (y ) obtained by replacing x by y in fX (x ). It is easily verified that μX = μY = , and

Thus Cov(X, Y) = . A negative covariance is reasonable here because more almonds in the can implies fewer cashews.

It might appear that the relationship in the insurance example is quite strong since Cov(X , Y ) = 1875, whereas Cov(X , Y ) = in the nut example would seem to imply quite a weak relationship. Unfortunately, the covariance has a serious defect that makes it impossible to interpret a computed value. In the insurance example, suppose we had expressed the deductible amount in cents rather than in dollars. Then 100X would replace X, 100Y would replace Y, and the resulting covariance would be Cov(100X , 100Y ) = (100)(100)Cov(X , Y ) = 18,750,000. If, on the other hand, the deductible amount had be. Ideally, the choice of units should have no effect on a measure of strength of relationship. This is achieved by scaling the covariance.

Correlation

DEFINITION

The correlation coefficient of X and Y , denoted by Corr(X , Y ), ρX ,Y , or just ρ, is defined by

Example 5.17

It is easily verified that in the insurance scenario of Example 5.15, E (X 2) = 36,250, = 36,250 − (175)2 = 5625, σX = 75, E (Y 2) = 22,500, = 6875, and σY = 82.92. This gives

The following proposition shows that p remedies the defect of Cov (X, Y ) and also suggests how to recognize the existence of a strong (linear) relationship.

PROPOSITION

1. If a and c are either both positive or both negative,

Corr(aX + b, cY + d ) = Corr(X , Y )

2. For any two rv’s X and Y, –1 ≤ Corr(X, Y ) ≤ 1.

Statement 1 says precisely that the correlation coefficient is not affected by a linear change in the units of measurement (if, say, X = temperature in °C, then 9X/5 + 32 = temperature in °F). According to Statement 2, the strongest possible positive relationship is evidenced by ρ = +1, whereas the strongest possible negative relationship corresponds to ρ = –1. The proof of the first statement is sketched in Exercise 35, and that of the second appears in Supplementary Exercise 87 at the end of the chapter. For descriptive purposes, the relationship will be described as strong if | ρ | ≥ .8, moderate if .5 ≤ | ρ | ≤ .8, and weak if | ρ | ≤ .5.

If we think of p (x, y ) or f (x, y ) as prescribing a mathematical model for how the two numerical variables X and Y are distributed in some population (height and weight, verbal SAT score and quantitative SAT score131 >Chapter 12, we will consider taking a sample of pairs (x_1, y_1),…, (xn, yn) from the population. The sample correlation coefficient r will then be defined and used to make inferences about ρ.

The correlation coefficient ρ is actually not a completely general measure of the strength of a relationship.

PROPOSITION

1. If X and Y are independent, then p = 0, but p = 0 does not imply independence.

2. p = 1 or –1 iff Y = aX + b for some numbers a and b with a ≠ 0.

This proposition says that ρ is a measure of the degree of linear relationship between X and Y, and only when the two variables are perfectly related in a linear manner will p be as positive or negative as it can be. A ρ less than 1 in absolute value indicates only that the relationship is not completely linear, but there may still be a very strong nonlinear relation. Also, ρ = 0 does not imply that X and Y are independent, but only that there is complete absence of a linear relationship. When ρ = 0, X and Y are said to be uncorrelated. Two variables could be uncorrelated yet highly dependent because there is a strong nonlinear relationship, so be careful not to conclude too much from knowing that ρ = 0.

Example 5.18 Let X and Y be discrete rv’s with joint pmf

The points that receive positive probability mass are identified on the (x, y ) coordinate system in Figure 5.5. It is evident from the figure that the value of X is completely determined by the value of Y and vice versa, so the two variables are completely dependent. However, by symmetry μX = μY = 0 and E (XY ) = (–4) + (–4) + (4) + (4) = 0, so Cov(X , Y ) = E (XY ) –μX · μY = 0 and thus ρX,Y = 0. Although there is perfect dependence, there is also complete absence of any linear relationship!

Figure 5.5 The population of pairs for Example 5.18

d of Exercise 7. Compute the expected revenue from a single trip.

27. Annie and Alvie have agreed to meet for lunch between noon (0:00 P.M.) and 1:00 P.M. Denote Annie’s arrival time by X, Alvie’s by Y, and suppose X and Y are independent with pdf’s

What is the expected amount of time that the one who arrives first must wait for the other person? [Hint: h (X, Y ) = |X –Y |.]

28. Show that if X and Y are independent rv’s, then E (XY ) = E (X ) · E (Y ). Then apply this in Exercise 25. [Hint: Consider the continuous case with f (x, y ) = fX (x ) · fY (y ).]

29. Compute the correlation coefficient ρ for X and Y of Example 5.16 (the covariance has already been computed).

30. a. Compute the covariance for X and Y in Exercise 22.

b. Compute ρ for X and Y in the same exercise.

31. a. Compute the covariance between X and Y in Exercise 9.

b. Compute the correlation coefficient ρ for this X and Y.

32. Reconsider the minicomputer component lifetimes X and Y as described in Exercise 12. Determine E (XY ). What can be said about Cov (X, Y ) and ρ?

33. Use the result of Exercise 28 to show that when X and Y are independent, Cov (X, Y ) = Corr (X, Y ) = 0.

34. a. Recalling the definition of σ 2 for a single rv X, write a formula that would be appropriate for computing the variance of a function h (X, Y ) of two random variables. [Hint: Remember that variance is just a special expected value.]

b. Use this formula to compute the variance of the recorded score h (X, Y ) [= max (X, Y) ] in part (b) of Exercise 22.

35. a. Use the rules of expected value to show that Cov(aX + b, cY+ d ) = ac Cov(X , Y ).

b. Use part (a) along with the rules of variance and standard deviation to show that Corr(aX + b, cY + d) = Corr(X , Y ) when a and c have the same sign.

c. What happens if a and c have opposite signs?

36. Show that if Y = aX + b (a ≠ 0), then Corr(X , Y ) = +1 or –1. Under what conditions will ρ = +1?

5.3 Statistics and Their Distributions

The observations in a single sample were denoted in Chapter 1 by x_1, x_2,…, xn. Consider selecting two different samples of size n from the same population distribution. The xis in the second sample will virtually always differ at least a bit from those in the first sample. For example, a first sample of n = 3 cars of a particular type might result in fuel efficiencies x_1 = 30.7, x_2 = 29.4, x_3 = 31.1, whereas a second sample may give x_1 = 28.8, x_2 = 30.0, and x_3 = 31.1. Before we obtain data, there is uncertainty about the value of each xi. Because of this uncertainty, before the data becomes available we view each observation as a random variable and denote the sample by X_1, X_2,…, Xn (uppercase letters for random variables).

This variation in observed values in turn implies that the value of any function of the sample observations—such as the sample mean, sample standard deviation, or sample fourth spread—also varies from sample to sample. That is, prior to obtaining x_1,…, xn, there is uncertainty as to the value of , the value of s, and so on.

Example 5.19

Suppose that material strength for a randomly selected specimen of a particular type has a Weibull distribution with parameter values α = 2 (shape) and β = 5 (scale). The corresponding density curve is shown in Figure 5.6. Formulas from Section 4.5 give

μ = E (X ) = 4.4311 = 4.1628 σ^2 = V (X ) = 5.365 σ = 2.316

The mean exceeds the median because of the distribution’s posmage">

We used MINITAB to generate six different samples, each with n = 10, from this distribution (material strengths for six different groups of ten specimens each). The results appear in Table 5.1, followed by the values of the sample mean, sample median, and sample standard deviation for each sample. Notice first that the ten observations in any particular sample are all different from those in any other sample. Second, the six values of the sample mean are all different from one another, as are the six values of the sample median and the six values of the sample standard deviation. The same is true of the sample 10% trimmed means, sample fourth spreads, and so on.

Table 5.1 Samples from the Weibull Distribution of Example 5.19

Furthermore, the value of the sample mean from any particular sample can be regarded as a point estimate (“point” because it is a single number, corresponding to a single point on the number line) of the population mean μ, whose value is known to be 4.4311. None of the estimates from these six samples is identical to what is being estimated. The estimates from the second and sixth samples are much too large, whereas the fifth sample gives a substantial underestimate. Similarly, the sample standard deviation gives a point estimate of the population standard deviation. All six of the resulting estimates are in error by at least a small amount.

In summary, the values of the individual sample observations vary from sample to sample, so in general the value of any quantity computed from sample data, and the value of a sample characteristic used as an estimate of the corresponding population characteristic, will virtually never coincide with what is being estimated.

DEFINITION

A statistic is any quantity whose value can be calculated from sample data. Prior to obtaining data, there is uncertainty as to what value of any particular statistic will result. Therefore, a statistic is a random variable and will be denoted by an uppercase letter; a lowercase letter is used to represent the calculated or observed value of the statistic.

Thus the sample mean, regarded as a statistic (before a sample has been selected or an experiment carried out), is denoted by ; the calculated value of this statistic is . Similarly, S represents the sample standard deviation thought of as a statistic, and its computed value is s. If samples of two different types of bricks are selected and the individual compressive strengths are denoted by X_1,…, Xm and Y_1,…, Yn, respectively, the {To ≤ 2} as

The mean and variance of the underlying exponential distribution are μ = 1/λ and σ^2 = 1 /λ^2. From Expressions (5.5) and (5.6), it can be verified that E () = 1 /λ, V(X) = 1/(2λ^2), E(To) = 2/λ, and V(To) = 2/λ^2. These results again suggest some general relationships between means and variances of ,To, and the underlying distribution.

Simulation Experiments

The second method of obtaining information about a statistic’s sampling distribution is to perform a simulation experiment. This method is usually used when a derivation via probability rules is too difficult or complicated to be carried out. Such an experiment is virtually always done with the aid of a computer. The following characteristics of an experiment must be specified:

1. The statistic of interest (, S, a particular trimmed mean, etc.)

2. The population distribution (normal with μ = 100 and σ = 15, uniform with lower limit A = 5 and upper limit B = 10, etc.)

3. The sample size n (e.g., n = 10 or n = 50)

4. The number of replications k (e.g., k = 500)

Then use a computer to obtain k different random samples, each of size n, from the designated population distribution. For each such sample, calculate the value of the statistic and construct a histogram of the k calculated values. This histogram gives the approximate sampling distribution of the statistic. The larger the value of k, the better the approximation will tend to be (the actual sampling distribution emerges as k → ∞ ). In practice, k = 500 or 1000 is usually enough if the statistic is “fairly simple.”

Example 5.22

The population distribution for our first simulation study is normal with μ = 8.25 and σ = .75, as pictured in Figure 5.10. [The article “Platelet Size in Myocardial Infarction” (British Med. J., 1983: 449–451) suggests this distribution for platelet volume in individuals with no history of serious heart problems.]

Normal distribution, with μ = 8.25 and σ = .75

We actually performed four different experiments, with 500 replications for each one. In the first experiment, 500 samples of n = 5 observations each were generated using MINITAB, and the sample sizes for the other three were n = 10, n = 20, and n = 30, respectively. The sample mean was calculated for each sample, and the resulting histograms of values appear in Figure 5.11.

Figure 5.11 Sample histograms for based on 500 samples, each consisting of n observations: (a) n = 5; (b) n = 10; (c) n = 20; (d) n = 30

The first thing to notice about the histograms is their shape. To a reasonable approximation, each of the four looks like a normal curve. The resemblance would be even more striking if each histogram had been based on many more than 500 values. Second, each histogram is centered approximately at 8.25, the mean of the population being sampled. Had the histograms been based on an unending sequence of values, their centers would have been exactly the population mean, 8.25.

The final aspect of the histograms to note is their spread relative to one another. The smaller the value of n, the greater the extent to which the sampling distribution spreads out about the mean value. This is why the histograms for n = 20 and n = 30 are based on narrower class intervals than those for the two smaller sample sizes. For the larger sample sizes, most of the values are quite close to 8.25. This is the effect of averaging. When n is small, a single unusual value can result in an value far from the center. With a larger sample size, any unusual x values, when averaged in with the other sample values, still tend to yield an value close to μ. Combining these insights yields a result that should appeal to your intuition: based on a large n tends to be closer to μ than does based on a small n.

Example 5.23

Consider a simulation experiment in which the population distribution is quite skewed. Figure 5.12 shows the density curve for lifetimes of a certain type of electronic control [this is actually a lognormal distribution with E (ln(X )) = 3 and Vame four sample sizes as in Example 5.22. The resulting histograms along with a normal probability plot from MINITAB for the 500 xX values based on n = 30 are shown in Figure 5.13.

Figure 5.12 Density curve for the simulation experiment of Example 5.23 [E (X ) = 21.7584, V (X ) = 82.1449]

Unlike the normal case, these histograms all differ in shape. In particular, they become progressively less skewed as the sample size n increases. The average of the 500 values for the four different sample sizes are all quite close to the mean value of the population distribution. If each histogram had been based on an unending sequence of values rather than just 500, all four would have been centered at exactly 21.7584. Thus different values of n change the shape but not the center of the sampling distribution of . Comparison of the four histograms in Figure 5.13 also shows that as n increases, the spread of the histograms decreases. Increasing n results in a greater degree of concentration about the population mean value and makes the histogram look more like a normal curve. The histogram of Figure 5.13(d) and the normal probability plot in Figure 5.13(e) provide convincing evidence that a sample size of n = 30 is sufficient to overcome the skewness of the population distribution and give an approximately normal sampling distribution.

Figure 5.13 Results of the simulation experiment of Example 5.23: (a) histogram for n = 5; (b) histogram for n = 10; (c) histogram for n = 20; (d) histogram for n = 30; (e) normal probability plot for n = 30 (from MINITAB)

EXERCISES Section 5.3 (37–45)

37. A particular brand of dishwasher soap is sold in three sizes: 25 oz, 40 oz, and 65 oz. Twenty percent of all purchasers select a 25-oz box, 50% select a 40-oz box, and the remaining 30% choose a 65-oz box. Let X_1 and X_2 denote the package sizes selected by two independently selected purchasers.

a. Determine the sampling distribution of , calculate E(), and compare tcted customer at a certain shipping facility. Suppose the distribution of X is as follows:

a. Consider a random sample of size n = 2 (two customers), and let be the sample mean number ofpackages shipped. Obtain the probability distribution of .

b. Refer to part (a) and calculate P( sampling distribution appear to be approximately normal?

45. Carry out a simulation experiment using a statistical computer package or other software to study the sampling distribution of when the population distribution is lognormal with E (ln (X )) = 3 and V (ln (X )) = 1. Consider the four sample sizes n = 10, 20, 30, and 50, and in each case use 500 replications. For which of these sample sizes does the sampling distribution appear to be approximately normal?

5.4 The Distribution of the Sample Mean

The importance of the sample mean springs from its use in drawing conclusions about the population mean μ. Some of the most frequently used inferential procedures are based on properties of the sampling distribution of . A preview of these properties appeared in the calculations and simulation experiments of the previous section, where we noted relationships between E () and μ and also among V(), σ 2, and n.

PROPOSITION

Let X_1, X_2,…, Xn be a random sample from a distribution with mean value μ and standard deviation σ. Then

1. E () = μ = μ

2.

In addition, with To = X_1 + … + Xn (the sample total), E (To) = n μ, V (To) = n σ^2, and σTo =

Proofs of these results are deferred to thejiext section. According to Result 1, the sampling (i.e., probability) distribution of is centered precisely at the mean of the population from which the sample has been selected. Result 2 shows that the distribution becomes more concentrated about μ as the sample size n increases. In marked contrast, the distribution of To becomes more spread out as n increases. Ah="0em">In a notched tensile fatigue test on a titanium specimen, the expected number of cycles to first acoustic emission (used to indicate crack initiation) is μ = 28,000, and the standard deviation of the number of cycles is σ = 5000. Let X_1, X_2,…, X_25 be a random sample of size 25, where each Xi is the number of cycles on a different randomly selected specimen. Then the expected value of the sample mean number of cycles until first emission is E () = μ = 28,000, and the expected total number of cycles for the_25 specimens is E(To) = n μ = 25(28,000) = 700,000. The standard deviations of and To are

If the sample size increases to n = 100, E() is unchanged, but σ = 500, half of its previous value (the sample size must be quadrupled to halve the standard deviation of ).

The Case of a Normal Population Distribution

The simulation experiment of Example 5.22 indicated that when the population distribution is normal, each histogram of values is well approximated by a normal curve.

PROPOSITION

Let X_1, X_2,…, Xn be a random sample from a normal distribution with mean μ and standard deviation σ. Then for any n, is normally distributed (with mean μ and standard deviation ), as is To (with mean n μ and standard deviation ).*

* A proof of the result for To when n = 2 is possible using the method in Example 5.21, but the details are messy. The general result is usually proved using a theoretical tool called a moment generating function. One of the chapter references can be consulted for more information.

We know everything there is to know about the and To distributions when the population distribution is normal. In particular, probabilities such as P (a ≤ ≤ b ) and P (c ≤ To ≤ d ) can be obtained simply by standardizing. Figure 5.14 illustrates the promuch less then 30 would suffice. For example, in the case of a uniform population distribution, the CLT gives a good approximation for n ≥ 12.

Other Applications of the Central Limit Theorem

The CLT can be used to justify the normal approximation to the binomial distribution discussed in Chapter 4. Recall that a binomial variable X is the number of successes in a binomial experiment consisting of n independent success/failure trials with p = P (S ) for any particular trial. Define new rv’s X_1, X_2,…, Xn by

Because the trials are independent and P(S) is constant from trial to trial, the Xis are iid (a random sample from a Bernoulli distribution). The CLT then implies that if n is sufficiently large, both the sum and the average of the Xi s have approximately normal distributions. When the Xi s are summed, a 1 is added for every S that occurs and a 0 for every F, so X_1 + … + Xn = X. The sample mean of the Xi s is X/n, the sample proportion of successes. That is, both X and X/n are approximately normal when n is large. The necessary sample size for this approximation depends on the value of p: When p is close to .5, the distribution of each Xi is reasonably symmetric (see Figure 5.17), whereas the distribution is quite skewed when p is near 0 or 1. Using the approximation only if both np ≥ 10 and n (1 –p ) ≥ 10 ensures that n is large enough to overcome any skewness in the underlying Bernoulli distribution.

Figure 5.17 Two Bernoulli distributions: (a) p 5 –4 (reasonably symmetric); (b) p 5 –1 (very skewed)

Recall from Section 4.5 that X has a lognormal distribution if ln(X) has a normal distribution.

PROPOSITION

Let X_1, X_2,…, Xn be a random sample from a distribution for which only positive values are possible [P (Xi > 0) = 1]. Then if n is sufficiently large, the product Y = X_1 X_2 Xn has approximately a lognormal distribution.

To verify this, note that

ln (Y ) = ln(X 1) + ln(X 2) + … + ln(Xn )

If grading times are independent and the instructor begins grading at 6:50 P.M. and grades continuously, what is the (approximate) probability that he is through grading before the 11:00 P.M. TV news begins?

b. If the sports report begins at 11:10, what is the probability that he misses part of the report if he waits until grading is done before turning on the TV?

50. The breaking strength of a rivet has a mean value of 10,000 psi and a standard deviation of 500 psi.

a. What is the probability that the sample mean breaking strength for a random sample of 40 rivets is between 9900 and 10,200?

b. If the sample size had been 15 rather than 40, could the probability requested in part (a) be calculated from the given information?

51. The time taken by a randomly selected applicant for a mortgage to fill out a certain form has a normal distribution with mean value 10 min and standard deviation 2 min. If five individuals fill out a form on one day and six on another, what is the probability that the sample average amount of time taken on each day is at most 11 min?

52. The lifetime of a certain type of battery is normally distributed with mean value 10 hours and standard deviation 1 hour. There are four batteries in a package. What lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only 5% of all packages?

53. Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.2.

a. If the distribution is normal, what is the probability that the sample mean hardness for a random sample of 9 pins is at least 51?

b. What is the (approximate) probability that the sample mean hardness for a random sample of 40 pins is at least 51?

54. Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65 and standard deviation .85 (suggested in “Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants,” Water Research, 1984: 1169-1174).

a. If a random sample of 25 specimens is selected, what is the probability that the sample average sediment density is at most 3.00? Between 2.65 and 3.00?

< The number of parking tickets issued in a certain city on any given weekday has a Poisson distribution with parameter λ = 50. What is the approximate probability that

a. Between 35 and 70 tickets are given out on a particular day? [Hint: When λ is large, a Poisson rv has approximately a normal distribution.]

b. The total number of tickets given out during a 5-day week is between 225 and 275?

56. A binary communication channel transmits a sequence of “bits” (0s and 1s). Suppose that for any particular bit transmitted, there is a 10% chance of a transmission error (a 0 becoming a 1 or a 1 becoming a 0). Assume that bit errors occur independently of one another.

a. Consider transmitting 1000 bits. What is the approximate probability that at most 125 transmission errors occur?

b. Suppose the same 1000-bit message is sent two different times independently of one another. What is the approximate probability that the number of errors in the first transmission is within 50 of the number of errors in the second?

57. Suppose the distribution of the time X (in hours) spent by students at a certain university on a particular project is gamma with parameters a = 50 and β = 2. Because a is large, it can be shown that X has approximately a normal distribution. Use this fact to compute the probability that a randomly selected student spends at most 125 hours on the project.

5.5 The Distribution of a Linear Combination

The sample mean and sample total To are special cases of a type of random variable that arises very frequently in statistical applications.

DEFINITION

Given a collection of n random variables X_1,…, Xn and n numerical constants a_1,…, an, the rv

is called a linear combination of the Xi s.

Taking a_1 = a_2 = … = an = 1 gives Y = X_1 + … + Xn = To, and a_1 = a_2 = … = an = yields Y =X_1 + … + ont face="Times New Roman">When the Xis form a random sample from a normal distribution, and To are both normally distributed. Here is a more general result concerning linear combinations.

PROPOSITION

If X_1, X_2,…, Xn are independent, normally distributed rv’s (with possibly different means and/or variances), then any linear combination of the Xis also has a normal distribution. In particular, the difference X_1—X_2 between two independent, normally distributed variables is itself normally distributed.

Example 5.30 (Example 5.28 continued)

The total revenue from the sale of the three grades of gasoline on a particular day was Y = 21.2X_1 + 21.35X_2 + 21.5X_3, and we calculated μY = 4125 and (assuming independence) σγ = 322.53. If the Xi s are normally distributed, the probability that revenue exceeds 4500 is

The CLT can also be generalized so it applies to certain linear combinations. Roughly speaking, if n is large and no individual term is likely to contribute too much to the overall value, then Y has approximately a normal distribution.

Proofs for the case n = 2

For the result concerning expected values, suppose that X_1 and X_2 are continuous with joint pdf f(x_1, x_2). Then

Summation replaces integration in the discrete case. The argument for the variance result does not require specifying whether either variable is discrete or continuous. Recalling that V (Y ) = E [(Y—μY )2],

The expression inside the braces is a linear combination of the variables Y_1 = (X_1—μ_1)2, Y_2 = (X_2—μ_2)2, and Y_3 = (X_1—μ_1)(X_2 –μ_2), so carrying the E operation through to the three terms gives a 1V(X_1) + a 2V (X_2) + 2a_1a_2 Cov(X_1, X_2) as required.

EXERCISES Section 5.5 (58–74)

58. A shipping company handles containers in three different sizes: (1) 27 ft^3 (3 × 3 × 3), (2) 125 ft^3, and (3) 512 ft^3. Let Xi (i = 1, 2, 3) denote the number of type i containers shipped during a given week. With μi = E (Xi) and = V (Xi), suppose that the mean values and standard deviations are as follows:

a. Assuming that X_1, X_2, X_3 are independent, calculate the expected value and variance of the total volume shipped. [Hint: Volume = 27X_1 + 125X_2 + 512X_3.]

b. Would your calculations necessarily be correct if the Xis were not independent? Explain.

59. Let X_1, X_2, and X_3 represent the times necessary to perform three successive repair tasks at a certain service facility. Suppose they are independent, normal rv’s with expected values μ_1, μ_2, and μ_3 and variances , , and , respectively.

a. If μ = μ_2 = μ_3 = 60 and , calculate P (X_1 + X_2 + X_3 ≤ 200). What is P (150 ≤ X_1 + X_2 + X_3 < 200)?

b. Using the μi s and σi s given in part (a), calculate P (55 ≤ ) and P (58 ≤ ≤ 62).

c. Using the μ s and σ s given in part (a), calculate P(–10 ≤ X_1 − .5X_2 − .5X_3 ≤ 5).

d. If μ_1 = 40, μ_2 = 50, μ_3 = 60, = 10, = 12, and = 14, calculate P (X_1 + X_2 + X_3 ≤ 160) and P (X_1 + X_2 ≥ 2 X_3).

X_1, X_2, X_3, X_4, and X_5 be the observed fuel efficiencies (mpg) for the five cars. Suppose these variables are independent and normally distributed with μ_1 = μ_2 = 20, μ_3 = μ_4 = μ_5 = 21, and σ 2 = 4 for the economy brand and 3.5 for the name brand. Define an rv Y by

so that Y is a measure of the difference in efficiency between economy gas and name-brand gas. Compute P (0 < Y ) and P (–1 ’ Y ’ 1). [Hint: Y = a_1X_1 + … + a_5X_5, with a_1 = 2, …, a_5 = –.]

61. Exercise 26 introduced random variables X and Y, the number of cars and buses, respectively, carried by a ferry on a single trip. The joint pmf of X and Y is given in the table in Exercise 7. It is readily verified that X and Y are independent.

a. Compute the expected value, variance, and standard deviation of the total number of vehicles on a single trip.

b. If each car is charged $3 and each bus $10, compute the expected value, variance, and standard deviation of the revenue resulting from a single trip.

62. Manufacture of a certain component requires three different machining operations. Machining time for each operation has a normal distribution, and the three times are independent of one another. The mean values are 15, 30, and 20 min, respectively, and the standard deviations are 1, 2, and 1.5 min, respectively. What is the probability that it takes at most 1 hour of machining time to produce a randomly selected component?

63. Refer to Exercise 3.

a. Calculate the covariance between X_1 = the number of customers in the express checkout and X_2 = the number of customers in the superexpress checkout.

b. Calculate V (X_1 + X_2). How does this compare to V (X_1) + V (X_2)?

64. Suppose your waiting time for a bus in the morning is uniformly distributed on [0, 8], whereas waiting time in the evenockquote>a. If you take the bus each morning and evening for a week, what is your total expected waiting time? [Hint: Define rv’ s X_1,…, X_10 and use a rule of expected value.]

b. What is the variance of your total waiting time?

c. What are the expected value and variance of the difference between morning and evening waiting times on a given day?

d. What are the expected value and variance of the difference between total morning waiting time and total evening waiting time for a particular week?

65. Suppose that when the pH of a certain chemical compound is 5.00, the pH measured by a randomly selected beginning chemistry student is a random variable with mean 5.00 and standard deviation .2. A large batch of the compound is subdivided and a sample given to each student in a morning lab and each student in an afternoon lab. Let the average pH as determined by the morning students and = the average pH as determined by the afternoon students.

a. If pH is a normal variable and there are 25 students in each lab, compute P(–.1 ≤ ≤ .1). [Hint : is a linear combination of normal variables, so is normally distributed. Compute μ and σ]

b. If there are 36 students in each lab, but pH determinations are not assumed normal, calculate (approximately) P (–.1 ≤ ≤ .1).

66. If two loads are applied to a cantilever beam as shown in the accompanying drawing, the bending moment at 0 due to the loads is a_1X_1 + a_2X_2.

a. Suppose that X_1 and X_2 are independent rv’s with means 2 and 4 kips, respectively, and standard deviations .5 and 1.0 kip, respectively. If a_1 = 5 ft and a_2 = 10 ft, what is the expected bending moment and what is the standard deviation of the bending moment?

b. If X_1 and X_2 are normally distributed, what is the probability the standard deviation of the total.

70. Consider a random sample of size n from a continuous distribution having median 0 so that the probability of any one observation being positive is .5. Disregarding the signs of the observations, rank them from smallest to largest in absolute value, and let W = the sum of the ranks of the observations having positive signs. For example, if the observations are -.3, + .7, +2.1, and -2.5, then the ranks of positive observations are 2 and 3, so W = 5. In Chapter 15, W will be called Wilcoxon’s signed-rank statistic. W can be represented as follows:

where the Yi s are independent Bernoulli rv’s, each with p = .5 (Yi = 1 corresponds to the observation with rank i being positive).

a. Determine E (Yi) and then E (W ) using the equation for W. [Hint: The first n positive integers sum to n (n + 1)/2.]

b. Determine V (Yi ) and then V (W ) [Hint: The sum of the squares of the first n positive integers can be expressed as n (n + 1)(2n + 1)/6.]

71. In Exercise 66, the weight of the beam itself contributes to the bending moment. Assume that the beam is of uniform thickness and density so that the resulting load is uniformly distributed on the beam. If the weight of the beam is random, the resulting load from the weight is also random; denote this load by W (kip-ft).

a. If the beam is 12 ft long, W has mean 1.5 and standard deviation .25, and the fixed loads are as described in part (a) of Exercise 66, what are the expected value and variance of the bending moment? [Hint: If the load due to the beam were w kip-ft, the contribution to the bending moment would be w x dx. ]

b. If all three variables (X_1, X_2, and W ) are normally distributed, what is the probability that the bending moment will be at most 200 kip-ft?

72. I have three errands to take care of in the Administration Building. Let Xi = the time that it takes for the ith errand (i = 1, 2, 3), and let X_4 = the total time in minutes that I spend walking to and from the building and between each errand. Suppos>X and Y independent? Justify your answer.

d. What is the expected total cost of the dinner for the two people?

e. Suppose that when a couple opens fortune cookies at the conclusion of the meal, they find the message “You will receive as a refund the difference between the cost of the more expensive and the less expensive meal that you have chosen.” How much does the restaurant expect to refund?

76. In cost estimation, the total cost of a project is the sum of component task costs. Each of these costs is a random variable with a probability distribution. It is customary to obtain information about the total cost distribution by adding together characteristics of the individual component cost distributions—this is called the “roll-up” procedure. For example, E(X_1 + … + Xn) = E(X_1) + … + E(Xn), so the roll-up procedure is valid for mean cost. Suppose that there are two component tasks and that X_1 and X_2 are independent, normally distributed random variables. Is the roll-up procedure valid for the 75th percentile? That is, is the 75th percentile of the distribution of X_1 + X_2 the same as the sum of the 75th percentiles of the two individual distributions? If not, what is the relationship between the percentile of the sum and the sum of percentiles? For what percentiles is the roll-up procedure valid in this case?

77. A health-food store stocks two different brands of a certain type of grain. Let X= the amount (lb) of brand A on hand and Y = the amount of brand B on hand. Suppose the joint pdf of X and Y is

a. Draw the region of positive density and determine the value of k.

b. Are X and Y independent? Answer by first deriving the marginal pdf of each variable.

c. Compute P (X + Y ≤ 25).

d. What is the expected total amount of this grain on hand?

e. Compute Cov (X, Y ) and Corr(X, Y).

f. What is the variance of the total amount of grain on hand?

78. Let X_1, X_2,…, Xn be random variableote>

c. Use the fact that V (W ) = 0 only if W is a constant to show that ρ = 1 only if Y = aX + b.

88. Suppose a randomly chosen individual’s verbal score X and quantitative score Y on a nationally administered aptitude examination have joint pdf

You are asked to provide a prediction t of the individual’s total score X + Y. The error of prediction is the mean squared error E[(X + Y –t)2]. What value of t minimizes the error of prediction?

89. a. Let X_1 have a chi-squared distribution with parameter v_1 (see Section 4.4), and let X_2 be independent of X_1 and have a chi-squared distribution with parameter v_2. Use the technique of Example 5.21 to show that X_1 + X_2 has a chi-squared distribution with parameter v_1 + v_2.

b. In Exercise 71 of Chapter 4, you were asked to show that if Z is a standard normal rv, then Z^2 has a chi-squared distribution with v = 1. Let Z_1, Z_2,…, Zn be n independent standard normal rv’s. What is the distribution of Justify your answer.

c. Let X_1,…, Xn be a random sample from a normal distribution with mean μ and variance σ^2. What is the distribution of the sum [(Xi—μ)/σ]2? Justify your answer.

90. a. Show that Cov(X, Y + Z) = Cov(X, Y ) + Cov(X, Z). b. Let X_1 and X_2 be quantitative and verbal scores on one aptitude exam, and let Y_1 and Y_2 be corresponding scores on another exam. If Cov(X_1, Y_1) = 5, Cov(X_1, Y_2) = 1, Cov(X_2, Y_1) = 2, and Cov(X_2, Y_2) = 8, what is the covariance between the two total scores X_1 + X_2 andY1 +Y2?

91. A rock specimen from a particular area is randomly selected and weighed two different times. Let W denote the actual weight and X_1 and X_2 the two measured weights. Then X_1 = W_1) = V(E_2) = .

a. Express ρ, the correlation coefficient between the two measured weights X_1 and X_2, in terms of , the variance of actual weight, and , the variance of measured weight.

b. Compute ρ when σW = 1 kg and σE = .01 kg.

92. Let A denote the percentage of one constituent in a randomly selected rock specimen, and let B denote the percentage of a second constituent in that same specimen. Suppose D and E are measurement errors in determining the values of A and B so that measured values are X = A + D and Y = B + E, respectively. Assume that measurement errors are independent of one another and of actual values.

a. Show that

Corr(X, Y ) = Corr(A , B ) · ) where X_1 and X_2 are replicate measurements on the value of A, and Y_1 and Y_2 are defined analogously with respect to B. What effect does the presence of measurement error have on the correlation?

b. What is the maximum value of Corr (X, Y ) when Corr(X_1, X_2) = .8100 and Corr(Y_1, Y_2) = .9025? Is this disturbing?

93. Let X_1,…, Xn be independent rv’s with mean values μ_1,…, μπ and variances ,…, . Consider a function h(x_1,…, xn), and use it to define a new rv Y = h(X_1,…, Xn). Under rather general conditions on the h function, if the σi s are all small relative to the corresponding μi s, it can be shown that E (Y ) ≈ h (μ_1,…, μn ) and

where each partial derivative is evaluated at (x_1,…, xn) =(μ_1,…, μn ). Suppose three resistors with resistances X_1, X_2, X_3 are connected in parallel across a battery with voltage X_4. Then by Ohm’s la>

b. Estimator = , estimate = = (27.94 + 27.98)/2 = 27.960

c. Estimator = [min(Xi ) + max(Xi )]/2 = the average of the two extreme lifetimes, estimate = [min(xi ) + max(xi )]/2 = (24.46 + 30.88)/2 = 27.670

d. Estimator = tr(10), the 10% trimmed mean (discard the smallest and largest 10% of the sample and then average),

Each one of the estimators (a)–(d) uses a different measure of the center of the sample to estimate μ. Which of the estimates is closest to the true value? We cannot answer this without knowing the true value. A question that can be answered is, “Which estimator, when used on other samples of Xi s, will tend to produce estimates closest to the true value?” We will shortly consider this type of question.

Example 6.3

In the near future there will be increasing interest in developing low-cost Mg-based alloys for various casting processes. It is therefore important to have practical ways of determining various mechanical properties of such alloys. The article “On the Development of a New Approach for the Determination of Yield Strength in Mg-based Alloys” (Light Metal Age, Oct. 1998: 50–53) proposed an ultrasonic method for this purpose. Consider the following sample of observations on elastic modulus (GPa) of AZ91D alloy specimens from a die-casting process:

44.2 43.9 44.7 44.2 44.0 43.8 44.6 43.1

Assume that these observations are the result of a random sample X_1, …, X_8from the population distribution of elastic modulus under such circumstances. We want to estimate the population variance σ^2. A natural estimator is the sample variance:

The corresponding estimate is

The estimate of σ would then be s = = .501.

An alternative estimator would result from using divi , X_2 , …, Xn be a random sample from a distribution with mean μ and variance σ^2. Then the estimator

is an unbiased estimator of σ^2.

Proof For any rv Y, V (Y ) = E (Y 2) –[E (Y )]2, so E (Y 2) = V (Y ) + [E (Y )]2. Applying this to

gives

The estimator that uses divisor n can be expressed as (n –1 )S 2/ n, so

This estimator is therefore not unbiased. The bias is (n –1 )σ^2/n –σ^2 = –σ^2/n . Because the bias is negative, the estimator with divisor n tends to underestimate σ^2, and this is why the divisor n –1 is preferred by many statisticians (though when n is large, the bias is small and there is little difference between the two).

Although S^2 is unbiased for σ^2, S is a biased estimator of σ (its bias is small unless n is quite small). However, there are other good reasons to use S as an estimator, especially when the population distribution is normal. These will become more apparent when we discuss confidence intervals and hypothesis testing in the next several chapters.

In Example 6.2, we proposed several different estimators for the mean μ of a normal distribution. If there were a unique unbiased estimator for μ, the estimation problem would be resolved by using that estimator. Unfortunately, this is not the case.

PROPOSITION

If X_1, X_2, …, Xn is a random sample from a distribution with mean μ, then is an unbiased estimator of μ. If in addition the distribution is continuous and symmetric, then and any trimmed mean are also unbiased estimators of μ.

The next example introduces another situation in which there are several unbiased estimators for a particular parameter.

Example 6.5

Under certain circumstances organic contaminants adhere readily to wafer surfaces and cause deterioration in semiconductor manufacturing devices. The paper “Ceramic Chemical Filter for Removal of Organic Contaminants” (J. of the Institute of Environmental Sciences and Technology, 2003: 59–65) discussed a recently developed alternative to conventional charcoal filters for removing organic airborne molecular contamination in cleanroom applications. One aspect of the investigation of filter performance involved studying how contaminant concentration in air related to concentration on a wafer surface after prolonged exposure. Consider the following representative data on x = DBP concentration in air and y = DBP concentration on a wafer surface after 4-hour exposure (both in μg/m^3, where DBP = dibutyl phthalate).

The authors comment that “DBP adhesion on the wafer surface was roughly proportional to the DBP concentration in air.” Figure 6.2 shows a plot of y versus x—i.e., of the (x y ) pairs.

Figure 6.2 Plot of the DBP data from Example 6.5

If y were exactly proportional to x, we would have y = βx for some value β, which says that the (x, y) points in the plot would lie exactly on a straight line with slope β passing through (0,0). But this is only approximately the case. So we now assume that for any fixed x, wafer DBP is a random variable Y having mean value βx. That is, we postulate that the mean value of Y is related to x by a line passing through (0,0) but that the observed value of Y will typically deviate from this line (this is referred to in the statistical literature as “regression through the origin”).

We now wish to estimate the slope parameter β. Consider the following three estimators:

The resulting estimates based on the given data are 1.3497, 1.1875, and 1.1222, respectively. So the estimate definitely depends on which estimator is used. If one of these three estimators were unbiased and the other two were biased, there would be a good case for using the unbiased one. But all three are unbiased; the the midpoint of the interval of positive density, so E (Xi) = θ/2. This implies that E () = θ/2, from which E (2) = θ. That is, the estimator 2 = 2 is unbiased for θ.

If X is uniformly distributed on the interval from A, B, then V (X ) = σ^2 = (B – A )2/12. Thus, in our situation, V (Xi ) = θ^2 /12, V () = σ^2 /n = θ^2/(12n), and V (2) = V (2) = 4V () = θ^2/(3n). The results of Exercise 32 can be used to show that V (1) = θ^2/[n (n + 2)]. The estimator 1 has smaller variance than does θ_2 if 3n < n (n + 2)—that is, if 0 < n^2 –n = n(n – 1). As long as n > 1, V (1) < V(2), so 1 is a better estimator than 2. More advanced methods can be used to show that 1 is the MVUE of θ—every other unbiased estimator of θ has variance that exceeds θ^2 /[n (n + 2)].

One of the triumphs of mathematical statistics has been the development of methodology for identifying the MVUE in a wide variety of situations. The most important result of this type for our purposes concerns estimating the mean μ of a normal distribution.

THEOREM

Let X_1, …, Xn be a random sample from a normal distribution with parameters μ and σ. Then the estimator μ = is the MVUE for μ.

Whenever we are convinced that the population being sampled is normal, the result says that should be used to estimate μ. In Example 6.2, then, our estimate would be = 27.793.

In some situations, it is possible to obtain an estimator with small bias that would be preferred to the best unbiased estimator. This is illustrated in Figure 6.4 does not suffer too much in comparison with the best procedure in any of the three situations.

More generally, recent research in statistics has established that when estimating a point of symmetry μ of a continuous probability distribution, a trimmed mean with trimming proportion 10% or 20% (from each end of the sample) produces reasonably behaved estimates over a very wide range of possible models. For this reason, a trimmed mean with small trimming percentage is said to be a robust estimator.

In some situations, the choice is not between two different estimators constructed from the same sample, but instead between estimators based on two different experiments.

Example 6.8

Suppose a certain type of component has a lifetime distribution that is exponential with parameter λ so that expected lifetime is μ = 1/λ. A sample of n such components is selected, and each is put into operation. If the experiment is continued until all n lifetimes, X_1, …, Xn, have been observed, then X is an unbiased estimator of μ.

In some experiments, though, the components are left in operation only until the time of the rth failure, where r < n. This procedure is referred to as censoring. Let Y_1 denote the time of the first failure (the minimum lifetime among the n components), Y_2 denote the time at which the second failure occurs (the second smallest lifetime), and so on. Since the experiment terminates at time Yr, the total accumulated lifetime at termination is

We now demonstrate that = Tr/r is an unbiased estimator for μ. To do so, we need two properties of exponential variables:

1. The memoryless property (see Section 4.4), which says that at any time point, remaining lifetime has the same exponential distribution as original lifetime.

2. If X_1, …, Xk are independent, each exponential with parameter λ, then min (X_1, …, Xk) is exponential with parameter k X and has expected value 1/(kX).

Since all n components last until Y_1, n – 1 last an additional Y_2 –Y_1, n – 2 an additional Y_3 –Y_2 amoub>3 –Y_2) + …

+ (n –r + l)(Yr –Yr–1)

But Y_1 is the minimum of n exponential variables, so E (Y_1) = 1/(nλ). Similarly, Y_2 –Y_1 is the smallest of the n – 1 remaining lifetimes, each exponential with parameter λ (by the memoryless property), so E (Y_2 –Y_1 = 1/[(n – 1)λ]. Continuing, E (Yi+1 –Yi) = 1/[(n –i )λ], so

Therefore, E (Tr/r ) = (1/r )E (Tr ) = (1/r ) · (r /λ) = 1/λ = μ, as claimed.

As an example, suppose 20 components are put on test and r = 10. Then if the first ten failure times are 11,15, 29, 33, 35,40,47, 55, 58, and 72, the estimate of μ is

The advantage of the experiment with censoring is that it terminates more quickly than the uncensored experiment. However, it can be shown that V (Tr/r ) = 1/(λ^2r ), which is larger than 1/(λ^2n), the variance of in the uncensored experiment.

Reporting a Point Estimate: The Standard Error

Besides reporting the value of a point estimate, some indication of its precision should be given. The usual measure of precision is the standard error of the estimator used.

DEFINITION

The standard error of an estimator is its standard deviation If the standard error itself involves unknown parameters whose values can be estimated, substitution of these estimates into σ yields the estimated standard error (estimated standard deviation) of the estimator. The estimated standard error can be denoted either by (the ^ over σ emphasizes that σ is being estimated) or by, s.

Example 6.9 (Example 6.2 continued)

Assuming that breakdown voltage is normally distributed, e best estimator of μ. If the value of σ is known to be 1.5, the standard error of is . If, as is usually the case, the value of σ is unknown, the estimate = s = 1.462 is substituted into σ to obtain the estimated standard error

Example 6.10 (Example 6.1 continued)

The standard error of = X/n is

Since p and q = 1 –p are unknown (else why estimate?), we substitute = x/n and = 1 –x/n into σ, yielding the estimated standard error . Alternatively, since the largest value of pq is attained when p = q = .5, an upper bound on the standard error is

When the point estimator has approximately a normal distribution, which will often be the case when n is large, then we can be reasonably confident that the true value of θ lies within approximately 2 standard errors (standard deviations) of . Thus if a sample of n = 36 component lifetimes gives = = 28.50 and s = 3.60, then = .60, so within 2 estimated standard errors of translates to the interval 28.50 ± (2)(.60) = (27.30, 29.70).

If is not necessarily approximately normal but is unbiased, then it can be shown that the estimate will deviate from θ by as much as 4 standard errors at most 6% of the time. We won introduced to address this problem. Suppose that the population pdf is f (x ; θ), a member of a particular parametric family, and that data x_1, x_2, …, xn gives = 21.7. We now use the computer to obtain “bootstrap samples” from the pdf f (x ; 21.7), and for each sample we calculate a “bootstrap estimate” *:

B = 100 or 200 is often used. Now let the sample mean of the bootstrap estimates. The bootstrap estimate of ’s standard error is now just the sample standard deviation of the

(In the bootstrap literature, B is often used in place of B – 1; for typical values of B, there is usually little difference between the resulting estimates.)

Example 6.11

A theoretical model suggests that X, the time to breakdown of an insulating fluid between electrodes at a particular voltage, has f (x ; λ) = λe –λx, an exponential distribution. A random sample of n = 10 breakdown times (min) gives the following data:

41.53 18.73 2.99 30.34 12.33 117.52 73.02 223.63 4.00 26.78

Since E (X ) = 1/λ, E () = 1/λ, so a reasonable estimate of λ is = 1/ = 1/55.087 = .018153. We then used a statistical computer package to obtain B = 100 bootstrap samples, each of size 10, from f (x ; .018153). The first such sample was 41.00, 109.70, 16.78, 6.31, 6.76, 5.62, 60.96, 78.81, 192.25, 27.61, from which = 545.8 and = 1/54.58 = .01832. The average of the 100 bootstrap estimates is * = .02153, and the sample standard deviation of these 100 estimates is s = .0091, the bootstrap estimate of ’s standard error. A histogram of the 100 was somewhat positively skewed, suggesting that the sampling distribution of also has this property.

Sometimes an investigator wishes to estimate a population characions.

d. Suppose a single beam and a single cylinder are randomly selected. Calculate a point estimate of the variance of the difference X – Y between beam strength and cylinder strength.

5. As an example of a situation in which several different statistics could reasonably be used to calculate a point estimate, consider a population of N invoices. Associated with each invoice is its “book value,” the recorded amount of that invoice. Let T denote the total book value, a known amount. Some of these book values are erroneous. An audit will be carried out by randomly selecting n invoices and determining the audited (correct) value for each one. Suppose that the sample gives the following results (in dollars).

Let

Y = sample mean book value

X = sample mean audited value

D = sample mean error

Propose three different statistics for estimating the total audited (i.e., correct) value—one involving just N and , another involving T, N, and , and the last involving T and /. If N = 5000 and T = 1,761,300, calculate the three corresponding point estimates. (The article “Statistical Models and Analysis in Auditing,” Statistical Science, 1989: 2-33). discusses properties of these estimators.)

6. Consider the accompanying observations on stream flow (1000 s of acre-feet) recorded at a station in Colorado for the period April 1-August 31 over a 31-year span (from an article in the 1974 volume of Water Resources Research).

An appropriate probability plot supports the use of the lognormal distribution (see Section 4.5) as a reasonable model for stream flow.

a. Estimate the parameters of the distribution. [Hint: Remember that X has a lognormal distribution with parameters μ and σ^2 if In(X ) is normally distributed with mean μ and variance σ^2.]

b. Use the estimates of part (a) to calculate an estimate of the expected value of stream flow. [Hint: What is E (X1 and n_2, respectively, of the two fertilizers. Show that the pooled (combined) estimator

is an unbiased estimator of σ^2.

13. Consider a random sample X_1, … , Xn from the pdf

f (x; θ) = .5(1 + θx ) –1 ≤ x ≤ 1

where –1 ≤ θ ≤ 1 (this distribution arises in particle physics). Show that = 3 is an unbiased estimator of θ . [Hint: First determine μ= E (X ) = E ().]

14. A sample of n captured Pandemonium jet fighters results in serial numbers x_1, x_2, x_3 , … , xn. The CIA knows that the aircraft were numbered consecutively at the factory starting with α and ending with β, so that the total number of planes manufactured is β –α + 1 (e.g., if α = 17 and β = 29, then 29 –17 + 1 = 13 planes having serial numbers 17, 18, 19, …, 28, 29 were manufactured). However, the CIA does not know the values of α or β. A CIA statistician suggests using the estimator max(Xi) –min(Xi) + 1 to estimate the total number of planes manufactured.

a. If n = 5, x_1 = 237, x_2 = 375, x_3 = 202, x_4 = 525, and x_5 = 418, what is the corresponding estimate?

b. Under what conditions on the sample will the value of the estimate be exactly equal to the true total number of planes? Will the estimate ever be larger than the true total? Do you think the estimator is unbiased for estimating β –α + 1? Explain in one or two sentences.

15. Let X_1, X_2 , … , Xn represent a random sample from a Rayleigh distribution with pdf

a. It can be shown that E(X^2) = 2θ. Use this fact to construct an unbiased estimator of θ based on meter to be estimated, the estimator is obtained by equating E(X) to . Since E(X) = l/λ for an exponential distribution, this gives l/λ = or λ = 1/. The moment estimator of λ is then λ = 1/.

Example 6.13

Let X_1, …, Xn be a random sample from a gamma distribution with parameters α and β. From Section 4.4, E(X) = αβ and E(X^2) = β^2Γ(α + 2)/Γ(α) = β^2(α + 1)α. The moment estimators of α and β are obtained by solving

Since α(α + 1)β^2 = α^2β^2 + αβ^2 and the first equation implies α^2β^2 = , the second equation becomes

Now dividing each side of this second equation by the corresponding side of the first equation and substituting back gives the estimators

To illustrate, the survival time data mentioned in Example 4.24 is

with = 113.5 and (1/20) = 14,087.8. The estimates are

These estimates of α and β differ from the values suggested by Gross and Clark because they used a different estimation technique.

Example 6.14

Let X_1, …, Xn be a random sample from a generalized negative binomial distribution with parameters r and p (Section 3.5). Since E(X) = r (l –p)/p and V(X) = r (1 –p)/p2 eventually gives

As an illustration, Reep, Pollard, and Benjamin (“Skill and Chance in Ball Games,” J. Royal Stat. Soc, 1971: 623-629) consider the negative binomial distribution as a model for the number of goals per game scored by National Hockey League teams. The data for 1966-1967 follows (420 games):

Then,

= Σxi/420 = [(0)(29) + (1)(71) + …+ (10)(3)]/420 = 2.98

and

Thus

Although r by definition must be positive, the denominator of could be negative, indicating that the negative binomial distribution is not appropriate (or that the moment estimator is flawed).

Maximum Likelihood Estimation

The method of maximum likelihood was first introduced by R. A. Fisher, a geneticist and statistician, in the 1920s. Most statisticians recommend this method, at least when the sample size is large, since the resulting estimators have certain desirable efficiency properties (see the proposition on page 249).

Example 6.15

A sample of ten new bike helmets manufactured by a certain company is obtained. Upon testing, it is found that the first, third, and tenth helmets are flawed, whereas the others are not. Let p = P (flawed helmet) and define X_1, …, X_10 by Xi = 1 if the ith helmet is flawed and zero otherwise. Then the observed xis are 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, so the joint pmf of the sample is

We now ask, “For what value of p is the observed sample most likely to have occurred?” That is, we wish to find the value of p that maximizes the pmf (6.4) or, equivalently, maximizes the natural log of (6.4).

which is a differentiable function of p, equating the derivative of (6.5) to zero gives the maximizing valued†

†This conclusion requires checking the second derivative, but the details are omitted.

where x is the observed number of successes (flawed helmets). The estimate of p is now . It is called the maximum likelihood estimate because for fixed x_1, …, x_10, it is the parameter value that maximizes the likelihood (joint pmf) of the observed sample.

Note that if we had been told only that among the ten helmets there were three that were flawed, Equation (6.4) would be replaced by the binomial pmf ()p^3(1 –p)7, which is also maximized for

DEFINITION

Let X_1, X_2, …, Xn have joint pmf or pdf

where the parameters θ_1 , … , θm have unknown values. When x_1,…, xn are the observed sample values and (6.6) is regarded as a function of 0_1, …, 0m , it is called the likelihood function. The maximum likelihood estimates (mle’s) θ_1,…, θm are those values of the θis that maximize the likelihood function, so that

f(x_1,…xn; 1,…, m ≥ f (x_1,…, xn; θ_1,…, θm) for all θ_1,…, θm

When the Xis are substituted in place of the xis, the maximum likelihood estimators result.

The likelihood function tells us how likely the observed sample is as a function of the possible parameter values. Maximizing the likelihood gives the parameter values for which the observed sample is most likely to have been generated—that is, the parameter values that “agree most closely” with the observed data.<*O|/p>

Example 6.16

Suppose X_1, X_2, …, Xn is a random sample from an exponential distribution with parameter λ. Because of independence, the likelihood function is a product of the individual pdf’s:

f(x_1, … , xn;λ) = (λe –λxn) = λneλΣxi

The ln(likelihood) is

In[f (x_1, …, xn; λ)] = n In(λ) –λΣxi

Equating (d /d λ)[In(likelihood)] to zero results in n /λ – Σxi = 0, or λ = n /Σxi = 1/ Thus the mle is = 1/; it is identical to the method of moments estimator [but it is not an unbiased estimator, since E (1/) ≠ 1/E()].

Example 6.17

Let X_1, …, Xn be a random sample from a normal distribution. The likelihood function is

SO

To find the maximizing values of μ and σ^2, we must take the partial derivatives of ln(f ) with respect to μ and σ^2, equate them to zero, and solve the resulting two equations. Omitting the details, the resulting mle’s are

The mle of σ^2 is not the unbiased estimator, so two different principles of estimation (unbiasedness and maximum likelihood) yield two different estimators.

Example 6.18

In Chapter 3, we discussed the use of the Poisson distribution for modeling the number of “events” that occur inm of the likelihood occurs at a point of discontinuity, but the figure shows that = max(Xi). Thus if my waiting times are 2.3, 3.7, 1.5, .4, and 3.2, then the mle is = 3.7.

Figure 6.5 The likelihood function for Example 6.22

Example 6.23

A method that is often used to estimate the size of a wildlife population involves performing a capture/recapture experiment. In this experiment, an initial sample of M animals is captured, each of these animals is tagged, and the animals are then returned to the population. After allowing enough time for the tagged individuals to mix into the population, another sample of size n is captured. With X = the number of tagged animals in the second sample, the objective is to use the observed x to estimate the population size N.

The parameter of interest is θ = N, which can assume only integer values, so even after determining the likelihood function (pmf of X here), using calculus to obtain N would present difficulties. If we think of a success as a previously tagged animal being recaptured, then sampling is without replacement from a population containing M successes and N –M failures, so that X is a hypergeometric rv and the likelihood function is

The integer-valued nature of N notwithstanding, it would be difficult to take the derivative of p(x; N). However, if we consider the ratio of p(x; N) to p(x; N –1), we have

This ratio is larger than 1 if and only if (iff) N < Mn/x. The value of N for which p(x; N) is maximized is therefore the largest integer less than Mn/x. If we use standard mathematical notation [r ] for the largest integer less than or equal to r , the mle of N is = [Mn/x]. As an illustration, if M = 200 fish are taken from a lake and tagged, subsequently n = 100 fish are recaptured, and among the 100 there are x = 11 tagged fish, then = [(200)(100)/11] = [1818.18] = 1818. The estimate is actually rather intuitive; x/n is the proportion of the recaptured sample that is tagged, whereas M/N is the proportion of the entire population that is tagged. The estimate is obtained by equating these two proportions (estimating a population proportion by a sample proportion).

Xn is a random sample from a pdf f (x ; θ) that is symmetric about θ but that the investigator is unsure of the form of the f function. It is then desirable to use an estimator that is robust—that is, one that performs well for a wide variety of underlying pdf’s. One such estimator is a trimmed mean. In recent years, statisticians have proposed another type of estimator, called an M-estimator, based on a generalization of maximum likelihood estimation. Instead of maximizing the log likelihood Σln[f (x ; θ)] for a specified f , one maximizes Σp(xi; θ). The “objective function” p is selected to yield an estimator with good robustness properties. The book by David Hoaglin et al. (see the bibliography) contains a good exposition on this subject.

EXERCISE Section 6.2 (20–30)

20. A random sample of n bike helmets manufactured by a certain company is selected. Let X = the number among the n that are flawed, and let p = P (flawed). Assume that only X is observed, rather than the sequence of S’s and F’s.

a. Derive the maximum likelihood estimator of p. If n = 20 and x = 3, what is the estimate?

b. Is the estimator of part (a) unbiased?

c. If n = 20 and x = 3, what is the mle of the probability (1 –p )5 that none of the next five helmets examined is flawed?

21. Let X have a Weibull distribution with parameters α and β, so

E(X) = β · Γ(1 + 1/α)

V(X) = β^2{Γ(1 + 2 /α) –[Γ(1 + 1/α)]2}

a. Based on a random sample X_1,…, Xn, write equations for the method of moments estimators of β and α. Show that, once the estimate of α has been obtained, the estimate of β can be found from a table of the gamma function and that the estimate of α is the solution to a complicated equation involving the gamma function.

b. If n = 20, = 28.0, and = 16,500, compute the estimates. [Hint: [Γ(1.2)]2/Γ(1.4) = .95.]

22. Let X denotee the pdf of X is

where –1 ≤ θ. A random sample of ten students yields data x_1 = .92, x_2 = .79, x_3 = .90, x_4 = .65, x_5 = .86, x_6 = .47, x_7 = .73, x_8 = .97, x_9 = .94, x_10 = .77.

a. Use the method of moments to obtain an estimator of θ, and then compute the estimate for this data.

b. Obtain the maximum likelihood estimator of θ, and then compute the estimate for the given data.

23. Two different computer systems are monitored for a total of n weeks. Let Xi denote the number of breakdowns of the first system during the ith week, and suppose the Xis are independent and drawn from a Poisson distribution with parameter λ_1. Similarly, let Yi denote the number of breakdowns of the second system during the ith week, and assume independence with each Yi Poisson with parameter λ_2. Derive the mle’s of λ_1, λ_2, and λ_1 –λ_2. [Hint: Using independence, write the joint pmf (likelihood) of the Xis and Yis together.]

24. Refer to Exercise 20. Instead of selecting n = 20 helmets to examine, suppose I examine helmets in succession until I have found r = 3 flawed ones. If the 20th helmet is the third flawed one (so that the number of helmets examined that were not flawed is x = 17), what is the mle of p? Is this the same as the estimate in Exercise 20? Why or why not? Is it the same as the estimate computed from the unbiased estimator of Exercise 17?

25. The shear strength of each of ten test spot welds is determined, yielding the following data (psi):

392 376 401 367 389 362 409 415 358 375

a. Assuming that shear strength is normally distributed, estimate the true average shear strength and standard deviation of shear strength using the method of maximum likelihood.

b. Again assuming a normal distribution, estimate the strength value below which 95% of all welds will have their strengths. [Hint: What is the 95th percentile in terms of μ and σ? Now use the invariance principle.]

26. Refer to Exercise 25. Suppose we decide to examine another test spot weld. Let X = shear strength of the weld. Use the given data to obtain the mle of P(X ≤ 400). [Hint: P(X ≤ 400) = Φ ((400 ̵="00681" alt="Image"> is unbiased, then MSE() = V(), but in general MSE() = V() + (bias)2. Consider the estimator 2 = KS^2, where S^2 = sample variance. What value of K minimizes the mean squared error of this estimator when the population distribution is normal? [Hint: It can be shown that

E[(S^2)2] = (n + 1)σ^4/(n –1)

In general, it is difficult to find to minimize MSE(), which is why we look only at unbiased estimators and minimize V().]

35. Let X_1, …, Xn be a random sample from a pdf that is symmetric about μ. An estimator for μ that has been found to perform well for a variety of underlying distributions is the Hodges-Lehmann estimator. To define it, first compute for each i ≤ j and each j = 1, 2, …, n the pairwise average i = (i + X )/2. Then the estimator is μ= the median of the Xi s. Compute the value of this estimate using the data of Exercise 44 of Chapter 1. [Hint: Construct a square table with the xis listed on the left margin and on top. Then compute averages on and above the diagonal.]

36. When the population distribution is normal, the statistic median{|X_1 – |, …, |Xn – |}/.6745 can be used to estimate σ. This estimator is more resistant to the effects of outliers (observations far from the bulk of the data) than is the sample standard deviation. Compute both the corresponding point estimate and s for the data of Example 6.2.

37. When the sample standard deviation S is based on a random sample from a normal population distribution, it can be shown that

Use this to obtain an unbiased estimator for σ of the form cS. What is c when n = 20?

38. Each of n specimens is to be weighed twice on the same scale. Let Xi and Yi denote the two observed weights for the ith specimen. Suppose Xi and Yi are independent of one another, e be to μ. An alternative to reporting a single sensible value for the parameter being estimated is to calculate and report an entire interval of plausible values—an interval estimate or confidence interval (CI). A confidence interval is always calculated by first selecting a confidence level, which is a measure of the degree of reliability of the interval. A confidence interval with a 95% confidence level for the true average breaking strength might have a lower limit of 9162.5 and an upper limit of 9482.9. Then at the 95% confidence level, any value of μ between 9162.5 and 9482.9 is plausible. A confidence level of 95% implies that 95% of all samples would give an interval that includes μ, or whatever other parameter is being estimated, and only 5% of all samples would yield an erroneous interval. The most frequently used confidence levels are 95%, 99%, and 90%. The higher the confidence level, the more strongly we believe that the value of the parameter being estimated lies within the interval (an interpretation of any particular confidence level will be given shortly).

Information about the precision of an interval estimate is conveyed by the width of the interval. If the confidence level is high and the resulting interval is quite narrow, our knowledge of the value of the parameter is reasonably precise. A very wide confidence interval, however, gives the message that there is a great deal of uncertainty concerning the value of what we are estimating. Figure 7.1 shows 95% confidence intervals for true average breaking strengths of two different brands of paper towels. One of these intervals suggests precise knowledge about μ, whereas the other suggests a very wide range of plausible values.

Figure 7.1 Confidence intervals indicating precise (brand 1) and imprecise (brand 2) information about μ

7.1 Basic Properties of Confidence Intervals

The basic concepts and properties of confidence intervals (CIs) are most easily introduced by first focusing on a simple, albeit somewhat unrealistic, problem situation. Suppose that the parameter of interest is a population mean μ and that

1. The population distribution is normal

2. The value of the population standard deviation σ is known

Normality of the population distribution is often a reasonable assumption. However, if the value of μ is unknown, it is implausible that the value of σ would be available (knowledge of a population’s center typically precedes information concerning spread). In later sections, we will develop methods based on less restrictive assumptions.

Example 7.1

Industrial engineers who specialize in ergonomics are concerned with designing workspace and devices operated by workers so as to achieve high productivity 220;Studies on Ergonomically Designed Alphanumeric Keyboards” (Human Factors, 1985: 175-187) reports on a study of preferred height for an experimental keyboard with large forearm-wrist support. A sample of n = 31 trained typists was selected, and the preferred keyboard height was determined for each typist. The resulting sample average preferred height was = 80.0 cm. Assuming that the preferred height is normally distributed with σ = 2.0 cm (a value suggested by data in the article), obtain a CI for μ, the true average preferred height for the population of all experienced typists.

The actual sample observations x_1, x_2, …, xn are assumed to be the result of a random sample X_1, …, Xn from a normal distribution with mean value μ and standard deviation σ. The results of Chapter 5 then imply that irrespective of the sample size n, the sample mean is normally distributed with expected value μ and standard deviation . Standardizing by first subtracting its expected value and then dividing by its standard deviation yields the standard normal variable

Because the area under the standard normal curve between –1.96 and 1.96 is .95,

Now let’s manipulate the inequalities inside the parentheses in (7.2) so that they appear in the equivalent form l < μ < u, where the endpoints l and u involve and . This is achieved through the following sequence of operations, each yielding inequalities equivalent to the original ones.

1. Multiply through by

2. Subtract from each term:

3. Multiply through by –1 to eliminate the minus sign in front of μ (which reverses the direction of each inequality):

that is,

The event inside the parentheses in (7.3) has a somewhat unfamiliar appearance; previously, the random quantity has appeared in the middle with constants on both ends, as in a ≤ Y ≤ b. In (7.3) the random quantity appears on the two ends, whereas the unknown constant μ appears in the middle. To interpret (7.3), think of a random interval having left endpoint – 1.96 • and right endpoint + 1.96 · . In interval notation, this becomes

The interval (7.4) is random because the two endpoints of the interval involve a random variable. It is centered at the sample mean and extends 1.96 to each side of . Thus the interval’s width is 2 • (1.96) • , which is not random; only the location of the interval (its midpoint ) is random (Figure 7.2). Now (7.3) can be paraphrased as “the probability is .95 that the random interval (7.4) includes or covers the true value of μ.” Before any experiment is performed and any data is gathered, it is quite likely that μ will lie inside the interval (7.4).

Figure 7.2 The random interval (7.4) centered at

DEFINITION

If after observing X_1 = x_1, X_2 = x_2, …, Xn = xn , we compute the observed sample mean and then substitute into (7.4) in place of , the resulting fixed interval is called a 95% confidence interval for μ. This CI can be expressed either as

is a 95% CI for μ

or as

with 95% confidence

A concise expression for the interval is ± 1.96 • , where – gives the left endpoint (lower limit) and + gives the right endpoint (upper limit).

n = (9.80)2 = 96.04

Since n must be an integer, a sample size of 97 is required.

The general formula for the sample size n necessary to ensure an interval width w is obtained from w = 2 • zα/2 • as

The smaller the desired width w, the larger n must be. In addition, n is an increasing function of σ (more population variability necessitates a larger sample size) and of the confidence level 100(1 –α) (as α decreases, zα/2 increases).

The half-width 1.96 of the 95% CI is sometimes called the bound on the error of estimation associated with a 95% confidence level. That is, with 95% confidence, the point estimate will be no farther than this from μ. Before obtaining data, an investigator may wish to determine a sample size for which a particular value of the bound is achieved. For example, with μ representing the average fuel efficiency (mpg) for all cars of a certain type, the objective of an investigation may be to estimate μ to within 1 mpg with 95% confidence. More generally, if we wish to estimate μ to within an amount B (the specified bound on the error of estimation) with 100(1 –α)% confidence, the necessary sample size results from replacing 2 /w by 1 /B in the formula in the preceding box.

Deriving a Confidence Interval

Let X_1, X_2, …, Xn denote the sample on which the CI for a parameter θ is to be based. Suppose a random variable satisfying the following two properties can be found:

1. The variable depends functionally on both X_1, …, Xn and θ.

2. The probability distribution of the variable does not depend on θ or on any other unknown parameters.

Let h(X_1, X_2, …, Xn; θ) denote this random variable. For example, if the population distribution is normal with known σ and θ = μ, the variable h(X_1, …, Xn; μ) = ( –μ)/() satisfies both properties; i the unknown percentiles in (7.7). Consult the Devore and Berk or Efron books cited in Chapter 6 for more information.

(EXERCISES Section 7.1 (1–11)

1. Consider a normal population distribution with the value of σ known.

a. What is the confidence level for the interval ± 2.81 ?

b. What is the confidence level for the interval ± 1.44?

c. What value of zα/2 in the CI formula (7.5) results in a confidence level of 99.7%?

d. Answer the question posed in part (c) for a confidence level of 75%.

2. Each of the following is a confidence interval for μ = true average (i.e., population mean) resonance frequency (Hz) for all tennis rackets of a certain type:

(114.4, 115.6)    (114.1, 115.9)

a. What is the value of the sample mean resonance frequency?

b. Both intervals were calculated from the same sample data. The confidence level for one of these intervals is 90% and for the other is 99%. Which of the intervals has the 90% confidence level, and why?

3. Suppose that a random sample of 50 bottles of a particular brand of cough syrup is selected and the alcohol content of each bottle is determined. Let μ denote the average alcohol content for the population of all bottles of the brand under study. Suppose that the resulting 95% confidence interval is (7.8, 9.4).

a. Would a 90% confidence interval calculated from this same sample have been narrower or wider than the given interval? Explain your reasoning.

b. Consider the following statement: There is a 95% chance that μ is between 7.8 and 9.4. Is this statement correct? Why or why not?

c. Consider the following statement: We can be highly confident thockquote>If n is sufficiently large, the standardized variable

has approximately a standard normal distribution. This implies that

is a large-sample confidence interval for μ with confidence level approximately 100(1 –α)%. This formula is valid regardless of the shape of the population distribution.

Generally speaking, n > 40 will be sufficient to justify the use of this interval. This is somewhat more conservative than the rule of thumb for the CLT because of the additional variability introduced by using S in place of σ.

Example 7.6

The alternating current (AC) breakdown voltage of an insulating liquid indicates its dielectric strength. The article “Testing Practices for the AC Breakdown Voltage Testing of Insulation Liquids” (IEEE Electrical Insulation Magazine, 1995: 21–26) gave the accompanying sample observations on breakdown voltage (kV) of a particular circuit under certain conditions.

A boxplot of the data (Figure 7.5) shows a high concentration in the middle half of the data (narrow box width). There is a single outlier at the upper end, but this value is actually a bit closer to the median (55) than is the smallest sample observation.

Figure 7.5 A boxplot for the breakdown voltage data from Example 7.6

Summary quantities include n = 48, Σxi = 2626, and = 144,950, from which = 54.7 and s = 5.23. The 95% confidence interval is then

That is,

53.2 < μ < 56.2

with a confidence level of approximately 95%. The interval is reasonably narrow, indicating that we have precisely estimated μ.

Unfortunately, the choice of sample size to yield a desired interval width is not as 952; does not involve θ but does involve at least one other unknown parameter. Let S be the estimate of σ obtained by using estimates in place of the unknown parameters (e.g., s/ estimates ). Under general conditions (essentially that s be close to σ for most samples), a valid CI is ± zα/2 • s. The large-sample interval ± za /2 · s / is an example.

Finally, suppose that σθ does involve the unknown θ. This is the case, for example, when θ = p, a population proportion. Then ( –θ)/σ = zα/2 can be difficult to solve. An approximate solution can often be obtained by replacing θ in σ by its estimate . This results in an estimated standard deviation sθ, and the corresponding interval is again ± Zθ/2 • s.

PROPOSITION

A confidence interval for a population proportion p with confidence level approximately 100(1 –α)% has

If the sample size is quite large, z^2 /(2n ) is negligible compared to z^2/(4n^2) under the square root is negligible compared to /n, and z^2/n is negligible compared to 1. Discarding these negligible terms gives approximate confidence limits

This is of the general form ± zα/2 of a large-sample interval suggested in the last subsection. For decades this latter interval has been recommended as long as the normal approximation for is justified. However, recent research has shown that the somewhat more complicated interval given in the proposition has an actual confidence level that tends to be closer to the nominal level than does the traditional interval (Agresti, Alan, and Coull, “Approximate Is Better Than ’Exact’ for Interval Estimation of a Binomial Proportion,” The American Statistician, 1998: 119-126). That is, if zα/2 = 1.96 is used, the confidence level for the “new” interval tends to be closer to 95% for almost all values of p than is the case for the traditional interval; this is also true for other confidence levels. In addition, Agresti and Coull state that the interval “can be recommended for use with nearly all sample sizes and parameter values,” so the conditions n ≥ 10 and n ≥ 10 need not be checked.

Example 7.8

The article “Repeatability and Reproducibility for Pass/Fail Data” (/. of Testing and Eval., 1997: 151-153) reported that in n = 48 trials in a particular laboratory, 16 resulted in ignition of a particular type of substrate by a lighted cigarette. Let p denote the long-run proportion of all such trials that would result in ignition. A point estimate for p is = 16/48 = .333. A confidence interval for p with a confidence level of approximately 95% is

.645 is .95,

Manipulating the inequality inside the parentheses to isolate μ on one side and replacing rv’s by calculated values gives the inequality μ > – 1.645s/; the expression on the right is the desired lower confidence bound. Starting with P(-1.645 < Z ) ≈ .95 and manipulating the inequality results in the upper confidence bound. A similar argument gives a one-sided bound associated with any other confidence level.

PROPOSITION

A large-sample upper confidence bound for μ is

and a large-sample lower confidence bound for μ is

A one-sided confidence bound for p results from replacing zα/2 by za and ± by either + or –in the CI formula (7.10) for p. In all cases the confidence level is approximately 100(1 –α)%.

Example 7.10 The slant shear test is the most widely accepted procedure for assessing the quality of a bond between a repair material and its concrete substrate. The article “Testing the Bond Between Repair Materials and Concrete Substrate” (ACI Materials J., 1996: 553-558) reported that in one particular investigation, a sample of 48 shear strength observations gave a sample mean strength of 17.17 N/mm^2 and a sample standard deviation of 3.28 N/mm^2. A lower confidence bound for true average shear strength μ with confidence level 95% is

That is, with a confidence level of 95%, the value of μ lies in the interval (16.39, °°).

(EXERCISES Section 7.2 (12–27)

12. A random sample of 110 lightning flashes in a certain region resulted in a sample average radar echo duration of .81 sec and a sample standard deviation of .34 sec (“Lightning Strikes to an Airplane in a Thunderstorm,” J. of Aircraft, 1984: 607-611). Calculate a 99% (two-sided) confidence interval for the true average echo duration μ, and interpret the resulting interval.

13. The article “Gas Cooking, Kitchen Ventilation, and Exposure to Combustion Products” (Indoor Air, 20reported that for a sample of 50 kitchens with gas cooking appliances monitored during a one-week period, the sample mean CO_2level (ppm) was 654.16, and the sample standard deviation was 164.43.

a. Calculate and interpret a 95% (two-sided) confidence interval for true average CO_2 level in the population of all homes from which the sample was selected.

b. Suppose the investigators had made a rough guess of 175 for the value of s before collecting data. What sample size would be necessary to obtain an interval width of 50 ppm for a confidence level of 95%?

14. The article “Evaluating Tunnel Kiln Performance” (Amer. Ceramic Soc. Bull., Aug. 1997: 59-63) gave the following summary information for fracture strengths (MPa) of n = 169 ceramic bars fired in a particular kiln: ) = 89.10, s = 3.73.

a. Calculate a (two-sided) confidence interval for true average fracture strength using a confidence level of 95%. Does it appear that true average fracture strength has been precisely estimated?

b. Suppose the investigators had believed a priori that the population standard deviation was about 4 MPa. Based on this supposition, how large a sample would have been required to estimate μ to within .5 MPa with 95% confidence?

15. Determine the confidence level for each of the following large-sample one-sided confidence bounds:

a. Upper bound: + .84s/

b. Lower bound: –2.05s/

c. Upper bound: + .67s/

16. The charge-to-tap time (min) for a carbon steel in one type of open hearth furnace was determined for each heat in a sample of size 46, resulting in a sample mean time of 382.1 and a sample standard deviation of 31.5. Calculate a 95% upper confidence bound for true average charge-to-tap time.

17. Exercise 1.13 gave a sample of ultimate tensile strength observations (ksi). Use the accompanying descriptive statistics output from MINITAB to calculate a 99% lower confidence bound for true average ultimate tensile strength, and interpret the result.

25. A state legislator wishes to survey residents of her district to see what proportion of the electorate is aware of her position on using state funds to pay for abortions.

a. What sample size is necessary if the 95% CI for p is to have width of at most .10 irrespective of p?

b. If the legislator has strong reason to believe that at least f of the electorate know of her position, how large a sample size would you recommend?

26. The superintendent of a large school district, having once had a course in probability and statistics, believes that the number of teachers absent on any given day has a Poisson distribution with parameter λ . Use the accompanying data on absences for 50 days to derive a large-sample CI for λ. [Hint: The mean and variance of a Poisson variable both equal λ, so

has approximately a standard normal distribution. Now proceed as in the derivation of the interval for p by making a probability statement (with probability 1 –α) and solving the resulting inequalities for λ (see the argument just after (7.10)).]

27. Reconsider the CI (7.10) for p, and focus on a confidence level of 95%. Show that the confidence limits agree quite well with those of the traditional interval (7.11) once two successes and two failures have been appended to the sample [i.e., (7.11) based on x + 2 S’s in n + 4 trials]. [Hint: 1.96 ≈ 2. Note: Agresti and Coull showed that this adjustment of the traditional interval also has actual confidence level close to the nominal level.]

7.3 Intervals Based on a Normal Population Distribution

The CI for μ presented in Section 7.2 is valid provided that n is large. The resulting interval can be used whatever the nature of the population distribution. The CLT cannot be invoked, however, when n is small. In this case, one way to proceed is to make a specific assumption about the form of the population distribution and then derive a CI tailored to that assumption. For example, we could develop a CI for μ when the population is described by a gamma distribution, another interval for the case of a Weibull population, and so on. Statistution is more frequently appropriate as a population model than is any other type of distribution, we will focus here on a CI for this situation.

ASSUMPTION

The population of interest is normal, so that X_1, …, Xn constitutes a random sample from a normal distribution with both μ and σ unknown.

The key result underlying the interval in Section 7.2 was that for large n, the rv Z = ( – μ)/(S/) has approximately a standard normal distribution. When n is small, S is no longer likely to be close to σ, so the variability in the distribution of Z arises from randomness in both the numerator and the denominator. This implies that the probability distribution of ( –μ )/(S/) will be more spread out than the standard normal distribution. The result on which inferences are based introduces a new family of probability distributions called the family of t distributions.

THEOREM

When is the mean of a random sample of size n from a normal distribution with mean μ, the rv

has a probability distribution called a t distribution with n – 1 degrees of freedom (df).

Properties of t Distributions

Before applying this theorem, a discussion of properties of t distributions is in order. Although the variable of interest is still ( –μ)/(S/), we now denote it by T to emphasize that it does not have a standard normal distribution when n is small. Recall that a normal distribution is governed by two parameters, the mean μ and the standard deviation σ. A t distribution is governed by only one parameter, called the number of degrees of freedom of the distribution, abbreviated by df. We denote this parameter by the Greek letter ν. Possible values of ν are the positive integers 1, 2, 3, .… Each different value of ν corresponds to a different t distribution.

For any fixed value of the parameter v, the density function that specifies the associated t curve has an even more complicated appearance than the normal density function. Fortunately, we need concern ourselves only with several of the more important features of these curves.

Properties of t Distributions

Let tν

The values of tα,ν exhibit regular behavior as we move across a row or down a column. For fixed ν, tα,ν increases as a decreases, since we must move farther to the right of zero to capture area a in the tail. For fixed α, as ν is increased (i.e., as we look down any particular column of the t table) the value of tα,ν decreases. This is because a larger value of ν implies a t distribution with smaller spread, so it is not necessary to go so far from zero to capture tail area α. Furthermore, tα,ν decreases more slowly as ν increases. Consequently, the table values are shown in increments of 2 between 30 df and 40 df and then jump to ν = 50, 60, 120, and finally ∞. Because t∞ is the standard normal curve, the familiar Zα values appear in the last row of the table. The rule of thumb suggested earlier for use of the large-sample CI (if n > 40) comes from the approximate equality of the standard normal and t distributions for ν ≥ 40.

The One-Sample t Confidence Interval

The standardized variable T has a t distribution with n –1 df, and the area under the corresponding t density curve between −tα/2n–1 and tα/2,n–1 is 1 –α (area α/2 lies in each tail), so

Expression (7.14) differs from expressions in previous sections in that T and tα/2,n–1 are used in place of Z and Zα/2 , but it can be manipulated in the same manner to obtain a confidence interval for μ.

PROPOSITION

Let and s be the sample mean and sample standard deviation computed from the results of a random sample from a normal population with mean μ. Then a 100(1 –α)% confidence interval for μ is

or, more compactly, ± tα/2,n–1 • s/.

An upper confidence bound for μ is

μ and σ, there is a good chance that the resulting interval will include less than 90% of the population values. Intuitively, to have an a priori 95% chance of the resulting interval including at least 90% of the population values, when and s are used in place of μ and σ, we should also replace 1.645 by some larger number. For example, when n = 20, the value 2.310 is such that we can be 95% confident that the interval ± 2.310s will include at least 90% of the fuel efficiency values in the population.

Let k be a number between 0 and 100. A tolerance interval for capturing at least k% of the values in a normal population distribution with a confidence level 95% has the form

± (tolerance critical value) · s

Tolerance critical values for k = 90, 95, and 99 in combination with various sample sizes are given in Appendix Table A.6. This table also includes critical values for a confidence level of 99% (these values are larger than the corresponding 95% values). Replacing ± by + gives an upper tolerance bound, and using –in place of ± results in a lower tolerance bound. Critical values for obtaining these one-sided bounds also appear in Appendix Table A.6.

Example 7.14

Let’s return to the modulus of elasticity data discussed in Example 7.11, where n = 16, = 14,532.5, s = 2055.67, and a normal probability plot of the data indicated that population normality was quite plausible. For a confidence level of 95%, a two-sided tolerance interval for capturing at least 95% of the modulus of elasticity values for specimens of lumber in the population sampled uses the tolerance critical value of 2.903. The resulting interval is

14,532.5 ± (2.903) (2055.67) = 14,532.5 ± 5967.6 = (8,564.9, 20,500.1)

We can be highly confident that at least 95% of all lumber specimens have modulus of elasticity values between 8,564.9 and 20,500.1.

The 95% CI for μ was (13,437.3, 15,627.7), and the 95% prediction interval for the modulus of elasticity of a single lumber specimen is (10,017.0, 19,048.0). Both the prediction interval and the tolerance interval are substantially wider than the confidence interval.

Intervals Based on Nonnormal Population Distributions

The one-sample t CI for μ is robust to small or even moderate departures from normality unless n is quite n level of 95% for the ultimate tensile strength of the next specimen selected.

41. A more extensive tabulation of t critical values than what appears in this book shows that for the t distribution with 20 df, the areas to the right of the values .687, .860, and 1.064 are .25, .20, and .15, respectively. What is the confidence level for each of the following three confidence intervals for the mean μ of a normal population distribution? Which of the three intervals would you recommend be used, and why?

a. ( –.687s, + 1.725s/)

b. ( –.860s, + 1.325s/)

c. ( –1.064s, + 1.064s/)

7.4 Confidence Intervals for the Variance and Standard Deviation of a Normal Population

Although inferences concerning a population variance σ^2 or standard deviation σ are usually of less interest than those about a mean or proportion, there are occasions when such procedures are needed. In the case of a normal population distribution, inferences are based on the following result concerning the sample variance S^2.

THEOREM

Let X_1, X_2, …, Xn be a random sample from a normal distribution with parameters μ and σ^2. Then the rv

has a chi-squared (χ^2) probability distribution with n –1 df.

As discussed in Sections 4.4 and 7.1, the chi-squared distribution is a continuous probability distribution with a single parameter ν, called the number of degrees of freedom, with possible values 1, 2, 3, .… The graphs of several χ^2 probability distribution functions (pdf’s) are illustrated in Figure 7.9. Each pdf f (x; ν) is positive only for x > 0, and each has a positive skew (long upper tail), though the distribution moves rightward and becomes more symmetric as ν increases. To specify inferential procedures that use the chi-squared distribution, we need notation analogous to that for a t critical value tα,ν .

corresponding limits in the interval for σ^2.

Example 7.15

The accompanying data on breakdown voltage of electrically stressed circuits was read from a normal probability plot that appeared in the article “Damage of Flexible Printed Wiring Boards Associated with Lightning-Induced Voltage Surges” (IEEE Transactions on Components, Hybrids, and Manuf. Tech., 1985: 214-220). The straightness of the plot gave strong support to the assumption that breakdown voltage is approximately normally distributed.

Let σ 2 denote the variance of the breakdown voltage distribution. The computed value of the sample variance is s^2 = 137,324.3, the point estimate of σ 2. With df = n –1 = 16, a 95% CI requires and . The interval is

Taking the square root of each endpoint yields (276.0, 564.0) as the 95% CI for σ. These intervals are quite wide, reflecting substantial variability in breakdown voltage in combination with a small sample size.

CIs for σ^2 and σ when the population distribution is not normal can be difficult to obtain, even when the sample size is large. For such cases, consult a knowledgeable statistician.

EXERCISES Section 7.4 (42–46)

42. Determine the values of the following quantities:

a.

b.

c.

d.

e.

f.

b. The 5th percentile of the chi-squared distribution with ν = 10

c. P (10.98 ≤ x 2 ≤ 36.78), where x 2 is a chi-squared rv with ν = 22

d. P (χ^2 < 14.611 or χ 2 > 37.652), where χ 2 is a chisquared rv with ν = 25

44. The amount of lateral expansion (mils) was determined for a sample of n = 9 pulsed-power gas metal arc welds used in LNG ship containment tanks. The resulting sample standard deviation was s = 2.81 mils. Assuming normality, derive a 95% CI for σ 2 and for σ.

45. The following observations were made on fracture toughness of a base plate of 18% nickel maraging steel [“Fracture Testing of Weldments,” ASTM Special Publ. No. 381, 1965: 328-356 (in ksi given in increasing order)]:

Calculate a 99% CI for the standard deviation of the fracture toughness distribution. Is this interval valid whatever the nature of the distribution? Explain.

46. The results of a Wagner turbidity test performed on 15 samples of standard Ottawa testing sand were (in microamperes)

a. Is it plausible that this sample was selected from a normal population distribution?

b. Calculate an upper confidence bound with confidence level 95% for the population standard deviation of turbidity.

SUPPLEMENTRY EXERCISE

47. Example 1.10 introduced the accompanying observations on bond strength.

a. Estimate true average bond strength in a way that conveys information about precision and reliability. [Hint: Σ Xi = 387.8 and = 4247.08.]

b. Calculate a 95% CI for the pr

56. Chronic exposure to asbestos fiber is a well-known health hazard. The article “The Acute Effects of Chrysotile Asbestos Exposure on Lung Function” (Environ. Research, 1978: 360-372) reports results of a study based on a sample of construction workers who had been exposed to asbestos over a prolonged period. Among the data given in the article were the following (ordered) values of pulmonary compliance (cm^3/cm H_2O) for each of 16 subjects 8 months after the exposure period (pulmonary compliance is a measure of lung elasticity, or how effectively the lungs are able to inhale and exhale):

a. Is it plausible that the population distribution is normal?

b. Compute a 95% CI for the true average pulmonary compliance after such exposure.

c. Calculate an interval that, with a confidence level of 95%, includes at least 95% of the pulmonary compliance values in the population distribution.

57. In Example 6.8, we introduced the concept of a censored experiment in which n components are put on test and the experiment terminates as soon as r of the components have failed. Suppose component lifetimes are independent, each having an exponential distribution with parameter λ . Let Y_1 denote the time at which the first failure occurs, Y_2 the time at which the second failure occurs, and so on, so that Tr = Y_1 + … + Yr + (n –r ) Yr is the total accumulated lifetime at termination. Then it can be shown that 2λTr has a chi-squared distribution with 2r df. Use this fact to develop a 100 (1 –α ) % CI formula for true average lifetime 1 /λ. Compute a 95% CI from the data in Example 6.8.

58. Let X_1, X_2, …, Xn be a random sample from a continuous probability distribution having median (so that P (Xi ≤ )= P (Xi ≥ ) = .5) .

a. Show that

P (min(Xi ) < max((Xi ) =

so that (min (Xi), max (Xi)) is a 100 (1 –α ) % confidenc>i)} is {max (Xi) ≤ }∪{min (Xi ) ≥ }. But max (Xi ) ≤ iff Xi ≤ for all i .]

b. For each of six normal male infants, the amount of the amino acid alanine (mg/100 mL) was determined while the infants were on an isoleucine-free diet, resulting in the following data:

2.84 3.54 2.80 1.44 2.94 2.70

Compute a 97% CI for the true median amount of alanine for infants on such a diet (“The Essential Amino Acid Requirements of Infants,” Amer. J. Nutrition, 1964: 322-330) .

c. Let X(2) denote the second smallest of the Xi s and x(n–1)denote the second largest of the xi s. What is the confidence coefficp>

The choice y = α /2 yields the usual interval derived in Section 7.2; if y ν α /2, this interval is not symmetric about . The width of this interval is w = s(zγ + zα –γ ) Show that w is minimized for the choice y = a /2, so that the symmetric interval is the shortest. [Hints: (a) By definition of zα , Φ (zα ) = 1 –α , so that zα = Φ-1 (1 –α ); (b) the relationship between the derivative of a function y = f (x ) and the inverse function x = f–1 (y) is (d/dy) f –1(y) = 1/f ′(x).]

61. Suppose X_1, X_2, …, Xn are observed values resulting from a random sample from a symmetric but possibly heavy-tailed distribution. Let and fs denote the sample median and fourth spread, respectively. Chapter 11 of Understanding Robust and Exploratory Data Analysis (see the bibliography in Chapter 6) suggests the following robust 95% CI for the population mean (point of symmetry):

The value of the quantity in parentheses is 2.10 for n = 10, 1.94 for n = 20, and 1.91 for n = 30. Compute this CI for the data of Exercise 45, and compare to the t CI appropriate for a normal population distribution.

62. a. Use the results of Example 7.5 to obtain a 95% lower confidence bound for the parameter λ of an exponential distribution, and calculate the bound based on the data given in the example.

b. If lifetime X has an exponential distribution, the probability that lifetime exceeds t is P (X > t) = e–λt Use the result of part (a) to obtain a 95% lower confidence bound for the probability that breakdown time exceeds 100 min.

Bibliography

DeGroot, Morris, and Mark Schervish, Probability and Statistics (3rd ed.), Addison-Wesley, Reading, MA, 2002. A very good exposition of the general principles of statistical inference.

Hahn, Gerald, and William Meeker, Statistical Intervals, Wiley, New York, 1991. Everything you ever wanted to know about

statistical intervals (confidence, predicd were defective. An engineer has suggested a change in the production process in the belief that it will result in a reduced defective rate. Let p denote the true proportion of defective boards resulting from the changed process. Then the research hypothesis, on which the burden of proof is placed, is the assertion that p < .10. Thus the alternative hypothesis is Ha : p < .10.

In our treatment of hypothesis testing, H_0 will always be stated as an equality claim. If θ denotes the parameter of interest, the null hypothesis will have the form H_0: θ = θ_0, where θ_0 is a specified number called the null value of the parameter (value claimed for θ by the null hypothesis). As an example, consider the circuit board situation just discussed. The suggested alternative hypothesis was Ha: p < .10, the claim that the defective rate is reduced by the process modification. A natural choice of H_0 in this situation is the claim that p ≥ .10, according to which the new process is either no better or worse than the one currently used. We will instead consider H_0: p = .10 versus Ha: p < .10. The rationale for using this simplified null hypothesis is that any reasonable decision procedure for deciding between H_0: p = .10 and Ha: p < .10 will also be reasonable for deciding between the claim that p ≥ .10 and Ha. The use of a simplified H_0 is preferred because it has certain technical benefits, which will be apparent shortly.

The alternative to the null hypothesis H_0: θ = θ_0 will look like one of the following three assertions: (1) Ha: θ > θ_0 (in which case the implicit null hypothesis is θ ≤ θ_0), (2) Ha: θ < θ_0 (so the implicit null hypothesis states that θ ≥ θ_0), or (3) Ha: θ ≠ θ_0 For example, let σ denote the standard deviation of the distribution of inside diameters (inches) for a certain type of metal sleeve. If the decision was made to use the sleeve unless sample evidence conclusively demonstrated that σ > .001, the appropriate hypotheses would be H_0: σ = .001 versus Ha: σ > .001. The number θ_0 that appears in both H_0 and Ha (separates the alternative from the null) is called the null value.

Test Procedures

A test procedure is a rule, based on sample data, for deciding whether to reject H_0. A test of H_0: p = .10 versus Ha: p < .10 in the circuit board problem might be based on examining a random sample of n = 200 boards. Let X denote the number of defective boards in the sample, a binomial random variable; x representmple 8.1

A certain type of automobile is known to sustain no visible damage 25% of the time in 10-mph crash tests. A modified bumper design has been proposed in an effort to increase this percentage. Let p denote the proportion of all 10-mph crashes with this new bumper that result in no visible damage. The hypotheses to be tested are H_0: p = .25 (no improvement) versus Ha: p > .25. The test will be based on an experiment involving n = 20 independent crashes with prototypes of the new design. Intuitively, H_0 should be rejected if a substantial number of the crashes show no damage. Consider the following test procedure:

Test statistic: X = the number of crashes with no visible damage

Rejection region: R_8 = {8, 9, 10, …, 19, 20}; that is, reject H_0 if x ≥ 8, where x is the observed value of the test statistic.

This rejection region is called upper-tailed because it consists only of large values of the test statistic.

When H_0 is true, X has a binomial probability distribution with n = 20 and p = .25. Then

α = P (type I error) = P(H_0 is rejected when it is true)

= P(X ≥ 8 when X ~ Bin(20, .25)) = 1 –B (7; 20, .25)

= 1 –.898 = .102

That is, when H_0 is actually true, roughly 10% of all experiments consisting of 20 crashes would result in H_0 being incorrectly rejected (a type I error).

In contrast to α , there is not a single β . Instead, there is a different β for each different p that exceeds .25. Thus there is a value of β for p = .3 (in which case X ~ Bin(20, .3)), another value of β for p = .5, and so on. For example,

β(.3) = P (type II error when p = .3)

= P (H_0 is not rejected when it is false because p = .3)

= P(X ≤ 7 when X ~ Bin(20, .3)) = B (7; 20, .3) = .772

When p is actually .3 rather than .25 (a “small”>0 is true: α (75), α (75.8), α (76.5), and so on. It is easily verified, though, that α (75) is the largest of all these type I error probabilities. Focusing on the boundary value amounts to working explicitly with the “worst case.”

The specification of a cutoff value for the rejection region in the examples just considered was somewhat arbitrary. Use of R_8 = {8, 9, …, 20} in Example 8.1 resulted in α = .102, β (.3) = .772, and β (.5) = .132. Many would think these error probabilities intolerably large. Perhaps they can be decreased by changing the cutoff value.

Example 8.3

(Example 8.1 continued)

Let us use the same experiment and test statistic X as previously described in the automobile bumper problem but now consider the rejection region R_9 = {9, 10, …, 20}. Since X still has a binomial distribution with parameters n = 20 and p ,

α = P(H_0 is rejected when p = .25)

= P(X ≥ 9 when X ~ Bin(20, .25)) = 1 –B (8; 20, .25) = .041

The type I error probability has been decreased by using the new rejection region. However, a price has been paid for this decrease:

β (.3) = P(H_0 is not rejected when p = .3)

= P(X ≤ 8 when X ~ Bin(20, .3)) = B (8; 20, .3) = .887

β (.5) = B (8; 20, .5) = .252

Both these β s are larger than the corresponding error probabilities .772 and .132 for the region R_8 . In retrospect, this is not surprising; α is computed by summing over probabilities of test statistic values in the rejection region , whereas β is the probability that X falls in the complement of the rejection region. Making the rejection region smaller must therefore decrease α while increasing β for any fixed alternative value of the parameter.

Example 8.4

(Example 8.2 continued)

The use of cutoff value c = 70.8 in the paint-drying example resulted in a very small value of α (.01) but rather large β s. Consider the same experiment and test statistic with the new rejection region ≤ 72. Because is still normally distributed with mean value μ = μ and σ = 1.8,

α = P (H_0 is rejected when it is true)

= P ( ≤ 72 when ~ N (75, 1.8^2))

= = Φ(–1.67) = .0475 ≈ .05

β (72) = P(H_0 is not rejected when μ = 72)

= P( > 72 when is a normal rv with mean 72 and standard deviation 1.8)

= 1 – = 1 –Φ(0) = .5

β (70) = = .1335 β (67) = .0027

The change in cutoff value has made the rejection region larger (it includes more values), resulting in a decrease in β for each fixed μ less than 75. However, α for this new region has increased from the previous value .01 to approximately .05. If a type I error probability this large can be tolerated, though, the second region (c = 72) is preferable to the first (c = 70.8) because of the smaller β s.

The results of these examples can be generalized in the following manner.

PROPOSITION

Suppose an experiment and a sample size are fixed and a test statistic is chosen. Then decreasing the size of the rejection region to obtain a smaller value of α results in a larger value of β for any particular parameter value consistent with Ha.

This proposition says that once the test statistic and n are fixed, there is no rejection region that will simultaneously make both α and all β s small. A region must be chosen to effect a compromise between α and β .

Because of the suggested guidproper choice of the hypotheses). The approach adhered to by most statistical practitioners is then to specify the largest value of α that can be tolerated and find a rejection region having that value of α rather than anything smaller. This makes β as small as possible subject to the bound on α. The resulting value of α is often referred to as the significance level of the test. Traditional levels of significance are .10, .05, and .01, though the level in any particular problem will depend on the seriousness of a type I error—the more serious this error, the smaller should be the significance level. The corresponding test procedure is called a level α test (e.g., a level .05 test or a level .01 test). A test with significance level α is one for which the type I error probability is controlled at the specified level.

Example 8.5

Consider the situation mentioned previously in which μ was the true average nicotine content of brand B cigarettes. The objective is to test H_0: μ = 1.5 versus Ha: μ > 1.5 based on a random sample X_1, X_2, …, X_32 of nicotine contents. Suppose the distribution of nicotine content is known to be normal with σ = .20. Then is normally distributed with mean value μ = μ and standard deviation σ = .20/ = .0354.

Rather than use itself as the test statistic, let’s standardize assuming that H_0 is true.

Z expresses the distance between and its expected value when H_0 is true as some number of standard deviations. For example, z = 3 results from an that is 3 standard deviations larger than we would have expected it to be were H_0 true.

Rejecting H_0 when “considerably” exceeds 1.5 is equivalent to rejecting H_0 when z “considerably” exceeds 0. That is, the form of the rejection region is z ≥ c . Let’s now determine c so that α = .05. When H_0 is true, Z has a standard normal distribution. Thus

α = P (type I error) = P (rejecting H_0 when H_0 is true)

= P(Z ≥ c when Z ~ N (0, 1))

The value c must capture upper-tail area .05 under the z curve. Either from H_0: μ = 150° versus Ha: μ > 150°. In the context of this situation, describe type I and type II errors. Which type of error would you consider more serious? Explain.

8. A regular type of laminate is currently being used by a manufacturer of circlude from the following sample data:

g. Reexpress the test procedure of part (b) in terms of the standardized test statistic Z = ( –10)/().

12. A new design for the braking system on a certain type of car has been proposed. For the current system, the true average braking distance at 40 mph under specified conditions is known to be 120 ft. It is proposed that the new design be implemented only if sample data strongly indicates a reduction in true average braking distance for the new design.

a. Define the parameter of interest and state the relevant hypotheses.

b. Suppose braking distance for the new system is normally distributed with σ = 10. Let denote the sample average braking distance for a random sample of 36 observations. Which of the following three rejection regions is appropriate: R_1 = {: ≥ 124.80}, R_2 = {: ≤ 115.20}, R_3 = {: either ≥ 125.13 or ≤ 114.87}?

c. What is the significance level for the appropriate region of part (b)? How would you change the region to obtain a test with α = .001?

d. What is the probability that the new design is not implemented when its true average braking distance is actually 115 ft and the appropriate region from part (b) is used?

e. Let Z = ( –120)/(). What is the significance level for the rejection region {z : z ≤ –2.33}? For the region {z : z ≤ –2.88}?

13. Let X_1, …, Xn denote a random sample from a normal population distribution with a known value of σ .

a. For testing the hypotheses H_0: μ = μ_0 versus Ha: μ > μ_0 (where μ_0 is a fixed number), show that the test with test statistic and rejection region ≥ μ_0 + 2.33 hasx="00055" alt="Image"> = 103, then the test statistic value is z = (103 –100)/2.0 = 1.5. That is, the observed value of is 1.5 standard deviations (of ) larger than what we expect it to be when H_0 is true. The statistic Z is a natural measure of the distance between , the estimator of μ , and its expected value when H_0 is true. If this distance is too great in a direction consistent with Ha, the null hypothesis should be rejected.

Suppose first that the alternative hypothesis has the form Ha: μ > μ_0. Then an value less than μ_0 certainly does not provide support for Ha. Such an corresponds to a negative value of z (since –μ_0 is negative and the divisor is positive). Similarly, an value that exceeds μ_0 by only a small amount (corresponding to z which is positive but small) does not suggest that H_0 should be rejected in favor of Ha. The rejection of H_0 is appropriate only when considerably exceeds μ_0—that is, when the z value is positive and large. In summary, the appropriate rejection region, based on the test statistic Z rather than , has the form z ≥ c .

As discussed in Section 8.1, the cutoff value c should be chosen to control the probability of a type I error at the desired level α . This is easily accomplished because the distribution of the test statistic Z when H_0 is true is the standard normal distribution (that’s why μ_0 was subtracted in standardizing). The required cutoff c is the z critical value that captures upper-tail area α under the z curve. As an example, let c = 1.645, the value that captures tail area .05 (z.05 = 1.645). Then,

α = P (type I error) = P (H_0 is rejected when H_0 is true)

= P(Z ≥ 1.645 when Z ~ N (0, 1)) = 1 –Φ(1.645) = .05

More generally, the rejection region z ≥ za has type I error probability α . The test procedure is upper-tailed because the rejection region consists only of large values of the test statistic.

Analogous reasoning for the alternative hypothesis Ha: μ < μ_0 suggests a rejection region of the form z ≤ c ="0em">

2. Determine the null value and state the null hypothesis.

3. State the appropriate alternative hypothesis.

4. Give the formula for the computed value of the test statistic (substituting the null value and the known values of any other parameters, but not those of any sample-based quantities).

5. State the rejection region for the selected significance level α .

6. Compute any necessary sample quantities, substitute into the formula for the test statistic value, and compute that value.

7. Decide whether H_0 should be rejected and state this conclusion in the problem context.

The formulation of hypotheses (Steps 2 and 3) should be done before examining the data.

Example 8.6

A manufacturer of sprinkler systems used for fire protection in office buildings claims that the true average system-activation temperature is 130°. A sample of n = 9 systems, when tested, yields a sample average activation temperature of 131.08°F. If the distribution of activation times is normal with standard deviation 1.5°F, does the data contradict the manufacturer’s claim at significance level α = .01?

1. Parameter of interest: μ = true average activation temperature.

2. Null hypothesis: H_0: μ = 130 (null value = μ_0 = 130).

3. Alternative hypothesis: Ha: μ ≠ 130 (a departure from the claimed value in either direction is of concern).

4. Test statistic value:

5. Rejection region: The form of Ha implies use of a two-tailed test with rejection region either z ≥ z.005 or z ≤ –z.005. From Section 4.3 or Appendix Table A.3, z.005 = 2.58, so we reject H0n . A parallel argument yields the necessary sample size for lower- and two-tailed tests as summarized in the next box.

where Φ(z ) = the standard normal cdf.

The sample size n for which a level α test also has β(μ′ ) = β at the alternative value μ′ is

Example 8.7

Let μ denote the true average tread life of a certain type of tire. Consider testing H_0: μ = 30,000 versus Ha: μ > 30,000 based on a sample of size n = 16 from a normal population distribution with σ = 1500. A test with α = .01 requires zα = z.01 = 2.33. The probability of making a type II error when μ = 31,000 is

Since z.1 = 1.28, the requirement that the level .01 test also have β (31,000) = .1 necessitates

The sample size must be an integer, so n = 30 tires should be used.

Case II: Large-Sample Tests

When the sample size is large, the z tests for case I are easily modified to yield valid test procedures without requiring either a normal population distribution or known σ . The key result was used in Chapter 7 to justify large-sample confidence intervals: A large n implies that the standardized variable

has approximately a standard normal distribution. Substitution of the null value μ_0 in place of μ yields the test statistic

which has approximately a standard normal distribution when H_0 is true. The use of rejection regions given previously for case I (e.g.ance level is approximately (rather than exactly) a . The rule of thumb n ≥ 40 will again be used to characterize a large sample size.

Example 8.8

A dynamic cone penetrometer (DCP) is used for measuring material resistance to penetration (mm/blow) as a cone is driven into pavement or subgrade. Suppose that for a particular application, it is required that the true average DCP value for a certain type of pavement be less than 30. The pavement will not be used unless there is conclusive evidence that the specification has been met. Let’s state and test the appropriate hypotheses using the following data (“Probabilistic Model for the Analysis of Dynamic Cone Penetrometer Test Values in Pavement Structure Evaluation,” J. of Testing and Evaluation , 1999: 7–14):

Figure 8.3 shows a descriptive summary obtained from MINITAB. The sample mean DCP is less than 30. However, there is a substantial amount of variation in the data (sample coefficient of variation = s/ = .4265), so the fact that the mean is less than the design specification cutoff may be a consequence just of sampling variability. Notice that the histogram does not resemble at all a normal curve (and a normal probability plot does not exhibit a linear pattern), but the large-sample z tests do not require a normal population distribution.

Figure 8.3 MINITAB descriptive summary for the DCP data of Example 8.8

Determination of β and the necessary sample size for these large-sample tests can be based either on specifying a plausible value of σ and using the case I formulas (even though s is used in the test) or on using the curves to be introduced shortly in connection with case III.

Case III: A Normal Population Distribution

When n is small, the Central Limit Theorem (CLT) can no longer be invoked to justify the use of a large-sample test. We faced this same difficulty in obtaining a small-sample confidence interval (CI) for μ in Chapter 7. Our approach here will be the same one used there: We will assume that the population distribution is at least approximately normal and describe test procedures whose validity rests on this assumption. If an investigator has good reason to believe that the population distribution is quite nonnormal, a distribution-free test from Chapter 15 can be used. Alternatively, a statistician can be consulted regarding procedures valid for specific families of population distributions other than the normal family. Or a bootstrap procedure can be developed.

The key result on which tests for a normal population mean are based was used in Chapter 7 to derive the one-sample t CI: If X_1, X_2, …, Xn is a random sample from a normal distribution, the standardized variable

has a t distribution with n –1 degrees of freedom (df). Consider testing H_0: μ = μ_0 against Ha: μ > μ_0 by using the test statistic . That is, the test statistic results from standardizing under the assumption that H_0 is true (using , the estimated standard deviation of , rather than ). When H_0 is true, the test statistic has a t distribution with n –1 df. Knowledge of the test statistic’s distribution when H_0 is true (the “null distribution”) allows us to construct a rejection region for which the type I error probability is controlled at the desired level. In particular, use of the upper-tail t critical value tα,n –1 to specify the rejection region t ≥ tα,n –1 implies that

P (type I error) = P(H_0 is rejected w; μ_0, n = 24, rejection region t ≥ –2.500

c. Ha: μ ≠ μ_0, n = 31, rejection region t ≥ 1.697 or t ≤ –1.697

17. Answer the following questions for the tire problem in Example 8.7.

a. If = 30,960 and a level α = .01 test is used, what is the decision?

b. If a level .01 test is used, what is β (30,500)?

c. If a level .01 test is used and it is also required that β (30,500) = .05, what sample size n is necessary?

d. If = 30,960, what is the smallest α at which H_0 can be rejected (based on n = 16)?

18. Reconsider the painc region is the average linear displacement (ALD). The paper gave the following ALD observations for a sample of 49 kidneys (units of pixel dimensions).

a. Summarize/describe the data.

b. Is it plausible that ALD is at least approximately normally distributed? Must normality be assumed prior to calculating a CI for true average ALD or testing hypotheses about true average ALD? Explain.

c. The authors commented that in most cases the ALD is better than or of the order of 1.0. Does the data in fact provide strong evidence for concluding that true average ALD under these circumstances is less than 1.0? Carry out an appropriate test of hypotheses.

d. Calculate an upper confidence bound for true average ALD using a confidence level of 95%, and interpret this bound.

28. Minor surgery on horses under field conditions requires a reliable short-term anesthetic producing good muscle relaxation, minimal cardiovascular and respiratory changes, and a quick, smooth recovery with minimal aftereffects so that horses can be left unattended. The article “A Field Trial of Ketamine Anesthesia in the Horse” (Equine Vet. J. , 1984: 176–179) reports that for a sample of n = 73 horses to which ketamine was administered under certain conditions, the sample average lateral recumbency (lying-down) time was 18.86 min and the standard deviation was 8.6 min. Does this data suggest that true average lateral recumbency time under these conditions is less than 20 min? Test the appropriate hypotheses at level of significance .10.

29. The amount of shaft wear (.0001 in.) after a fixed mileage was determined for each of n = 8 internal combustion engines having copper lead as a bearing material, resulting in = 3.72 and s = 1.25.

a. Assuming that the distribution of shaft wear is normal with mean μ , use the t test at level .05 to test H_0: μ = 3.50 versus Ha: μ > 3.50.

b. Using σ = 1.25, what is the type II error probability β(μ′ ) of the test for the alternative μ ′= 4.00?

30. The recommended daily dietary allowance for zinc among males older than age 50 years is 15 mg/day. The article “Nutrient Intakes and Dietary Patterns of Older Americans: A National Study” (J. Gerontology , 1992: M145-150) reports the following summary data on intake for a sample of males age 65–74 years: n = 115, = 11.3, and s = 6.43. Does this data indicate that average daily zinc intake in the population of 8211;74 falls below the recommended allowance?

31. In an experiment designed to measure the time necessary for an inspector’s eyes to become used to the reduced amount of light necessary for penetrant inspection, the sample average time for n = 9 inspectors was 6.32 sec and the sample standard deviation was 1.65 sec. It has previously been assumed that the average adaptation time was at least 7 sec. Assuming adaptation time to be normally distributed, does the data contradict prior belief? Use the t test with α = .1.

32. A sample of 12 radon detectors of a certain type was selected, and each was exposed to 100 pCi/L of radon. The resulting readings were as follows:

a. Does this data suggest that the population mean reading under these conditions differs from 100? State and test the appropriate hypotheses using α = .05.

b. Suppose that prior to the experiment, a value of σ = 7.5 had been assumed. How many determinations would then have been appropriate to obtain β = .10 for the alternative μ = 95?

33. Show that for any Δ> 0, when the population distribution is normal and σ is known, the two-tailed test satisfies β (μ_0 –Δ) = β (μ_0 + Δ), so that β (μ′) is symmetric about μ_0

34. For a fixed alternative value μ ′, show that β (μ′) → 0 as n → ∞ for either a one-tailed or a two-tailed z test in the case of a normal population distribution with known σ .

8.3 Tests Concerning a Population Proportion

Let p denote the proportion of individuals or objects in a population who possess a specified property (e.g., cars with manual transmissions or smokers who smoke a filter cigarette). If an individual or object with the property is labeled a success (S ), then p is the population proportion of successes. Tests concerning p will be based on a random sample of size n from the population. Provided that n is small relative to the population size, X (the number of S ’s in the sample) has (approximately) a binomial distribution. Furthermore, if n itself is large, both X and the estimator = X/n are approximately normally distributed. We first consider large-sample tests based on this latter fact and then turn to the small-sample case that directly uses the binomial distribution.

Large-Sample Tests

Large-sample tests concerning p are a special case of the more general large-sample procedures for a parameter θ . Let be an estimator of θ that is (at least approximately) unbiased and has approximately a normal distribution. The null hypothesis has the form H_0: θ = θ_0, where θ_0 denotes a number (the null value) appropriate to the problem context. Suppose that when H_0 is true, the standard deviation of , σ, involves no unknown parameters. For example, if θ = μ and = ,, which involves no unknown parameters only if the value of σ is known. A large-sample test statistic results from standardizing under the assumption that H_0 is true (so that E() = θ_0):

If the alternative hypothesis is Ha: θ > θ_0, an upper-tailed test whose significance level is approximately α is specified by the rejection region z ≥ zα . The other two alternatives, Ha: θ < θ_0 and Ha: θ ≠ θ_0, are tested using a lower-tailed z test and a two-tailed z test, respectively.

In the case θ = p , σ will not involve any unknown parameters when H_0 is true, but this is atypical. When σ does involve unknown parameters, it is often possible to use an estimated standard deviation S in place of σ and still have Z approximately normally distributed when H_0 is true (because when n is large, S ≈ σ for most samples). The large-sample test of the previous section furnishes an example of this: Because σ is usually unknown, we use in place of in the denominator of z .

The estimator = X/n is unbiased (E() = p ), has approximately a normal distribution, and its standard deviation is . These facts were used in Section 7.2 to obtain a confidence interval for ="center">

has approximately a standard normal distribution. If the alternative hypothesis is Ha: p > p_0 and the upper-tailed rejection region z ≥ zα is used, then

P (type I error) = P (H_0 is rejected when it is true)

= P (Z ≥ zα when Z has approximately a standard normal distribution) ≈ α

Thus the desired level of significance α is attained by using the critical value that captures area α in the upper tail of the z curve. Rejection regions for the other two alternative hypotheses, lower-tailed for Ha: p < p_0 and two-tailed for Ha: p ≠ p_0, are justified in an analogous manner.

Null hypothesis: H_0: p = p_0

Test statistic value:

These test procedures are valid provided that np_0 ≥ 10 and n (1 –p_0) ≥ 10.

Example 8.11

Recent information suggests that obesity is an increasing problem in America among all age groups. The Associated Press (October 9, 2002) reported that 1276 individuals in a sample of 4115 adults were found to be obese (a body mass index exceeding 30; this index is a measure of weight relative to height). A 1998 survey based on people’s own assessment revealed that 20% of adult Americans considered themselves obese. Does the most recent data suggest that the true proportion of adults who are obese is more than 1.5 times the percentage from the self-assessment survey? Let’s carry out a test of hypotheses using a significance level of .10.

1. p = the proportion of all American adults who are obese.

2. Saying that the current percentage is 1.5 times the self-assessment percentage is equivalent to the assertion that the current percentage is 30%, from which we have the null hypothesis as H_0: p = .30.

3. The phrase “more than” in the problem description 15(.7) > 10, the large-sample z test can certainly be used. The test statistic value is

5. The form of Ha implies that an upper-tailed test is appropriate: reject H_0 if z > z.10 = 1.28.

6. = 1276/4115 = .310, from which we have = .010/.0071 = 1.40.

7. Since 1.40 exceeds the critical value 1.28, z lies in the rejection region. This justifies rejecting the null hypothesis. Using a significance level of .10, it does appear that more than 30% of American adults are obese.

β and Sample Size Determination When H_0 is true, the test statistic Z has approximately a standard normal distribution. Now suppose that H_0 is not true and that p = p′ . Then Z still has approximately a normal distribution (because it is a linear function of ), but its mean value and variance are no longer 0 and 1, respectively. Instead,

The probability of a type II error for an upper-tailed test is β (p ′) = P (Z < zα when p = p′ ). This can be computed by using the given mean and variance to standardize and then referring to the standard normal cdf. In addition, if it is desired that the level α test also have β (p ′) = β for a specified value of β , this equation can be solved for the necessary n as in Section 8.2. General expressions for β (p ′) and n are given in the accompanying box.

The sample size n for which the level α test also satisfies β(p′ ) = β is

Example 8.12

A package-delivery service advertises that at least 90% of all packages brought to its office by 9 A.M . for delivery in the same city are delivered by noon that day. Let p denote the true proportion of such packages that are delivered as advertised and consider the hypotheses H_0: p = .9 versus Ha: p < .9/em> ′)

= P (X < c when X ~ Bin(n , p ′)) = B (c –1; n , p ′)

That is, β (p ′) is the result of a straightforward binomial probability calculation. The sample size n necessary to ensure that a level α test also has specified β at a particular alternative value p′ must be determined by trial and error using the binomial cdf.

Test procedures for Ha: p < p_0 and for Ha: p ≠ p_0 are constructed in a similar manner. In the former case, the appropriate rejection region has the form x ≤ c (a lower-tailed test). The critical value c is the largest number satisfying B (c ; n , p_0) ≤ α . The rejection region when the alternative hypothesis is Ha: p ≠ p_0 consists of both large and small x values.

Example 8.13

A plastics manufacturer has developed a new type of plastic trash can and proposes to sell them with an unconditional 6-year warranty. To see whether this is economically feasible, 20 prototype cans are subjected to an accelerated life test to simulate 6 years of use. The proposed warranty will be modified only if the sample data strongly suggests that fewer than 90% of such cans would survive the 6-year period. Let p denote the proportion of all cans that survive the accelerated test. The relevant hypotheses are H_0: p = .9 versus Ha: p < .9. A decision will be based on the test statistic X , the number among the 20 that survive. If the desired significance level is α = .05, c must satisfy B (c ; 20, .9) ≤ .05. From Appendix Table A.1, B (15; 20, .9) = .043, whereas B (16; 20, .9) = .133. The appropriate rejection region is therefore x ≤ 15. If the accelerated test results in x = 14, H_0 would be rejected in favor of Ha, necessitating a modification of the proposed warranty. The probability of a type II error for the alternative value p′ = .8 is

β (.8) = P(H 0 is not rejected when X ~ Bin(20, .8))

= P (X ≥ 16 when X ~ Bin(20, .8))

= 1 –B (15; 20, .8) = 1 –.370 = .630

That is, when p = .8, 63% of all samples consisting of n = 20 cans would result in H_0 being incorrectly not rejected. ndex="00007" alt="Image">

EXERCISES Section 8.3 (35–44)

35. State DMV records indicate that of all vehicles undergoing emissions testing during the previous year, 70% passed on the first try. A random sample of 200 cars tested in a particular county during the current year yields 124 that passed on the initial test. Does this suggest that the true proportion for this county during the current year differs from the previous statewide proportion? Test the relevant hypotheses using α = .05.

36. A manufacturer of nickel-hydrogen batteries randomly selects 100 nickel plates for test cells, cycles them a specified number of times, and determines that 14 of the plates have blistered.

a. Does this provide compelling evidence for concluding that more than 10% of all plates blister under such circumstances? State and test the appropriate hypotheses using a significance level of .05. In reaching your conclusion, what type of error might you have committed?

b. If it is really the case that 15% of all plates blister under these circumstances and a sample size of 100 is used, how likely is it that the null hypothesis of part (a) will not be rejected by the level .05 test? Answer this question for a sample size of 200.

c. How many plates would have to be tested to have β (.15) = .10 for the test of part (a)?

37. A random sample of 150 recent donations at a certain blood bank reveals that 82 were type A blood. Does this suggest that the actual percentage of type A donations differs from 40%, the percentage of the population having type A blood? Carry out a test of the appropriate hypotheses using a significance level of .01. Would your conclusion have been different if a significance level of .05 had been used?

38. It is known that roughly 2/3 of all human beings have a dominant right foot or eye. Is there also right-sided dominance in kissing behavior? The article “Human Behavior: Adult Persistence of Head-Turning Asymmetry” (Nature , 2003: 771) reported that in a random sample of 124 kissing couples, both people in 80 of the couples tended to lean more to the right than to the left.

a. If 2/3 of all kissing couples exhibit this right-leaning behavior, what is the probability that the number in a sample of 124 who do so differs from the expected value by at least as much as what was actually observed?

a. Among the 1000 books searched, 15 were misshelved or unlocatable. Test the relevant hypotheses and advise the librarian what to do (use α = .05).

b. If the true proportion of misshelved and lost books is actually .01, what is the probability that the inventory will be (unnecessarily) taken?

c. If the true proportion is .05, what is the probability that the inventory will be postponed?

40. The article “Statistical Evidence of Discrimination” (J. Amer. Stat. Assoc. , 1982: 773–783) discusses the court case Swain v. Alabama (1965), in which it was alleged that there was discrimination against blacks in grand jury selection. Census data suggested that 25% of those eligible for grand jury service were black, yet a random sample of 1050 called to appear for possible duty yielded only 177 blacks. Using a level .01 test, does this data argue strongly for a conclusion of discrimination?

41. A plan for an executive traveler’s club has been developed by an airline on the premise that 5% of its current customers would qualify for membership. A random sample of 500 customers yielded 40 who would qualify.

a. Using this data, test at level .01 the null hypothesis that the company’s premise is correct against the alternative that it is not correct.

b. What is the probability that when the test of part (a) is used, the company’s premise will be judged correct when in fact 10% of all current customers qualify?

42. Each of a group of 20 intermediate tennis players is given two rackets, one having nylon strings and the other synthetic gut strings. After several weeks of playing with the two rackets, each player will be asked to state a preference for one of the two types of strings. Let p denote the proportion of all such players who would prefer gut to nylon, and let X be the number of players in the sample who prefer gut. Because gut strings are more expensive, consider the null hypothesis that at most 50% of all such players prefer gut. We simplify this to H_0: p = .5, planning to reject H_0 only if sample evidence strongly favors gut strings.

a. Which of the rejection regions {15, 16, 17, 18, 19, 20}>

b. What is the probability of a type I error for the chosen region of part (a)? Does the region specify a level .05 test? Is it the best level .05 test?

c. If 60% of all enthusiasts prefer gut, calculate the probability of a type II error using the appropriate region from part (a). Repeat if 80% of all enthusiasts prefer gut.

d. If 13 out of the 20 players prefer gut, should H_0 be rejected using a significance level of .10?

43. A manufacturer of plumbing fixtures has developed a new type of washerless faucet. Let p = P (a randomly selected faucet of this type will develop a leak within 2 years under normal use). The manufacturer has decided to proceed with production unless it can be determined that p is too large; the borderline acceptable value of p is specified as .10. The manufacturer decides to subject n of these faucets to accelerated testing (approximating 2 years of normal use). With X = the number among the n faucets that leak before the test concludes, production will commence unless the observed X is too large. It is decided that if p = .10, the probability of not proceeding should be at most .10, whereas if p = .30 the probability of proceeding should be at most .10. Can n = 10 be used? n = 20? n = 25? What is the appropriate rejection region for the chosen n , and what are the actual error probabilities when this region is used?

44. Scientists think that robots will play a crucial role in factories in the next several decades. Suppose that in an experiment to determine whether the use of robots to weave computer cables is feasible, a robot was used to assemble 500 cables. The cables were examined and there were 15 defectives. If human assemblers have a defect rate of .035 (3.5%), does this data support the hypothesis that the proportion of defectives is lower for robots than humans? Use a .01 significance level.

8.4 P-Values

One way to report the result of a hypothesis-testing analysis is to simply say whether the null hypothesis was rejected at a specified level of significance. Thus an investigator might state that H_0 was rejected at level of significance .05 or that use of a level .01 test resulted in not rejecting H_0. This type of statement is somewhat inadequate because it says nothing about whether the conclusion was a very close call or quite clearcut. A related difficulty is that such a report imposes the specified significance level on other decision makers. In many decision situations, individuals may have different views concerning the consequences of a type I or type II error. Each individual would then want to select his or her own significance level—some selecting α = .05, others .01, and so on—and reach a conclusion accordingly. This could result in some individuals rejecting H_0 while others conclude that the data does not shman">Example 8.14

The true average time to initial relief of pain for a best-selling pain reliever is known to be 10 min. Let μ denote the true average time to relief for a company’s newly developed reliever. The company wishes to produce and market this reliever only if it provides quicker relief than the best-seller, so wishes to test H_0: μ = 10 versus Ha: μ < 10. Only if experimental evidence leads to rejection of H_0 will the new reliever be introduced. After weighing the relative seriousness of each type of error, a single level of significance must be agreed on and a decision—to reject H_0 and introduce the reliever or not to do so—made at that level.

Suppose the new reliever has been introduced. The company supports its claim of quicker relief by stating that, based on an analysis of experimental data, H_0: μ = 10 was rejected in favor of Ha: μ < 10 using level of significance α = .10. Any individuals contemplating a switch to this new reliever would naturally want to reach their own conclusions concerning the validity of the claim. Individuals who are satisfied with the best-seller would view a type I error (concluding that the new product provides quicker relief when it actually does not) as serious so might wish to use α = .05, .01, or even smaller levels. Unfortunately, the nature of the company’s statement prevents an individual decision maker from reaching a conclusion at such a level. The company has imposed its own choice of significance level on others. The report could have been done in a manner that allowed each individual flexibility in drawing a conclusion at a personally selected a .

A P-value conveys much information about the strength of evidence against H_0 α and allows an individual decision maker to draw a conclusion at any specified level a . Before we give a general definition, consider how the conclusion in a hypothesis-testing problem depends on the selected level a .

Example 8.15

The nicotine content problem discussed in Example 8.5 involved testing H_0: μ = 1.5 versus Ha: μ > 1.5. Because of the inequality in Ha, the rejection region is upper-tailed, with H_0 rejected if z ≥ zα . Suppose z = 2.10. The accompanying table displays the rejection region for each of four different α s along with the resulting conclusion.

Figure 8.5 Relationship between α and tail area capturem>α is to draw a picture like that of Figure 8.6. The calculation of the P -value depends on whether the test is upper-, lower-, or two-tailed. However, once it has been calculated, the comparison with α does not depend on which type of test was used.

Figure 8.6 Comparing α and the P-value: (a) reject H_0 when α lies here; (b) do not reject H_0 when α lies here

Example 8.16 (Example 8.14 continued)

Suppose that when data from an experiment involving the new pain reliever was analyzed, the P -value for testing H_0: μ = 10 versus Ha: μ < 10 was calculated as .0384. Since α = .05 is larger than the P -value (.05 lies in the interval (a) of Figure 8.6), H_0 would be rejected by anyone carrying out the test at level .05. However, at level .01, H_0 would not be rejected because .01 is smaller than the smallest level (.0384) at which H_0 can be rejected.

The most widely used statistical computer packages automatically include a P -value when a hypothesis-testing analysis is performed. A conclusion can then be drawn directly from the output, without reference to a table of critical values.

A useful alternative definition equivalent to the one just given is as follows:

DEFINITION

The P -value is the probability, calculated assuming H_0 is true, of obtaining a test statistic value at least as contradictory to H_0 as the value that actually resulted. The smaller the P -value, the more contradictory is the data to H_0.

Thus if z = 2.10 for an upper-tailed z test, P -value = P (Z ≥ 2.10 when H_0 is true) = 1 –Φ(2.10) = .0179, as before. Beware: The P -value is not the probability that H_0 is true, nor is it an error probability!

P-Values for z Tests

The P -value for a z test (one based on a test statistic whose distribution when H_0 is true is at least approximately standard normal) is easily determined from the information in Appendix Table A.3. Consider an upper-tailed test and let z denote the computed value of the test statistic Z . The null hypothesis is rejected if z ≥ zα , and the P -value is the en <€smallest α for which this is the case. Since zα increases as α decreases, the P -value is the value of α for which z = zα . That is, the P -value is just the area captured by the computed value z in the upper tail of the standard normal curve. The corresponding cumulative area is Φ(z ), so in this case P -value = 1 –Φ(z ).

An analogous argument for a lower-tailed test shows that the P -value is the area captured by the computed value z in the lower tail of the standard normal curve. More care must be exercised in the case of a two-tailed test. Suppose first that z is positive. Then the P -value is the value of α satisfying z = zα/2 (i.e., computed z = upper-tail critical value). This says that the area captured in the upper tail is half the P -value, so that P -value = 2[1 –Φ(z )]. If z is negative, the P -value is the α for which z = –zα /2, or, equivalently, –z = zα /2, so P -value = 2[1 –Φ(–z )]. Since –z = |z | when z is negative, P -value = 2[1 –Φ(|z |)] for either positive or negative z .

Each of these is the probability of getting a value at least as extreme as what was obtained (assuming H_0 true). The three cases are illustrated in Figure 8.7.

The next example illustrates the use of the P -value approach to hypothesis testing by means of a sequence of steps modified from our previously recommended sequence.

Figure 8.7 Determination of the P -value for a z test

Example 8.17

The target thickness for silicon wafers used in a certain type of integrated circuit is 245 μ m. A sample of 50 wafers is obtained and the thickness of each one is determined, resulting in a sample mean thickness of 246.18 μ m and a sample standard deviation of 3.60 (m. Does this data suggest that true average wafer thickness is something other than the target value?

1. Parameter of interest: μ = true average wafer thickness

2. Null hypothesis: H_0: μ = 245

3. Alternative hypothesis: Ha: μ ≠ 245

5. Calculation of test statistic value:

6. Determination of P -value: Because the test is two-tailed,

P -value = 2(1 –Φ(2.32)) = .0204

7. Conclusion: Using a significance level of .01, H_0 would not be rejected since .0204 > .01. At this significance level, there is insufficient evidence to conclude that true average thickness differs from the target value.

P-Values for t Tests

Just as the P -value for a z test is a z curve area, the P -value for a t test will be a t curve area. Figure 8.8 on the next page illustrates the three different cases. The number of df for the one-sample t test is n –1.

Figure 8.8 P -values for t tests

The table of t critical values used previously for confidence and prediction intervals doesn’t contain enough information about any particular t distribution to allow for accurate determination of desired areas. So we have included another t table in Appendix Table A.8, one that contains a tabulation of upper-tail t curve areas. Each different column of the table is for a different number of df, and the rows are for calculated values of the test statistic t ranging from 0.0 to 4.0 in increments of .1. For example, the number .074 appears at the intersection of the 1.6 row and the 8 df column, so the area under the 8 df curve to the right of 1.6 (an upper-tail area) is .074. Because t curves are symmetric, .074 is also the area under the 8 df curve to the left of -1.6 (a lower-tail area).

Suppose, for example, that a test of H_0: μ = 100 versus Ha: μ > 100 is based on the 8 df t distribution. If the calculated value of the test statistic is t = 1.6, then the P -value for this upper-tailed test is .074. Because .074 exceeds .05, we would not be able to reject H_0 at a significance level of .05. If the alternative hypothesis is Ha: μ < 100 and a test based on 20 df yields t = –3.2, then Appendix Table A.8 shows that the P -value is the captured lower-tail area .0"-2em">
50. The paint used to make lines on roads must reflect enough light to be clearly visible at night. Let μ denote the true average reflectometer reading for a new type of paint under consideration. A test of "-2em">
58. A certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of a writing machine) is at least 10 hours. A random sample of 18 pens is selected, the writing lifetime of each is determined, and a normal probability plot of the resulting data supports the use of a one-sample t test.

a. What hypotheses should be tested if the investigators believe a priori that the design specification has been satisfied?

b. What conclusion is appropriate if the hypotheses of part (a) are tested, t = –2.3, and α = .05?

c. What conclusion is appropriate if the hypotheses of part (a) are tested, t = –1.8, and α = .01?

d. What should be concluded if the hypotheses of part (a) are tested and t = –3.6?

59. A spectrophotometer used for measuring CO concentration [ppm (parts per million) by volume] is checked for accuracy by taking readings on a manufactured gas (called span gas) in which the CO concentration is very precisely controlled at 70 ppm. If the readings suggest that the spectrophotometer is not working properly, it will have to be recalibrated. Assume that if it is properly calibrated, measured concentration for span gas samples is normally distributed. On the basis of the six readings—85, 77, 82, 68, 72, and 69—is recalibration necessary? Carry out a test of the relevant hypotheses using the P -value approach with α = .05.

60. The relative conductivity of a semiconductor device is determined by the amount of impurity “doped” into the device during its manufacture. A silicon diode to be used for a specific purpose requires an average cut-on voltage of .60 V, and if this is not achieved, the amount of impurity must be adjusted. A sample of diodes was selected and the cut-on voltage was determined. The accompanying SAS output resulted from a request to test the appropriate hypotheses.

[Note: SAS explicitly tests H_0: μ = 0, so to test H_0: μ = .60, the null value .60 must be subtracted from each xi; the reported mean is then the average of the (xi –.60) values. Also, SAS’s P -value is always for a two-tailed test.] What would be concluded for a significance level of .01? .05? .10?

8.5 Some Comments on Selecting a Test

Once the experimenter hted to a large audience, rejection of H_0 at level .05 will be much more convincing if the observed value of the test statistic greatly exceeds the 5% critical value than if it barely exceeds that value. This is precisely what led to the notion of P -value as a way of reporting significance without imposing a particular α on others who might wish to draw their own conclusions.

Even if a P -value is included in a summary of results, however, there may be difficulty in interpreting this value and in making a decision. This is because a small P -value, which would ordinarily indicate statistical significance in that it would strongly suggest rejection of H_0 in favor of Ha, may be the result of a large sample size in combination with a departure from H_0 that has little practical significance. In many experimental situations, only departures from H_0 of large magnitude would be worthy of detection, whereas a small departure from H_0 would have little practical significance.

Consider as an example testing H_0: μ = 100 versus Ha: μ > 100 where μ is the mean of a normal population with σ = 10. Suppose a true value of μ = 101 would not represent a serious departure from H_0 in the sense that not rejecting H_0 when μ = 101 would be a relatively inexpensive error. For a reasonably large sample size n , this μ would lead to an value near 101, so we would not want this sample evidence to argue strongly for rejection of H_0 when = 101 is observed. For various sample sizes, Table 8.1 records both the P -value when = 101 and also the probability of not rejecting H_0 at level .01 when μ = 101.

The second column in Table 8.1 shows that even for moderately large sample sizes, the P -value of = 101 argues very strongly for rejection of H_0, whereas the observed ξ itself suggests that in practical terms the true value of μ differs little from the null value μ_0 = 100. The third column points out that even when there is little practical difference between the true μ and the null value, for a fixed level of significance a large sample size will almost always lead to rejection of the null hypothesis at that level. To summarize, one must be especially careful in interpreting evidence when the sample size is large, since any small departure from H_0 will almost surely be detected by a test, yet such a departure may have little practical significance .

Table 8.1 An Illustration of the Effect of Sample Size on P -values and β

A practical limitation on the use of the likelihood ratio principle is that, to construct the likelihood ratio test statistic, the form of the probability distribution from which the sample comes must be specified. To derive the t test from the likelihood ratio principle, the investigator must assume a normal pdf. If an investigator is willing to assume that the distribution is symmetric but does not want to be specific about its exact form (such as normal, uniform, or Cauchy), then the principle fails because there is no way to write a joint pdf simultaneously valid for all symmetric distributions. In Chapter 15, we will present several distribution-free test procedures, so called because the probability of a type I error is controlled simultaneously for many different underlying distributions. These procedures are useful when the investigator has limited knowledge of the underlying distribution. We shall also say more about issues 3 and 4 listed at the outset of this section.

EXERCISES Section 8.5 (61–62)

61. Reconsider the paint-drying problem discussed in Example 8.2. The hypotheses were H_0: μ = 75 versus Ha: μ < 75, with σ assumed to have value 9.0. Consider the alternative value μ = 74, which in the context of the problem would presumably not be a practically significant departure from H_0.

a. For a level .01 test, compute β at this alternative for sample sizes n = 100, 900, and 2500.

b. If the observed value of is = 74, what can you say about the resulting P -value when n = 2500? Is the data statistically significant at any of the standard values of α ?

c. Would you really want to use a sample size of 2500 along with a level .01 test (disregarding the cost of such an experiment)? Explain.

62. Consider the large-sample level .01 test in Section 8.3 for testing H_0: p = .2 against Ha: p > .2.

a. For the alternative value p = .21, compute β (.21) for sample sizes n = 100, 2500, 10,000, 40,000, and 90,000.

b. For = x/n = .21, compute the P -value when n = 100, 2500, 10,000, and 40,000.

c. In most situations, would it be reasonable to use a level .01 test in conjunction with a sample size of 40,000? Why or why not?

SUPPLEMENTARY EXERCISES (63–85)

63. A sample of 50 lenses used in eyeglasses yields a sample mean thickness of 3.05 σm and a sample standard deviation of .34 mm. The desired true average thickness of such lenses is 3.20 mm. Does the data strongly suggest that the true average thickness of such lenses is something other than what is desired? Test using α = .05.

64. In Exercise 63, suppose the experimenter had believed before collecting the data that the value of σ was approximately .30. If the experimenter wished the probability of a type II error to be .05 when μ = 3.00, was a sample size 50 unnecessarily large?

65. It is specified that a certain type of iron should contain .85 gm of silicon per 100 gm of iron (.85%). The silicon content of each of 25 randomly selected iron specimens was determined, and the accompanying MINITAB output resulted from a test of the appropriate hypotheses.

a. What hypotheses were tested?

b. What conclusion would be reached for a significance level of .05, and why? Answer the same question for a significance level of .10.

66. One method for straightening wire before coiling it to make a spring is called “roller straightening.” The article “The Effect of Roller and Spinner Wire Straightening on Coiling Performance and Wire Properties” (Springs , 1987: 27–28) reports on the tensile properties of wire. Suppose a sample of 16 wires is selected and each is tested to determine tensile strength (N/mm^2). The resulting sample mean and standard deviation are 2160 and 30, respectively.

a. The mean tensile strength for springs made using spinner straightening is 2150 N/mm^2. What hypotheses should be tested to determine whether the mean tensile strength for the roller method exceeds 2150?

b. Assuming that the tensile strength distribution is approximately normal, what test statistic would you use to test the hypotheses in part (a)?

c. What is the value of the test statistic for this data?

d. What is the P -value for the value of the test statistic computed in part (c)?

e. For a level .05 test, what conclusion would you reach “A Rapid Method to Determine Total Phosphorus in Soils” (Soil Sci. Amer. J. , 1988: 1301–1304). Suppose a sample of 11 soil specimens, each with a true phosphorus content of 548 mg/kg, is analyzed using the new method. The resulting sample mean and standard deviation for phosphorus level are 587 and 10, respectively.

a. Is there evidence that the mean phosphorus level reported by the new method differs significantly from the true value of 548 mg/kg? Use α = .05.

b. What assumptions must you make for the test in part (a) to be appropriate?

68. The article “Orchard Floor Management Utilizing Soil-Applied Coal Dust for Frost Protection” (Agri. and Forest Meteorology , 1988: 71–82) reports the following values for soil heat flux of eight plots covered with coal dust.

The mean soil heat flux for plots covered only with grass is 29.0. Assuming that the heat-flux distribution is approximately normal, does the data suggest that the coal dust is effective in increasing the mean heat flux over that for grass? Test the appropriate hypotheses using α = .05.

69. The article “Caffeine Knowledge, Attitudes, and Consumption in Adult Women” (J. of Nutrition Educ. , 1992: 179–184) reports the following summary data on daily caffeine consumption for a sample of adult women: n = 47, = 215 mg, s = 235 mg, and range = 5–1176.

a. Does it appear plausible that the population distribution of daily caffeine consumption is normal? Is it necessary to assume a normal population distribution to test hypotheses about the value of the population mean consumption? Explain your reasoning.

b. Suppose it had previously been believed that mean consumption was at most 200 mg. Does the given data contradict this prior belief? Test the appropriate hypotheses at significance level .10 and include a P -value in your analysis.

70. The accompanying output resulted when MINITAB was used to test the appropriate hypotheses about true average activation time based on the data in Exercise 57. Use this information to reach a conclusion at significance level .05 and also at level .01.

71. The true average breaking strength of ceramic insulators of a certain type is supposed to be at least 10 psi. They will be used for a particular application unl="Times New Roman">76. To test the ability of auto mechanics to identify simple engine problems, an automobile with a single such problem was taken in turn to 72 different car repair facilities. Only 42 of the 72 mechanics who worked on the car correctly identified the problem. Does this strongly indicate that the true proportion of mechanics who could identify this problem is less than .75? Compute the P -value and reach a conclusion accordingly.

77. When X_1, X_2, …, Xn are independent Poisson variables, each with parameter λ, and n is large, the sample mean has approximately a normal distribution with μ = E() = λ and σ 2 = V () = λ/n . This implies that

has approximately a standard normal distribution. For testing H_0: λ = λ_0, we can replace λ by λ_0 in the equation for Z to obtain a test statistic. This statistic is actually preferred to the large-sample statistic with denominator (when the Xi s are Poisson) because it is tailored explicitly to the Poisson assumption. If the number of requests for consulting received by a certain statistician during a 5-day work week has a Poisson distribution and the total number of consulting requests during a 36-week period is 160, does this suggest that the true average number of weekly requests exceeds 4.0? Test using α = .02.

78. An article in the Nov. 11, 2005, issue of the San Luis Obispo Tribune reported that researchers making random purchases at California Wal-Mart stores found scanners coming up with the wrong price 8.3% of the time. Suppose this was based on 200 purchases. The National Institute for Standards and Technology says that in the long run at most two out of every 100 items should have incorrectly scanned prices.

a. Develop a test procedure with a significance level of (approximately) .05, and then carry out the test to decide whether the NIST benchmark is not satisfied.

b. For the test procedure you employed in (a), what is the probability of deciding that the NIST benchmark has been satisfied when in fact the mistake rate is 5%?

79. A hot-tub manufacturer advertises that with its heating equipment, a temperature of 100°F can be achieved in at most 15 min. A random sample of 32 tubs is selected, and the time necessary to achieve a 100°F temperature is determined for each tub. The sample average time and sample standard deviation are 17.5 min and 2.2 min, respectively. Does this data cast doubt on the company’s claim? Compute the P -value and use it to reach a conclusion atard. One such hypothesis would state that μ_1 –μ_2 = 0, that is, that μ_1 = μ_2. Alternatively, it may be appropriate to estimate μ_1 –μ_2 by computing a 95% CI. Such inferences are based on a sample of strength observations for each type of fiberboard.

Basic Assumptions

1. X_1, X_2, …, Xm is a random sample from a population with mean μ_1 and variance .

2. Y_1, Y_2, …, Yn is a random sample from a population with mean μ_2 and variance .

3. The X and Y samples are independent of one another.

The natural estimator of μ_1 –μ_2 is , the difference between the corresponding sample means. The test statistic results from standardizing this estimator, so we need expressions for the expected value and standard deviation of .

PROPOSITION

The expected value of is μ_1 –μ_2, so is an unbiased estimator of μ_1 –μ_2. The standard deviation of is

Proof Both these results depend on the rules of expected value and variance presented in Chapter 5. Since the expected value of a difference is the difference of expected values,

E () = E () –E () = μ_1 –μ_2

Because the X and Y samples are independent, and are independent quantities, so the variance of the difference is the sum of V () and V ():

If we think of μ_1 –μ_2 as a parameter θ , then its estimator is with standard deviation σ§ given by the proposition. When and both have known values, the test statistic will have the form ( –null value)σ; this form of a test statistic was used in several one-sample problems in the previous chapter. The sample variances must be used to estimate σ when and are unknown.

Test Procedures for Normal Populations with Known Variances

In Chapters 7 and 8, the first CI and test procedure for a population mean μ were based on the assumption that the population distribution was normal with the value of the population variance σ 2 known to the investigator. Similarly, we first assume here that both population distributions are normal and that the values of both and are known. Situations in which one or both of these assumptions can be dispensed with will be presented shortly.

Because the population distributions are normal, both and have normal distributions. This implies that is normally distributed, with expected value μ_1 –μ_2 and standard deviation given in the foregoing proposition. Standardizing gives the standard normal variable

In a hypothesis-testing problem, the null hypothesis will state that μ_1 –μ_2 has a specified value. Denoting this null value by Δ_0, we have H_0: μ_1 μ_2 = Δ_0. Often Δ_0 = 0, in which case H_0 says that μ_1 = μ_2. A test statistic results from replacing μ_1 –μ_2 in Expression (9.1) by the null value Δ_0. The test statistic Z is obtained by standardizing under the assumption that H_0 is true, so it has a standard normal distribution in this case. Consider the alternative hypothesis Ha: μ_1 –μ_2 > Δ_0. A value that considerably exceeds Δ_0 (the expected value of when H_0 is true) provides evidence against H0z. Thus H_0 should be rejected in favor of H a if z is greater than or equal to an appropriately chosen critical value. Because the test statistic Z has a standard normal distribution when H_0 is true, the upper-tailed rejection region z ≤ zα gives a test with significance level (type I error probability) α. Rejection regions for Ha: μ_1 –μ_2 < Δ_0 and Ha: μ_1 –μ_2 ≠ Δ_0 that yield tests with desired significance level α are lower-tailed and two-tailed, respectively.

Null hypothesis: H_0: μ_1 –μ_2 = Δ_0

Test statistic value:

Because these are z tests, a P -value is computed as it was for the z tests in Chapter 8 (e.g., P -value = 1 –Φ(z ) for an upper-tailed test).

Example 9.1

Analysis of a random sample consisting of m = 20 specimens of cold-rolled steel to determine yield strengths resulted in a sample average strength of = 29.8 ksi. A second random sample of n = 25 two-sided galvanized steel specimens gave a sample average strength of = 34.7 ksi. Assuming that the two yield-strength distributions are normal with σ_1 = 4.0 and σ_2 = 5.0 (suggested by a graph in the article “Zinc-Coated Sheet Steel: An Overview,” Automotive Engr., Dec. 1984: 39-43), does the data indicate that the corresponding true average yield strengths μ_1 and μ_2 are different? Let’s carry out a test at significance level α = .01.

1. The parameter of interest is μ_1 –μ_2, the difference between the true average strengths for the two types of steel.

2. The null hypothesis is H_0: μ_1 –μ_2 = 0.

3. The alternative hypothesis is Ha: μ_1 –μ_2 ≠ 0; if Ha is true, then μ_1 and μLet μ_1 denote the true average number of years lived beyond graduation for physicians in full-time practice, and let μ 2 denote the same quantity for physicians with academic affiliations. Assume the 125 and 90 physicians to be random samples from populations 1 and 2, respectively (which may not be reasonable if there is reason to believe that Harvard graduates have special characteristics that differentiate them from all other physicians—in this case inferences would be restricted just to the “Harvard populations”). The letter from which the data was taken gave no information about variances, so for illustration assume that σ_1 = 14.6 and σ_2 = 14.4. The hypotheses are H_0: μ_1 –μ_2 = 0 versus Ha: μ_1 –μ_2 > 0, so Δ_0 is zero. The computed value of the test statistic is

The P -value for an upper-tailed test is 1 –Φ(2.85) = .0022. At significance level .01, H_0 is rejected (because α > P -value) in favor of the conclusion that μ_1 –μ_2 > 0 (μ_1 > μ_2). This is consistent with the information reported in the letter.

This data resulted from a retrospective observational study; the investigator did not start out by selecting a sample of doctors and assigning some to the “academic affiliation” treatment and the others to the “full-time practice” treatment, but instead identified members of the two groups by looking backward in time (through obituaries!) to past records. Can the statistically significant result here really be attributed to a difference in the type of medical practice after graduation, or is there some other underlying factor (e.g., age at graduation, exercise regimens, etc.) that might also furnish a plausible explanation for the difference? Observational studies have been used to argue for a causal link between smoking and lung cancer. There are many studies that show that the incidence of lung cancer is significantly higher among smokers than among nonsmokers. However, individuals had decided whether to become smokers long before investigators arrived on the scene, and factors in making this decision may have played a causal role in the contraction of lung cancer.

A randomized controlled experiment results when investigators assign subjects to the two treatments in a random fashion. When statistical significance is observed in such an experiment, the investigator and other interested parties will have more confidence in the conclusion that the difference in response has been caused by a difference in treatments. A very famous example of this type of experiment and conclusion is the Salk polio vaccine experiment described in Section 9.4. These issues are discussed at greater length in the (nonmathematical) books by Moore and by Freedman et al., listed in the Chapter 1 reight="0em" width="2em">The probability of a type II error is easily calculated when both population distributions are normal with known values of σ_1 and σ_2. Consider the case in which the alternative hypothesis is Ha: μ_1 –μ_2 > Δ_0 Let Δ′ denote a value of μ_1 –μ_2 that exceeds Δ_0 (a value for which H_0 is false). The upper-tailed rejection region z ≥ zα can be reexpressed in the form ≥ Δ_0 + zα. Thus the probability of a type II error when μ_1 –μ_2 = Δ′ is

β(Δ′) = P (not rejecting H_0 when μ_1 –μ_2 = Δ′)

= P ( <Δ_0 + zα when μ_1 –μ_2 = Δ′)

When μ_1 –μ_2 = Δ′, is normally distributed with mean value Δ′ and standard deviation (the same standard deviation as when H_0 is true); using these values to standardize the inequality in parentheses gives β .

Example 9.3 (Example 9.1 continued)

Suppose that when μ_1 and μ_2 (the true average yield strengths for the two types of steel) differ by as much as 5, the probability of detecting such a departure from H_0 should be .90. Does a level .01 test with sample sizes m = 20 and n = 25 satisfy this condition? The value of σ for these sample sizes (the denominator of z ) was previously calculated as 1.34. The probability of a type II error for the two-tailed level .01 test when μ_1 –μ_2 = Δ′= 5 is

It is easy to verify that β (–5) = .1251 also (because the rejection region is symmetric). Thus the probability of detecting such a departure is 1 –β (5) = .8749. Because this is somewhat less than .9, slightly larger sample sizes should be used.

As in Chapter 8, sample sizes m and n can be determined that will satisfy b>μ_2 = Δ′) = a specified β. For an upper-tailed test, equating the previous expression for β(Δ′) to the specified value of β gives

When the two sample sizes are equal, this equation yields

These expressions are also correct for a lower-tailed test, whereas α is replaced by α /2 for a two-tailed test.

Large-Sample Tests

The assumptions of normal population distributions and known values of σ_1 and σ_2 are unnecessary when both sample sizes are large. In this case, the Central Limit Theorem guarantees that has approximately a normal distribution regardless of the underlying population distributions. Furthermore, using and in place of and in Expression (9.1) gives a variable whose distribution is approximately standard normal:

A large-sample test statistic results from replacing μ_1 –μ_2 by Δ_0, the expected value of when H_0 is true. This statistic Z then has approximately a standard normal distribution when H_0 is true. Level α tests are obtained by using z critical values exactly as before.

Use of the test statistic value

along with the previously stated upper-, lower-, and two-tailed rejection regions based on z critical values gives large-sample tests whose significance levels are approximately α. These tests are usually appropriate if both m > 40 and n > 40. A P -value is computed exactly as it was for our earlier z tests.

Example 9.4

In selecting a sulphur concrete for roadway construction in regions that experience heavy frost, it is important that the chosen concrete have a low value of thermal conductivity to minimize subsequent damage due to changing temperatures. Suppose two types of concrete, a graded aggregate and a no-fines aggregate, are being considered for a certain road. Table 9.1 summarizes data from an experiment carried out to compare the two types of concrete. Does this information suggest that true average conductivity for the graded concrete exceeds that for the no-fines concrete? Let’s use a test with α = .01ce level is approximately 100(1 –α)%. Furthermore, use of the sample variances and in the standardized variable Z yields a valid interval in which and replace and .

Provided that m and n are both large, a CI for μ_1 –μ_2 with a confidence level of approximately 100(1 –a )% is

where –gives the lower limit and + the upper limit of the interval. An upper or lower confidence bound can also be calculated by retaining the appropriate sign (+ or –) and replacing zα/2 by zα.

Our standard rule of thumb for characterizing sample sizes as large is m > 40 and n > 40.

Example 9.5

An experiment carried out to study various characteristics of anchor bolts resulted in 78 observations on shear strength (kip) of 3/8-in. diameter bolts and 88 observations on strength of 1/2-in. diameter bolts. Summary quantities from MINITAB follow, and a comparative boxplot is presented in Figure 9.1. The sample sizes, sample means, and sample standard deviations agree with values given in the article “Ultimate Load Capacities of Expansion Anchor Bolts” (J. Energy Engr., 1993: 139–158). The summaries suggest that the main difference between the two samples is in where they are centered.

Figure 9.1 A comparative boxplot of the shear strength data

Let’s now calculate a confidence interval for the difference between true average shear strength for 3/8-in. bolts (μ_1) and true average shear strength for 1/2-in. bolts (μ_2) using a confidence level of 95%:

That is, with 95% confidence, –3.34 < μ_1 –μ_2 < –2.44. We can therefore be highly confident that the true average shear strength for the 1/2-in. bolts exceeds that for the 3/8-in. bolts by between 2.44 kip and 3.34 kip. Notice that if we relabel so that μ_1 refers to 1/2-in. bolts and μ_2 to 3/8-in. bolts, the confidence interval is now centered at +2.89 and the value .45 is still subtracted and added to div height="0em">

If the variances and are at least approximately known and the investigator uses equal sample sizes, then the common sample size n that yields a 100(1 –α)% interval of width w is

which will generally have to be rounded up to an integer.

EXERCISES Section 9.1 (1–16)

1. An article in the November 1983 Consumer Reports compared various types of batteries. The average lifetimes of Duracell Alkaline AA batteries and Eveready Energizer Alkaline AA batteries were given as 4.1 hours and 4.5 hours, respectively. Suppose these are the population average lifetimes.

a. Let be the sample average lifetime of 100 Duracell batteries and be the sample average lifetime of 100 Eveready batteries. What is the mean value of (i.e., where is the distribution of centered)? How does your answer depend on the specified sample sizes?

b. Suppose the population standard deviations of lifetime are 1.8 hours for Duracell batteries and 2.0 hours for Eveready batteries. With the sample sizes given in part (a), what is the variance of the statistic , and what is its standard deviation?

c. For the sample sizes given in part (a), draw a picture of the approximate distribution curve of (include a measurement scale on the horizontal axis). Would the shape of the curve necessarily be the same for sample sizes of 10 batteries of each type? Explain.

2. Let μ_1 and μ_2 denote true average tread lives for two competing brands of size P205/65R15 radial tires. Test H_0: μ_1 –μ_2 = 0 versus Ha: μ_1 –μ_2 ≠ 0 at level .05 using the following data: m = 45, = 42,500, s_1 = 2200, n = 45, = 40,400, and s_2 = 1900.

3. Let μ_1 denote true average tread life for a premium brand of P205/65R15 radial tire and let μ_2 denote the true average tread life for an economy brand of the same size. Test H_0: μ_1 –μ_2 = 5000 versus Ha: μ_1level .01 using the following data: m = 45, = 42,500, s_1 = 2200, n = 45, = 36,800, and s_2 = 1500.

4. a. Use the data of Exercise 2 to compute a 95% CI for μ_1 –μ_2. Does the resulting interval suggest that μ_1 –μ_2 h has been precisely estimated?

b. Use the data of Exercise 3 to compute a 95% upper confidence bound for μ_1 –μ_2.

5. Persons having Reynaud’s syndrome are apt to suffer a sudden impairment of blood circulation in fingers and toes. In an experiment to study the extent of this impairment, each subject immersed a forefinger in water and the resulting heat output (cal/cm^2/min) was measured. For m = 10 subjects with the syndrome, the average heat output was = .64, and for n = 10 nonsufferers, the average output was 2.05. Let μ_1 and μ_2 denote the true average heat outputs for the two types of subjects. Assume that the two distributions of heat output are normal with σ_1 = .2 and σ_2 = .4.

a. Consider testing H_0: μ_1 –μ_2 = –1.0 versus Ha: μ_1 –μ_2 < –1.0 at level .01. Describe in words what Ha says, and then carry out the test.

b. Compute the P -value for the value of Z obtained in part (a).

c. What is the probability of a type II error when the actual difference between μ_1 and μ_2 is μ_1 –μ_2 = –1.2?

d. Assuming that m = n, what sample sizes are required to ensure that β = .1 when μ_1 –μ_2 = –1.2?

6. An experiment to compare the tension bond strength of polymer latex modified mortar (Portland cement mortar to which polymer latex emulsions have been added during mixing) to that of unmodified mortar resulted in = 18.12 kgf/cm^2 for the modified mortar (m = 40) and = 16.87 kgf/cm^2 for the unmodified mortar (n = 32). Let μ_1 and μ_2 be the true average tension bond strengths for the modified and unmodified mortars, respectively. Assume that the bond strength distributions are both normal.

a= 1.6 and σ_2 = 1.4, test H_0: μ_1 –μ_2 = 0 versus Ha: μ_1 –μ_2 > 0 at level .01.

b. Compute the probability of a type II error for the test of part (a) when μ_1 –μ_2 = 1.

c. Suppose the investigator decided to use a level .05 test and wished β = .10 when μ_1 –μ_2 = 1. If m = 40, what value of n is necessary?

d. How would the analysis and conclusion of part (a) change if σ_1 and σ_2 were unknown but s_1 = 1.6 and s_2 = 1.4?

7. Are male college students more easily bored than their female counterparts? This question was examined in the article “Boredom in Young Adults—Gender and Cultural Comparisons” (J. of Cross-Cultural Psych., 1991: 209–223). The authors administered a scale called the Boredom Proneness Scale to 97 male and 148 female U.S. college students. Does the accompanying data support the research hypothesis that the mean Boredom Proneness Rating is higher for men than for women? Test the appropriate hypotheses using a .05 significance level.

8. Tensile strength tests were carried out on two different grades of wire rod (“Fluidized Bed Patenting of Wire Rods,” Wire J., June 1977: 56-61), resulting in the accompanying data.

a. Does the data provide compelling evidence for concluding that true average strength for the 1078 grade exceeds that for the 1064 grade by more than 10 kg/mm^2? Test the appropriate hypotheses using the P -value approach.

b. Estimate the difference between true average strengths for the two grades in a way that provides information about precision and reliability.

9. The article “Evaluation of a Ventilation Strategy to Prevent Barotrauma in Patients at High Risk for Acute Respiratory Distress Syndrome” (New Engl. J. of Med., 1998: 355-358) reported on an experiment in which 120 patients with similar clinical features were randomly divided into a control group and a treatment group, each consisting of 60 patients. The sample mean ICU stay (days) and sample standard deviation for the treatment group were 19.9 and 39.1>2 with confidence level 100(1—α)% is then

A one-sided confidence bound can be calculated as described earlier.

The two-sample t test for testing H_0: μ_1 –μ_2 = Δ_0 is as follows:

Test statistic value:

A P-value can be computed as described in Section 8.4 for the one-sample t test.

Example 9.6

The void volume within a textile fabric affects comfort, flammability, and insulation properties. Permeability of a fabric refers to the accessibility of void space to the flow of a gas or liquid. The article “The Relationship Between Porosity and Air Permeability of Woven Textile Fabrics” (J. of Testing and Eval., 1997: 108-114) gave summary information on air permeability (cm^3/cm^2/sec) for a number of different fabric types. Consider the following data on two different types of plain-weave fabric:

Assuming that the porosity distributions for both types of fabric are normal, let’s calculate a confidence interval for the difference between true average porosity for the cotton fabric and that for the acetate fabric, using a 95% confidence level. Before the appropriate t critical value can be selected, df must be determined:

Thus we use ν = 9; Appendix Table A.5 gives t.0259 = 2.262. The resulting interval is

With a high degree of confidence, we can say that true average porosity for triacetate fabric specimens exceeds that for cotton specimens by between 81.80 and 87.06 cm^3/cm^2/sec.

Example 9.7

The deterioration of many municipal pipeline networks across the country is a growing concern. One technology proposed for pipeline rehabilitation uses a flexible liner threaded through existing pipe. The article “Effect of Welding on a High-Density Polyethylene Liner” (J. of Materials in Civil Engr., 1996: 94-100) reported the following data on tensile strength e="Times New Roman">Table A.8 shows that the area under the 15 df t curve to the right of 1.8 is .046, so the P -value for a lower-tailed test is also .046. The following MINITAB output summarizes all the computations:

Twosample T for nofusion vs fused

7. Using a significance level of .05, we can barely reject the null hypothesis in favor of the alternative hypothesis, confirming the conclusion stated in the article. However, someone demanding more compelling evidence might select α = .01, a level for which H_0 cannot be rejected.

If the question posed had been whether fusing increased true average strength by more than 100 psi, then the relevant hypotheses would have been H_0: μ_1 –μ_2 = –100 versus Ha: μ_1 –μ_2 < –100; that is, the null value would have been Δ_0 = –100.

Pooled t Procedures

Alternatives to the two-sample t procedures just described result from assuming not only that the two population distributions are normal but also that they have equal variances ( = ). That is, the two population distribution curves are assumed normal with equal spreads, the only possible difference between them being where they are centered.

Let σ 2 denote the common population variance. Then standardizing gives

which has a standard normal distribution. Before this variable can be used as a basis for making inferences about μ_1 –μ_2, the common variance must be estimated from sample data. One estimator of σ^2 is , the variance of the m observations in the first sample, and another is , the variance of the second sample. Intuitively, a better estimator than either individual sample variance results from combining the two sample variances. A first thought might be to use , the ordinary average of the two sample variances. However, if m > n, then the first sample contains more information about σ^2 than does the second sample, and an analogous comment applies if m < n . The following weighted average of the two sample variances, called the pooled (i.e., combined) estimator of σ 2 , adjusts for any difference between the two sample sgn="center">

The first sample contributes m –1 degrees of freedom to the estimate of σ 2, and the second sample contributes n –1 df, for a total of m + n –2 df. Statistical theory says that if replaces σ 2 in the expression for Z , the resulting standardized variable has a t distribution based on m + n –2 df. In the same way that earlier standardized variables were used as a basis for deriving confidence intervals and test procedures, this t variable immediately leads to the pooled t CI for estimating μ_1 –μ_2 and the pooled t test for testing hypotheses about a difference between means.

In the past, many statisticians recommended these pooled t procedures over the two-sample t procedures. The pooled t test, for example, can be derived from the likelihood ratio principle, whereas the two-sample t test is not a likelihood ratio test. Furthermore, the significance level for the pooled t test is exact, whereas it is only approximate for the two-sample t test. However, recent research has shown that although the pooled t test does outperform the two-sample t test by a bit (smaller β s for the same α) when , the former test can easily lead to erroneous conclusions if applied when the variances are different. Analogous comments apply to the behavior of the two confidence intervals. That is, the pooled t procedures are not robust to violations of the equal variance assumption.

It has been suggested that one could carry out a preliminary test of and use a pooled t procedure if this null hypothesis is not rejected. Unfortunately, the usual “F test” of equal variances (Section 9.5) is quite sensitive to the assumption of normal population distributions, much more so than t procedures. We therefore recommend the conservative approach of using two-sample t procedures unless there is really compelling evidence for doing otherwise, particularly when the two sample sizes are different.

Type II Error Probabilities

Determining type II error probabilities (or equivalently, power = 1 –β ) for the two-sample t test is complicated. There does not appear to be any simple way to use the β curves of Appendix Table A.17. The most recent version of MINITAB (Version 14) will calculate power for the pooled t test but not for the two-sample t test. However, the UCLA Statistics Department homepage (http://www.stat.ucla.edu) permits access to a power calculator that will do this. For example, we specified m = 10, n = 8, σ 1 = 300, σ 2 = 225 (these are the sample sizes for Example 9.7, whose sample standard deviations are somewhat smaller than these values of σ 1 and σed on an experiment involving a bird-feeder table, time-lapse video, and artificial foods. Information was collected for two different bird species at both the experimental location and at a natural vineyard setting. Consider the following data on time (sec) spent on a single visit to the location.

a. Calculate an upper confidence bound for the true average time that blackbirds spend on a single visit at the experimental location.

b. Does it appear that true average time spent by blackbirds at the experimental location exceeds the true average time birds of this type spend at the natural location? Carry out a test of appropriate hypotheses.

c. Estimate the difference between the true average time blackbirds spend at the natural location and true average time that silvereyes spend at the natural location, and do so in a way that conveys information about reliability and precision.

[Note: The sample medians reported in the article all seemed significantly smaller than the means, suggesting substantial population distribution skewness. The authors actually used the distribution-free test procedure presented in Section 2 of Chapter 15.]

25. Low-back pain (LBP) is a serious health problem in many industrial settings. The article “Isodynamic Evaluation of Trunk Muscles and Low-Back Pain Among Workers in a Steel Factory” (Ergonomics, 1995: 2107-2117) reported the accompanying summary data on lateral range of motion (degrees) for a sample of workers without a history of LBP and another sample with a history of this malady.

Calculate a 90% confidence interval for the difference between population mean extent of lateral motion for the two conditions. Does the interval suggest that population mean lateral motion differs for the two conditions? Is the message different if a confidence level of 95% is used?

26. The article “The Influence of Corrosion Inhibitor and Surface Abrasion on the Failure of Aluminum-Wired Twist-on Connections” (IEEE Trans. on Components, Hybrids, and Manuf. Tech., 1984: 20–25) reported data on potential drop measurements for one sample of connectors wired with alloy aluminum and another sample wired with EC aluminum. Does the accompanying SAS output suggest that the true average potential drop for alloy connections (type 1) is higher than that for EC connections (as stated in the article)? Carry out the appropriate test using a significance level of .01. In reaching your conclusion, what type of erro="0em" width="-2em">
27. Tennis elbow is thought to be aggravated by the impact experienced when hitting the ball. The article “Forces on the Hand in the Tennis One-Handed Backhand” (Intl. J. of Sport Biomechanics, 1991: 282-292) reported the force (N) on the hand just after impact on a one-handed backhand drive for six advanced players and for eight intermediate players.

In their analysis of the data, the authors assumed that both force distributions were normal. Calculate a 95% CI for the difference between true average force for advanced players (μ_1) and true average force for intermediate players (μ_2). Does your interval provide compelling evidence for concluding that the two μ s are different? Would you have reached the same conclusion by calculating a CI for μ_2 –μ_1(i.e., by reversing the 1 and 2 labels on the two types of players)? Explain.

28. As the population ages, there is increasing concern about accident-related injuries to the elderly. The article “Age and Gender Differences in Single-Step Recovery from a Forward Fall” (J. of Gerontology, 1999: M44-M50) reported on an experiment in which the maximum lean angle—the furthest a subject is able to lean and still recover in one step—was determined for both a sample of younger females (21-29 years) and a sample of older females (67–81 years). The following observations are consistent with summary data given in the article:

Does the data suggest that true average maximum lean angle for older females is more than 10 degrees smaller than it is for younger females? State and test the relevant hypotheses at significance level .10 by obtaining a P -value.

29. The article “Effect of Internal Gas Pressure on the Compression Strength of Beverage Cans and Plastic Bottles” (J. Testing and Evaluation, 1993: 129-131) includes the accompanying data on compression strength (lb) for a sample of 12-oz aluminum cans filled with strawberry drink and another sample filled with cola. Does the data suggest that the extra carbonation of cola results in a higher average compression strength? Base your answer on a P -value. What assumptions are necessary for your analysis?

30. The article “Flexure of Concrete Beams Reinforced with Advanced Composite Orthogrids” (J. of Aerospace Engr., 1997: 7-15) gave the accompanying data on ultimate load (kN) for two different types of beams="0em" align="center">

Figure 9.6 A boxplot of the differences in Example 9.9

Let’s now use the recommended sequence of steps to test the appropriate hypotheses.

1. Let μD denote the true average difference between elevation time before the change in work conditions and time after the change.

2. H_0: μD = 0 (there is no difference between true average time before the change and true average time after the change)

3. Ha : μD ≠ 0

4.

5. n = 16, ∑di = 108, = 1746, from which = 6.75, sD = 8.234, and

6. Appendix Table A.8 shows that the area to the right of 3.3 under the t curve with 15 df is .002. The inequality in Ha implies that a two-tailed test is appropriate, so the P -value is approximately 2(.002) = .004 (MINITAB gives .0051).

7. Since .004 < .01, the null hypothesis can be rejected at either significance level .05 or .01. It does appear that the true average difference between times is something other than zero; that is, true average time after the change is different from that before the change.

When the number of pairs is large, the assumption of a normal difference distribution is not necessary. The CLT validates the resulting z test.

A Confidence Interval for μD

In the same way that the t CI for a single population mean μ is bacharacteristics thought to influence responses. For example, in a medical experiment to compare the efficacy of two drugs for lowering blood pressure, the experimenter's budget might allow for the treatment of 20 patients. If 10 patients are randomly selected for treatment with the first drug and another 10 independently selected for treatment with the second drug, an independent-samples experiment results.

However, the experimenter, knowing that blood pressure is influenced by age and weight, might decide to create pairs of patients so that within each of the resulting 10 pairs, age and weight were approximately equal (though there might be sizable differences between pairs). Then each drug would be given to a different patient within each pair for a total of 10 observations on each drug.

Without this matching (or “blocking”), one drug might appear to outperform the other just because patients in one sample were lighter and younger and thus more susceptible to a decrease in blood pressure than the heavier and older patients in the second sample. However, there is a price to be paid for pairing—a smaller number of degrees of freedom for the paired analysis—so we must ask when one type of experiment should be preferred to the other.

There is no straightforward and precise answer to this question, but there are some useful guidelines. If we have a choice between two t tests that are both valid (and carried out at the same level of significance a), we should prefer the test that has the larger number of degrees of freedom. The reason for this is that a larger number of degrees of freedom means smaller β for any fixed alternative value of the parameter or parameters. That is, for a fixed type I error probability, the probability of a type II error is decreased by increasing degrees of freedom.

However, if the experimental units are quite heterogeneous in their responses, it will be difficult to detect small but significant differences between two treatments. This is essentially what happened in the data set in Example 9.8; for both “treatments” (bottom water and surface water), there is great between-location variability, which tends to mask differences in treatments within locations. If there is a high pos-itive correlation within experimental units or subjects, the variance of = will be much smaller than the unpaired variance. Because of this reduced variance, it will be easier to detect a difference with paired samples than with independent samples. The pros and cons of pairing can now be summarized as follows.

1. If there is great heterogeneity between experimental units and a large correlation within experimental units (large positive ρ), then the loss in degrees of freedom will be compensated for by the increased precision associated with pairing, so a paired experiment is preferable to an independent-samples experiment.

2. If the experimental units are relatively homogeneous and the correlation within pairindex="01113" alt="Image">, and ρ will not usually be known very precisely, so an investigator will be required to make an educated guess as to whether Situation 1 or 2 obtains. In general, if the number of observations that can be obtained is large, then a loss in degrees of freedom (e.g., from 40 to 20) will not be serious; but if the number is small, then the loss (say, from 16 to 8) because of pairing may be serious if not compensated for by increased precision. Similar considerations apply when choosing between the two types of experiments to estimate μ_1 –μ_2 with a confidence interval.

EXERCISES Section 9.3 (36–46)

36. Consider the accompanying data on breaking load (kg/25 σm width) for various fabrics in both an unabraded condition and an abraded condition (“The Effect of Wet Abrasive Wear on the Tensile Properties of Cotton and Polyester-Cotton Fabrics,” J. Testing and Evaluation , 1993: 84-93). Use the paired t test, as did the authors of the cited article, to test H_0: μD = 0 versus Ha: μD > 0 at significance level .01.

37. Hexavalent chromium has been identified as an inhalation carcinogen and an air toxin of concern in a number of different locales. The article “Airborne Hexavalent Chromium in Southwestern Ontario” (J. of Air and Waste Mgmnt. Assoc. , 1997: 905-910) gave the accompanying data on both indoor and outdoor concentration (nanograms/m^3) for a sample of houses selected from a certain region.

a. Calculate a confidence interval for the population mean difference between indoor and outdoor concentrations using a confidence level of 95%, and interpret the resulting interval.

b. If a 34th house were to be randomly selected from the population, between what values would you predict the difference in concentrations to lie?

38. Concrete specimens with varying height-to-diameter ratios cut from various positions on the original cylinder were obtained both from a normal-strength concrete mix and from a high-strength mix. The peak stress (MPa) wvel of .05. [Note: The appropriate normal probability plot shows some curvature but not enough to cast substantial doubt on a normality assumption.]

b. Calculate an upper confidence bound using a 95% confidence level for the true average difference between TBBMC during postweaning and during lactation.

c. Does the (incorrect) use of the two-sample t test to test the hypotheses suggested in (a) lead to the same conclusion that you obtained there? Explain.

41. In an experiment designed to study the effects of illumination level on task performance (“Performance of Complex Tasks Under Different Levels of Illumination,” J. Illuminating Eng. , 1976: 235-242), subjects were required to insert a fine-tipped probe into the eyeholes of ten needles in rapid succession both for a low light level with a black background and a higher level with a white background. Each data value is the time (sec) required to complete the task.

Does the data indicate that the higher level of illumination yields a decrease of more than 5 sec in true average task completion time? Test the appropriate hypotheses using the P -value approach.

42. It has been estimated that between 1945 and 1971, as many as 2 million children were born to mothers treated with di-ethylstilbestrol (DES), a nonsteroidal estrogen recommended for pregnancy maintenance. The FDA banned this drug in 1971 because research indicated a link with the incidence of cervical cancer. The article “Effects of Prenatal Exposure to Diethylstilbestrol (DES) on Hemispheric Laterality and Spatial Ability in Human Males” (Hormones and Behavior , 1992: 62-75) discussed a study in which 10 males exposed to DES and their unexposed brothers underwent various tests. This is the summary data on the results of a spatial ability test: = 12.6 (exposed), = 13.7, and standard error of mean difference = .5. Test at level .05 to see whether exposure is associated with reduced spatial ability by obtaining the P -value.

43. Cushing's disease is characterized by muscular weakness due to adrenal or pituitary dysfunction. To provide effective treatment, it is important to detect childhood Cushing's disease as early as possible. Age at onset of symptoms and age at diagnosis for 15 children suffering from the disease were given in the article “Treatment of Cushing's Disease in Childhood and Adolescence by Transphenoidal Microadenomectomy” (New Engl. J. of Med. , 1984: 889). Here are the values of the differences between age at onset of symptoms an>a. Does the accompanying normal probability plot cast strong doubt on the approximate normality of the population distribution of differences?

b. Calculate a lower 95% confidence bound for the population mean difference, and interpret the resulting bound.

c. Suppose the (age at diagnosis) –(age at onset) differences had been calculated. What would be a 95% upper confidence bound for the corresponding population mean difference?

44. Example 7.11 gave data on the modulus of elasticity obtained 1 minute after loading in a certain configuration. The cited article also gave the values of modulus of elasticity obtained 4 weeks after loading for the same lumber specimens. The data is presented here.

Calculate and interpret an upper confidence bound for the true average difference between 1-minute modulus and 4-week modulus; first check the plausibility of any necessary assumptions.

45. The paper “Slender High-Strength RC Columns Under Eccentric Compression” (Magazine of Concrete Res. , 2005: 361 -370) gave the accompanying data on cylinder strength (MPa) for various types of columns cured under both moist conditions and laboratory drying conditions.

a. Estimate the difference in true average strength under the two drying conditions in a way that conveys information about reliability and precision, and interpret the estimate. What does the estimate suggest about how true average strength under moist drying conditions compares to that under laboratory drying conditions?

b. Check the plausibility of any assumptions that underlie your analysis of (a).

46. Construct a paired data set for which t = ∞, so that the data is highly significant when the correct analysis is used, yet t for the two-sample t test is quite near zero, so the incorrect analysis yields an insignificant result.

9.4 Inferences Concerning a Difference Between Population Proportions

Having presented methods for comparing the means of two different populations, we now turn attention to the comparison of two population proportions. Regard an individual or object as a success S if he/she/it processes some characteristic of interest (someone who graduated from college, a refrigerator with an icemaker, etc.). Let

1 –2 also has approximately a normal distribution. Standardizing 1 –2 yields a variable Z whose distribution is approximately standard normal:

A Large-Sample Test Procedure

Analogously to the hypotheses for μ_1 –μ_2, the most general null hypothesis an investigator might consider would be of the form H_0: p_1 –p_2 = Δ_0, where Δ_0 is again a specified number. Although for population means the case Δ_0 Φ 0 presented no difficulties, for population proportions the cases Δ_0 = 0 and Δ_0 Φ 0 must be considered separately. Since the vast majority of actual problems of this sort involve Δ0 = 0 (i.e., the null hypothesis p_1 = p_2), we will concentrate on this case. When H_0: p_1 –p_2 = 0 is true, let p denote the common value of p_1 and p_2 (and similarly for q ). Then the standardized variable

has approximately a standard normal distribution when H 0 is true. However, this Z cannot serve as a test statistic because the value of p is unknown—H_0 asserts only that there is a common value of p , but does not say what that value is. A test statistic results from replacing p and q in (9.4) by appropriate estimators.

Assuming that p_1 = p_2 = p , instead of separate samples of size m and n from two different populations (two different binomial distributions), we really have a single sample of size m + n from one population with proportion p. The total number of individuals in this combined sample having the characteristic of interest is X + Y . The estimator of p is then

The second expression for p shows that it is actually a weighted average of estimators 1 and 2 obtained from the two samples. Using and = 1 – in place of p and q in (9.4) gives a test statistic having approximately a standard normal distribution when H_0 is true.

Null hypothesis: H_0: p_1 –p_2 = 0

Test statistic value (large samples):

Here the determination of β is a bit more cumbersome than it was for other large-sample tests. The reason is that the denominator of Z is an estimate of the standard deviation of –2, assuming that p_1 = p_2 = p . When H_0 is false, 1 –2 must be restandardized using

The form of σ implies that β is not a function of just p_1 –p_2, so we denote it by β (p_1, p_2).

where = (mp_1 + np_2)/(m + n ), = (mq_1 + nq_2)/(m + n ), and σ is given by (9.6).

Proof For the upper-tailed test (Ha: p_1 –p_2 > 0),

When m and n are both large,

= (m_1 + n2)/(m + n) ≈ (mp_1 + np_2)/(m + n) =

and ≈ , which yields the previous (approximate) expression for β(p_1, p_2).

Alternatively, for specified p_1, p_2 with p_1 –p_2 = d , the sample sizes necessary to achieve β(p_1, p_2) = β can be determined. For example, for the upper-tailed test, we equate –Zβ to the argument of Φ(•) (i.e., what's inside the parentheses) in the foregoing box. If m = n , there is a simple expression for the common value.

For the case m = n , the level α test has type II error probability β at the alternative values p_1, p_2 with p_1 –p_2 = d when

Inferential Methods

A test procedure for hypotheses concerning the ratio as well as a CI for this ratio are based on the following result.

THEOREM

Let X_1, …, Xm be a random sample from a normal distribution with variance , let Y_1, …, Yn be another random sample (independent of the Xis) from a normal distribution with variance , and let and denote the two sample variances. Then the rv

has an F distribution with v_1 = m –1 and v_2 = n –1.

This theorem results from combining (9.8) with the fact that the variables and each have a chi-squared distribution with m –1 and n –1 df, respectively (see Section 7.4). Because F involves a ratio rather than a difference, the test statistic is the ratio of sample variances. The claim that is then rejected if the ratio differs by too much from 1.

The F Test for Equality of Variances

Null hypothesis:

Test statistic value:

Since critical values are tabled only for α = .10, .05, .01, and .001, the two-tailed test can be performed only at levels .20, .10, .02, and .002. Other F critical values can be obtained from statistical software.

Example 9.14

On the basis of data reported in a 1979 article in the Journal of Gerontology (“Serum Ferritin in anistribution in the elderly had a smaller variance than in the younger adults. (Serum ferritin is used in diagnosing iron deficiency.) For a sample of 28 elderly men, the sample standard deviation of serum ferritin (mg/L) was s_1 = 52.6; for 26 young men, the sample standard deviation was s_2 = 84.2. Does this data support the conclusion as applied to men?

Let and denote the variance of the serum ferritin distributions for elderly men and young men, respectively. We wish to test versus . At level .01, H_0 will be rejected if f ≤ F.99,27,25. To obtain the critical value, we need F.01,25,27. From Appendix Table A.9, F.01,25,27 = 2.54, so F.99,27,25 = 1/2.54 = .394. The computed value of F is (52.6)2/(84.2)2 = .390. Since .390 ≤ .394, H_0 is rejected at level .01 in favor of Ha, so variability does appear to be greater in young men than in elderly men.

P-Values for F Tests

Recall that the P -value for an upper-tailed t test is the area under the relevant t curve (the one with appropriate df) to the right of the calculated t. In the same way, the P-value for an upper-tailed F test is the area under the F curve with appropriate numerator and denominator df to the right of the calculated f. Figure 9.9 illustrates this for a test based on v_1 = 4 and v_2 = 6.

Figure 9.9 A P -value for an upper-tailed F test

Tabulation of F curve upper-tail areas is much more cumbersome than for t curves because two df's are involved. For each combination of v_1 and v_2, our F table gives only the four critical values that capture areas .10, .05, .01, and .001. Figure 9.10 shows what can be said about the P -value depending on where f falls relative to the four critical values.

Figure 9.10 Obtaining P-value information from the F table for an upper-tailed F test

For example, for a test with v_1 = 4 and v_2 = 6,

f = 5.70 ⇒ .01 < P -value < .05

f = 25.03 ⇒ P -value < .001

Only if f equals a tabulated value do we obtain an exact P -value (e.g., if f = 4.53, then P-value = .05). Once we know that .01 < P -value < .05, H_0 would be rejected at a significance level of .05 but not at a level of .01. When P-value < .001, H_0 should be rejected at any reasonable significance level.

The F tests discussed in succeeding chapters will all be upper-tailed. If, however, a lower-tailed F test is appropriate, then (9.9) should be used to obtain lower-tailed critical values so that a bound or bounds on the P -value can be established. In the case of a two-tailed test, the bound or bounds from a one-tailed test should be multiplied by 2. For example, if f = 5.82 when v_1 = 4 and v_2 = 6, then since 5.82 falls between the .05 and .01 critical values, 2(.01) < P-value < 2(.05), giving .02 < P-value < .10. H_0 would then be rejected if α = .10 but not if α = .01. In this case, we cannot say from our table what conclusion is appropriate when α = .05 (since we don't know whether the P -value is smaller or larger than this). However, statistical software shows that the area to the right of 5.82 under this F curve is .029, so the P-value is .058 and the null hypothesis should therefore not be rejected at level .05 (.058 is the smallest α for which H_0 can be rejected and our chosen α is smaller than this). Various statistical software packages will, of course, provide an exact P -value for any F test.

A Confidence Interval for σ_1//σ_2

The CI for is based on replacing F in the probability statement

P (F 1-α/2,v_1,v_2 < F < Fα/2,v_1,v_2) = 1 –α

by the F variable (9.10) and manipulating the inequalities to isolate . An interval for σ_1/σ_2 results from taking the square root of each limit. The details are left for an exercise.

EXERCISES Section 9.5 (57–64)

57. Obtain or compute the following quantities:

a. F .05,5,8 b. F .05,8,5 c. F .95,5,8 d. F .95,8,5

e. The 99th percentile of the F distribution with v_1 = 10, v_2 = 12

f. The 1s cattle grazing in a nonanomalous area (“An Investigation into Copper Deficiency in Cattle in the Southern Pennines,” J. Agricultural Soc. Cambridge , 1972: 157-163), resulting in s_1 = 21.5 (m = 48) for the anomalous condition and s_2 = 19.45 (n = 45) for the nonanomalous condition. Test for the equality versus inequality of population variances at significance level .10 by using the P-value approach.

63. The article “Enhancement of Compressive Properties of Failed Concrete Cylinders with Polymer Impregnation” (J. Testing and Evaluation , 1977: 333-337) reports the following data on impregnated compressive modulus (psi X 10^6) when two different polymers were used to repair cracks in failed concrete.

Obtain a 90% CI for the ratio of variances by first using the method suggested in the text to obtain a general confidence interval formula.

64. Reconsider the data of Example 9.6, and calculate a 95% upper confidence bound for the ratio of the standard deviation of the triacetate porosity distribution to that of the cotton porosity distribution.

SUPPLEMENTARY EXERCISES (65–93)

65. The accompanying summary data on compression strength (lb) for 12 X 10 X 8 in. boxes appeared in the article “Compression of Single-Wall Corrugated Shipping Containers Using Fixed and Floating Test Platens” (J. Testing and Evaluation , 1992: 318-320). The authors stated that “the difference between the compression strength using fixed and floating platen method was found to be small compared to normal variation in compression strength between identical boxes.” Do you agree? Is your analysis predicated on any assumptions?

66. The authors of the article “Dynamics of Canopy Structure and Light Interception in Pinus elliotti , North Florida” (Ecological Monographs , 1991: 33-51) planned an experiment to determine the effect of fertilizer on a measure of leaf area. A number of plots were available for the study, and half were selected at random to be fertilized. To ensure that the plots to receive the fertilizer and the control plots were similar, before beginning the experiment tree density (the number of trees per hectare) was recorded for eight plots to be fertilized and eight control plots, resulting in the given data. MINITAB output follows.

a. Construct a comparative boxplot and com1em">
b. Would you conclude that there is a significant difference in the mean tree density for fertilizer and control plots? Use α = .05.

c. Interpret the given confidence interval.

67. Is the response rate for questionnaires affected by including some sort of incentive to respond along with the questionnaire? In one experiment, 110 questionnaires with no incentive resulted in 75 being returned, whereas 98 questionnaires that included a chance to win a lottery yielded 66 responses (“Charities, No; Lotteries, No; Cash, Yes,” Public Opinion Quarterly , 1996: 542-562). Does this data suggest that including an incentive increases the likelihood of a response? State and test the relevant hypotheses at significance level .10 by using the P -value method.

68. The accompanying data was obtained in a study to evaluate the liquefaction potential at a proposed nuclear power station (“Cyclic Strengths Compared for Two Sampling Techniques,” J. Geotechnical Division, Am. Soc. Civil Engrs. Proceedings , 1981: 563-576). Before cyclic strength testing, soil samples were gathered using both a pitcher tube method and a block method, resulting in the following observed values of dry density (lb/ft^3):

Calculate and interpret a 95% CI for the difference between true average dry densities for the two sampling methods.

69. The article “Quantitative MRI and Electrophysiology of Preoperative Carpal Tunnel Syndrome in a Female Population” (Ergonomics , 1997: 642-649) reported that (-473.3, 1691.9) was a large-sample 95% confidence interval for the difference between true average thenar muscle volume (mm^3) for sufferers of carpal tunnel syndrome and true average volume for nonsufferers. Calculate a 90% confidence interval for this difference.

70. The following summary data on bending strength (lb-in/in) of joints is taken from the article “Bending Strength of Corner Joints Constructed with Injection Molded Splines” (Forest Products J. , April, 1997: 89-92).

a. Calculate a 95% lower confidence bound for true average strength of joints with a side coating.

b. Calculate a 95% lower prediction bound for the strength of a sinecindex="01193" alt="Image"> and , the standard deviations of the two log concentration distributions, are not equal, would μ_1 and μ_2, the means of the concentration distributions, be the same if ? Explain your reasoning.

77. Long-term exposure of textile workers to cotton dust released during processing can result in substantial health problems, so textile researchers have been investigating methods that will result in reduced risks while preserving important fabric properties. The accompanying data on roving cohesion strength (kN • m/kg) for specimens produced at five different twist multiples is from the article “Heat Treatment of Cotton: Effect on Endotoxin Content, Fiber and Yarn Properties, and Processability” (Textile Research J. , 1996: 727-738).

The authors of the cited article stated that strength for heated specimens appeared to be slightly higher on average than for the control specimens. Is the difference statistically significant? State and test the relevant hypotheses using α = .05 by calculating the P -value.

78. The accompanying summary data on the ratio of strength to cross-sectional area for knee extensors is taken from the article “Knee Extensor and Knee Flexor Strength: Cross-Sectional Area Ratios in Young and Elderly Men” (J. of Gerontology , 1992: M204-M210).

Does this data suggest that the true average ratio for young men exceeds that for elderly men? Carry out a test of appropriate hypotheses using α = .05. Be sure to state any assumptions necessary for your analysis.

79. The accompanying data on response time appeared in the article “The Extinguishment of Fires Using Low-Flow Water Hose Streams—Part II” (Fire Technology , 1991: 291-320).

The authors analyzed the data with the pooled t test. Does the use of this test appear justified? [Hint: Check for normality. The normal scores for n = 8 are -1.53, -.89, -.49, -.15, .15, .49, .89, and 1.53.]

80. Acrylic bone cement is commonly used in total joint arthro-plasty as a grout that allows for the smooth transfer of loads from a metal prosthesis to bone structure. The paper “Validation of the Small-Punch Test as a Technique for Characterizing the Mechanical Properties of Acrylic Bone Cement” (J. of Engr. in Med. , 2006: 11-21) gave the following data on breaking forcev height="0em">

Assume that all population distributions are normal.

a. Estimate true average breaking force in a dry medium at 37° in a way that conveys information about reliability and precision, and interpret your estimate.

b. Estimate the difference between true average breaking force in a dry medium at 37° and true average force at the same temperature in a wet medium, and do so in a way that conveys information about precision and reliability. Then interpret your estimate.

c. Is there strong evidence for concluding that true average force in a dry medium at the higher temperature exceeds that at the lower temperature by more than 100 N?

81. In an experiment to compare bearing strengths of pegs inserted in two different types of mounts, a sample of 14 observations on stress limit for red oak mounts resulted in a sample mean and sample standard deviation of 8.48 MPa and .79 MPa, respectively, whereas a sample of 12 observations when Douglas fir mounts were used gave a mean of 9.36 and a standard deviation of 1.52 (“Bearing Strength of White Oak Pegs in Red Oak and Douglas Fir Timbers,” J. of Testing and Evaluation , 1998, 109-114). Consider testing whether or not true average stress limits are identical for the two types of mounts. Compare df's and P-values for the unpooled and pooled t tests.

82. How does energy intake compare to energy expenditure? One aspect of this issue was considered in the article “Measurement of Total Energy Expenditure by the Doubly Labelled Water Method in Professional Soccer Players” (J. of Sports Sciences , 2002: 391-397), which contained the accompanying data (MJ/day).

Test to see whether there is a significant difference between intake and expenditure. Does the conclusion depend on whether a significance level of .05, .01, or .001 is used?

83. An experimenter wishes to obtain a CI for the difference between true average breaking strength for cables manufactured by company I and by company II. Suppose breaking strength is normally distributed for both types of cable with σ_1 = 30 psi and σ_2 = 20 psi.

a. If costs dictate that the sample size for the type I cable should be three times the sample size for the type II cable, how many observations are required if the 99% CI is to be no wider than 20 psi?

b. Suppose a total of 400 observations is to be made. How many of the observations should be made on type I cable samples if the width of the resulou think of real problems in which the difference μ_1 –μ_2 = 1 has little practical significance? Would sample sizes of n = 10,000 be desirable in such problems?

88. The following data refers to airborne bacteria count (number of colonies/ft^3) both for m = 8 carpeted hospital rooms and for n = 8 uncarpeted rooms (“Microbial Air Sampling in a Carpeted Hospital,” J. Environmental Health , 1968: 405). Does there appear to be a difference in true average bacteria count between carpeted and uncarpeted rooms?

Suppose you later learned that all carpeted rooms were in a veterans' hospital, whereas all uncarpeted rooms were in a children's hospital. Would you be able to assess the effect of carpeting? Comment.

89. Researchers sent 5000 resumes in response to job ads that appeared in the Boston Globe and Chicago Tribune. The resumes were identical except that 2500 of them had “white sounding” first names, such as Brett and Emily, whereas the other 2500 had “black sounding” names such as Tamika and Rasheed. The resumes of the first type elicited 250 responses and the resumes of the second type only 167 responses (these numbers are very consistent with information that appeared in a Jan. 15, 2003, report by the Associated Press). Does this data strongly suggest that a resume with a “black” name is less likely to result in a response than is a resume with a “white” name?

90. McNemar's test, developed in Exercise 54, can also be used when individuals are paired (matched) to yield n pairs and then one member of each pair is given treatment 1 and the other is given treatment 2. Then X_1 is the number of pairs in which both treatments were successful, and similarly for X_2, X_3, and X_4. The test statistic for testing equal efficacy of the two treatments is given by , which has approximately a standard normal distribution when H_0 is true. Use this to test whether the drug ergotamine is effective in the treatment of migraine headaches.

The data is fictitious, but the conclusion agrees with that in the article “Controlled Clinical Trial of Ergotamine Tar-trate” (British Med. J. , 1970: 325-327).

91. The article “Evaluating Variability in Filling Operations” (Food Tech. , 1984: 51-55) describes two different filling operations used in a ground-beef packing plant. Both filling operations were set to fill packages with 1400 g of ground beef. In a random sample of ied

μ_1 = the mean of population I or the true average response when treatment I is applied

The relevant hypotheses are

H_0 : μ_1 = μ_2 = …= μI

versus

Ha: at least two of the μi s are different

If I = 4, H_0 is true only if all four μi s are identical. Ha would be true, for example, if μ_1 = μ_2 ≠ μ_3 = μ_4, if μ_1 = μ_3 = μ_4 ≠ μ_2, or if all four μ, s differ from one another.

A test of these hypotheses requires that we have available a random sample from each population or treatment.

Example 10.1

The article “Compression of Single-Wall Corrugated Shipping Containers Using Fixed and Floating Test Platens” (J . Testing and Evaluation, 1992: 318–320) describes an experiment in which several different types of boxes were compared with respect to compression strength (lb). Table 10.1 presents the results of a single-factor ANOVA experiment involving I = 4 types of boxes (the sample means and standard deviations are in good agreement with values given in the article).

Table 10.1 The Data and Summary Quantities for Example 10.1

With μi denoting the true average compression strength for boxes of type i (i = 1, 2, 3, 4), the null hypothesis is H_0: μ_1 = μ_2 = μ_3 = μ_4. Figure 10.1 (a) shows a comparative boxplot for the four samples. There is a substantial amount of overlap among observations on the first three types of boxes, but compression strengths for the fourth type appear considerably smaller than for the other types. This suggests that H_0 is not true. The comparative boxplot in Figure 10.1 (b) is based on adding 120 to each observation in the fourth sample (giving mean 682.02 and the same standard deviation) and leaving the other observations unaltered. It is no long1em" width="0em">If H_0 is true, the J observations in each sample come from a normal population distribution with the same mean value μ, in which case the sample means 1..…, I. should be reasonably close. The test procedure is based on comparing a measure of differences among the i .s (“between-samples” variation) to a measure of variation calculated from within each of the samples.

DEFINITION

Mean square for treatments is given by

and mean square for error is

The test statistic for single-factor ANOVA is F = MSTr/MSE.

The terminology “mean square” will be explained shortly. Notice that uppercase s and s^2’s are used, so MSTr and MSE are defined as statistics. We will follow tradition and also use MSTr and MSE (rather than mstr and mse) to denote the calculated values of these statistics. Each assesses variation within a particular sample, so MSE is a measure of within-samples variation.

What kind of value of F provides evidence for or against H_0? If H_0 is true (all μi s are equal), the values of the individual sample means should be close to one another and therefore close to the grand mean, resulting in a relatively small value of MSTr. However, if the μi s are quite different, some i. s should differ quite a bit from… So the value of MSTr is affected by the status of H 0(true or false). This is not the case with MSE, because the s depend only on the underlying value of σ^2 and not on where the various distributions are centered. The following box presents an important property of E (MSTr) and E (MSE), the expected values of these two statistics.

PROPOSITION

When H_0 is true,

E (MSTr) = E (MSE) = σ^2

whereas when H_0 is false,

E(MSTr) > E(MSE) = σ 2

That c assumptions of this section are satisfied, F has an F distribution with v_1 = I –1 and v_2 = I(J –1). With f denoting the computed value of F, the rejection region f ≥ Fα,I –1,I(J –1) then specifies a test with significance level α. Refer to Section 9.5 to see how P -value information for F tests is obtained.

The rationale for v_1 = I –1 is that although MSTr is based on the I deviations 1. .., …, 1. –Σ(i. –) = 0, so only I –1 of these are freely determined. Because each sample contributes J –1 df to MSE and these samples are independent, v 2 = (J –1) + … + (J –1) = I (J –1).

Example 10.2 (Example 10.1 continued)

The values of I and J for the strength data are 4 and 6, respectively, so numerator df = I –1 = 3 and denominator df = I (J –1) = 20. At significance level .05, H_0: μ_1 = μ_2 = μ_3 = μ_4 will be rejected in favor of the conclusion that at least two μi. s are different if f ≥ F.05,3,20 = 3.10. The grand mean is = ΣΣ xij /(IJ ) = 682.50

MSTr = [(713.00 –682.50)2 + (756.93 –682.50)2 + (698.07 –682.50)2 + (562.02 –682.50)2] = 42,455.86

MSE = [(46.55)2 + (40.34)2 + (37.20)2 + (39.87)2] = 1691.92

f = MSTr/MSE = 42,455.86/1691.92 = 25.09

Since 25.09 ≥ 3.10, H_0 is resoundingly rejected at significance level .05. True average compression strength does appear to depend on box type. In fact, P -value = area under F curve to the right of 25.09 = .000. H_0 would be rejected at any reasonable significance level.

Sums of Squares

The introduction of sums of squares facilitates developing an intuitive appreciation for the rationale underlying single-factor and multifactor ANOVAs. Let xi. represent the sum (not the average, since there is nem" width="2em">Once SSTr and SSE are computed, each is divided by its associated df to obtain a mean square (mean in the sense of average). Then F is the ratio of the two mean squares.

The computations are often summarized in a tabular format, called an ANOVA table, as displayed in Table 10.2. Tables produced by statistical software customarily include a P -value column to the right of f

Table 10.1 An ANOVA Table

Example 10.3

The accompanying data resulted from an experiment comparing the degree of soiling for fabric copolymerized with three different mixtures of methacrylic acid (similar data appeared in the article “Chemical Factors Affecting Soiling and Soil Release from Cotton DP Fabric,” American Dyestuff Reporter, 1983: 25-30).

Let μi denote the true average degree of soiling when mixture i is used (i = 1, 2, 3). The null hypothesis H_0: μ_1 = μ_2 = μ_3 states that the true average degree of soiling is identical for the three mixtures. We will carry out a test at significance level .01 to see whether H_0 should be rejected in favor of the assertion that true average degree of soiling is not the same for all mixtures. Since I –1 = 2 and I (J –1) = 12, the F critical value for the rejection region is F.01,2,12 = 6.93. Squaring each of the 15 observations and summing gives = (.56)2 + (1.12)2 + … + (.93)2 = 12.1351. The values of the three sums of squares are

SST = 12.1351 –(13.25)2/15 = 12.1351 –11.7042 = .4309

SSTr = [(4.59)2 + (3.97)2 + (4.69)2] –11.7042

= 11.7650 –11.7042 = .0608

SSE = .4309 –.0608 = .3701

The remaining computations are summarized in the accompanying ANOVA table. Because f = .99 is not at least F.01,2,12 = 6.93, H_0 is not rejected at significance level .01. The mixtures appear to be indistinguishable with respect to degree of soiling (F.10,2,12 = 2.81 ⇒ P -value > .10).

= 911.91. Use the ANOVA F test at level .05 to decide whether there are any differences between true average flight times for the four treatments.

5. Consider the following summary data on the modulus of elasticity (× 10^6 psi) for lumber of three different grades (in close agreement with values in the article “Bending Strength and Stiffness of Second-Growth Douglas-Fir Dimension Lumber” (Forest Products J., 1991: 35—13), except that the sample sizes there were larger):

Use this data and a significance level of .01 to test the null hypothesis of no difference in mean modulus of elasticity for the three grades.

6. The article “Origin of Precambrian Iron Formations” (Econ. Geology, 1964: 1025-1057) reports the following data on total Fe for four types of iron formation (1 = carbonate, 2 = silicate, 3 = magnetite, 4 = hematite).

Carry out an analysis of variance F test at significance level .01, and summarize the results in an ANOVA table.

7. In an experiment to investigate the performance of four different brands of spark plugs intended for use on a 125-cc two-stroke motorcycle, five plugs of each brand were tested and the number of miles (at a constant speed) until failure was observed. The partial ANOVA table for the data is given here. Fill in the missing entries, state the relevant hypotheses, and carry out a test by obtaining as much information as you can about the P -value.

8. A study of the properties of metal plate-connected trusses used for roof support (“Modeling Joints Made with Light-Gauge Metal Connector Plates,” Forest Products J., 1979: 39–14) yielded the following observations on axial stiffness index (kips/in.) for plate lengths 4, 6, 8, 10, and 12 in:

Does variation in plate length have any effect on true average axial stiffness? State and test the relevant hypotheses using analysis of variance with α = .01. Display your results in an ANOVA table. [Hint: = 5,241, 420.79.]

9. Six samples of each of four types of cereal grain grown in a certgnificantly different. Following the same line of reasoning for each of the other intervals, we end up being able to judge for each pair of μ s whether or not they differ significantly from one another.

The procedures based on this idea differ in the method used to calculate the various CIs. Here we present a popular method that controls the simultaneous confidence level for all I (I –1)/2 intervals calculated.

Tukey’s Procedure (the T Method)

Tukey’s procedure involves the use of another probability distribution called the Studentized range distribution. The distribution depends on two parameters: a numerator df m and a denominator df v . Let Qα,m,v denote the upper-tail α critical value of the Studentized range distribution with m numerator df and v denominator df (analogous to Fα,v1,v2 ). Values of Qα,m,v are given in Appendix Table A.10.

PROPOSITION

With probability 1 –α ,

for every i and j (i = 1, …, I and j = 1, …, I) with i < j.

Notice that numerator df for the appropriate Qa critical value is I , the number of population or treatment means being compared, and not I –1 as in the F test. When the computed i., j., and MSE are substituted into (10.4), the result is a collection of confidence intervals with simultaneous confidence level 100(1 –α )%, for all pair-wise differences of the form μi –μj with i < j . Each interval that does not include 0 yields the conclusion that the corresponding values of μi and μj differ significantly from one another.

Since we are not really interested in the lower and upper limits of the various intervals but only in which include 0 and which do not, much of the arithmetic associated with (10.4) can be avoided. The following box gives details and describes how differences can be identified visually using an “underscoring pattern.”

The T Method for Identifying Significantly Different μi’s

Select a, extract Qα,I,I (J –1) from Appendix Table A.10, and calculate w = Qα,I,I (J –1) • . Then list the sample means in increasing order and underline those pairs that dme line corresponds to a pair of population or treatment means that are judged significantly different.

Suppose, for example, that I = 5 and that

2. < 5. < 4. < 1. < 3.

Then

1. Consider first the smallest mean 2. If 5 –2 ≥ w , proceed to Step 2. However, if 5 –2 < w , connect these first two means with a line segment. Then if possible extend this line segment even further to the right to the largest i that differs from 2. by less than w (so the line may connect two, three, or even more means).

2. Now move to 5 and again extend a line segment to the largest i. to its right that differs from 5. by less than w (it may not be possible to draw this line, or alternatively it may underscore just two means, or three, or even all four remaining means).

3. Continue by moving to 4 and repeating, and then finally move to 1.

To summarize, starting from each mean in the ordered list, a line segment is extended as far to the right as possible as long as the difference between the means is smaller than w. It is easily verified that a particular interval of the form (10.4) will contain 0 if and only if the corresponding pair of sample means is underscored by the same line segment.

Example 10.4

An experiment was carried out to compare five different brands of automobile oil filters with respect to their ability to capture foreign material. Let μi denote the true average amount of material captured by brand i filters (i = 1, …, 5) under controlled conditions. A sample of nine filters of each brand was used, resulting in the following sample mean amounts: 1. = 14.5, 2. = 13.8, 3. = 13.3, Table 10.3 ANOVA Table for Example 10.4

Since F.05,4,40 = 2.61, H_0 is rejected (decisively) at level .05. We now use Tukey’s procedure to look for significant differences among the μi s. From Appendix Table A.10, Q.05,5,40 = 4.04 (the second subscript on Q is I and not I –1 as in F ), so w = = .4. After arranging the five sample means in increasing order, the two smallest can be connected by a line segment because they differ by less than .4. However, this segment cannot be extended further to the right since 13.8 –13.1 = .7 ≥ .4. Moving one mean to the right, the pair3. and 2. cannot be underscored because these means differ by more than .4. Again moving to the right, the next mean, 13.8, cannot be connected to any further to the right. The last two means can be underscored with the same line segment.

Thus brands 1 and 4 are not significantly different from one another, but are significantly higher than the other three brands in their true average contents. Brand 2 is significantly better than 3 and 5 but worse than 1 and 4, and brands 3 and 5 do not differ significantly.

If 2. = 14.15 rather than 13.8 with the same computed w, then the configuration of underscored means would be

Example 10.5

A biologist wished to study the effects of ethanol on sleep time. A sample of 20 rats, matched for age and other characteristics, was selected, and each rat was given an oral injection having a particular concentration of ethanol per body weight. The rapid eye movement (REM) sleep time for each rat was then recorded for a 24-hour period, with the following results:

Does the data indicate that the true average REM sleep time depends on the concentration of ethanol? (This example is based on an experiment reported in “Relationship of Ethanol Blood Level to REM and Non-REM Sleep Time and Distribution in the Rat,” Life Sciences, 1978: 839-846.)

The i. s differ rather substantially from one another, but there is also a great deal of variability within each sample, so to answer the question precisely we must carry oo the probability that the second interval will include p. Because the two samples are selected independently of one another, the probability that both intervals will include the values of the respective parameters is (.95)(.95) = (.95)2 ≈ .90. Thus the simultaneous or joint confidence level for the two intervals is roughly 90%—if pairs of intervals are calculated over and over again from independent samples, in the long run roughly 90% of the time the first interval will capture μ and the second will include p. Similarly, if three CIs are calculated based on independent samples, the simultaneous confidence level will be 100(.95)3% ≈ 86%. Clearly, as the number of intervals increases, the simultaneous confidence level that all intervals capture their respective parameters will decrease.

Now suppose that we want to maintain the simultaneous confidence level at 95%. Then for two independent samples, the individual confidence level for each would have to be ≈ 97.5%. The larger the number of intervals, the higher the individual confidence level would have to be to maintain the 95% simultaneous level.

The tricky thing about the Tukey intervals is that they are not based on independent samples—MSE appears in every one, and various intervals share the same i. s (e.g., in the case I = 4, three different intervals all use 1). This implies that there is no straightforward probability argument for ascertaining the simultaneous confidence level from the individual confidence levels. Nevertheless, it can be shown that if Q_05 is used, the simultaneous confidence level is controlled at 95%, whereas using Q.01 gives a simultaneous 99% level. To obtain a 95% simultaneous level, the individual level for each interval must be considerably larger than 95%. Said in a slightly different way, to obtain a 5% experimentwise or family error rate, the individual or per-comparison error rate for each interval must be considerably smaller than .05. MINITAB asks the user to specify the family error rate (e.g., 5%) and then includes on output the individual error rate (see Exercise 16).

Confidence Intervals for Other Parametric Functions

In some situations, a CI is desired for a function of the μi s more complicated than a difference μi –μj Let θ = Σciμi , where the ci s are constants. One such function is , which in the context of Example 10.4 measures the difference between the group consisting of the first two brands and that of the last three brands. Because the Xij s are normally distributed with E (Xij ) = μi and V(Xij ) = σ^2, = Σici σ 2 by MSE and forming results in a t variable ( –θ)/, which can be manipulated to obtain the following 100(1 –α )% confidence interval for Σiμi:

Example 10.6 (Example 10.4 continued)

The parametric function for comparing the first two (store) brands of oil filter with the last three (national) brands is , from which

With = (1. + 2.) – (3. + 4. + 5.) = 5.83 and MSE = .088, a 95% interval is

Sometimes an experiment is carried out to compare each of several “new” treatments to a control treatment. In such situations, a multiple comparisons technique called Dunnett’s method is appropriate.

EXERCISES Section 10.2 (11–21)

11. An experiment to compare the spreading rates of five different brands of yellow interior latex paint available in a particular area used 4 gallons (J = 4) of each paint. The sample average spreading rates (ft^2/gal) for the five brands were 1. = 462.0, 2. = 512.8, 3. = 437.5, 4. = 469.3, and 5. = 532.1. The computed value of F was found to be significant at level α = .05. With MSE = 272.8, use Tukey’s procedure to investigate significant differences in the true average spreading rates between brands.

12. In Exercise 11, suppose 3. = 427.5. Now which true average spreading rates differ significantly from one another? Be sure to use the method of underscoring to illustrate your conclusions, and write a paragraph summarizing your results.

13. Repeatge">3. = 427.5.

14. Use Tukey’s procedure on the data in Exercise 4 to identify differences in true average flight times among the four types of mosquitos.

15. Use Tukey’s procedure on the data of Exercise 6 to identify differences in true average total Fe among the four types of formations (use MSE = 15.64).

16. Reconsider the axial stiffness data given in Exercise 8.

a. Is it plausible that the variances of the five axial stiffness index distributions are identical? Explain.

b. Use the output (without reference to our F table) to test the relevant hypotheses.

c. Use the Tukey intervals given in the output to determine which means differ, and construct the corresponding underscoring pattern.

17. Refer to Exercise 5. Compute a 95% t CI for θ =

18. Consider the accompanying data on plant growth after the application of different types of growth hormone.

a. Perform an F test at level α = .05.

b. What happens when Tukey’s procedure is applied?

19. Consider a single-factor ANOVA experiment in which I = 3, J = 5, 1. = 10, 2. = 12, and 3. = 20. Find a value of SSE for which f > F.05,2,12, so that H_0: μ_1 = μ_2 = μ_3 is rejected, yet when Tukey’s procedure is applied none of the s can be said to differ significantly from one another.

20.est and Tukey’s procedure?

21. The article “The Effect of Enzyme Inducing Agents on the Survival Times of Rats Exposed to Lethal Levels of Nitrogen Dioxide” (Toxicology and Applied Pharmacology, 1978: 169-174) reports the following data on survival times for rats exposed to nitrogen dioxide (70 ppm) via different injection regimens. There were J = 14 rats in each group.

a. Test the null hypothesis that true average survival time does not depend on injection regimen against the alternative that there is some dependence on injection regimen using α = .01.

b. Suppose that 100(1 –α)% CIs for k different parametric functions are computed from the same ANOVA data set. Then it is easily verified that the simultaneous confidence level is at least 100(1 –kα )%. Compute CIs with simultaneous confidence level at least 98% for and

10.3 More on Single-Factor ANOVA

We now briefly consider some additional issues relating to single-factor ANOVA. These include an alternative description of the model parameters, β for the F test, the relationship of the test to procedures previously considered, data transformation, a random effects model, and formulas for the case of unequal sample sizes.

The ANOVA Model

The assumptions of single-factor ANOVA can be described succinctly by means of the “model equation”

Xij = μi + εij

where εij represents a random deviation from the population or true treatment mean μi The εij s are assumed to be independent, normally distributed rv’s (implying that the Xij s are also) with E (εij ) = 0 [so that E (Xij = μi ] and V (εij = σ 2 [from which V (Xij ) = σ^2 for every i and j ]. An alternative description of single-factor ANOVA will give added insight and suggest appropriate generalizations to models involving more than one factor. Define a parameter μ by

and the parameters α_1, …, α_1 by

<#956;i –μ (i = 1, …, I )

Then the treatment mean μi can be written as μ + αi , where μ represents the true average overall response in the experiment, and αi is the effect, measured as a departure from μ , due to the ith treatment. Whereas we initially had I parameters, we now have I + 1 (μ , αi , …, αI ). However, because Σαi = 0 μ_0 (the average departure from the overall mean response is zero), only I of these new parameters are independently determined, so there are as many independent parameters as there were before. In terms of μ and the αi s, the model becomes

Xij μ + μ + αi + εij (i = 1, …, I , j = 1, …, J )

In Chapter 11, we will develop analogous models for multifactor ANOVA. The claim that the μi s are identical is equivalent to the equality of the αi s, and because Σαi = 0, the null hypothesis becomes

H_0: α_1 = α_2 = … = αI = 0

In Section 10.1, it was stated that MSTr is an unbiased estimator of σ_2 when H_0 is true but otherwise tends to overestimate σ_2. More precisely,

When H_0 is true, = 0 so E(MSTr) = σ 2 (MSE is unbiased whether or not H_0 is true). If is used as a measure of the extent to which H_0 is false, then a larger value of will result in a greater tendency for MSTr to overestimate σ 2. In the next chapter, formulas for expected mean squares for multifactor models will be used to suggest how to form F ratios to test various hypotheses.

Proof of the Formula for E (MSTr) For any rv Y , E (Y 2) = V (Y ) + [E (Y )]2, so

The result then follows from the relationship MSTr = SSTr /(I- 1).

β for the F Test
Consider a set of parameter values α_1, α_2, …, αI for which H_0 is not true. The probability of a type II error, β, is the probability that H_0 is not rejected when that set is the set of true values. One might think that β would have to be determined separately for each different configuration of ais. Fortunately, since β for the F test depends on the αi s and σ^2 only through , it can be simultaneously evaluated for many different alternatives. For example, = 4 for each of the following sets of ais for which H_0 is false, so β is identical for all three alternatives:

1. α_1 = –1, α_2 = –1, α_3 = 1, α_4 = 1

2. α_1 = – =, α_2 = , α_3 = 0, α_4 = 0

3. α_1 = –, α_2 = –, α_3 = , α_4 =

The quantity J is called the noncentrality parameter for one-way ANOVA (because when H_0 is false the test statistic has a noncentral F distribution with this as one of its parameters), and β is a decreasing function of the value of this parameter. Thus, for fixed values of σ 2 and J , the null hypothesis is more likely to be rejected for alternatives far from H_0 (large ) than for alternatives close to H_0. For a fixed value of , β decreases as the sample size J on each treatment increases, and it increases as the variance σ^2 increases (since greater underlying variability makes it more difficult to detect any given departure from H_0).

Figure 10.5 Power curves for the ANOVA F test (v, = 3)

Because hand computation of β and sample size determination for the F test are quite difficult (as in the case of t tests), statisticians have constructed sets of curves from which β can be obtained. Sare package has a function that calculates the cumulative area under a noncentral F curve (inputs Fα, numerator df, denominator df, and φ^2), and this area is β. Version 14 of MINITAB does this and also something rather different. The user is asked to specify the maximum difference between μi s rather than the individual means. For example, we might wish to calculate the power of the test when I = 4, μ_1 = 100, μ_2 = 101, μ_3 = 102, and μ_4 = 106. Then the maximum difference is 106 –100 = 6. However, the power depends not only on this maximum difference but on the values of all the μi s. In this situation MINITAB calculates the smallest possible value of power subject to μ_1 = 100 and μ_4 = 106, which occurs when the two other μ s are both halfway between 100 and 106. If this power is .85, then we can say that the power is at least .85 and β is at most .15 when the two most extreme μ s are separated by 6 (the common sample size, α, and σ must also be specified). The software will also determine the necessary common sample size if maximum difference and minimum power are specified.

Relationship of the F Test to the t Test

When the number of treatments or populations is I = 2, all formulas and results connected with the F test still make sense, so ANOVA can be used to test H_0: μ_1 = μ_2 versus Ha: μ_1 ≠ μ_2. In this case, a two-tailed, two-sample t test can also be used. In Section 9.3, we mentioned the pooled t test, which requires equal variances, as an alternative to the two-sample t procedure. It can be shown that the single-factor ANOVA F test and the two-tailed pooled t test are equivalent; for any given data set, the P -values for the two tests will be identical, so the same conclusion will be reached by either test.

The two-sample t test is more flexible than the F test when I = 2 for two reasons. First, it is valid without the assumption that σ 1 = σ 2; second, it can be used to test Ha: μ_1 > μ_2 (an upper-tailed t test) or Ha: μ_1 < μ_2 as well as Ha: μ_1 ≠ μ_2. In the case of I ≥ 3, there is unfortunately no general test procedure known to have good properties without assuming equal variances.

Unequal Sample Sizes

When the sample sizes from each population or treatment are not equal, let J_1, J_2, …, JI denote the I sample sizes and let n = Σi Ji denote the total number of observations. The accompanying box gives ANOVA formulas and the test procedure.

Test statistic value:

Rejection region:f ≥ Fα,I –1,n –I

Example 10.8

The article “On the Development of a New Approach for the Determination of Yield Strength in Mg-based Alloys” (Light Metal Age, Oct., 1998: 51-53) presented the following data on elastic modulus (GPa) obtained by a new ultrasonic method for specimens of a certain alloy produced using three different casting processes.

Let μ_1, μ_2, and μ_3 denote the true average elastic moduli for the three different processes under the given circumstances. The relevant hypotheses are H_0: μ_1 = μ_2 = μ_3 versus Ha: at least two of the μi s are different. The test statistic is, of course, F = MSTr/MSE, based on I –1 = 2 numerator df and n –I = 22 –3 = 19 denominator df. Relevant quantities include

The remaining computations are displayed in the accompanying ANOVA table. Since F.001,2,19 = 10.16 < 12.56 = f , the P -value is smaller than .001. Thus the null hypothesis should be rejected at any reasonable significance level; there is compelling evidence for concluding that true average elastic modulus somehow depends on which casting process is used.

There is more controversy among statisticians regarding which multiple comparisons procedure to use when sample sizes are unequal than there is in the case of equal sample sizes. The procedure that we present here is recommended in the excellent book Beyond ANOVA: Basics of Applied Statistics (see the chapter bibliography) for use when the I sample sizes J_1, J_2, …, JI are reasonably close to one another (̶ex="00055" alt="image">i ––j ––wij ≤ μi –μj. ≤ i. –j. –+ wij

for every i and j (i = 1, …, I and j = 1, …, I ) with i ≠ j .

The simultaneous confidence level 100(1 –α ) is only approximate rather than exact as it is with equal sample sizes. The underscoring method can still be used, but now the wij factor used to decide whether i and j. can be connected will depend on J and Jj.

Example 10.9 (Example 10.8 continued)

The sample sizes for the elastic modulus data were J_1 = 8, J_2 = 8, J_3 = 6, and I = 3, n –I = 19, MSE = .316. A simultaneous confidence level of approximately 95% requires Q.05,3,19 = 3.59, from which

Since 1. –2. = 44.71 –44.06 = .65 < w_12, μ_1 and μ_2 are judged not significantly different. The accompanying underscoring scheme shows that μ_1 and μ_3 appear to differ significantly, as do μ_2 and μ_3.

Data Transformation

The use of ANOVA methods can be invalidated by substantial differences in the variances (which until now have been assumed equal with common value σ 2). It sometimes happens that V (Xij) = = g (μi ), a known function of μi (so that when H_0 is false, the variances are not equal). For example, if Xij has a Poisson distribution with parameter λi (approximately normal if λi ≥ 10), then μi = λi and = λi so g (μi ) = μi is the known function. In such cases, one can ofttely equal variances (while leaving the transformed variables approximately normal), and then the F test can be used on the transformed observations. The key idea in choosing a transformation h (•) is that often V [h (Xij )] ≈ V (Xij ) • [h ′(μi )]2 = g (μi ) • [h ′(μi )]2. We wish to find the function h(•) for which g (μi ) • [h ′(μi )]2 = c (a constant) for every i .

PROPOSITION

If V (Xij ) = g (μi ), a known function of μi then a transformation h (Xij ) that “stabilizes the variance” so that V [h (Xij )] is approximately the same for each i is given by h (x ) ∝ ∫ [g (x )]–1/2 dx.

In the Poisson case, g (x ) = x , so h (x ) should be proportional to ∫ x 1/2 dx = 2x1/2. Thus Poisson data should be transformed to before the analysis.

A Random Effects Model

The single-factor problems considered so far have all been assumed to be examples of a fixed effects ANOVA model. By this we mean that the chosen levels of the factor under study are the only ones considered relevant by the experimenter. The single-factor fixed effects model is

where the Εij s are random and both μ and the αi s are fixed parameters whose values are unknown.

In some single-factor problems, the particular levels studied by the experimenter are chosen, either by design or through sampling, from a large population of levels. For example, to study the effects on task performance time of using different operators on a particular machine, a sample of five operators might be chosen from a large pool of operators. Similarly, the effect of soil pH on the yield of maize plants might be studied by using soils with four specific pH values chosen from among the many possible pH levels. When the levels used are selected at random from a larger population of possible levels, the factor is said to be random rather than fixed, and the fixed effects model (10.6) is no longer appropriate. An analogous random effects model is obtained by replacing the fixed αi s in (10.6) by random variables. The resulting model description is

all Ai s and Εij s normally distributed and independent of one another.

The condition E (Ai ) = 0 in (10.7) is similar to the condition Σαi = 0 in (10.6); it states that the expected or average effect of the ith level measured as a departure from μ is zero.

For the random effects model (10.7), the hypothesis of no effects due to different levels is H_0: = 0, which says that different levels of the factor contribute nothing to variability of the response. Although the hypotheses in the single-factor fixed and random effects models are different, they are tested in exactly the same way, by forming F = MSTr/MSE and rejecting H_0 if f ≥ Fα,I –1,n –I . This can be justified intuitively by noting that E (MSE) = σ 2 (as for fixed effects), whereas

where J_1, J_2, …, JI are the sample sizes and n = ΣJi . The factor in parentheses on the right side of (10.8) is nonnegative, so again E (MSTr) = σ^2 if H_0 is true and E(MSTr) > σ 2 if H_0 is false.

Example 10.10

The study of nondestructive forces and stresses in materials furnishes important information for efficient engineering design. The article “Zero-Force Travel-Time Parameters for Ultrasonic Head-Waves in Railroad Rail” (Materials Evaluation, 1985: 854-858) reports on a study of travel time for a certain type of wave that results from longitudinal stress of rails used for railroad track. Three measurements were made on each of six rails randomly selected from a population of rails. The investigators used random effects ANOVA to decide whether some variation in travel time could be attributed to “between-rail variability.” The data is given in the accompanying table (each value, in nanoseconds, resulted from subtracting 36.1 μ s from the original observation) along with the derived ANOVA table. The value of the F ratio is highly significant, so H_0: = 0 is rejected in favor of the conclusion that differences between rails is a source of travel-time variability.

EXERCISES Section 10.3 (22–34)

22. The following data refers to yield of tomatoes (kg/plot) for four different levels of salinity; salinity level here refers to electrical conductivity (EC), where the chosen levels were EC = 1.6, 3.8, 6.0, and 10.2 nmhos/cm:

Use the F test at level α = .05 to test for any differences in true average yield due to the different salinity levels.

23. Apply the modified Tukey’s method to the data in Exercise 22 to identify significant differences among the μi s.

24. The following partial ANOVA table is taken from the article “Perception of Spatial Incongruity” (J. Nervous and Mental Disease, 1961: 222) in which the abilities of three different groups to identify a perceptual incongruity were assessed and compared. All individuals in the experiment had been hospitalized to undergo psychiatric treatment. There were 21 individuals in the depressive group, 32 individuals in the functional “other” group, and 21 individuals in the braindamaged group. Complete the ANOVA table and carry out the F test at level α = .01.

25. Lipids provide much of the dietary energy in the bodies of infants and young children. There is a growing interest in the quality of the dietary lipid supply during infancy as a major determinant of growth, visual and neural development, and long-term health. The article “Essential Fat Requirements of Preterm Infants” (Amer. J. of Clinical Nutrition , 2000: 245S–250S) reported the following data on total polyunsaturated fats (%) for infants who were randomized to four different feeding regimens: breast milk, corn-oil-based formula, soy-oil-based formula, or soy-andmarine-oil-based formula:

a. What assumptions must be made about the four total polyunsaturated fat distributions before carrying out a single-factor ANOVA to decide whether there are any differences in true average fat content?

b. Carry out the test suggested in part (a). What can be said about the P -value?

26. Samples of six different brands of diet/imitation margarine were analyzed to determine the level of physiologically active polyunsaturated fatty acids (PAPFUA, in percentages), resulting in the following data:

(The preceding numbers are fictitious, but the sample means agree with data reported in the Janu is E(MSTr) when H_0 is true? [This expectation is correct if Σji = 0 is replaced by the restriction Σ αi = 0 (or any other single linear restriction on the αi s used to reduce the model to I independent parameters), but Σji = 0 simplifies the algebra and yields natural estimates for the model parameters (in particular, i = i –..).]

30. Reconsider Example 10.7 involving an investigation of the effects of different heat treatments on the yield point of steel ingots.

a. If J = 8 and σ = 1, what is β for a level .05 F test when μ_1 = μ_2 , μ_3 = μ_1 –1, and μ_4 = μ_1 + 1?

b. For the alternative of part (a), what value of J is necessary to obtain β = .05?

c. If there are I = 5 heat treatments, J = 10, and σ = 1, what is β for the level .05 F test when four of the μi s are equal and the fifth differs by 1 from the other four?

31. When sample sizes are not equal, the noncentrality parameter is = and Referring to Exercise 22, what is the power of the test when μ_2 = μ_3 , μ_1 = μ_2 –σ, and μ_4 = μ_2 + σ?

32. In an experiment to compare the quality of four different brands of reel-to-reel recording tape, five 2400–ft reels of each brand (A–D) were selected and the number of flaws in each reel was determined.

It is believed that the number of flaws has approximately a Poisson distribution for each brand. Analyze the data at level .01 to see whether the expected number of flaws per reel is the same for each brand.

33. Suppose that Xij is a binomial variable with parameters n and pi (so approximately normal when npi ≤ 5 and nqi ≤ 5). Then since μi = npi , V (Xij ) = = npi (1 –pi ) = μi (1 –μi /n ). How should the Xij s be transformed so as to stabild.Hint: g(μi ) = μi (1 –μi /n ).]

Bibliography

Miller, Rupert, Beyond ANOVA: The Basics of Applied Statistics, Wiley, New York, 1986. An excellent source of information about assumption checking and alternative methods of analysis.

Montgomery, Douglas, Design and Analysis of Experiments (5th ed.), Wiley, New York, 2001. A very up-to-date presentation of ANOVA models and methodology.

Neter, John, William Wasserman, and Michael Kutner, Applied Linear Statistical Models (4th ed.), Irwin, Homewood, IL, 1996. The second half of this book contains a very well-presented survey of ANOVA; the level is comparable to that of the present text, but the discussion is more comprehensive, making the book an excellent reference.

Ott, R. Lyman and Michael Longnecker. An Introduction to Statistical Methods and Data Analysis (5th ed.), Duxbury Press, Belmont, CA, 2001. Includes several chapters on ANOVA methodology that can profitably be read by students desiring a very nonmathematical exposition; there is a good chapter on various multiple comparison methods.

11 Multifactor Analysis of Variance

INTRODUCTION

In the previous chapter, we used the analysis of variance (ANOVA) to test for equality of either I different population means or the true average responses associated with I different levels of a single factor (alternatively referred to as I different treatments). In many experimental situations, there are two or more factors that are of simultaneous interest. This chapter extends the methods of Chapter 10 to investigate such multifactor situations.

In the first two sections, we concentrate on the case of two factors of interest. We will use I to denote the number of levels of the first factor (A ) and J to denote the number of levels of the second factor (B ). Then there are IJ possible combinations consisting of one level of factor A and one of factor B ; each such combination is called a treatment, so there are IJ different treatments. The number of observations made on treatment (i , j ) will be denoted by Kij. In Section 11.1, we consider Kij = 1. An important special case of this type is a randomized block design, in which a single factor A is of primary interest but another factor, “blocks,” is created to control for extraneous variability in experimental units or subjects. In Section 11.2, we focus on the case Kij = K > 1, and mention briefly the difficd at level i and the jth column contains the observed values when factor B is held at level j. In the erasable-pen experiment of Example 11.1, the number of levels of factor A is I = 3, the number of levels of factor B is J = 4, x_13 = .48, x_22 = .14, and so on.

Whereas in single-factor ANOVA we were interested only in row means and the grand mean, here we are interested also in column means. Let

with observed values i., .j, and ... Totals rather than averages are denoted by omitting the horizontal bar (so x.j = ∑ixij, etc.). Intuitively, to see whether there is any effect due to the levels of factor A, we should compare the observed i.s with one another, and information about the different levels of factor B should come from the .js.

The Fixed Effects Model

Proceeding by analogy to single-factor ANOVA, one’s first inclination in specifying a model is to let μij = the true average response when factor A is at level i and factor B at level j, giving IJ mean parameters. Then let

Xij = μij + ∈ij

where ∈ij is the random amount by which the observed value differs from its expectation and the ∈ijs are assumed normal and independent with common variance σ 2. Unfortunately, there is no valid test procedure for this choice of parameters. The reason is that under the alternative hypothesis of interest, the μijs are free to take on any values whatsoever, whereas σ 2 can be any value greater than zero, so that there are IJ + 1 freely varying parameters. But there are only IJ observations, so after using each xij as an estimate of μij, there is no way to estimate σ 2.

To rectify this problem of a model having more parameters than observed values, we must specify a model that is realistic yet involves relatively few parameters.

Assume the existence of I parameters α_1, α_2, …, αI and J parameters β_1, β_2, …, βJ such that

I + J + 1 model parameters, so if I ≥ 3 and J ≥ 3, then there will be fewer parameters than observations (in fact, we will shortly modify (11.2) so that even I = 2 and/or J = 2 will be accommodated).

The model specified in (11.1) and (11.2) is called an additive model because each mean response μij is the sum of an effect due to factor A at level i (αi) and an effect due to factor B at level j (βj). The difference between mean responses for factor A at level i and level i′ when B is held at level j is μij –μi′j. When the model is additive,

μij –μi′j = (αi + βj) –(αi′ + βj) = αi –αi′

which is independent of the level j of the second factor. A similar result holds for μij –μij′. Thus additivity means that the difference in mean responses for two levels of one of the factors is the same for all levels of the other factor. Figure 11.1(a) shows a set of mean responses that satisfy the condition of additivity, and Figure 11.1(b) shows a nonadditive configuration of mean responses.

Figure 11.1 Mean responses for two types of model: (a) additive; (b) nonadditive

Example 11.2 (Example 11.1 continued)

When we plot the observed xijs in a manner analogous to that of Figure 11.1, we get the result shown in Figure 11.2. Although there is some “crossing over” in the observed xijs, the configuration is reasonably representative of what would be expected under additivity with just one observation per treatment.

Figure 11.2 Plot of data from Example 11.1

Expression (11.2) is not quite the final model description because the αis and βjs are not uniquely determined. Following are two different configurations of the αis and βjs that yield the same additive μij s.

By subtracting any constant c from all αis and adding The ANOVA calculations are summarized in Table 11.3.

Table 11.3 ANOVA Table for Example 11.5

Since F.05,4,12 = 3.26 and fA = 95.57 ≥ 3.26, H_0 is rejected in favor of Ha, and we conclude that power consumption does depend on the brand of humidifier. To identify significantly different brands, we use Tukey’s procedure. Q.05,5,12 = 4.51 and .

The underscoring indicates that the brands can be divided into three groups with respect to power consumption.

Because the block factor is of secondary interest, F.05,3,12 is not needed, though the computed value of FB is clearly highly significant. Figure 11.4 shows SAS output for this data. Notice that in the first part of the ANOVA table, the sums of squares (SS’s) for treatments (brands) and blocks (humidity levels) are combined into a single “model” SS.

Figure 11.4 SAS output for power consumption data

In many experimental situations in which treatments are to be applied to subjects, a single subject can receive all I of the treatments. Blocking is then often done on the subjects themselves to control for variability between subjects; each subject is then said to act as its own control. Social scientists sometimes refer to such experiments as repeated-measures designs. The “units” within a block are then the different “instances” of treatment application. Similarly, blocks are often taken as different time periods, locations, or observers.

Example 11.6

The data in Table 11.4 is from the article “Compounding of Discriminative Stimuli from the Same and Different Sensory Modalities” (J. Experimental Analysis Behavior, 1971: 337–342). Rat response was maintained by fixed interval schedules of reinforcement in the presence of a tone or two separate lights. The lights were either of moderate (L1) or low intensity (L2). Observations are given as the mean number of responses emitted by each subject during single and compound stimuli presentations over a 4-day period.

Table 11.4 Response Data for Example 11.6

In many experiments, the actual levels of a factor used in the experiment, rather than being the only ones of interest to the experimenter, have been selected from a much larger population of possible levels of the factor. In a two-factor situation, when this is the case for both factors, a random effects model is appropriate. The case in which the levels of one factor are the only ones of interest and the levels of the other factor are selected from a population of levels leads to a mixed effects model. The two-factor random effects model when Kij = 1 is

Xij = μ + Ai + Bj + ∈ij (i = 1, …, I, j = 1, …, J )

where the A, s, Bj s, and ∈ijs are all independent, normally distributed rv’s with mean 0 and variances , , and σ 2, respectively. The hypotheses of interest are then H0A: (level of factor A does not contribute to variation in the response) versus HaA: and H0B: versus HaB: . Whereas E(MSE) = σ 2 as before, the expected mean squares for factors A and B are now

Thus when H0A (H0B) is true, FA (FB ) is still a ratio of two unbiased estimators of σ^2. It can be shown that a level α test for H0A versus HaA still rejects HOA if fA ≥ Fα,I–1,(I–1)(J–1), and, similarly, the same procedure as before is used to decide between H0B and HaB.

For the case in which factor A is fixed and factor B is random, the mixed model is

Xij = μ + αi + Bj + ∈ij (i = 1, …, I, j = 1, …, J )

where ∑αi = 0 and the Bjs and ∈ijs are normally distributed with mean 0 and variances and σ 2, respectively. Now the two null hypotheses are

H0A: α_1 = … = αI = 0 and H0B: = 0

with expected mean squares

The test procedures for H0A versus HaA m> are exactly as before. For example, in the analysis of the color change data in Example 11.1, if the four wash treatments were randomly selected, then because fB = 11.05 and F.05,3,6 = 4.76, H0B: is rejected in favor of HaB: . An estimate of the “variance component” is then given by (MSB –MSE)II = .0485.

Summarizing, when Kij = 1, although the hypotheses and expected mean squares differ from the case of both effects fixed, the test procedures are identical.

EXERCISES Section 11.1 (1–15)

1. The number of miles of useful tread wear (in 1000 s) was determined for tires of each of five different makes of sub-compact car (factor A, with I = 5) in combination with each of four different brands of radial tires (factor B , with J = 4), resulting in IJ = 20 observations. The values SSA = 30.6, SSB = 44.1, and SSE = 59.2 were then computed. Assume that an additive model is appropriate.

a. Test H_0: α_1 = α_2 = α_3 = α_4 = α_5 = 0 (no differences in true average tire lifetime due to makes of cars) versus Ha: at least one αi ≠ 0 using a level .05 test.

b. H_0: β_1 = β_2 = β_3 = β_4 = 0 (no differences in true average tire lifetime due to brands of tires) versus Ha: at least one βj ≠ 0 using a level .05 test.

2. Four different coatings are being considered for corrosion protection of metal pipe. The pipe will be buried in three different types of soil. To investigate whether the amount of corrosion depends either on the coating or on the type of soil, 12 pieces of pipe are selected. Each piece is coated with one of the four coatings and buried in one of the three types of soil for a fixed time, after which the amount of corrosion (depth of maximum pits, in .0001 in.) is determined. The data appears in the table.

a. Assuming the validity of the additive model, carry out the ANOVA analysis using an ANOVA table to see whether the amount of corrosion depends on either the type of coating used or the type of soil. Use α = .05.

b. Compute , and

3. The article “Adiabatic Humidification of Air with Waterft^2 on °F) as a function of gas rate (factor A ) and liquid rate (factor B ).

a. After constructing an ANOVA table, test at level .01 both the hypothesis of no gas-rate effect against the appropriate alternative and the hypothesis of no liquid-rate effect against the appropriate alternative.

b. Use Tukey’s procedure to investigate differences in expected heat transfer coefficient due to different gas rates.

c. Repeat part (b) for liquid rates.

4. In an experiment to see whether the amount of coverage of light-blue interior latex paint depends either on the brand of paint or on the brand of roller used, 1 gallon of each of four brands of paint was applied using each of three brands of roller, resulting in the following data (number of square feet covered).

a. Construct the ANOVA table. [Hint: The computations can be expedited by subtracting 400 (or any other convenient number) from each observation. This does not affect the final results.]

b. State and test hypotheses appropriate for deciding whether paint brand has any effect on coverage. Use α = .05.

c. Repeat part (b) for brand of roller.

d. Use Tukey’s method to identify significant differences among brands. Is there one brand that seems clearly preferable to the others?

5. In an experiment to assess the effect of the angle of pull on the force required to cause separation in electrical connectors, four different angles (factor A ) were used and each of a sample of five connectors (factor B ) was pulled once at each angle (“A Mixed Model Factorial Experiment in Testing Electrical Connectors,” Industrial Quality Control, 1960: 12–16). The data appears in the accompanying table.

Does the data suggest that true average separation force is affected by the angle of pull? State and test the appropriate hypotheses at level .01 by first constructing an ANOVA table (SST = 396.13, SSA = 58.16, and SSB = 246.97).

6. A particular county employs three assessors who are responsible for determining the value of residential property in the county. Tee whether these assessors differ systematically in their assessments, 5 houses are selected, and each assessor is asked to determine the market value of each house. With factor A denoting assessors (I = 3) and factor B denoting houses (J = 5), suppose SSA = 11.7, SSB = 113.5, and SSE = 25.6.

a. Test H_0: α_1 = α_2 = α_3 = 0 at level .05. (H_0 states that there are no systematic differences among assessors.)

b. Explain why a randomized block experiment with only 5 houses was used rather than a one-way ANOVA experiment involving a total of 15 different houses with each assessor asked to assess 5 different houses (a different group of 5 for each assessor).

7. The article “Rate of Stuttering Adaptation Under Two Electro-Shock Conditions” (Behavior Research Therapy, 1967: 49–54) gives adaptation scores for three different treatments: (1) no shock, (2) shock following each stuttered word, and (3) shock during each moment of stuttering. These treatments were used on each of 18 stutterers, resulting in SST = 3476.00, SSTr = 28.78, and SSBl = 2977.67.

a. Construct the ANOVA table and test at level .05 to see whether true average adaptation score depends on the treatment given.

b. Judging from the F ratio for subjects (factor B ), do you think that blocking on subjects was effective in this experiment? Explain.

8. The paper “Exercise Thermoregulation and Hyperprolactinaemia” (Ergonomics, 2005: 1547–1557) discussed how various aspects of exercise capacity might depend on the temperature of the environment. The accompanying data on body mass loss (kg) after exercising on a semi-recumbent cycle ergometer in three different ambient temperatures (6°C, 18°C, and 30°C) was provided by the paper’s authors.

a. Does temperature affect true average body mass loss? Carry out a test using a significance level of .01 (as did the authors of the cited paper).

b. Investigate significant differences among the temperatures.

c. The residuals are .20, .30, –.40, –.07, .30, .00, .03, –.20, –.14, .13, .23, –.27, –.04, .03, –.27, –.04, .33, –.10, –.33, –.53, .67, .11, –.33, .27, .01, –.13, .24. Use these as a basis for investigating the plausibility of the assumptions that underlie your analysis in (a).

9. The article “ of observations per “cell” (for each combination of levels) is constant.

Parameters and Hypotheses for the Fixed Effects Model

Rather than use the μijs themselves as model parameters, it is usual to use an equivalent set that reveals more clearly the role of interaction. Let

Thus μ is the expected response averaged over all levels of both factors (the true grand mean), μi, is the expected response averaged over levels of the second factor when the first factor A is held at level i, and similarly for μ.j.

DEFINITION

from which

The model is additive if and only if all γijs = 0. The γijs are referred to as the interaction parameters. The αis are called the main effects for factor A , and the βjs are the main effects for factor B . Although there are I αi s, J βj s, and IJ γijs in addition to μ, the conditions ∑αi = 0, ∑βj = 0, ∑jγij = 0 for any i, and ∑jγij = 0 for any j [all by virtue of (11.7) and (11.8)], imply that only IJ of these new parameters are independently determined: μ , I—1 of the αi s, J— 1 of the βj s, and (I –1)(J –1) of the γij s.

There are now three sets of hypotheses that will be considered:

H0AB: γij = 0 for all i , j versus HaAB: at least one γij ≠ 0

H0A: α_1 = … = αI = 0 versus HaA: at least one αi ≠ 0

H0B: β_1 = … = βJ = 0 versus HaB: at least one βj ≠ 0

The no-interaction hypothesis H0AB is usually tested first. If H0AB is not rejected, then the other two hypotheses can be tested to see whether the main effects are significant. If H0AB is rejected and H0A is then tested and not rejected, the resulting model μij = μ + βj + γij dos not lend itself to straightforward interpretation. In such a case, it is best to construct a picture similar to that of Figure 11.1(b) to try to visualize the way in which the factors interact.

The Model and Test Procedures

We now use triple subscripts for both random variables and observed values, with Xijk and xijk referring to the kth observation (replication) when factor A is at level i and factor B is at level j. The model is then

i = 1, …, I, j = 1, …, J, k = 1, …, K

where the ∈ijs are independent and normally distributed, each with mean 0 and variance σ^2.

Again a dot in place of a subscript denotes summation over all values of that subscript and a horizontal bar indicates averaging. Thus Xij. is the total of all K observations made for factor A at level i and factor B at level j [all observations in the (i , j )th cell], and ij., is the average of these K observations.

To test the hypotheses of interest, we again employ sums of squares:

DEFINITION

The fundamental identity is

SST = SSA + SSB + SSAB + SSE

SSAB is referred to as interaction sum of squares.

Total variation is thus partitioned into four pieces: unexplained (SSE—which would be present whether or not any of the three null hypotheses was true) and three pieces that may be explained by the truth or falsity of the three H_0s. Each of four mean squares is defined by MS = SS/df. The expected mean squares suggest that each set of hypotheses should be tested using the appropriate ratio of mean squares with MSE in the denominator:

E(MSE) = σ 2

Figure 11.5 Interaction plot from MINITAB for the tomato yield data

To verify plausibility of the normality and constant variance assumptions we can construct plots similar to those of Section 11.1. Define the predicted values (fitted values) to be the cell means, ijk = ij, so the residuals, the differences between the observations and predicted values, are xijk –ij.. The normal plot of the residuals is Figure 11.6(a), and the plot of the residuals against the fitted values is Figure 11.6(b). The normal plot is sufficiently straight that there should be no concern about the normality assumption. The plot of residuals against predicted values has a fairly uniform vertical spread, so there is no cause for concern about the constant variance assumption.

Figure 11.6 Plots from MINITAB to verify assumptions for Example 11.7

Multiple Comparisons

When the no-interaction hypothesis H0AB is not rejected and at least one of the two main effect null hypotheses is rejected, Tukey’s method can be used to identify significant differences in levels. For identifying differences among the αi’s when H0A is rejected,

1. Obtain QαI ,IJ (K –1), where the second subscript I identifies the number of levels being compared and the third subscript refers to the number of degrees of freedom for error.

2. Compute , where JK is the number of observations averaged to obtain each of the i...’s compared in Step 3.

3. Order the i..’s from smallest to largest and, as before, underscore all pairs that differ by less than w. Pairs not underscored correspond to significantly different levels of factor A.

To identifyem" width="0em">Example 11.8 (Example 11.7 continued)

For factor A (varieties), I = 3, so with α = .01 and IJ(K –1) = 24, Q.01,3,24 = 4.55. Then , so ordering and underscoring gives

The Harvester and Ife varieties do not appear to differ significantly from one another in effect on true average yield, but both differ from the Pusa variety.

For factor B (density), J = 4 so Q.01,4,24 = 4.91 and .

Thus with experimentwise error rate .01, which is quite conservative, only the lowest density appears to differ significantly from all others. Even with α = .05 (so that w = 1.64), densities 2 and 3 cannot be judged significantly different from one another in their effect on yield.

Models with Mixed and Random Effects

In some problems, the levels of either factor may have been chosen from a large population of possible levels, so that the effects contributed by the factor are random rather than fixed. As in Section 11.1, if both factors contribute random effects, the model is referred to as a random effects model, whereas if one factor is fixed and the other is random, a mixed effects model results. We will now consider the analysis for a mixed effects model in which factor A (rows) is the fixed factor and factor B (columns) is the random factor. The case in which both factors are random is dealt with in Exercise 26.

DEFINITION

The mixed effects model when factor A is fixed and factor B is random is

Xijk = μ + αi + Bj + Gij + ∈ijk

i = 1, …, I, j = 1, …, J, k = 1, …, K

Here μ and αi’s are constants with ∑αi = 0, and the Bj’s, Gij’s, and ∈ijk’s are independent, normally distributed random variables with expected value 0 and variances , andomly selected from a much larger population, so factor B is random. The relevant null hypotheses are

MINITAB output appears in Figure 11.7. The P -value column in the ANOVA table indicates that the latter two null hypotheses should be rejected at significance level .05. Different casing materials by themselves do not appear to affect vibration, but interaction between material and supplier is a significant source of variation in vibration.

Figure 11.7 Output from MINITAB’s balanced ANOVA option for the data of Example 11.9

When at least two of the Kij’s are unequal, the ANOVA computations are much more complex than for the case Kij = K, and there is controversy as to which test procedures should be used. One of the chapter references can be consulted for more information.

EXERCISES Section 11.2 (16–26)

16. In an experiment to assess the effects of curing time (factor A ) and type of mix (factor B ) on the compressive strength of hardened cement cubes, three different curing times were used in combination with four different mixes, with three observations obtained for each of the 12 curing time–mix combinations. The resulting sums of squares were computed to be SSA = 30,763.0, SSB = 34,185.6, SSE = 97,436.8, and SST = 205,966.6.

a. Construct an ANOVA table.

b. Test at level .05 the null hypothesis H0AB: all γij’s = 0 (no interaction of factors) against HaAB: at least one γij ≠ 0.

c. Test at level .05 the null hypothesis H0A: α_1 = α_2 = α_3 = 0 (factor A main effects are absent) against HaA: at least one αi ≠ 0.

d. Test H0B: β_1 = β_2 = β_3 = β_4 = 0 versus HaB: at least one βj ≠ 0 using a level .05 test.

e. The values of the i..’s were 1.. = 4010.88, 2.. = 4029.10, and 3.. = 3960.02. Use Tukehe three curing times.

17. The article “Towards Improving the Properties of Plaster Moulds and Castings” (J. Engr. Manuf., 1991: 265–269) describes several ANOVAs carried out to study how the amount of carbon fiber and sand additions affect various characteristics of the molding process. Here we give data on casting hardness and on wet-mold strength.

a. An ANOVA for wet-mold strength gives SS-Sand = 705, SSFiber = 1278, SSE = 843, and SST = 3105. Test for the presence of any effects using α = .05.

b. Carry out an ANOVA on the casting hardness observations using α = .05.

c. Plot sample mean hardness against sand percentage for different levels of carbon fiber. Is the plot consistent with your analysis in part (b)?

18. The accompanying data resulted from an experiment to investigate whether yield from a certain chemical process depended either on the formulation of a particular input or on mixer speed.

A statistical computer package gave SS(Form) = 2253.44, SS(Speed) = 230.81, SS(Form*Speed) = 18.58, and SSE = 71.87.

a. Does there appear to be interaction between the factors?

b. Does yield appear to depend on either formulation or speed?

c. Calculate estimates of the main effects.

d. The fitted values are , and the residuals are . Verify that the residuals are .23, –.87, .63, 4.50, –1.20, –3.30, –2.03, 1.97, .07, –1.10, –.30, 1.40, .67, –1.23, .57, –3.43, –.13, and 3.57.

e. Construct a normal probability plot from the residuals given in part (d). Do the ∈ijk’s appear to be normally distributed?

19. The accompanying data table gives observations on total acidity of coal samples of three different types, with determinations made using three different concentrations of ethanolic NaOH (“Chemistry of Brown Coals,”are examples of 3 × 3, 4 × 4, and 5 × 5 Latin squares. There are 12 different 3 × 3 Latin squares, and the number of different N × N Latin squares increases rapidly with N (e.g., every permutation of rows of a given Latin square yields a Latin square, and similarly for column permutations). It is recommended that the square actually used in a particular experiment be chosen at random from the set of all possible squares of the desired dimension; for further details, consult one of the chapter references.

Figure 11.9 Examples of Latin squares

The letter N will be used to denote the common value of I , J , and K . Then a complete layout with one observation per combination would require N 3 observations, whereas a Latin square requires only N 2 observations. Once a particular square has been chosen, the value of k (the level of factor C ) is completely determined by the values of i and j . To emphasize this, we use xij(k) to denote the observed value when the three factors are at levels i , j, and k, respectively, with k taking on only one value for each i , j pair. The model is then

Xij(k) = μ + αi + βj + δk + ∈ij(k) i, j, k = 1, …, N

where ∑αi = ∑βj = ∑δk = 0 and the ∈ij(k)s are independent and normally distributed with mean 0 and variance σ^2.

We employ the following notation for totals and averages:

Note that although Xi.. previously suggested a double summation, now it corresponds to a single sum over all j (and the associated values of k ).

DEFINITION

Sums of squares for a Latin square experiment are

Each mean square is, of course, the ratio SS/df. For testing H0C: δ_1 = δ_2 = … δN = 0, the test statistic value is fC = MSC/MSE, with H0C rejected if fC ≥ Fα,N –1,(N –1)(N–2). The other two main effect null hypotheses are also rejected if the corresponding F ratio exceeds Fα,N-1,(N-1)(N-2).

If any of the null hypotheses is rejected, significant differences can be identified by using Tukey’s procedure. Aing w = QαN,(N-1)(N-2) · , pairs of sample means (the i.. s, .j. s, or ..k s) differing by more than w correspond to significant differences between associated factor effects (the αi s, βj s, or δk s).

The hypothesis H 0C is frequently the one of central interest. A Latin square design is used to control for extraneous variation in the A and B factors, as was done by a randomized block design for the case of a single extraneous factor. Thus in the product sales example mentioned previously, variation due to both stores and time periods is controlled by a Latin square design, enabling an investigator to test for the presence of effects due to different product display configurations.

Example 11.11

In an experiment to investigate the effect of relative humidity on abrasion resistance of leather cut from a rectangular pattern (“The Abrasion of Leather,” J. Inter. Soc. Leather Trades’ Chemists, 1946: 287), a 6 × 6 Latin square was used to control for possible variability due to row and column position in the pattern. The six levels of relative humidity studied were 1 = 25%, 2 = 37%, 3 = 50%, 4 = 62%, 5 = 75%, and 6 = 87%, with the following results:

Also, x..1 = 46.10, x..2 = 40.59, x..3 = 39.56, x..4 = 35.86, x..5 = 32.23, x..6 = 32.64, x… = 226.98. Further computations are summarized in Table 11.9.

Table 11.9 ANOVA Table for Example 11.11

Since F.05,5,20 = 2.71 and 26.89 ≥ 2.71, H0C is rejected in favor of the hypothesis that relative humidity does on average affect abrasion resistance.

To apply Tukey’s procedure, . Ordering the ..ks and underscoring yields

In particular, the lowest relative humidity appears to result in a true average abrasion resistance significantly higher than for any other relative humidity studied.

EXERCISES Section 11.3 (27–37)

a. Construct the ANOVA table.

b. Use appropriate F tests to show that none of the F ratios for two- or three-factor interactions is significant at level .05.

c. Which main effects appear significant?

d. With x..1. = 8242, x..2. = 9732, and x..3. = 11,210, use Tukey’s procedure to identify significant differences among the levels of factor C.

28. To see whether thrust force in drilling is affected by drilling speed (A ), feed rate (B ), or material used (C ), an experiment using four speeds, three rates, and two materials was performed, with two samples (L = 2) drilled at each combination of levels of the three factors. Sums of squares were calculated as follows: SSA = 19,149.73, SSB = 2,589,047.62, SSC = 157,437.52, SSAB = 53,238.21, SSAC = 9033.73, SSBC = 91,880.04, SSE = 56,819.50, and SST = 2,983,164.81. Construct the ANOVA table and identify significant interactions using α = .01. Is there any single factor that appears to have no effect on thrust force? (In other words, does any factor appear nonsignificant in every effect in which it appears?)

29. The article “An Analysis of Variance Applied to Screw Machines” (Industrial Quality Control, 1956: 8–9) describes an experiment to investigate how the length of steel bars was affected by time of day (A ), heat treatment applied (B ), and screw machine used (C ). The three times were 8:00 A.M ., 11:00 A.M ., and 3:00 P.M ., and there were two treatments and four machines (a 3 × 2 × 4 factorial experiment), resulting in the accompanying data [coded as 1000(length –4.380), which does not affect the analysis].

Sums of squares include SSAB = 1.646, SSAC = 71.021, SSBC = 1.542, SSE = 447.500, and SST = 1037.833.Figure 11.12 Confounding ABC in a 2^3 experiment

In the absence of replications, the data from such an experiment would usually be analyzed by assuming that there were no two-factor interactions (additivity) and using SSE = SSAB + SSAC + SSBC with 3 df to test for the presence of main effects. Alternatively, a normal probability plot of effect contrasts or effect parameter estimates could be examined. Most frequently, though, there are replications when just three factors are being studied. Suppose there are u replicates, resulting in a total of 2r · u blocks in the experiment. Then after subtracting from SST all sums of squares associated with effects not confounded with blocks (computed using Yates’s method), the block sum of squares is computed using the 2r · u block totals and then subtracted to yield SSE (so there are 2r · u –1 df for blocks).

Example 11.15

The article “Factorial Experiments in Pilot Plant Studies” (Industrial and Eng. Chemistry, 1951: 1300–1306) reports the results of an experiment to assess the effects of reactor temperature (A ), gas throughput (B ), and concentration of active constituent (C ) on strength of the product solution (measured in arbitrary units) in a recirculation unit. Two blocks were used, with the ABC effect confounded with blocks, and there were two replications, resulting in the data in Figure 11.13. The four block × replication totals are 288, 212, 88, and 220, with a grand total of 808, so

Figure 11.13 Data for Example 11.15

The other sums of squares are computed by Yates’s method using the eight experimental condition totals, resulting in the ANOVA table given as Table 11.14. By comparison with F.05,1,6 = 5.99, we conclude that only the main effects for A and C differ significantly from zero.

Table 11.14 ANOVA Table for Example 11.15

Confounding Using More Than Two Blocks

In the case r = 2 (four blocks), three effects are confounded with blocks. The experimenter first chooses two defining effects to be confounded. For example, in temperature, B = amount of material B, C = solvent volume, D = condensation time, and E = amount of material E. The response variable was the yield of the chemical process. The chosen defining contrasts were ACE and BDE, with generalized interaction (ACE )(BDE ) = ABCD. The remaining 28 main effects and interactions can now be partitioned into seven groups of four effects each such that the effects within a group cannot be assessed separately. For example, the generalized interactions of A with the nonestimable effects are (A )(ACE ) = CE, (A)(BDE ) = ABDE, and (A)(ABCD ) = BCD, so one alias group is {A , CE , ABDE , BCD }. The complete set

Analysis of a Fractional Replicate

Once the defining contrasts have been chosen for a quarter-replicate, they are used as in the discussion of confounding to divide the 2p treatment conditions into four groups of 2p –2 conditions each. Then any one of the four groups is selected as the set of conditions for which data will be collected. Similar comments apply to a 1/2r replicate of a 2p factorial experiment.

Having made observations for the selected treatment combinations, a table of signs similar to Table 11.10 is constructed. The table contains a row only for each of the treatment combinations actually observed rather than the full 2p rows, and there is a single column for each alias group (since each effect in the group would have the same set of signs for the treatment conditions selected for observation). The signs in each column indicate as usual how contrasts for the various sums of squares are computed. Yates’s method can also be used, but the rule for arranging observed conditions in standard order must be modified.

The difficult part of a fractional replication analysis typically involves deciding what to use for error sum of squares. Since there will usually be no replication (though one could observe, e.g., two replicates of a quarter-replicate), some effect sums of squares must be pooled to obtain an error sum of squares. In a half-replicate of a 2^8 experiment, for example, an alias structure can be chosen so that the eight main effects and 28 two-factor interactions are each confounded only with higher-order interactions and that there are an additional 27 alias groups involving only higher-order interactions. Assuming the absence of higher-order interaction effects, the resulting 27 sums of squares can then be added to yield an error sum of squares, allowing 1-df tests for all main effects and two-factor interactions. However, in many cases tests for main effects can be obtained only by pooling some or all of the sums of squares associated with alias groups involving two-factor interactions, and the corresponding two-factor interactions cannot be investigated.

Figure 11.15 displays a normal probability plot of the effect contrasts. All but two of the points fall quite close to a straight line. The discrepant points correspond

Figure 11.15 Normal probability plot of contrasts from Example 11.18

to effects E = wire braid type and G = wire diameter, suggesting that these two factors are the only ones that affect the amount of shrinkage.

The subjects of factorial experimentation, confounding, and fractional replication encompass many models and techniques we have not discussed. For more information, the chapter references should be consulted.

EXERCISES Section 11.4 (38–49)

38. The accompanying data resulted from an experiment to study the nature of dependence of welding current on the three factors: welding voltage, wire feed speed, and tip-to-workpiece distance. There were two levels of each factor (a 2^3 experiment) with two replications per combination of levels (the averages across replications agree with values given in the article “A Study on Prediction of Welding Current in Gas Metal Arc Welding,” J. Engr. Manuf., 1991: 64–69). The first two given numbers are for the treatment (1), the next two for a , and so on in standard order: 200.0, 204.2, 215.5, 219.5, 272.7, 276.9, 299.5, 302.7, 166.6, 172.6, 186.4, 192.0, 232.6, 240.8, 253.4, 261.6.

a. Verify that the sums of squares are as given in the accompanying ANOVA table from MINITAB.

b. Which effects appear to be important, and why?

39. The accompanying data resulted from a 2^3 experiment with three replications per combination of treatments designed to study the effects of concentration of detergent (A ), concentration of sodium carbonate (B ), and concentration of sodium carboxymethyl cellulose (C ) on cleaning ability of a solution in washing tests (a larger number indicates better cleaning ability than a smaller number).

b. Use the cell totals along with Yates’s method to compute the effect contrasts and sums of squares. Then construct an ANOVA table and test all appropriate hypotheses using α = .05.

40. In a study of processes used to remove impurities from cellulose goods (“Optimization of Rope-Range Bleaching of Cellulosic Fabrics,” Textile Research J., 1976: 493–496), the following data resulted from a 2^4 experiment involving the desizing process. The four factors were enzyme concentration (A ), pH (B ), temperature (C ), and time (D).

a. Use Yates’s algorithm to obtain sums of squares and the ANOVA table.

b. Do there appear to be any second-, third-, or fourth-order interaction effects present? Explain your reasoning. Which main effects appear to be significant?

41. In Exercise 39, suppose a low water temperature has been used to obtain the data. The entire experiment is then repeated with a higher water temperature to obtain the following data. Use Yates’s algorithm on the entire set of 48 observations to obtain the sums of squares and ANOVA table, and then test appropriate hypotheses at level .05.

42. The following data on power consumption in electric furnace heats (kW consumed per ton of melted product) resulted from a 2^4 factorial experiment with three replicates (“Studies on a 10-cwt Arc Furnace,” J. Iron and Steel Institute, 1956: 22). The factors were nature of roof A (low, high), power setting B (low, high), scrap used C (tube, plate), and charge D (700 lb, 1000 lb).

Construct the ANOVA table, and test all hypotheses of interest using α = .01.

43. The article “Statistical Design and Analysis of Qualification Test Program for a Small Rocket Engine” (Industrial Quality Control, 1964: 14–18) presents data from an experiment to assess the effects of vibration (A ), temperature cycling (B ), altitude cycling (C ), and temperature for altitude cycling and firing (D ) on thrust duration. A subset of the data is given here. (In the article, there were four levels of D rather than just two.) Use the Yates method to obtain sums of squares and the ANOVA table. Then assume tha height="0em" width="-1em">
The data from an unreplicated 2^4 experiment is listed in the next table.

a. Calculate estimates of all main effects and two-factor interaction effects for this experiment.

b. Create a probability plot of the effects. Which effects appear to be important?

56. Factorial designs have been used in forestry to assess the effects of various factors on the growth behavior of trees. In one such experiment, researchers thought that healthy spruce seedlings should bud sooner than diseased spruce seedlings (“Practical Analysis of Factorial Experiments in Forestry,” Canadian J. of Forestry, 1995: 446–461). In addition, before planting, seedlings were also exposed to three levels of pH to see whether this factor has an effect on virus uptake into the root system. The following table shows data from a 2 × 3 experiment to study both factors:

The response variable is an average rating of five buds from a seedling. The ratings are 0 (bud not broken), 1 (bud partially expanded), and 2 (bud fully expanded). Analyze this data.

57. One property of automobile air bags that contributes to their ability to absorb energy is the permeability (ft^3/ft^2/min) of the woven material used to construct the air bags. Understanding how permeability is influenced by various factors is important for increasing the effectiveness of air bags. In one study, the effects of three factors, each at three levels, were studied (“Analysis of Fabrics used in Passive Restraint Systems—Airbags,” J. of the Textile Institute, 1996: 554–571):

A (Temperature): 8°C, 50°C, 75°C

B (Fabric denier): 420-D, 630-D, 840-D

C (Air pressure): 17.2 kPa, 34.4 kPa, 103.4 kPa

Analyze this data and state your conclusions (assume that all factors are fixed).

58. AThe objective of this section is to develop a linear probabilistic model.

* The slope of a line is the change in y for a 1-unit increase in x . For example, if y = –3x + 10, then y decreases by 3 when x increases by 1, so the slope is –3. The y -intercept is the height at which the line crosses the vertical axis and is obtained by setting x = 0 in the equation.

If the two variables are not deterministically related, then for a fixed value of x , the value of the second variable is random. For example, if we are investigating the relationship between age of child and size of vocabulary and decide to select a child of age x = 5.0 years, then before the selection is made, vocabulary size is a random variable Y . After a particular 5-year-old child has been selected and tested, a vocabulary of 2000 words may result. We would then say that the observed value of Y associated with fixing x = 5.0 was y = 2000.

More generally, the variable whose value is fixed by the experimenter will be denoted by x and will be called the independent, predictor, or explanatory variable. For fixed x , the second variable will be random; we denote this random variable and its observed value by Y and y, respectively, and refer to it as the dependent or response variable.

Usually observations will be made for a number of settings of the independent variable. Let x_1, x_2, …, xn denote values of the independent variable for which observations are made, and let Yi and yi, respectively denote the random variable and observed value associated with x . The available bivariate data then consists of the n pairs (x_1, y_1), (x_2, y_2), …, (xn, yn). A first step in regression analysis involving two variables is to construct a scatter plot of the observed data. In such a plot, each (x , y ) is represented as a point plotted on a two-dimensional coordinate system.

Example 12.1

Visual and musculoskeletal problems associated with the use of visual display terminals (VDTs) have become rather common in recent years. Some researchers have focused on vertical gaze direction as a source of eye strain and irritation. This direction is known to be closely related to ocular surface area (OSA), so a method of measuring OSA is needed. The accompanying representative data on y = OSA (cm^2) and x = width of the palprebal fissure (i.e., the horizontal width of the eye opening, in cm) is from the article “Analysis of Ocular Surface Area for Comfortable VDT Workstation Layout” (Ergonomics, 1996: 877–884). The order in which observations were obtained was not given, so for convenience they are listed in increasing order of x values.

That is, even though we expected Y to decrease when x increases by 1 unit, it is not unlikely that the observed Y at x + 1 will be larger than the observed Y at x .

EXERCISES Section 12.1 (1–11)

1. The efficiency ratio for a steel specimen immersed in a phosphating tank is the weight of the phosphate coating divided by the metal loss (both in mg/ft^2). The article “Statistical Process Control of a Phosphate Coating Line” (Wire J. Intl., May, 1997: 78–81) gave the accompanying data on tank temperature (x ) and efficiency ratio (y ).

a. Construct stem-and-leaf displays of both temperature and efficiency ratio, and comment on interesting features.

b. Is the value of efficiency ratio completely and uniquely determined by tank temperature? Explain your reasoning.

c. Construct a scatter plot of the data. Does it appear that efficiency ratio could be very well predicted by the value of temperature? Explain your reasoning.

2. The article “Exhaust Emissions from Four-Stroke Lawn Mower Engines” (J. of the Air and Water Mgmnt. Assoc., 1997: 945–952) reported data from a study in which both a baseline gasoline mixture and a reformulated gasoline were used. Consider the following observations on age (yr) and NOX emissions (g/kWh):

Construct scatter plots of NOx emissions versus age. What appears to be the nature of the relationship between these two variables? [Note: The authors of the cited article commented on the relationship.]

3. Bivariate data often arises from the use of two different techniques to measure the same quantity. As an example, the accompanying observations on x = hydrogen concentration (ppm) using a gas chromatography method and y = concentration using a new sensor method were read from a graph in the article “A New Method to Measure the Diffusible Hydrogen Content in Steel Weldments Using a Polymer Electrolyte-Based Hydrogen Sensor” (Welding Res., July 1997: 251sRblockquote width="-0em">Construct a scatter plot. Does there appear to be a very strong relationship between the two types of concentration measurements? Do the two methods appear to be measuring roughly the same quantity? Explain your reasoning.

4. A study to assess the capability of subsurface flow wetland systems to remove biochemical oxygen demand (BOD) and various other chemical constituents resulted in the accompanying data on x = BOD mass loading (kg/ha/d) and y = BOD mass removal (kg/ha/d) (“Subsurface Flow Wetlands—A Performance Evaluation,” Water Envir. Res., 1995: 244–247).

a. Construct boxplots of both mass loading and mass removal, and comment on any interesting features.

b. Construct a scatter plot of the data, and comment on any interesting features.

5. The article “Objective Measurement of the Stretchability of Mozzarella Cheese” (J. of Texture Studies, 1992: 185–194) reported on an experiment to investigate how the behavior of mozzarella cheese varied with temperature. Consider the accompanying data on x = temperature and y = elongation (%) at failure of the cheese. [Note: The researchers were Italian and used real mozzarella cheese, not the poor cousin widely available in the United States.]

a. Construct a scatter plot in which the axes intersect at (0, 0). Mark 0, 20, 40, 60, 80, and 100 on the horizontal axis and 0, 50, 100, 150, 200, and 250 on the vertical axis.

b. Construct a scatter plot in which the axes intersect at (55, 100), as was done in the cited article. Does this plot seem preferable to the one in part (a)? Explain your reasoning.

c. What do the plots of parts (a) and (b) suggest about the nature of the relationship between the two variables?

6. One factor in the development of tennis elbow, a malady that strikes fear in the hearts of all serious tennis players, is the impact-induced vibration of the racket-and-arm system at ball contact. It is well known that the likelihood of getting tennis elbow depends on various properties of the racket used.

Consider the scatter plot of x = racket resonance frequency (Hz) and y = sum of peak-to-peak acceleration (a characteristic of arm vibration, in m/sec/sec) for n = 23 different rackets (“Transfer of Tennis R>

c. What is the expected flow rate for a pressure drop of 10 in.? A drop of 15 in.?

d. Suppose σ = .025 and consider a pressure drop of 10 in. What is the probability that the observed value of flow rate will exceed .835? That observed flow rate will exceed .840?

e. What is the probability that an observation on flow rate when pressure drop is 10 in. will exceed an observation on flow rate made when pressure drop is 11 in.?

10. Suppose the expected cost of a production run is related to the size of the run by the equation y = 4000 + 10x . Let Y denote an observation on the cost of a run. If the variables size and cost are related according to the simple linear regression model, could it be the case that P (Y > 5500 when x = 100) = .05 and P (Y > 6500 when x = 200) = .10? Explain.

11. Suppose that in a certain chemical process the reaction time y (hr) is related to the temperature (°F) in the chamber in which the reaction takes place according to the simple linear regression model with equation y = 5.00 –.01x and σ = .075.

a. What is the expected change in reaction time for a 1°F increase in temperature? For a 10°F increase in temperature?

b. What is the expected reaction time when temperature is 200°F? When temperature is 250°F?

c. Suppose five observations are made independently on reaction time, each one for a temperature of 250°F. What is the probability that all five times are between 2.4 and 2.6 hr?

d. What is the probability that two independently observed reaction times for temperatures 1° apart are such that the time at the higher temperature exceeds the time at the lower temperature?

12.2 Estimating Model Parameters

We will assume in this and the next several sections that the variables x and y are related according to the simple linear regression model. The values of β_0, β_1, and σ 2 will almost never be known to an investigator. Instead, sample data consisting of n observed pairs (x_1, y_1), …, (xn, yn) will be available, from which the model parameters and the true regression line itself can be estimated. These observations are assumed to have been obtained independently of one another. That is, yise, if is large in magnitude, rounding will affect the final answer. We emphasize that before 1 and 0 are computed, a scatter plot should be examined to see whether a linear probabilistic model is plausible. If the points do not tend to cluster about a straight line with roughly the same degree of spread for all x , other models should be investigated. In practice, plots and regression calculations are usually done by using a statistical computer package.

Example 12.4

No-fines concrete, made from a uniformly graded coarse aggregate and a cement–water paste, is beneficial in areas prone to excessive rainfall because of its excellent drainage properties. The article “Pavement Thickness Design for No-Fines Concrete Parking Lots” (J. of Transportation Engr., 1995: 476–484) employed a least squares analysis in studying how y = porosity (%) is related to x = unit weight (pcf) in concrete specimens. Consider the following representative data, displayed in a tabular format convenient for calculating the values of the summary statistics:

Thus = 109.34, = 19.986667, and

0 = 19.986667 –(-.90473066)(109.34) = 118.909917 ≈ 118.91

We estimate that the expected change in porosity associated with a 1-pcf increase in unit weight is –.905% (a decrease of .905%). The equation of the estimated regression line (least squares line) is then y = 118.91 –.905x. Figure 12.8, generated by the statistical computer package S-Plus, shows that the least squares line provides an excellent summary of the relationship between the two variables.

Figure 12.8 A scatter plot of the data in Example 12.4 with the least squares line superimposed, from S-Plus

The estimated regression line can immediately be used for two different purposes. For a fixed x value x *, 0 + 1x * (the height of the line above x *) gives either (1) a point estimate of the expected value of Y when x = x* or (2) a point prediction of the Y value that will result from a single new observation made at x = x*.

Example 12.5

Refer to the un |it weight–porosity data in the previous example. A point estimate for true average porosity for all specimens whose unit weight is 110 is

Y.110 = 0 + 1(110) = 118.91 –.905(110) = 19.4%

If a single specimen whose unit weight is 110 pcf is to be selected, 19.4% is also a point prediction for the porosity of this specimen.

The least squares line should not be used to make a prediction for an x value much beyond the range of the data, such as x = 90 or x = 135 in Example 12.4. The danger of extrapolation is that the fitted relationship (a line here) may not be valid for such x values. (In the foregoing example, x = 135 gives = –3.3, a patently ridiculous value of porosity, but extrapolation will not always result in such inconsistencies.)

Estimating σ^2 and σ

The parameter σ 2 determines the amount of variability inherent in the regression model. A large value of σ 2 will lead to observed (xi, yi)s that are quite spread out about the true regression line, whereas when σ 2 is small the observed points will tend to fall very close to the true line (see Figure 12.9). An estimate of σ 2 will be used in confidence interval (CI) formulas and hypothesis-testing procedures presented in the next two sections. Because the equation of the true line is unknown, the estimate is based on the extent to which the sample observations deviate from the estimated line. Many large deviations (residuals) suggest a large value of σ 2, whereas deviations all of which are small in magnitude suggest that σ 2 is small.

Figure 12.9 Typical sample for σ^2: (a) small; (b) large

DEFINITION

The fitted (or predicted) values y_1, 2, …, n are obtained by successively substituting x_1, …, xn into the equation of the estimated regression line: 1 = 0 + 1x_1, i = 72.958547 + .04103377xi:

1 = 72.958547 + .04103377(125.3) ≈ 78.100, y_1 –1 ≈ –.200, etc.

A positive residual corresponds to a point in the scatter plot that lies above the graph of the least squares line, whereas a negative residual results from a point lying below the line. All predicted values (fits) and residuals appear in the accompanying table.

In much the same way that the deviations from the mean in a one-sample situation were combined to obtain the estimate s 2 = ∑(xi –)2/(n –1), the estimate of σ 2 in regression analysis is based on squaring and summing the residuals. We will continue to use the symbol s 2 for this estimated variance, so don’t confuse it with our previous s^2.

DEFINITION

The error sum of squares, (equivalently, residual sum of squares) denoted by SSE, is

SSE = ∑(yi –i)2 = ∑ [yi –(0 + 1xi)]2

and the estimate of σ 2 is

The divisor n –2 in s^2 is the number of degrees of freedom (df) associated with the estimate (or, equivalently, with error sum of squares). This is because to obtain s 2, the two parameters β_0 and β_1 must first be estimated, which results in a loss of 2 df (just as μ had to be estimated in one-sample problems, resulting in an estimated variance based on n –1 df). Replacing each yi in the formula for s 2 by the rv Yi gives the estimator S 2. It can be shown that S 2 is an unbiased estimator for σ 2 (though the estimator S is not unbiased for σ).

Example 12.7 (Example 12.6 continued)

The residuals for the fil–.188)2 +... + (1.099)2 = 7.968

The estimate of &#ockquote>

a. Does a scatter plot of the data support the use of the simple linear regression model?

b. Calculate point estimates of the slope and intercept of the population regression line.

c. Calculate a point estimate of the true average runoff volume when rainfall volume is 50.

d. Calculate a point estimate of the standard deviation σ.

e. What proportion of the observed variation in runoff volume can be attributed to the simple linear regression relationship between runoff and rainfall?

17. A regression of y = calcium content (g/L) on x = dissolved material (mg/cm^2) was reported in the article “Use of Fly Ash or Silica Fume to Increase the Resistance of Concrete to Feed Acids” (Magazine of Concrete Research, 1997: 337–344). The equation of the estimated regression line was y = 3.678 + .144x , with r 2 = .860, based on n = 23.

a. Interpret the estimated slope .144 and the coefficient of determination .860.

b. Calculate a point estimate of the true average calcium content when the amount of dissolved material is 50 mg/cm^2.

c. The value of total sum of squares was SST = 320.398. Calculate an estimate of the error standard deviation σ in the simple linear regression model.

18. The following summary statistics were obtained from a study that used regression analysis to investigate the relationship between pavement deflection and surface temperature of the pavement at various locations on a state highway. Here x = temperature (°F) and y = deflection adjustment factor (y > 0):

(Many more than 15 observations were made in the study; the reference is “Flexible Pavement Evaluation and Rehabilitation,” Transportation Eng. J., 1977: 75–85.)

a. Compute 1, 0, and the equation of the estimated regression line. Gactor when temperature is increased by 1°F?

c. Suppose temperature were measured in °C rather than in °F. What would be the estimated regression line? Answer part (b) for an increase of 1°C. [Hint: °F = (9/5)°C + 32; now substitute for the “old x” in terms of the “new x.”]

d. If a 200°F surface temperature were within the realm of possibility, would you use the estimated line of part (a) to predict deflection factor for this temperature? Why or why not?

19. The following data is representative of that reported in the article “An Experimental Correlation of Oxides of Nitrogen Emissions from Power Boilers Based on Field Data” (J. Eng. for Power, July 1973: 165–170), with x = burner area liberation rate (MBtu/hr-ft^2) and y = NOx emission rate (ppm):

a. Assuming that the simple linear regression model is valid, obtain the least squares estimate of the true regression line.

b. What is the estimate of expected NOx emission rate when burner area liberation rate equals 225?

c. Estimate the amount by which you expect NOx emission rate to change when burner area liberation rate is decreased by 50.

d. Would you use the estimated regression line to predict emission rate for a liberation rate of 500? Why or why not?

20. A number of studies have shown lichens (certain plants composed of an alga and a fungus) to be excellent bioindicators of air pollution. The article “The Epiphytic Lichen Hypogymnia Physodes as a Biomonitor of Atmospheric Nitrogen and Sulphur Deposition in Norway” (Envir. Monitoring and Assessment, 1993: 27–47) gives the following data (read from a graph) on x = wet deposition (g N/m^2) and y = lichen N (% dry weight):

The author used simple linear regression to analyze the data. Use the accompanying MINITAB output to answer the following questions:

a. What are the least squares estimates of β_0 and Liom

There is clearly variation in values of the estimated slope and estimated intercept, as well as the estimated standard deviation. The equation of the least squares line thus varies from one sample to the next. Figure 12.13 shows a dotplot of the estimated slopes as well as graphs of the true regression line and the 20 sample regression lines.

Figure 12.13 Simulation results from Example 12.10: (a) dotplot of estimated slopes; (b) graphs of the true regression line and 20 least squares lines (from S-Plus)

The slope β_1 of the population regression line is the true average change in the dependent variable y associated with a 1-unit increase in the independent variable x. The slope of the least squares line, 1, gives a point estimate of β_1. In the same way that a confidence interval for μ and procedures for testing hypotheses about μ were based on properties of the sampling distribution of , further inferences about β_1 are based on thinking of 1 as a statistic and investigating its sampling distribution.

The values of the xi s are assumed to be chosen before the experiment is performed, so only the Yi s are random. The estimators (statistics, and thus random variables) for β_0 and β_1 are obtained by replacing yi by Yi in (12.2) and (12.3):

Similarly, the estimator for σ 2 results from replacing each yi in the formula for s 2by the rv Yi:

The denominator of l, Sxx = ∑(xi –)2, depends only on the xis and not on the Yi s, so it is a constant. Then because ∑(xi –) = ∑ (xi –) = · 0 = 0, the slope estimator can be written as

That is, ear function of the independent rv’s Yl, Y_2, …, Yn, each of which is normally distributed. Invoking properties of a linear function of random variables discussed in Section 5.5 leads to the following results.

1. The mean value of 1 is E (1) = μ_1 = 1, so 1 is an unbiased estimator of β_1 (the distribution of β_1 is always centered at the value of β_1).

2. The variance and standard deviation of β_1 are

where . Replacing σ by its estimate s gives an estimate for σ_1 (the estimated standard deviation, i.e., estimated standard error, of 1:

(This estimate can also be denoted by σ_1.)

3. The estimator 1 has a normal distribution (because it is a linear function of independent normal rv’s).

According to (12.4), the variance of 1 equals the variance σ 2 of the random error term—or, equivalently, of any Yi—divided by . Because is a measure of how spread out the xi s are about , we conclude that making observations at xi values that are quite spread out results in a more precise estimator of the slope parameter (smaller variance of 1), whereas values of xi all close to one another imply a highly variable estimator. Of course, if the xi s are spread out too far, a linear model may not be appropriate throughout the range of observation.

Many inferential procedures discussed previously were based on standardizing an estimator by first subtracting its mean value and then dividing by its estimated standard deviation. In particular, test procedures and a CI for the mean μ of a normal population utilized the fact that the standardized variable —that is, —had a t distribution with n –1 df. A similar result here prondex="01622" alt="Image">

The values of the summary statistics required for calculation of the least squares estimates are

from which Sxy = -372.404, Sxx = 405.836, 1 = -.917622, 0 = 126.248889, SST = 454.1693, SSE = 112.4432, and r 2 = 1 –112.4432I454.1693 = .752. Roughly 75% of the observed variation in density can be attributed to the simple linear regression model relationship between density and air content. Error df is 15 –2 = 13, giving s^2 = 112.4432I13 = 8.6495 and s = 2.941.

The estimated standard deviation of 1 is

The t critical value for a confidence level of 95% is t.025,13 = 2.160. The confidence interval is

-.918 ± (2.160)(.1460) = -.918 ± .315 = (-1.233, -.603)

With a high degree of confidence, we estimate that an average decrease in density of between .603 lb/ft^3 and 1.233 lb/ft^3 is associated with a 1% increase in air content (at least for air content values between roughly 5% and 25%, corresponding to the x values in our sample). The interval is reasonably narrow, indicating that the slope of the population line has been precisely estimated. Notice that the interval includes only negative values, so we can be quite confident of the tendency for density to decrease as air content increases.

Looking at the SAS output of Figure 12.15, we find the value of s_1 under Parameter Estimates as the second number in the Standard Error column. All of the widely used statistical packages include this estimated standard error in output. There is also an estimated standard error for the statistic 0 from which a confidence interval for the intercept β_0 of the population regression line can be calculated.

Figure 12.15 SAS output for the data of Example 12.11

Hypothesis-Testing Procedures

As before, the null hypothesis in a test about β_1 will be an equality statement. The null value (value of β_1 claimed true by the null hypothesis) will be denoted by β_10 (read “beta one nought,he test statistic results from replacing β_1 in the standardized variable T by the null value β_10—that is, from standardizing the estimator of β_1 under the assumption that H_0 is true. The test statistic thus has a t distribution with n –2 df when H_0 is true, so the type I error probability is controlled at the desired level α by using an appropriate t critical value.

The most commonly encountered pair of hypotheses about β_1 is H_0: β_1 = 0 versus Ha: β_1 ≠ 0. When this null hypothesis is true, μY.x = β_0 independent of x , so knowledge of x gives no information about the value of the dependent variable. A test of these two hypotheses is often referred to as the model utility test in simple linear regression. Unless n is quite small, H_0 will be rejected and the utility of the model confirmed precisely when r 2 is reasonably large. The simple linear regression model should not be used for further inferences (estimates of mean value or predictions of future values) unless the model utility test results in rejection of H_0 for a suitably small α.

Null hypothesis: H_0: β_1 = β_10

A P -value based on n - 2 df can be calculated just as was done previously for t tests in Chapters 8 and 9.

The model utility test is the test of H_0: β_1 = 0 versus Ha: β_1 ≠ 0, in which case the test statistic value is the t ratio

Example 12.12

The cleanliness of molten aluminum metal or alloy prior to casting is determined mainly by the hydrogen and inclusion content of the melt. The article “Effect of Melt Cleanliness on the Properties of an A1-10 Wt Pct Si-10 Vol Pct SiC(p) Composite” (Metallurgical Trans., 1993: 1631–1645) reports on a study in which various tensile properties were related to x = volume fraction of oxides/inclusions (%). Here we present data (read from a graph) on y = elongation (%) of test bars. The authors state that the scatter plot shows a linear relationship and give the equation of the least squares line. Let’s use the MINITAB output of Figure 12.16 to carry out the model utility test at significance level α = .01.

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Figure 12.16 MINITAB output for Example 12.12

The parameter of interest is β_1, the expected change in percentage elongation associated with a 1% increase in the volume fraction of oxides/inclusions. H_0: β_1 = 0 will be rejected in favor of Ha: β_1 ≠ 0 if the t ratio satisfies either t ≥ tα/2,n-2 = t.005,18 = 2.878 or t ≤ -2.878. From Figure 12.16, 1 = -.64884, = .05840, and

Clearly, -11.11 ≤ –2.878, so H_0 is resoundingly rejected. Alternatively, the P- value is twice the area captured under the 18 df t curve to the left of –11.11. MINITAB gives P -value = .000, so H_0 should be rejected at any reasonable α. This confirmation of the utility of the simple linear regression model gives us license to calculate various estimates and predictions as described in Section 12.4.

Regression and ANOVA

The decomposition of the total sum of squares ∑(yi –)2 into a part SSE, which measures unexplained variation, and a part SSR, which measures variation explained by the linear relationship, is strongly reminiscent of one-way ANOVA. In fact, the null hypothesis H_0: β_1 = 0 can be tested against Ha: β_1 ≠ 0 by constructing an ANOVA table (Table 12.2) and rejecting H_0 if f ≥ Fα,1,n − 2.

Table 12.2 ANOVA Table for Simple Linear Regression

The F test gives exactly the same result as the model utility t test because t^2 = f and Virtually all computer packages that have regression options include such an ANOVA table in the output. For example, Figure 12.15 shows SAS output for the mortar data of Example 12.11. The ANOVA table at ght="1em" width="0em">EXERCISES Section 12.3 (30–43)

30. Reconsider the situation described in Exercise 7, in which x = accelerated strength of concrete and y = 28-day cured strength. Suppose the simple linear regression model is valid for x between 1000 and 4000 and that β_1 = 1.25 and σ = 350. Consider an experiment in which n = 7, and the x values at which observations are made are x_1 = 1000, x_2 = 1500, x_3 = 2000, x_4 = 2500, x_5 = 3000, x_6 = 3500, and x_7 = 4000.

a. Calculate , the standard deviation of 1.

b. What is the probability that the estimated slope based on such observations will be between 1.00 and 1.50?

c. Suppose it is also possible to make a single observation at each of the n = 11 values x_1 = 2000, x_2 = 2100, …, x_11 = 3000. If a major objective is to estimate β_1 as accurately as possible, would the experiment with n = 11 be preferable to the one with n = 7?

31. Reconsider the summary quantities given in Exercise 18 for the regression of y = deflection factor on x = temperature.

a. Compute the estimated standard deviation .

b. Calculate a 95% CI for β_1, the expected change in deflection factor associated with a 1°F increase in temperature.

32. Exercise 16 of Section 12.2 gave data on x = rainfall volume and y = runoff volume (both in m^3). Use the accompanying MINITAB output to decide whether there is a useful linear relationship between rainfall and runoff, and then calculate a confidence interval for the true average change in runoff volume associated with a 1-m^3 increase in rainfall volume.

33. Exercise 15 of Section 12.2 included MINITAB output for a regression of flexural strength of concrete beams on modulus of elasticity.

a. Use the output to calculate a confidence interval with a confidence level of 95% for the slope β_1 of kquote>

b. Suppose it had previously been believed that when modulus of elasticity increased by 1 GPa, the associated true average change in flexural strength would be at most .1 MPa. Does the sample data contradict this belief? State and test the relevant hypotheses.

34. Refer to the MINITAB output of Exercise 20, in which x = wet deposition and y = lichen N (%).

a. Carry out the model utility test at level .01, using the rejection region approach.

b. Repeat part (a) using the P -value approach.

c. Suppose it had previously been believed that when wet deposition increases by .1 g N/m^2, the associated change in expected lichen N is at least .15%. Carry out a test of hypotheses at level .01 to decide whether the data contradicts this prior belief.

35. How does lateral acceleration—side forces experienced in turns that are largely under driver control—affect nausea as perceived by bus passengers? The article “Motion Sickness in Public Road Transport: The Effect of Driver, Route, and Vehicle” (Ergonomics, 1999: 1646–1664) reported data on x = motion sickness dose (calculated in accordance with a British standard for evaluating similar motion at sea) and y = reported nausea (%). Relevant summary quantities are

Values of dose in the sample ranged from 6.0 to 17.6.

a. Assuming that the simple linear regression model is valid for relating these two variables (this is supported by the raw data), calculate and interpret an estimate of the slope parameter that conveys information about the precision and reliability of estimation.

b. Does it appear that there is a useful linear relationship between these two variables? Answer the question by employing the P -value approach.

c. Would it be sensible to use the simple linear regression model as a basis for predicting % nausea when dose = 5.0? Explain your reasoning.

d. When MINITAB was used to fit the simple linear regression model to the raw data, the observation (6.0, 2.50) was flagged as possibly having a substantial impact on the fit. Eliminate this observation from the sample and recalculate the estimate of/strong> Mist (airborne droplets or aerosols) is generated when metal-removing fluids are used in machining operations to cool and lubricate the tool and workpiece. Mist generation is a concern to OSHA, which has recently lowered substantially the workplace standard. The article “Variables Affecting Mist Generaton from Metal Removal Fluids” (Lubrication Engr., 2002: 10–17) gave the accompanying data on x = fluid flow velocity for a 5% soluble oil (cm/sec) and y = the extent of mist droplets having diameters smaller than 10 μm (mg/m^3):

a. The investigators performed a simple linear regression analysis to relate the two variables. Does a scatter plot of the data support this strategy?

b. What proportion of observed variation in mist can be attributed to the simple linear regression relationship between velocity and mist?

c. The investigators were particularly interested in the impact on mist of increasing velocity from 100 to 1000 (a factor of 10 corresponding to the difference between the smallest and largest x values in the sample). When x increases in this way, is there substantial evidence that the true average increase in y is less than .6?

d. Estimate the true average change in mist associated with a 1 cm/sec increase in velocity, and do so in a way that conveys information about precision and reliability.

37. Magnetic resonance imaging (MRI) is well established as a tool for measuring blood velocities and volume flows. The article “Correlation Analysis of Stenotic Aortic Valve Flow Patterns Using Phase Contrast MRI,” referenced in Exercise 1.67, proposed using this methodology for determination of valve area in patients with aortic stenosis. The accompanying data on peak velocity (m/s) from scans of 23 patients in two different planes was read from a graph in the cited paper.

a. Does there appear to be a difference between true average velocity in the two different planes? Carry out an appropriate test of hypotheses (as did the authors of the article).

b. The authors of the article also regressed Level--velocity against Level–velocity. The resulting estimated intercept and slope are .14701 and .65393, with corresponding estimated standard errors .07877 and .05947, coefficient of determination .852, and s = .110673. The article included a comment that this regression showed evidence of a strong linear relationship but a regression slope well below 1. Do you agree?

38. Refer to the data on x = liberation rate and y = NOx emission rate given in Exercise 19. = 402, , s = 3.725, and 1 = 1.7035. Consider testing at level .01 H_0: β_1 = 1, which states that the expected increase in % BOD removal is 1 when filter application temperature increases by 1°C, against the alternative Ha: β_1 > 1. Determine P(type II error) when β′1 = 2, σ = 4.

12.4 Inferences Concerning μY·x* and the Prediction of Future Y Values

Let x * denote a specified value of the independent variable x . Once the estimates 0 and 1 have been calculated, can be regarded either as a point estimate of (the expected or true average value of Y when x = x *) or as a prediction of the Y value that will result from a single observation made when x = x *. The point estimate or prediction by itself gives no information concerning how precisely . has been estimated or Y has been predicted. This can be remedied by developing a CI for and a prediction interval (PI) for a single Y value.

Before we obtain sample data, both 0 and 1 are subject to sampling variability—that is, they are both statistics whose values will vary from sample to sample. Suppose, for example, that β_0 = 50 and β_1 = 2. Then a first sample of (x , y ) pairs might give 0 = 52.35, 1 = 1.895, a second sample might result in 0 = 46.52, 1 = 2.056, and so on. It follows that = itself varies in value from sample to sample, so it is a statistic. If the intercept and slope of the population line are the aforementioned values 50 and 2, respectively, and x * = 10, then this statistic is trying to estimate the value 50 + 2(10) = 70. The estimate from a first sample might be 52.35 + 1.895(10) = 71.30, from a second sample might be 46.52 + 2.056(10) = 67.08, and so on. In the same way that a confidence interval for β_1 was based on properties of the sampling distribution of 1, a confidence interval for a mean y value in regression is based on properties of the sampling distribution of the statistic .

Substitution of the expressions for 0 and 1 into 0 + 1x * followed by some algebraic manipulation leads to the representation of 0 + 1x * as a linear function of the Yi s:

The coefficients d_1, d_2, …, dn in this linear function involve the x, s and x *, all of which are fixed. Application of the rules of Section 5.5 to this linear function gives the following properties.

Let = , where x * is some fixed value of x . Then

1. The mean value of is

2. The variance of is

and the standard deviation σ is the square root of this expression. The estimated standard deviation of , denoted by S or S, results from replacing σ by its estimate s:

3. has a normal distribution.

The variance of is smallest when x * = and increases as x * moves away from in either direction. Thus the estimator of is more precise when x * is near the center of the xis than when it is far from the x values at which observations have been made. This will imply that both the CI and PI are narrower for an x * near than for an x * far from . Most statistical computer packages will provide both and S for any specified x * upon request.

Inferences Concerning

Just as inferential procedures for β_1 were based on the t variable obtained by standardizing β_1, a t variable obtained by standardizing leads to a CI and test procedures here.

THEOREM

The variable

has a t distribution with n –2 df.

As for β_1 in the previous section, a probability statement involving this standardized variable can be manipulated to yield a confidence interval for .

This CI is centered at the point estimate for and extends out to each side by an amount that depends on the confidence level and on the extent of variability in the estimator on which the point estimate is based.

Example 12.13

Corrosion of steel reinforcing bars is the most important durability problem for reinforced concrete structures. Carbonation of concrete results from a chemical reaction that lowers the pH value by enough to initiate corrosion of the rebar. Representative data on x = carbonation depth (mm) and y = strength (MPa) for a sample of core specimens taken from a particular building follow (read from a plot in the article “The Carbonation of Concrete Structures in the Tropical Environment of Singapore,” Magazine of Concrete Res., 1996: 293–300).

Figure 12.17 MINITAB scatter plot with confidence intervals and prediction intervals for the data of Example 12.13

A scatter plot of the data (see Figure 12.17) gives strong support to use of the simple linear regression model. Relevant quantities are as follows:

Let’s now calculate a confidence interval, using a 95% confidence level, for the mean strength for all core specimens having a carbonation depth of 45 mm—that is, a confidence interval for β_0 + β_1(45). The interval is centered at

= 0 + 1(45) = 27.18 –.2976(45) = 13.79

The estimated standard deviation of the statistic is

The 16 df t critical value for a 95% confidence level is 2.120, from which w0)(.7582) = 13.79 ± 1.61 = (12.18, 15.40)

The narrowness of this interval suggests that we have reasonably precise information about the mean value being estimated. Remember that if we recalculated this interval for sample after sample, in the long run about 95% of the calculated intervals would include β_0 + β_1(45). We can only hope that this mean value lies in the single interval that we have calculated.

Figure 12.18 shows MINITAB output resulting from a request to fit the simple linear regression model and calculate confidence intervals for the mean value of strength at depths of 45 σm and 35 mm. The intervals are at the bottom of the output; note that the second interval is narrower than the first, because 35 is much closer to than is 45. Figure 12.17 shows (1) curves corresponding to the confidence limits for each different x value and (2) prediction limits, to be discussed shortly. Notice how the curves get farther and farther apart as x moves away from :

Figure 12.18 MINITAB regression output for the data of Example 12.13

In some situations, a CI is desired not just for a single x value but for two or more x values. Suppose an investigator wishes a CI both for μY·v and for μY·w , where ν and w are two different values of the independent variable. It is tempting to compute the interval (12.6) first for x = v and then for x = w . Suppose we use α = .05 in each computation to get two 95% intervals. Then if the variables involved in computing the two intervals were independent of one another, the joint confidence coefficient would be (.95) · (.95) ≈ .90.

However, the intervals are not independent because the same 0, 1, and S are used in each. We therefore cannot assert that the joint confidence level for the two intervals is exactly 90%. It can be shown, though, that if the 100(1 –α )% CI (12.6) is computed both for x = v and for x = w to obtain joint CIs for μY·v and μY·w, then the joint confidence level on the resulting pair of intervals is at least 100(1 –2α)%. In particular, using α = .05 results in a joint confidence level of at least 90%, whereas using α = .01 results in at least 98% confidence. For example, in Example 12.13 a 95% CI for μY·45 was (12.185, 15.401) and a 95% CI for μY·35 was (15.330, 18.207). The simultaneous or joint confidence level for the two sus CIs rests on a probability result called the Bonferroni inequality, so the joint CIs are referred to as Bonferroni intervals. The method is easily generalized to yield joint intervals for k different μY·x S. Using the interval (12.6) separately first for x = , then for x = , …, and finally for x = yields a set of k CIs for which the joint or simultaneous confidence level is guaranteed to be at least 100(1 –kα)%.

Tests of hypotheses about are based on the test statistic T obtained by replacing in the numerator of (12.5) by the null value μ_0. For example, H_0: β_0 + β_1(45) = 15 in Example 12.13 says that when carbonation depth is 45, expected (i.e., true average) strength is 15. The test statistic value is then , and the test is upper-, lower-, or two-tailed according to the inequality in Ha.

A Prediction Interval for a Future Value of Y

Analogous to the CI (12.6) for , one frequently wishes to obtain an interval of plausible values for the value of Y associated with some future observation when the independent variable has value x*. For instance, in the example in which vocabulary size y is related to the age x of a child, for x = 6 years (12.6) would provide a CI for the true average vocabulary size of all 6-year-old children. Alternatively, we might wish an interval of plausible values for the vocabulary size of a particular 6-year-old child.

A CI refers to a parameter, or population characteristic, whose value is fixed but unknown to us. In contrast, a future value of Y is not a parameter but instead a random variable; for this reason we refer to an interval of plausible values for a future Y as a prediction interval rather than a confidence interval. The error of estimation is –(), a difference between a fixed (but unknown) quantity and a random variable. The error of prediction is Y –(), a difference between two random variables. There is thus more uncertainty in prediction than in estimation, so a PI will be wider than a CI. Because the future value Y is independent of the observed Yi s,

Furthermore, because E (Y ) = and E () = , the expected value of the prediction error is E (Y –()) = 0. It can then be shown that the standardized variable

tuting this T into the probability statement and manipulating to isolate Y between the two inequalities yields the following interval.

A 100(1 –α)% PI for a future Y observation to be made when x = x* is

The interpretation of the prediction level 100(1 –α)% is identical to that of previous confidence levels—if (12.7) is used repeatedly, in the long run the resulting intervals will actually contain the observed y values 100(1 –α)% of the time. Notice that the 1 underneath the initial square root symbol makes the PI (12.7) wider than the CI (12.6), though the intervals are both centered at . Also, as n → ∞, the width of the CI approaches 0, whereas the width of the PI does not (because even with perfect knowledge of β_0 and β_1, there will still be uncertainty in prediction).

Example 12.14

Let’s return to the carbonation depth-strength data of Example 12.13 and calculate a 95% prediction interval for a strength value that would result from selecting a single core specimen whose carbonation depth is 45 mm. Relevant quantities from that example are

For a prediction level of 95% based on n –2 = 16 df, the t critical value is 2.120, exactly what we previously used for a 95% confidence level. The prediction interval is then

Plausible values for a single observation on strength when depth is 45 σm are (at the 95% prediction level) between 7.51 MPa and 20.07 MPa. The 95% confidence interval for mean strength when depth is 45 was (12.18, 15.40). The prediction interval is much wider than this because of the extra (2.8640)2 under the square root. Figure 12.18, the MINITAB output in Example 12.13, shows this interval as well as the confidence interval.

The Bonferroni technique can be employed as in the case of confidence intervals. If a 100(1 –α)% PI is calculated for each of k different values of x , the simultaneous or joint prediction level for all k intervals is at least 100(1 –kα)%.

EXERCISES Section 12.4 (44–56)

44. Fitting the simple linear regression model to the n = 27 observations on x = modulus of elasticity and y = flexural strength given in Exercise 15 of Section 12.2 resulted >

b. Obtain an interval estimate for mean Fermi level position when concentration is .50, and interpret your estimate.

c. Obtain an interval of plausible values for position resulting from a single observation to be made when concentration is .50, interpret your interval, and compare to the interval of (b).

d. Obtain simultaneous CIs for expected position when concentration is .3, .5, and .7; the joint confidence level should be at least 97%.

51. Refer to Example 12.12 in which x = volume fraction of oxides/inclusions and y = % elongation.

a. MINITAB gave = .0311 and = .0352. Why is the former estimated standard deviation smaller than the latter one?

b. Use the MINITAB output from the example to calculate a 95% CI for expected % elongation when volume fraction = .40.

c. Use the MINITAB output to calculate a 95% PI for a single value of % elongation to be observed when volume fraction = 1.20.

52. Plasma etching is essential to the fine-line pattern transfer in current semiconductor processes. The article “Ion Beam-Assisted Etching of Aluminum with Chlorine” (J. Electrochem. Soc., 1985: 2010–2012) gives the accompanying data (read from a graph) on chlorine flow (x , in SCCM) through a nozzle used in the etching mechanism and etch rate (y, in 100 A/min).

The summary statistics are = 24.0, = 312.5, = 70.50, = 902.25, = 11,626.75, 0 = 6.448718, 1 = 10.602564.

a. Does the simple linear regression model specify a useful relationship between chlorine flow and etch rate?

b. Estimate the true average change in etch rate associated with a 1-SCCM increase in flow rate using a 95% confidence interval, and interpret the interval.

d. Calculate a 95% PI for a single future observation on etch rate to be made when flow = 3.0. Is the prediction likely to be accurate?

e. Would the 95% CI and PI when flow = 2.5 be wider or narrower than the corresponding intervals of parts (c) and (d)? Answer without actually computing the intervals.

f. Would you recommend calculating a 95% PI for a flow of 6.0? Explain.

53. Consider the following four intervals based on the data of Example 12.4 (Section 12.2):

a. A 95% CI for mean porosity when unit weight is 110

b. A 95% PI for porosity when unit weight is 110

c. A 95% CI for mean porosity when unit weight is 115

d. A 95% PI for porosity when unit weight is 115

Without computing any of these intervals, what can be said about their widths relative to one another?

54. The decline of water supplies in certain areas of the United States has created the need for increased understanding of relationships between economic factors such as crop yield and hydrologic and soil factors. The article “Variability of Soil Water Properties and Crop Yield in a Sloped Watershed” (Water Resources Bull., 1988: 281–288) gives data on grain sorghum yield (y, in g/m-row) and distance upslope (x, in m) on a sloping watershed. Selected observations are given in the accompanying table.

a. Construct a scatter plot. Does the simple linear regression model appear to be plausible?

b. Carry out a test of model utility.

c. Estimate true average yield when distance upslope is 75 by giving an interval of plausible values.

55. Verify that V (0 + 1xem" width="0em">

from which

Properties of r

The most important properties of r are as follows:

1. The value of r does not depend on which of the two variables under study is labeled x and which is labeled y .

2. The value of r is independent of the units in which x and y are measured.

3. –1 ≤ r ≤ 1

4. r = 1 if and only if (iff) all (xi, yi) pairs lie on a straight line with positive slope, and r = –1 iff all (xi, yi ) pairs lie on a straight line with negative slope.

5. The square of the sample correlation coefficient gives the value of the coefficient of determination that would result from fitting the simple linear regression model—in symbols, (r)2 = r 2.

Property 1 stands in marked contrast to what happens in regression analysis, where virtually all quantities of interest (the estimated slope, estimated y -intercept, s 2, etc.) depend on which of the two variables is treated as the dependent variable. However, Property 5 shows that the proportion of variation in the dependent variable explained by fitting the simple linear regression model does not depend on which variable plays this role.

Property 2 is equivalent to saying that r is unchanged if each xi is replaced by cxi and if each yi is replaced by dyi (a change in the scale of measurement), as well as if each xi is replaced by xi—a and yi by yi—b (which changes the location of zero on the measurement axis). This implies, for example, that r is the same whether temperature is measured in mg recindex="01694" alt="Image">

Example 12.16

In some locations, there is a strong association between concentrations of two different pollutants. The article “The Carbon Component of the Los Angeles Aerosol: Source Apportionment and Contributions to the Visibility Budget” (J. Air Pollution Control Fed., 1984: 643–650) reports the accompanying data on ozone concentration x (ppm) and secondary carbon concentration y (μ/m^3).

The summary quantities are n = 16, = 1.656, = 170.6, = .196912, = 20.0397, and = 2253.56, from which

The point estimate of the population correlation coefficient ρ between ozone concentration and secondary carbon concentration is = r = .716.

The small-sample intervals and test procedures presented in Chapters 7-9 were based on an assumption of population normality. To test hypotheses about ρ, we must make an analogous assumption about the distribution of pairs of (x , y ) values in the population. We are now assuming that both X and Y are random, whereas much of our regression work focused on x fixed by the experimenter.

ASSUMPTION

The joint probability distribution of (X , Y ) is specified by

where μ_1 and σ_1 are the mean and standard deviation of X, and μ_2 and σ_2 are the mean and standard deviation of Y; f (x , y ) is called the bivariate normal probability distribution.

The bivariate normal distribution is obviously rather complicated, but for our purposes we need only a passing acquaintance with several of its properties. The surface determined by f (x , y ) lies entirely above the x, y plane [f (x , y ) ≥ 0] and has a three-dimensional bell- or mound-shaped appearance, as illustrated in Figure 12.21Scand. J. Work and Environ. Health, 1996: 381–386) gave data on x = manganese concentration in blood (ppb) and y = concentration (μmol/L) of malondialdehyde, which is a stable product of lipid peroxidation, both for a sample of 22 workers exposed to manganese and for a control sample of 45 individuals. The value of r for the control sample was .29, from which

The corresponding P -value for a two-tailed t test based on 43 df is roughly .052 (the cited article reported only that P -value > .05). We would not want to reject the assertion that ρ = 0 at either significance level .01 or .05. For the sample of exposed workers, r = .83 and t ≈ 6.7, clear evidence that there is a linear relationship in the entire population of exposed workers from which the sample was selected.

Because ρ measures the extent to which there is a linear relationship between the two variables in the population, the null hypothesis H_0: ρ = 0 states that there is no such population relationship. In Section 12.3, we used the t-ratio to test for a linear relationship between the two variables in the context of regression analysis. It turns out that the two test procedures are completely equivalent because . When interest lies only in assessing the strength of any linear relationship rather than in fitting a model and using it to estimate or predict, the test statistic formula just presented requires fewer computations than does the t -ratio.

Other Inferences Concerning ρ

The procedure for testing H_0: ρ = ρ_0 when ρ_0 ≠ 0 is not equivalent to any procedure from regression analysis. The test statistic is based on a transformation of R called the Fisher transformation.

PROPOSITION

When (X_1, Y_1), …, (Xn, Yn) is a sample from a bivariate normal distribution, the rv

has approximately a normal distribution with mean and variance

The rationale for the transformation is to obtain a function of R that has a variance independent

where c_1 and c_2 are the left and right endpoints, respectively, of the interval (12.11).

Example 12.19

The article “A Study of a Partial Nutrient Removal System for Wastewater Treatment Plants” (Water Research, 1972: 1389–1397) reports on a method of nitrogen removal that involves the treatment of the supernatant from an aerobic digester. Both the influent total nitrogen x (mg/L) and the percentage y of nitrogen removed were recorded for 20 days, with resulting summary statistics = 285.90, = 4409.55, = 690.30, = 29,040.29, and = 10,818.56. The sample correlation coefficient between influent nitrogen and percentage nitrogen removed is r = .733, giving v = .935. With n = 20, a 95% confidence interval for μv , is (.935 – = (.460, 1.410) = (c_1, c_2). The 95% interval for ρ is

In Chapter 5, we cautioned that a large value of the correlation coefficient (near 1 or –1) implies only association and not causation. This applies to both ρ and r.

EXERCISES Section 12.5 (57–67)

57. The article “Behavioural Effects of Mobile Telephone Use During Simulated Driving” (Ergonomics, 1995: 2536–2562) reported that for a sample of 20 experimental subjects, the sample correlation coefficient for x = age and y = time since the subject had acquired a driving license (yr) was .97. Why do you think the value of r is so close to 1? (The article’s authors gave an explanation.)

58. The Turbine Oil Oxidation Test (TOST) and the Rotating Bomb Oxidation Test (RBOT) are two different procedures for evaluating the oxidation stability of steam turbine oils. The article “Dependence of Oxidation Stability of Steam Turbine Oil on Base Oil Composition” (J. of the Society of Tribologists and Lubrication Engrs., Oct. 1997: 19–24) reported the accompanying observations on x = TOST time (hr) and y = RBOT time (min) for 12 oil specimens.

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61. The authors of the paper “Objective Effects of a Six Months’ Endurance and Strength Training Program in Outpatients with Congestive Heart Failure” (Medicine and Science in Sports and Exercise, 1999: 1102–1107) presented a correlation analysis to investigate the relationship between maximal lac-tate level x and muscular endurance y. The accompanying data was read from a plot in the paper.

Sxx = 36.9839, Syy = 2,628,930.357, Sxy = 7377.704. A scatter plot shows a linear pattern.

a. Test to see whether there is a positive correlation between maximal lactate level and muscular endurance in the population from which this data was selected.

b. If a regression analysis were to be carried out to predict endurance from lactate level, what proportion of observed variation in endurance could be attributed to the approximate linear relationship? Answer the analogous question if regression is used to predict lactate level from endurance—and answer both questions without doing any regression calculations.

62. Hydrogen content is conjectured to be an important factor in porosity of aluminum alloy castings. The article “The Reduced Pressure Test as a Measuring Tool in the Evaluation of Porosity/Hydrogen Content in A1–7 Wt Pct Si-10 Vol Pct SiC(p) Metal Matrix Composite” (Metallurgical Trans, 1993: 1857–1868) gives the accompanying data on x = content and y = gas porosity for one particular measurement technique.

MINITAB gives the following output in response to a CORRELATION command:

a. Test at level .05 to see whether the population correlation coefficient differs from 0.

b. If a simple linear regression analysis had been carried out, what percentage of observed variation in porosity could be attributed to the model relationship?

63. Physical properties of six flame-retardant fabric samples were investigated in the article “Sensory and Physical Properties of Inherently Flame-Retardant Fnce level to determine whether a linear relationship exists between stiffness x (mg-cm) and thickness y (mm). Is the result of the test surprising in light of the value of r?

64. The article “Increases in Steroid Binding Globulins Induced by Tamoxifen in Patients with Carcinoma of the Breast” (J. Endocrinology, 1978: 219–226) reports data on the effects of the drug tamoxifen on change in the level of cortisol-binding globulin (CBG) of patients during treatment. With age = x and ΔCBG = y , summary values are n = 26, ∑ xi = 1613, ∑(xi—)2 = 3756.96, = 281.9, ∑(yi—)2 = 465.34, and = 16,731.

a. Compute a 90% CI for the true correlation coefficient ρ.

b. Test H_0: ρ = –.5 versus Ha: ρ < –.5 at level .05.

c. In a regression analysis of y on x , what proportion of variation in change of cortisol-binding globulin level could be explained by variation in patient age within the sample?

d. If you decide to perform a regression analysis with age as the dependent variable, what proportion of variation in age is explainable by variation in ΔCBG?

65. The article “Chronological Trend in Blood Lead Levels” (N. Engl. J. Med., 1983: 1373–1377) gives the following data on y = average blood lead level of white children age 6 months to 5 years and x = amount of lead used in gasoline production (in 1000 tons) for ten 6-month periods:

a. Construct separate normal probability plots for x and y. Do you think it is reasonable to assume that the (x, y) pairs are from a bivariate normal population?

b. Does the data provide sufficient evidence to conclude that there is a linear relationship between blood lead level and the amount of lead used in gasoline production? Use α = .01.

66. Consider a time series—that is, a sequence of observations Xm>1, x_2, …, xn. Suppose that the series shows no upward or downward trend over time. An investigator will frequently want to know just how strongly values in the series separated by a specified number of time units are related. The lag-one sample autocorrelation coefficient r_1 is just the value of the sample correlation coefficient r for the pairs (x_1, x_2), (x_2, x_3), …, (xn–1, xn), that is, pairs of values separated by one time unit. Similarly, the lag-two sample autocorrelation coefficient r_2 is r for the n – 2 pairs (x_1, x_3), (x_2, x_4), …, (xn–2, xn).

a. Calculate the values of r_1, r_2, and r_3 for the temperature data from Exercise 82 of Chapter 1, and comment.

b. Analogous to the population correlation coefficient ρ, let ρ_1, ρ_2, … denote the theoretical or long-run autocorrelation coefficients at the various lags. If all these ρ s are 0, there is no (linear) relationship at any lag. In this case, if n is large, each Ri has approximately a normal distribution with mean 0 and standard deviation 1/Vn and different Ris are almost independent. Thus H_0: ρi = 0 can be rejected at a significance level of approximately .05 if either ri ≥ 2/ or ri ≤ –2/. If n = 100 and r_1 = .16, r_2 = -.09, and r_3 = –.15, is there any evidence of theoretical autocorrelation at the first three lags?

c. If you are simultaneously testing the null hypothesis in part (b) for more than one lag, why might you want to increase the cutoff constant 2 in the rejection region?

67. A sample of n = 500 (x, y) pairs was collected and a test of H_0: ρ = 0 versus Ha: ρ ≠ 0 was carried out. The resulting P -value was computed to be .00032.

a. What conclusion would be appropriate at level of significance .001?

b. Does this small P -value indicate that there is a very strong linear relationship between x and y (a value of ρ that differs considerably from 0)? Explain.

c. Now suppose a sample of n = 10,000 (x, y) pairs resulted in r = .022. Test H_0: ρ = 0 versus Ha: ρ ≠ 0 at level .05. Is the result statistically significant?sis.

SUPPLEMENTARY EXERCISES (68–87)

68. The appraisal of a warehouse can appear straightforward compared to other appraisal assignments. A warehouse appraisal involves comparing a building that is primarily an open shell to other such buildings. However, there are still a number of warehouse attributes that are plausibly related to appraised value. The article “Challenges In Appraising ’Simple’ Warehouse Properties” (Donald Sonneman, The Appraisal Journal, April 2001, 174–178) gave the accompanying data on truss height (ft), which determines how high stored goods can be stacked, and sale price ($) per square foot.

a. Is it the case that truss height and sale price are “deterministically” related—i.e., that sale price is determined completely and uniquely by truss height? [Hint: Look at the data.]

b. Construct a scatterplot of the data. What does it suggest?

c. Determine the equation of the least squares line.

d. Give a point prediction of price when truss height is 27 ft, and calculate the corresponding residual.

e. What percentage of observed variation in sale price can be attributed to the approximate linear relationship between truss height and price?

69. Refer to the previous exercise, which gave data on truss heights for a sample of warehouses and the corresponding sale prices.

a. Estimate the true average change in sale price associated with a one-foot increase in truss height, and do so in a way that conveys information about the precision of estimation.

b. Estimate the true average sale price for all warehouses having a truss height of 25 ft, and do so in a way that conveys information about the precision of estimation.

c. Predict the sale price for a single warehouse whose truss height is 25 ft, and do so in a way that conveys information about the precision of prediction. How does this prediction compare to the estimate of (b)?

d. Without calculating any intervals, how would the width of a 95% prediction interval for sale price when truss height is 25 ft compare to the width of a 95% interval when height is 30 ft? Expla the sample correlation coefficient.

70. Forensic scientists are often interested in making a measurement of some sort on a body (alive or dead) and then using that as a basis for inferring something about the age of the body. Consider the accompanying data on age (yr) and % D-aspertic acid (hereafter %DAA) from a particular tooth (“An Improved Method for Age at Death Determination from the Measurements of D-Aspertic Acid in Dental Collagen,” Archaeometry, 1990: 61–70.)

Suppose a tooth from another individual has 2.01%DAA. Might it be the case that the individual is younger than 22? This question was relevant to whether or not the individual could receive a life sentence for murder.

A seemingly sensible strategy is to regress age on %DAA and then compute a PI for age when %DAA = 2.01. However, it is more natural here to regard age as the independent variable x and %DAA as the dependent variable y, so the regression model is %DAA = β_0 + ∈. After estimating the regression coefficients, we can substitute y * = 2.01 into the estimated equation and then solve for a prediction of age . This “inverse” use of the regression line is called “calibration.” A PI for age with prediction level approximately 100(1—α)% is ± tα/2,n –2 · SE where

Calculate this PI for y* = 2.01 and then address the question posed earlier.

71. The accompanying data on x = diesel oil consumption rate measured by the drain–weigh method and y = rate measured by the CI-trace method, both in g/hr, was read from a graph in the article “A New Measurement Method of Diesel Engine Oil Consumption Rate” (J. Society Auto Engr., 1985: 28–33).

a. Assuming that x and y are related by the simple linear regression model, carry out a test to decide whether it is plausible that on average the change in the rate measured by the CI-trace method is identical to the change in the rate measured by the drain–weigh method.

b. Calculate and interpret the value of the sample correlation coefficient.

72. The SAS output on the next page is based on data from the article “Evidence for and the Rate of Denitrification in the Arabian Sea” (<#8221; (J. of the Air and Waste Mgmt. Assoc., 1998: 16–25). The response variable is ΔNOy and the explanatory variable is ΔCO.

a. Fit an appropriate model to the data and judge the utility of the model.

b. Predict the value of ΔNOy that would result from making one more observation when ΔCO is 400, and do so in a way that conveys information about precision and reliability. Does it appear that ΔNOy can be accurately predicted? Explain.

SAS output for Exercise 73

c. The largest value of ΔCO is much greater than the other values. Does this observation appear to have had a substantial impact on the fitted equation?

75. An investigation was carried out to study the relationship between speed (ft/sec) and stride rate (number of steps taken/sec) among female marathon runners. Resulting summary quantities included n = 11, ∑(speed) = 205.4, ∑(speed)2 = 3880.08, ∑(rate) = 35.16, ∑(rate)2 = 112.681, and ∑(speed)(rate) = 660.130.

a. Calculate the equation of the least squares line that you would use to predict stride rate from speed.

b. Calculate the equation of the least squares line that you would use to predict speed from stride rate.

c. Calculate the coefficient of determination for the regression of stride rate on speed of part (a) and for the regression of speed on stride rate of part (b). How are these related?

76. “Mode-mixity” refers to how much of crack propagation is attributable to the three conventional fracture modes of opening, sliding, and tearing. For plane problems, only the first two modes are present, and the mode-mixity angle is a measure of the extent to which propagation is due to sliding as opposed to opening. The article “Increasing Allowable Flight Loads by Improved Structural Modeling” (AIAA J., 2006: 376–381) gave the following data on x = mode-mixity angle (degrees) and y = fracture toughness (N/m) for sandwich panels use in aircraft construction.

V () and a CI for . Taking x * = 0 gives and a CI for β_0. Use the data of Example 12.11 to calculate the estimated standard deviation of 0 and a 95% CI for the y -intercept of the true regression line.

79. Show that SSE = Syy –1Sxy, which gives an alternative computational formula for SSE.

80. Suppose that x and y are positive variables and that a sample of n pairs results in r ≈ 1. If the sample correlation coefficient is computed for the (x , y 2) pairs, will the resulting value also be approximately 1? Explain.

81. Let sx and sy denote the sample standard deviations of the observed x ’s and y ’s, respectively [so = ∑(xi –)2/(n –1) and similarly for ].

a. Show that an alternative expression for the estimated regression line y = is

b. This expression for the regression line can be interpreted as follows. Suppose r = .5. What then is the predicted y for an x that lies 1 SD (sx units) above the mean of the xi s? If r were 1, the prediction would be for y to lie 1 SD above its mean , but since r = .5, we predict a y that is only .5 SD (.5sy unit) above y . Using the data in Exercise 64 for a patient whose age is 1 SD below the average age in the sample, by how many standard deviations is the patient’s predicted ΔCBG above or below the average ΔCBG for the sample?

82. Verify that the t statistic for testing H_0: β_1 = 0 in Section 12.3 is identical to the t statistic in Section 12.5 for testing H_0: ρ = 0.

83. Use the formula for computing SSE to verify that r^2 = 1 –SSE/SST.

84. In biofiltration of wastewater, air discharged from a treatment facility is passed through a damp porous membrane that causes contaminants to dissolve in water and be transformed into harmless products. The accompanying data on x = inlet temperature (°C) and y = removal efficien, New York, 1999. The most comprehensive and authoritative book on regression analysis currently in print.

Neter, John, Michael Kutner, Christopher Nachtsheim, and William Wasserman, Applied Linear Statistical Models (4th ed.), Irwin, Homewood, IL, 1996. The first 15 chapters constitute an extremely readable and informative survey of regression analysis.

13 Nonlinear and Multiple Regression

INTRODUCTION

The probabilistic model studied in Chapter 12 specified that the observed value of the dependent variable Y deviated from the linear regression function μγ.X = β_0 + β_1x by a random amount. Here we consider two ways of generalizing the simple linear regression model. The first way is to replace β_0 + β_1X by a nonlinear function of x, and the second is to use a regression function involving more than a single independent variable. After fitting a regression function of the chosen form to the given data, it is of course important to have methods available for making inferences about the parameters of the chosen model. Before these methods are used, though, the data analyst should first assess the validity of the chosen model. In Section 13.1, we discuss methods, based primarily on a graphical analysis of the residuals (observed minus predicted y ’s), for checking the aptness of a fitted model.

In Section 13.2, we consider nonlinear regression functions of a single independent variable X that are ”intrinsically linear.” By this we mean that it is possible to transform one or both of the variables so that the relationship between the new variables is linear. An alternative class of nonlinear relations is obtained by using polynomial regression functions of the form μγ.X = β_0 + β_1X + β_2X^2 + ••• + βkXk; these polynomial models are the subject of Section 13.3. Multiple regression analysis involves building models for relating y to two or more independent variables. The focus in Section 13.4 is on interpretation of various multiple regression models and on understanding and using the regression output from various statistical computer packages. The last section of the chapter surveys some extensions and pitfalls of multiple regression modeling.

13.1 Aptness of the Model and Model Checking

A plot of the observed pairs (xi, yi) is a necessary first step in deciding on the form of a mathematical relationship between x and y. It is possible to fit many functions other than a linear one (y = b_0 + b_1x) to the data, using either the principle of lan">Figure 13.1 presents a scatter plot of the data and the four plots just recommended. The plot of versus y confirms the impression given by r 2 that x is effective in predicting y and also indicates that there is no observed y for which the predicted value is terribly far off the mark. Both residual plots show no unusual pattern or discrepant values. There is one standardized residual slightly outside the interval (-2, 2), but this is not surprising in a sample of size 14. The normal probability plot of the standardized residuals is reasonably straight. In summary, the plots leave us with no qualms about either the appropriateness of a simple linear relationship or the fit to the given data.

Figure 13.1 Plots for the data from Example 13.1

Difficulties and Remedies

Although we hope that our analysis will yield plots like those of Figure 13.1, quite frequently the plots will suggest one or more of the following difficulties:

1. A nonlinear probabilistic relationship between x and y is appropriate.

2. The variance of ∈ (and of Y ) is not a constant σ^2 but depends on x.

3. The selected model fits the data well except for a very few discrepant or outlying data values, which may have greatly influenced the choice of the best-fit function.

4. The error term ∈ does not have a normal distribution.

5. When the subscript i indicates the time order of the observations, the ∈is exhibit dependence over time.

6. One or more relevant independent variables have been omitted from the model.

Figure 13.2 presents residual plots corresponding to items 1-3, 5, and 6. In Chapter 4, we discussed patterns in normal probability plots that cast doubt on the assumption of an underlying normal distribution. Notice that the residuals from the data in Figure 13.2(d) with the circled point included would not by themselves necessarily suggest further analysis, yet when a new line is fit with that point deleted, the new line differs considerably from the original line.e is a single (or very few) data point(s) with independent variable value(s) far removed from the remainder of the data.

Figure 13.2 Plots that indicate abnormality in data: (a) nonlinear relationship; (b) nonconstant variance; (c) discrepant observation; (d) observation with large influence; (e) dependence in errors; (f) variable omitted

We now indicate briefly what remedies are available for the types of difficulties. For a more comprehensive discussion, one or more of the references on regression analysis should be consulted. If the residual plot looks something like that of Figure 13.2(a), exhibiting a curved pattern, then a nonlinear function of x may be fit.

The residual plot of Figure 13.2(b) suggests that, although a straight-line relationship may be reasonable, the assumption that V(Yi) = σ^2 for each i is of doubtful validity. When the assumptions of Chapter 12 are valid, it can be shown that among all unbiased estimators of β_0 and β_1, the ordinary least squares estimators have minimum variance. These estimators give equal weight to each (xi, Yi). If the variance of Y increases with x, then Yis for large xi should be given less weight than those with small xi. This suggests that β_0 and β_1 should be estimated by minimizing

where the wi s are weights that decrease with increasing xi. Minimization of Expression (13.4) yields weighted least squares estimates. For example, if the standard deviation of Y is proportional to x (for x > 0)—that is, V(Y) = kx^2—then it can be shown that the weights wi = 1/ yield best estimators of β_0 and β_1. The books by John Neter et al. and by S. Chatterjee and Bertram Price contain more detail (see the chapter bibliography). Weighted least squares is used quite frequently by econome-tricians (economists who use statistical methods) to estimate parameters.

When plots or other evidence suggest that the data set contains outliers or points having large influence on the resulting fit, one possible approach is to omit these outlying points and recompute the estimated regression equation. This would certainly be correct if it were found that the outliers resulted from errors in recording data values or experimental errors. If no assignable cause can be found for the outliers, it is still desirable to report the estimated equation both with and without outliers omitted. Yet another approach is to retain possible outliers but to use an estimation principle that puts relatively less weight on outlying values than does the principle of least squares. One such principle is MAD (minimize absolute deviations), which selects 0 and 1 to minimize ∑ | yi −> )|. Unlike the estimates of least squares, there are no nice formulas for the MAD estimates; their values must be found by using an iterative computational procedure. Such procedures are also used when it is suspected that the ∈is have a distribution that is not normal but instead has ”heavy tails” (making it much more likely than for the normal distribution that discrepant values will enter the sample); robust regression procedures are those that produce reliable estimates for a wide variety of underlying error distributions. Least squares estimators are not robust in the same way that the sample mean is not a robust estimator for μ.

When a plot suggests time dependence in the error terms, an appropriate analysis may involve a transformation of the y’s or else a model explicitly including a time variable. Lastly, a plot such as that of Figure 13.2(f), which shows a pattern in the residuals when plotted against an omitted variable, suggests that a multiple regression model that includes the previously omitted variable should be considered.

EXERCISES Section 13.1 (1–14)

1. Suppose the variables x = commuting distance and y = commuting time are related according to the simple linear regression model with σ = 10.

a. If n = 5 observations are made at the x values x_1 = 5, x_2 = 10, x_3 = 15, x_4 = 20, and x_5 = 25, calculate the standard deviations of the five corresponding residuals.

b. Repeat part (a) for x_1 = 5, x_2 = 10, x_3 = 15, x_4 = 20, and x_5 = 50.

c. What do the results of parts (a) and (b) imply about the deviation of the estimated line from the observation made at the largest sampled x value?

2. The x values and standardized residuals for the chlorine flow/etch rate data of Exercise 52 (Section 12.4) are displayed in the accompanying table. Construct a standardized residual plot and comment on its appearance.

3. Example 12.6 presented the residuals from a simple linear regression of moisture content y on filtration rate x.

a. Plot the residuals against x. Does the resulting plot suggest that a straight-line regression function is a reasonable choice of mouals. Is ≈ ei/s for i = 1, …, n, or are the s not close to being proportional to the ei s?

c. Plot the standardized residuals against x. Does the plot differ significantly in general appearance from the plot of part (a)?

4. Wear resistance of certain nuclear reactor components made of Zircaloy-2 is partly determined by properties of the oxide layer. The following data appears in an article that proposed a new nondestructive testing method to monitor thickness of the layer (“Monitoring of Oxide Layer Thickness on Zircaloy-2 by the Eddy Current Test Method,” J. of Testing and Eval., 1987: 333-336). The variables are x = oxide-layer thickness (μm) and y = eddy-current response (arbitrary units).

a. The authors summarized the relationship by giving the equation of the least squares line as y = 20.6 − .047x. Calculate and plot the residuals against x and then comment on the appropriateness of the simple linear regression model.

b. Use s = .7921 to calculate the standardized residuals from a simple linear regression. Construct a standardized residual plot and comment. Also construct a normal probability plot and comment.

5. As the air temperature drops, river water becomes supercooled and ice crystals form. Such ice can significantly affect the hydraulics of a river. The article ”Laboratory Study of Anchor Ice Growth” (J. of Cold Regions Engr., 2001: 60-66) described an experiment in which ice thickness (mm) was studied as a function of elapsed time (hr) under specified conditions. The following data was read from a graph in the article: n = 33; x = .17, .33, .50, .67, …, 5.50; y = .50, 1.25, 1.50, 2.75, 3.50, 4.75, 5.75, 5.60, 7.00, 8.00, 8.25, 9.50, 10.50, 11.00, 10.75, 12.50, 12.25, 13.25, 15.50, 15.00, 15.25, 16.25, 17.25, 18.00, 18.25, 18.15, 20.25, 19.50, 20.00, 20.50, 20.60, 20.50, 19.80.

a. The r^2 value resulting from a least squares fit is .977. Interpret this value and comment on the appropriateness of assuming an approximate linear relationship.

b. The residuals, listed in the same order as the x values, are

Plot the residuals against elapsed time. What does the plot suggest?
e the same x values, so these values are listed only once Frank Anscombe, ”Graphs in Statistical Analysis,” Amer. Statistician, 1973: 17-21):

For each of these four data sets, the values of the summary statistics ∑xi, , ∑yi, , and c0507-02.gifxiyi are virtually identical, so all quantities computed from these five will be essentially identical for the four sets—the least squares line (y = 3 + .5x), SSE, s^2, r^2, t intervals, t statistics, and so on. The summary statistics provide no way of distinguishing among the four data sets. Based on a scatter plot and a residual plot for each set, comment on the appropriateness or inappro-priateness of fitting a straight-line model; include in your comments any specific suggestions for how a ”straight-line analysis” might be modified or qualified.

10. a. Show that ei = 0 when the ei s are the residuals from a simple linear regression.

b. Are the residuals from a simple linear regression independent of one another, positively correlated, or negatively correlated? Explain.

c. Show that xiei = 0 for the residuals from a simple linear regression. (This result along with part (a) shows that there are two linear restrictions on the ei s, resulting in a loss of 2 df when the squared residuals are used to estimate σ^2.)

d. Is it true that = 0? Give a proof or a counter example.

11. a. Express the ith residual Yi —i (where i = 0 + 1xi) in the form ∑ciYi, a linear function of the Yj s. Then use rules of variance to verify that V(Yi—i) is given by Expression (13.2).

b. It can be shown that i and Yi—i (the ith predicted value and residual) are independent of one another. Use this fact, the relation Yi = i + (Yi — i), and the expression for V() from Section 12.4 to again verify Expression (13.2).

c. As xi moves farther away from , what happens to V(i) and to V(Yi—i)?

12. a. Could a linear regression result in residuals 23, -27, 5, 17, -8, 9, and 15? Why or why not?

b. Could a linear regression result in residuals 23, -27, 5, 17, -8, -12, and 2 corresponding to x values 3, -4, 8, 12, -14, -20, and 25? Why or why not? [Hint: See Exercise 10.]

13. Recall that 0 + 1x has a normal distribution with expected value β_0 + β_1x and variance

so that

has a standard normal distribution. If is substituted for σ, the resulting variable has a t distribution with n— 2 df. By analogy, what is the distribution of any particular standardized residual? If n = 25, what is the probability that a particular standardized residual falls outside the interval (-2.50, 2.50)?

14. If there is at least one x value at which more than one observation has been made, there is a formal test procedure for testing

H_0: μγ.x = β_0 + β_1x for some values β_0, β_1 (the true regression function is linear)

versus

Ha: H_0 is not true (the true regression function is not linear)

Suppose observations are made at x_1, x_2, …, xc. Let Y_11, Y_12, …, Y1n_1 denote the n_1 observations when x = x_1; …; Yc1, Yc2, …, Ycnc denote the nc observations when x = xc. With n = ∑ni (the total number of observations), SSE has n— 2 df. We break SSE into two pieces, SSPE (pure error) and SSLF (lack of fit), as follows:

c. Y = α + β log(x) + ∈, so that x’= log(x) immediately linearizes the model.

d. Y = α + β • 1/x + ∈, so that x’= 1/x yields a linear model.

The additive exponential and power models, Y = αeβx + ∈ and Y = α xβ + ∈, are not intrinsically linear. Notice that both (a) and (b) require a transformation on Y and, as a result, a transformation on the error variable ∈. In fact, if ∈ has a lognormal distribution (see Chapter 4) with E(∈) = eσ^2/2 and V(∈) = τ^2 independent of x, then the transformed models for both (a) and (b) will satisfy all the assumptions of Chapter 12 regarding the linear probabilistic model; this in turn implies that all inferences for the parameters of the transformed model based on these assumptions will be valid. If σ^2 is small, μγ.x ≈ αeβx in (a) or αxβ in (b).

The major advantage of an intrinsically linear model is that the parameters β_0 and β_1 of the transformed model can be immediately estimated using the principle of least squares simply by substituting x’ and y’ into the estimating formulas:

Parameters of the original nonlinear model can then be estimated by transforming back 0 and/or 1 if necessary. Once a prediction interval for y’ when x’= x’* has been calculated, reversing the transformation gives a PI for y itself. In cases (a) and (b), when σ^2 is small, an approximate CI for μY.x* results from taking antilogs of the limits in the CI for β_0 + β_1X’* (strictly speaking, taking antilogs gives a CI for the median of the Y distribution, i.e., for Y.x*. Because the lognormal distribution is positively skewed, μ > ; the two are approximately equal if σ^2 is close to 0.)

Example 13.3

Taylor’s equation m>y = αxβ. The article “The Effect of Experimental Error on the Determination of Optimum Metal Cutting Conditions” (J. Eng. for Industry, 1967: 315-322) observes that the relationship is not exact (deterministic) and that the parameters α and β must be estimated from data. Thus an appropriate model is the multiplicative power model Y = α • xβ • ∈, which the author fit to the accompanying data consisting of 12 carbide tool life observations (Table 13.2). In addition to the x, y, x’, and y’ values, the predicted transformed values (’) and the predicted values on the original scale (, after transforming back) are given.

Table 13.2 Data for Example 13.3

The summary statistics for fitting a straight line to the transformed data are = 74.41200, = 26.22601, 2 = 461.75874, 2 = 67.74609, and = 160.84601, so

The estimated values of α and β, the parameters of the power function model, are = 1 = -5.3996 and = e_0 = 3.094491530 • 10^15. Thus the estimated regression function is Y•X ≈ 3.094491530 • 10^15 • x-53996. To recapture Taylor’s (estimated) equation, set y = 3.094491530 • 10^15 • x-5.3996, whence, xy .185 = 740.

Figure 13.4(a) gives a plot of the standardized residuals from the linear regression using transformed variables (for which r 2 = .922); there is no apparent pattern in the plot, though one standardized residual is a bit large, and the residuals look as they should for a simple linear regression. Figure 13.4(b) pictures a plot of y versus y, which indicates satisfactory predictions on the original scale.

To obtain a confidence interval for median tool life when cutting time is 500, we transform x = 500 to x′ = 6.21461. Then 0 + 1x’ = 2.1120, and a 95% CI for β_0 + β_1(6.21461) is (from Y•500 is then obtained by taking antilogs: (e1.928, e2.296) = (6.876, 9.930). It is easily checked that for the transformed data s^2 = 2 ≈ .081. Because this is quite small, (6.876, 9.930) is an approximate interval for μY•500.

Figure 13.4 (a) Standardized residuals versus x’ from Example 13.3; (b) y versus y from Example 13.3

Example 13.4

In the article ”Ethylene Synthesis in Lettuce Seeds: Its Physiological Significance” (Plant Physiology, 1972: 719-722), ethylene content of lettuce seeds (y, in nL/g dry wt) was studied as a function of exposure time (x, in min) to an ethylene absorbent. Figure 13.5 presents both a scatter plot of the data and a plot of the residuals generated from a linear regression of y on x. Both plots show a strong curved pattern, suggesting that a transformation to achieve linearity is appropriate. In addition, a linear regression gives negative predictions for x= 90 and x= 100.

Figure 13.5 (a) Scatter plot; (b) residual plot from linear regression for the data in Example 13.4

The author did not give any argument for a theoretical model, but his plot of y’= ln(y) versus x shows a strong linear relationship, suggesting that an exponential function will provide a good fit to the data. Table 13.3 shows the data values and other information from a linear regression of y’ on x. The estimates of parameters of

Table 13.3 Data for Example 13.4

the linear model are 1 = -.0323 and 0 = 5.941, with r 2 = .995. The estimated regression function for the exponential model is Y•x ≈ e_0 • e_1x = 380.32e-.0323x. The predicted values as is usually not feasible, so it is desirable to have a method for estimating weight from various characteristics of an animal that can be easily determined. MINITAB has a stored data set consisting of various characteristics for a sample of n = 143 wild bears. Figure 13.7(a) displays a scatter plot of y = weight versus x= distance around the chest (chest girth). At first glance, it looks as though a single line obtained from ordinary least squares would effectively summarize the pattern. Figure 13.7(b) shows the LOWESS curve produced by MINITAB using a span of 50% (the fit at (x*, y*) is determined by the closest 50% of the sample). The curve appears to consist of two straight line segments joined together above approximately x = 38. The steeper line is to the right of 38, indicating that weight tends to increase more rapidly as girth does for girths exceeding 38 in.

Figure 13.7 A MINITAB scatter plot and LOWESS curve for the bear weight data

It is complicated to make other inferences (e.g., obtain a CI for a mean y value) based on this general type of regression model. The bootstrap technique mentioned earlier can be used for this purpose.

Logistic Regression

The simple linear regression model is appropriate for relating a quantitative response variable y to a quantitative predictor x. Suppose that y is a dichotomous variable with possible values 1 and 0 corresponding to success and failure. Let p = P(S) = P(y = 1). Frequently, the value of p will depend on the value of some quantitative variable x. For example, the probability that a car needs warranty service of a certain kind might well depend on the car’s mileage, or the probability of avoiding an infection of a certain type might depend on the dosage in an inoculation. Instead of using just the symbol p for the success probability, we now use p(x) to emphasize the dependence of this probability on the value of x. The simple linear regression equation Y = β_0 + β_1x + ∈ is no longer appropriate, for taking the mean value on each side of the equation gives

μY.x = 1 • p(x) + 0 • (1—p(x)) = p(x) = β_0 + β_1X

Whereas p(x) is a probability and therefore must be between 0 and 1, β_0 + β_1x need not be in this range.

Instead of letting the mean value of y be a linear function of x, we now consider a model in which some function of the mean value of y is a linear function of x. In other words, we allow p(x) to be a function of β_0 + β_1x rather than β_0 + β_1x itself. A function that has been found quite useful in many applications is the logit fs was the case for Y = β_0 + β_1x + ∈, where V(Y) = σ^2]? Explain your reasoning. Draw a picture of a prototype scatter plot resulting from this model. Answer the same questions for the power model Y = αxβ • ∈.

24. Kyphosis refers to severe forward flexion of the spine following corrective spinal surgery. A study carried out to determine risk factors for kyphosis reported the accompanying ages (months) for 40 subjects at the time of the operation; the first 18 subjects did have kyphosis and the remaining 22 did not.

Use the accompanying MINITAB logistic regression output to decide whether age appears to have a significant impact on the presence of kyphosis.

25. The following data resulted from a study commissioned by a large management consulting company to investigate the relationship between amount of job experience (months) for a junior consultant and the likelihood of the consultant being able to perform a certain complex task.

Interpret the accompanying MINITAB logistic regression output, and sketch a graph of the estimated probability of task performance as a function of experience.

Logistic regression table for Exercise 24

Logistic regression table for Exercise 25

13.3 Polynomial Regression

The nonlinear yet intrinsically linear models of Section 13.2 involved functions of the independent variable x that were either strictly increasing or strictly decreasing. In many situations, either theoretical reasoning or else a scatter plot of the data suggests that the true regression function μY.x has one or more peaks or valleys—that is, at least one relative minimum or maximum. In such cases, a polynomial function y = β_0 + β_1x +••• + βkxk may provide a satisfactory approximation to the true regression function.

DEFINITION

The k th-degree polynomial regression model equation is

All standard statistical computer packages will automatically solve the equations in (13.10) and provide the estimates as well as much other information.*

* We will see in Section 13.4 that polynomial regression is a special case of multiple regression, so a command appropriate for this latter task is generally used.

Example 13.7

The article ”Residual Stresses and Adhesion of Thermal Spray Coatings” (Surface Engineering, 2005: 35–40) considered the relationship between the thickness (μηι) of NiCrAl coatings deposited on stainless steel substrate and corresponding bond strength (MPa). The following data was read from a plot in the paper:

The scatter plot in Figure 13.11(a) supports the choice of the quadratic regression model. Figure 13.11(b) contains MINITAB output from a fit of this model. The estimated regression coefficients are

0 = 14.521 1 = .04323 2 = -.00006001

from which the estimated regression function is

y = 14.521 + .04323x - .00006001x^2

Substitution of the successive x values 220, 220, …, 860, and 860 into this function gives the predicted values 1 = 21.128, …, 20 = 7.321, and the residuals y_1 − 1 = 2.872, …, y_20 − 20 =-4.521 result from subtraction. Figure 13.12 shows a plot of the standardized residuals versus y and also a normal probability plot of the standardized residuals, both of which validate the quadratic model.

Figure 13.11 (a) Scatter plot of data from Example 13.7 (b) MINITAB output from fit of quadratic model

Figure 13.12 Diagnostic plots for quadratic model fit to data of Example 13.7

Example 13.8

SSE and SST are typically found on computer output in an ANOVA table. Fig-(Example 13.7 ure 13.11(b) gives SSE = 181.71 and SST = 825.00, for the bond strength data, continued) from which R^2 = 1 − 181.71/825.00 = .780 (alternatively, R^2 = SSR/SST = 643.29/825.00 = .780). Thus 78.0% of the observed variation in bond strength can be attributed to the model relationship. Adjusted R^2 = .754, only a small downward change in R^2. The estimates of σ 2 and σ are

Besides computing R^2 and adjusted R^2, one should examine the usual diagnostic plots to determine whether model assumptions are valid or whether modification may be appropriate.

Statistical Intervals and Test Procedures

Because the yi s appear in the normal equations (13.10) only on the right-hand side and in a linear fashion, the resulting estimates 0, …, k are themselves linear functions of the yi s. Thus the estimators are linear functions of the Yi s, so each i has a normal distribution. It can also be shown that each i is an unbiased estimator of i.

Let σi. denote the standard deviation of the estimator i. This standard deviation has the form

Fortunately, the expression in braces has been programmed into all of the most frequently used statistical computer packages. The estimated standard deviation of i, Si, results from substituting s in place of σ in the expression for σi. These estimated standard deviations s_0, s_1,…, and Sk appear on output from all the aforementioned statistical packages. Let ight="0em" align="center">

has a t distribution based on η − (k + 1) df. This leads to the following inferential procedures.

A 100(1 − α)% CI for βi, the coefficient of xi in the polynomial regression function, is

i ± tα/2,n-(k+1) • Si

A test of H_0: β,i = βi0 is based on the t statistic value

The test is based on η − (k + 1) df and is upper-, lower-, or two-tailed according to whether the inequality in Ha is >, <, or Φ .

A point estimate of μY.x—that is, of β_0 + β_1x +••• + βkX k—is Y.x = 0 + 1X +•••+ kxk. The estimated standard deviation of the corresponding estimator is rather complicated. Many computer packages will give this estimated standard deviation for any x value when requested to do so by a user. This, along with an appropriate standardized t variable, can be used to justify the following procedures.

Let x* denote a specified value of x . A 100(1 − α)% CI for μY.x* is

With = 0 + 1x* + •••+ βk(x*)k, denoting the calculated value of for the given data and S denoting the estimated standard deviation of the statistic , the formula for the CI is much like the one in the case of simple linear regression:

± tα/2,n-(k+1) • S

A 100(1 − α)% PI for a future y value to be observed when x = x* is

Example 13.9 (Example 13.8 continued)

Figure 13.11(b) shows that 2 =-.00006001 and S_2 = .00001786 (from the SE Coef column at the top of the output). The null hypothesis H_0: β_2 = 0 says that as long as the linear predictor x is retained in the model, the quadratic predictor x^2 provides no additional useful information. The relevant alternative is Ha: β_2 Φ 0 and the test statistic is T = 2/S_2,with computed value -3.36. The test is based on n − (k + 1) = 17 df. At significance level .05, the null hypothesis is rejected because -3.36 < -2.110 = − t.025, 17. Inclusion of the quadratic predictor is justified. The same conclusion results from comparing the reported P-value .004 to the chosen significance level .05.

The output in Figure 13.11(b) also contains estimation and prediction information both for x = 500 and for x = 800. In particular, for x = 500,

= 0 + 1(500) + 2(500)2 = Fit = 21.136

S = estimated SD of = SE Fit = 1.167

from which a 95% CI for mean strength when thickness = 500 is 21.136 ± (2.110)(1.167) = (18.67, 23.60). A 95% PI for the strength resulting from a single bond when thickness = 500 is 21.136 ± (2.110)[(3.27)2 + (1.167)2]½ = (13.81, 28.46). As before the PI is substantially wider than the CI because s is large compared to SE Fit.

Centering x Values

For the quadratic model with regression function βY.x = β_0 + β_1x + β_2x^2, the parameters β_0, β_1, and β_2 characterize the behavior of the function near x = 0. For example, β_0 is the height at which the regression function crosses the vertical axis x = 0, whereas β_1 is the first derivative of the function at x = 0 (instantaneous rate of change of βY.x at x = 0). If the xi s all lie far from 0, we may not have precise information about the values of these parameters. Let = the average of the xi s for which observations are to be takener">

In the model (13.14), , and the parameters now describe the behavior of the regression function near the center of the data.

To estimate the parameters of (13.14), we simply subtract from each xi to obtain = xi - , and then use the s in place of the xi s. An important benefit of this is that the coefficients of b_0, …, bk in the normal equations (13.10) will be of much smaller magnitude than would be the case were the original xi s used. When the system is solved by computer, this centering protects against any round-off error that may result.

Example 13.10

The article ”A Method for Improving the Accuracy of Polynomial Regression Analysis” (/. Quality Technology, 1971: 149-155) reports the following data on x = cure temperature (°F) and y = ultimate shear strength of a rubber compound (psi), with = 297.13:

A computer analysis yielded the results shown in Table 13.4.

Table 13.4 Estimated Coefficients and Standard Deviations for Example 13.10

The estimated regression function using the original model is y =-26,219.64 + 189.21 x- .3312 x^2, whereas for the centered model the function is y= 759.36 - 7.61(x − 297.13) − .3312(x − 297.13)2. These estimated functions are identical; the only difference is that different parameters have been estimated for the two models. The estimated standard deviations indicate clearly that and have been more accurately estimated than β_0 and β_1. The quadratic parameters are identical , as can be seen by comparing the x^2 term in (13.14) with the original model. We emphasize again that a major benefit of centering is the gain in computational accuracy, not only in quadratic but also in higher-degree models.

The book by Neter et al., listed in the chapter bibliography, is a good source for more information about polynomial regression.

Section 13.3 (26–35)

26. In addition to a linear regression of true density on moisture content, the article cited in Exercise 6 considered a quadratic regression of bulk density versus moisture content. Data from a graph in the article follows, along with a MINITAB output from the quadratic fit.

a. Does a scatter plot of the data appear consistent with the quadratic regression model?

b. What proportion of observed variation in density can be attributed to the model relationship?

c. Does the quadratic model appear to be useful? Carry out a test at significance level .05.

d. The last line of output is from a request for estimation and prediction information when moisture content is 14. Calculate a 99% PI for density when moisture content is 14.

e. Does the quadratic predictor appear to provide useful information? Test the appropriate hypotheses at significance level .05.

27. The following data on y = glucose concentration (g/L) and x = fermentation time (days) for a particular blend of malt liquor was read from a scatter plot in the article ’’Improving Fermentation Productivity with Reverse Osmosis” (Food Tech., 1984: 92-96):

a. Verify that a scatter plot of the data is consistent with the choice of a quadratic regression model.

b. The estimated quadratic regression equation is y = 84.482 − 15.875x + 1.7679x^2. Predict the value of glucose concentration for a fermentation time of 6 days, and compute the corresponding residual.

c. Using SSE = 61.77, what proportion of observed variation can be attributed to the quadratic regression relationship?

d. The n = 8 standardized residuals based on the quadratic model are 1.91, -1.95, -.25, .58, .90, .04, -.66, and .20. Construct a plot of the standardized residuals versus x and a normal probability plot. Do the plots exhibit any troublesome features?

e. The estimated standard deviation of Y.6—that is, >2(36)—is 1.69. Compute a 95% CI for μY.6.

f. Compute a 95% PI for a glucose concentration observation made after 6 days of fermentation time.

28. The viscosity (y) of an oil was measured by a cone and plate viscometer at six different cone speeds (x). It was assumed that a quadratic regression model was appropriate, and the estimated regression function resulting from the n = 6 observations was

y = -113.0937 + 3.3684x − .01780x^2

a. Estimate μY.75 , the expected viscosity when speed is 75 rpm.

b. What viscosity would you predict for a cone speed of 60 rpm?

c. If = 8386.43, ∑yi = 210.70, ∑xiyi = 17,002.00, and yi = 1,419,780, compute SSE [= − 0∑yi − 1∑xiyi − 2 yi], s 2, and s.

d. From part (c), SST = 8386.43 − (210.70)2/6 = 987.35. Using SSE computed in part (c), what is the computed value of R^2?

e. If the estimated standard deviation of 2 is s_2 = .00226, test H_0: β_2 = 0 versus Ha: β_2 Φ 0 at level .01, and interpret the result.

29. High-alumina refractory castables have been extensively investigated in recent years because of their significant advantages over other refractory brick of the same class—lower production and application costs, versatility, and performance at high temperatures. The accompanying data on x = viscosity (MPa • s) and y = free-flow (%) was read from a graph in the article ”Processing of Zero-Cement Self-Flow Alumina Castables” (The Amer. Ceramic Soc. Bull., 1998: 60-66):

The authors of the cited paper related these two variables using a quadratic regression model. The estimated regression function is y = -295.96 + 2.1885x − .0031662x^2.

a. Compute the predicted values and residuals, and then SSE and s^2.

b. Compute and interpret the coefficient of multiple determination.

c. The estimated SD of 2 is s_2 = .0004835. Does the quadratic predictor belong in the regression model?

d. The estimated SD of 1 is .4050. Use this and the information in (c) to obtain joint CIs for the linear and quadratic regression coefficients with a joint confidence level of (at least) 95%.

e. The estimated SD of Y.400 is 1.198. Calculate a 95% CI for true average free-flow when viscosity = 400 and also a 95% PI for free-flow resulting from a single observation made when viscosity = 400, and compare the intervals.

30. The article ”A Simulation-Based Evaluation of Three Cropping Systems on Cracking-Clay Soils in a Summer Rainfall Environment” (Agricultural Meteorology, 1976: 211-229) proposes a quadratic model for the relationship between water supply index (x) and farm wheat yield (y). Representative data and the resulting MINITAB output follow:

a. Interpret the value of the coefficient of multiple determination.

b. Calculate a 95% CI for the coefficient of the quadratic predictor.

c. The estimated standard deviation of 0 + 1x + 2x^2 when x = 2.5 is 53.5. Test H_0: μY•2.5 = 1500 versus Ha: μY•2.5 < 1500 using α = .01.

d. Obtain a 95% PI for wheat yield when the water supply index is 2.5 by using the information given in part (c).

31. The accompanying data was obtained from a study of a certain method for preparing pure alcohol from refinery streams (“Direct Hydration of Olefins,” Industrial and Eng. Chemistry, 1961: 209-211). The independent variable x is volume hourly space velocity, and the dependent variable y is the amount of conversion of iso-butylene.

a. Assuming that a quadratic probabilistic model is appropriate, estimate the regression function.

b. Determine the predicted values and residuals, and construct a residual plot. Does the plot look roughly as expected when the quadratic model is correct? Does the plot indicate that any observation has had a great influence on the fit? Does a scatter plot identify a point having large influence? If so, which point?

c. Obtain s^2 and R^2. Does the quadratic model provide a good fit to the data?

d. In Exercise 11, it was noted that the predicted value j and the residual Yj —j are independent of one another, so that σ^2 = V(Yj) = V(j) + V(Yj —j). A computer printout gives the estimated standard deviations of the predicted values as .955, .955, .712, .777, .777, .777, and 1.407. Use these values along with s^2 to compute the estimated standard deviation of each residual. Then compute the standardized residuals and plot them against x. Does the plot look much like the plot of part (b)? Suppose you had standardized the residuals using just s in the denominator. Would the resulting values differ much from the correct values?

e. Using information given in part (d), compute a 90% PI for isobutylene conversion when volume hourly space velocity is 4.

32. The following data is a subset of data obtained in an experiment to study the relationship between soil pH x and y = A1. Concentration/EC (“Root Responses of Three Gramineae Species to Soil Acidity in an Oxisol and an Ultisol,” Soil Science, 1973: 295-302):

A cubic model was proposed in the article, but the version of MINITAB used by the author of the present text refused to include the x^3 term in the model, stating that ”x^3 is highly correlated with other predictor variables.” To remedy this, = 4.3456 was subtracted from each x value to yield x’ = x—. A cubic regression was then requested to fit the model having regression function

y = exchangeable acidity/total cation exchange capacity (“Exchangeable Acidity in Unburnt Colliery Spoil,” Nature, 1969: 161):

Standardizing the independent variable x to obtain x’ = (x—)/sx and fitting the regression function y = + x’ + (x’)2 yielded the accompanying computer output.

a. Estimate μY .50.

b. Compute the value of the coefficient of multiple determination. (See Exercise 28(c).)

c. What is the estimated regression function 0 + 1x + 2x^2 using the unstandardized variable x?

d. What is the estimated standard deviation of β_2 computed in part (c)?

e. Carry out a test using the standardized estimates to decide whether the quadratic term should be retained in the model. Repeat using the unstandardized estimates. Do your conclusions differ?

35. The article ”The Respiration in Air and in Water of the Limpets Patella caerulea and Patella lusitanica” (Comp. Biochemistry and Physiology, 1975: 407—U1) proposed a simple power model for the relationship between respiration rate y and temperature x for P. caerulea in air. However, a plot of ln(y) versus x exhibits a curved pattern. Fit the quadratic power model Y = αeβx+γx^2 • ∈ to the accompanying data.

13.4 Multiple Regression Analysis

In multiple regression, the objective is to build a probabilistic model that relates a dependent variable y to more than one independent or predictor variable. Let k represent the number of predictor variables (k ≥ 2) and denote these predictors by x_1, x_2, …, xk. For example, in attempting to predict the selling prer of rooms.

DEFINITION

The general additive multiple regression model equation is

where E(∈) = 0 and V(∈) = σ^2. In addition, for purposes of testing hypotheses and calculating CIs or PIs, it is assumed that ∈ is normally distributed.

Let , , …, be particular values of x_1, …, xk. Then (13.15) implies that

Thus just as β_0 + β_1x describes the mean Y value as a function of x in simple linear regression, the true (or population) regression function β_0 + β_1x_1 + •••+ βkxk gives the expected value of Y as a function of x_1, …, xk. The βi s are the true (or population) regression coefficients. The regression coefficient β_1 is interpreted as the expected change in Y associated with a 1-unit increase in x_1 while x_2, …, xk are held fixed. Analogous interpretations hold for β_2, …, βk.

Models with Interaction and Quadratic Predictors

If an investigator has obtained observations on y, x_1, and x_2, one possible model is Y = β_0 + β_1x_1 + β_2x_2 + ∈. However, other models can be constructed by forming predictors that are mathematical functions of x_1 and/or x_2. For example, with and x_4 = x_1x_2, the model

Y = β_0 + β_1x_1 + β_2x_2 + β_3x_3 + β_4x_4 + ∈

has the general form of (13.15). In general, it is not only permissible for some predictors to be mathematical functions of others but also often highly desirable in the sense that the resulting model may be much more successful in explaining variation in y than any model without such predictors. This discussion also shows that polynomial regiv height= Mression is indeed a special case of multiple regression. For example, the quadratic model Y = β_0 + β_1x + β_2x^2 + ∈ has the form of (13.15) with k = 2, x_1 = x, and x_2 = x^2.

For the case of two independent variables, x 1 and x 2, there are four useful multiple regression models.

1. The first-order model:

Y = β_0 + β_1x_1 + β_2x_2 + ∈

2. The second-order no-interaction model:

3. The model with first-order predictors and interaction:

Y = β_0 + β_1x_1 + β_2x_2 + β_3x_1x + ∈

4. The complete second-order or full quadratic model:

Understanding the differences among these models is an important first step in building realistic regression models from the independent variables under study.

The first-order model is the most straightforward generalization of simple linear regression. It states that for a fixed value of either variable, the expected value of Y is a linear function of the other variable and that the expected change in Y for a unit increase in x_1 (x_2) is β_1 (β_2) independent of the level of x_2 (x_1). Thus if we graph the regression function as a function of x_1 for several different values of x_2, we obtain as contours of the regression function a collection of parallel lines, as pictured in Figure 13.13(a). The function y = β_0 + β_1x_1 + β_2x_2 specifies a plane in three-dimensional space; the firstorder model says that each observed value of the dependent variable corresponds to a point which deviates verAccording to the second-order no-interaction model, if x_2 is fixed, the expected change in Y for a 1-unit increase in x_1 is

Because this expected change does not depend on x_2, the contours of the regression function for different values of x_2 are still parallel to one another. However, the dependence of the expected change on the value of x_1 means that the contours are now curves rather than straight lines. This is pictured in Figure 13.13(b). In this case, the regression surface is no longer a plane in three-dimensional space but is instead a curved surface.

The contours of the regression function for the first-order interaction model are nonparallel straight lines. This is because the expected change in Y when x_1 is increased by 1 is

β_0 + β_1(x_1 + 1) + β_2x_2 + β_3(x_1 + 1)x_2—(β_0 + β_1x_1 + β_2x_2 + β_3x_1_2) = β_1 + β_3x_2

This expected change depends on the value of x 2, so each contour line must have a different slope, as in Figure 13.13(c). The word interaction reflects the fact that an expected change in Y when one variable increases in value depends on the value of the other variable.

Finally, for the complete second-order model, the expected change in Y when x_2 is held fixed while x_1 is increased by 1 unit is β_1 + β_3 + 2β_3x_1+ β_5x_2, which is a function of both x_1 and x_2. This implies that the contours of the regression function are both curved and not parallel to one another, as illustrated in Figure 13.13(d).

Figure 13.13 Contours of four different regression functions

Similar considerations apply to models constructed from more than two independent variables. In general, the presence of interaction terms in the model implies that the expected change in Y depends not only on the variable being increased or decreased but also on the values of some of the fixed variables. As in ANOVA, it is possible to have higher-way interaction terms (e.g., x_1x_2x_3), making model interpretation more difficult.

Models with Predictors for Categorical Variables

Thus far we have explicitly considered the inclusion of only quantitative (numerical) predictor variables in a multiple regression model. Using simple numerical coding, qualitative (categorical) variables, such as bearing material (aluminum or copper/ lead) or type of wood (pine, oak, or walnut), can also be incorporated into a model. Let’s first focus on the case of a dichotomous variable, one with just two possible categories—male or female, U.S. or foreign manufacture, and so on. With any such variable, we associate a dummy or indicator variable x whose possible values 0 and 1 indicate which category is relevant for any particular observation.

Example 13.11

The article ”Estimating Urban Travel Times: A Comparative Study” (Trans. Res., 1980: 173-175) described a study relating the dependent variable y = travel time between locations in a certain city and the independent variable x 2 = distance between locations. Two types of vehicles, passenger cars and trucks, were used in the study. Let

One possible multiple regression model is

Y = β_0 + β_1x_1 + β_2x_2 + ∈

The mean value of travel time depends on whether a vehicle is a car or a truck:

The coefficient β_1 is the difference in mean times between trucks and cars with distance held fixed; if β_1 > 0, on average it will take trucks longer to traverse any particular distance than it will for cars.

A second possibility is a model with an interaction predictor:

Y = β_0 + β_1x_1 + β_2x_2 + β_3x_1x_1 + ∈

Now the mean times for the two types of vehicles are

Figure 13.14 Regression fudiv height="1em">

For each model, the graph of the mean time versus distance is a straight line for either type of vehicle, as illustrated in Figure 13.14. The two lines are parallel for the first (no-interaction) model, but in general they will have different slopes when the second model is correct. For this latter model, the change in mean travel time associated with a 1-mile increase in distance depends on which type of vehicle is involved—the two variables ”vehicle type” and ”travel time” interact. Indeed, data collected by the authors of the cited article suggested the presence of interaction.

You might think that the way to handle a three-category situation is to define a single numerical variable with coded values such as 0, 1, and 2 corresponding to the three categories. This is incorrect, because it imposes an ordering on the categories that is not necessarily implied by the problem context. The correct approach to incorporating three categories is to define two different dummy variables. Suppose, for example, that y is the lifetime of a certain cutting tool, x 1 is cutting speed, and that there are three brands of tool being investigated. Then let

When an observation on a brand A tool is made, x_2 = 1 and x_3 = 0, whereas for a brand B tool, x_2 = 0 and x_3 = 1. An observation made on a brand C tool has x_2 = x_3 = 0, and it is not possible that x_2 = x_3 = 1 because a tool cannot simultaneously be both brand A and brand B. The no-interaction model would have only the predictors x_1, x_2, and x_3. The following interaction model allows the mean change in lifetime associated with a 1-unit increase in speed to depend on the brand of tool:

Y = β_0 + β_1x_1 + β_2x_2 + β 3x_3 + β_4x_1x_2 + β 5x_1x_3 + ∈

Construction of a picture like Figure 13.14 with a graph for each of the three possible (x_2, x_3) pairs gives three nonparallel lines (unless β_4 = β_5 = 0).

More generally, incorporating a categorical variable with c possible categories into a multiple regression model requires the use of c - 1 indicator variables (e.g., five brands of tools would necessitate using four indicator variables). Thus even one categorical variable can add many predictors tneering Problem Solving by Stephen Vardeman, an excellent exposition of the territory covered by our book, albeit at a somewhat higher level.

A statistical computer package gave the following least squares estimates:

0 = -37.48 1 = .2117 2 = .4983 3 = .1297 4 = .2583

Thus we estimate that .1297 gm is the average change in strength associated with a 1-degree increase in temperature when the other three predictors are held fixed; the other estimated coefficients are interpreted in a similar manner.

The estimated regression equation is

y = -37.48 + .2117x_1 + .4983x_2 + .1297x_3 + .2583x_4

A point prediction of strength resulting from a force of 35 gm, power of 75 mW, temperature of 200° degrees, and time of 20 msec is

= -37.48 + (.2117)(35) + (.4983)(75) + (.1297)(200) + (.2583)(20)

= 38.41 gm

This is also a point estimate of the mean value of strength for the specified values of force, power, temperature, and time.

2 and R^2

Substituting the values of the predictors from the successive observations into the equation for the estimated regression gives the predicted or fitted values 1, 2, …, n. For example, since the values of the four predictors for the last observation in Example 13.12 are 35, 75, 200, and 20, respectively, the corresponding predicted value is 30 = 38.41. The residuals are the differences y_1 − 1, …, yn - n. The last residual in Example 13.12 is 40.3 − 38.41 = 1.89. The closer the residuals are to 0, the better the job our estimated equation is doing in predicting the y values corresponding to values of the predictors in our sample.

SSE = ∑(yj, − j,)2 = ∑[yj - (0 + 1x1j + •••+ kxkj)]2

An efficient calculation formula for SSE is employed by most statistical computer packages. Because k + 1 parameters (β_0, β_1, …, βk) have been estimated, k + 1 df are lost, so n − (k + 1) df is associated with SSE, and

With SST = ∑(yi - )2, the proportion of total variation explained by the multiple regression model is R^2 = 1 - SSE/SST, the coefficient of multiple determination. As in polynomial regression, R^2 is often adjusted for the number of parameters in the model by the formula

The positive square root of the coefficient of multiple determination is called the multiple correlation coefficient R. It can be shown that R is the sample correlation coefficient r between the observed y s and the predictedy s (i.e., using xj = j in the formula for r results in r = R).

Example 13.13

Investigators carried out a study to see how various characteristics of concrete are influenced by x_1 = % limestone powder and x_2 = water-cement ratio, resulting in the accompanying data (“Durability of Concrete with Addition of Limestone Powder,” Magazine of Concrete Research, 1996: 131-137).

Consider first compressive strength as the dependent variable y . Fitting the first-order model results in

y = 84.82 + .1643 x_1—79.67x_2 SSE = 72.25 (df = 6) R 2 = .741

whereas including an interaction predictor gives

y = 6.22 + 5.779x_1 + 51.33x_2—9.357x_1>Based on this latter fit, a prediction for compressive strength when % limestone = 14 and water-cement ratio = .60 is

= 6.22 + 5.779(14) + 51.33(.60)—9.357(8.4) = 39.32

Fitting the full quadratic relationship results in virtually no change in the R 2 value. However, when the dependent variable is adsorbability, the following results are obtained: R 2 = .747 when just two predictors are used, .802 when the interaction predictor is added, and .889 when the five predictors for the full quadratic relationship are used.

In general, 1 can be interpreted as an estimate of the average change in y associated with a 1-unit increase in xi while values of all other predictors are held fixed. Sometimes, though, it is difficult or even impossible to increase the value of one predictor while holding all others fixed. In such situations, there is an alternative interpretation of the estimated regression coefficients. For concreteness, suppose that k = 2, and let 1 denote the estimate of β_1 in the regression of y on the two predictors x_1 and x_2. Then

1. Regress y against just x_2 (a simple linear regression) and denote the resulting residuals by g_1, g_2, …, gn . These residuals represent variation in y after removing or adjusting for the effects of x_2.

2. Regress x_1 against x_2 (that is, regard x_1 as the dependent variable and x_2 as the independent variable in this simple linear regression), and denote the residuals by f_1, …, fn . These residuals represent variation in x_1 after removing or adjusting for the effects of x_2.

Now consider plotting the residuals from the first regression against those from the second; that is, plot the pairs (f_1, g_1), …, (fn , gn ). The result is called a partial residual plot or adjusted residual plot . If a regression line is fit to the points in this plot, the slope turns out to be exactly 1 (furthermore, the residuals from this line are exactly the residuals e_1, …, en from the multiple regression of y on x_1 and x_2). Thus 1 can be interpreted as the estimated changes in y associated with a 1-unit increase in x_1 after removing or adjusting for the effects of any other model predictors. The same interpretation holds for gressed against all predictors other than x_1 in Step 1 and x_1 is regressed against the other k—1 predictors in Step 2).

As an example, suppose that y is the sale price of an apartment building and that the predictors are number of apartments, age, lot size, number of parking spaces, and gross building area (ft^2). It may not be reasonable to increase the number of apartments without also increasing gross area. However, if 5 = 16.00, then we can say that a $16 increase in sale price is associated with each extra square foot of gross area after adjusting for the effects of the other four predictors.

A Model Utility Test

With multivariate data, there is no preliminary picture analogous to a scatter plot to indicate whether a particular multiple regression model will be judged useful. The value of R 2 certainly communicates a preliminary message, but this value is sometimes deceptive because it can be greatly inflated by using a large number of predictors (large k ) relative to the sample size n (this is the rationale behind adjusting R 2).

The model utility test in simple linear regression involved the null hypothesis H_0: β_1 = 0, according to which there is no useful relation between y and the single predictor x . Here we consider the assertion that β_1 = 0, β_2 = 0, …, βk = 0, which says that there is no useful relationship between y and any of the k predictors. If at least one of these β s is not 0, the corresponding predictor(s) is (are) useful. The test is based on a statistic that has a particular F distribution when H_0 is true.

Null hypothesis: H_0: β_1 = β_2 = ··· = βk = 0

Alternative hypothesis: Ha: at least one βi ≠ 0 (i = 1, …, k )

where SSR = regression sum of squares = SST –SSE

Rejection region for a level α test: f ≥ Fa, k, n (k + 1)

Except for a constant multiple, the test statistic here is R 2/(1—R 2), the ratio of explained to unexplained variation. If the proportion of explained variation is high relative to unexplained, we would naturally want to reject H_0 and confirm the utility of the model. However, if k is lar s is quite complicated, but all standard regression computer packages compute and show the si . Inferences concerning a single βi are based on the standardized variable

which has a t distribution with n—(k + 1) df.

The point estimate of , the expected value of Y when . The estimated standard deviation of the corresponding estimator is again a complicated expression involving the sample xij s. However, the better statistical computer packages will calculate it on request. Inferences about are based on standardizing its estimator to obtain a t variable having n—(k + 1) df.

1. A 100(1—α )% CI for βi, the coefficient of xi in the regression function, is

i ± ta /2, n –(k + 1) · St

2. A test for H_0: βi = βi 0 uses the t statistic value t = (i —βi 0)/si based on n—(k + 1) df. The test is upper-, lower-, or two-tailed according to whether Ha contains the inequality >, <, or ≠.

3. A 100(1—α )% CI for is

± ta /2, n –(k + 1) · {estimated SD of } = ± ta /2, n –(k + 1) · S

where is the statistic and y is the calculated value of .

4. A 100(1—α )% PI for a future y value is

± ta /2, n –(k + 1) · {d-order predictors justified? That is, should the reduced model consisting of just the predictors x_1, x_2, x_3, and x_4(l = 4) be used? Output resulting from fitting the two models follows:

The hypotheses to be tested are

H_0: β_5 = β_6 = · · · = β_14 = 0

versus

Ha: at least one among β_5, …, β_14 is not 0

With k = 14 and l = 4, the F critical value for a test with α = .01 is F.01,10,15 = 3.80. The test statistic value is

Since 4.36 ≥ 3.80, H_0 is rejected. We conclude that the appropriate model should include at least one of the second-order predictors.

Assessing Model Adequacy

The standardized residuals in multiple regression result from dividing each residual by its estimated standard deviation; the formula for these standard deviations is substantially more complicated than in the case of simple linear regression. We recommend a normal probability plot of the standardized residuals as a basis for validating the normality assumption. Plots of the standardized residuals versus each predictor and versus y should show no discernible pattern. Adjusted residual plots can also be helpful in this endeavor. The book by Neter et al. is an extremely useful reference.

Example 13.17

Figure 13.16 shows a normal probability plot of the standardized residuals for the adsorption data and fitted model given in Example 13.15. The straightness of the plot casts little doubt on the assumption that the random deviation e is normally distributed.

Figure 13.16 A normal probability plot of the standardized residuals for the data and model of Example 13.15

Figure 13.17 shows the other suggested plots for the adsorption data. Given that there are only 13 observations in the data set, there is not much evidence of a pattern in any of the first three plots other than randomness. The point at the bottom of each of these three plots corresp?

e. When x_1 = 10 and x_2 = 25, = 42.253 and S = .350. Calculate and interpret a confidence interval and a prediction interval.

f. Including an interaction predictor x_3 gives 3 = -.04304 and S_3.01773. At significance level .01, should this predictor be retained?

45. The article “Analysis of the Modeling Methodologies for Predicting the Strength of Air-Jet Spun Yarns” (Textile Res. J ., 1997: 39–44) reported on a study carried out to relate yarn tenacity (y , in g/tex) to yarn count (x_1, in tex), percentage polyester (x_2), first nozzle pressure (x_3, in kg/cm^2), and second nozzle pressure (x_4, in kg/cm^2). The estimate of the constant term in the corresponding multiple regression equation was 6.121. The estimated coefficients for the four predictors were –.082, .113, .256, and –.219, respectively, and the coefficient of multiple determination was .946.

SPSS output for Exercise 44

a. Assuming that the sample size was n = 25, state and test the appropriate hypotheses to decide whether the fitted model specifies a useful linear relationship between the dependent variable and at least one of the four model predictors.

b. Again using n = 25, calculate the value of adjusted R 2.

c. Calculate a 99% confidence interval for true mean yarn tenacity when yarn count is 16.5, yarn contains 50% polyester, first nozzle pressure is 3, and second nozzle pressure is 5 if the estimated standard deviation of predicted tenacity under these circumstances is .350.

46. A regression analysis carried out to relate y = repair time for a water filtration system (hr) to x_1 = elapsed time since the previous service (months) and x_2 = type of repair (1 if electrical and 0 if mechanical) yielded the following model based on n = 12 observations: y = .950 + .400x_1 + 1.250x_2. In addition, SST = 12.72, SSE = 2.09, and S_2 = .312.

a. Does there appear to be a useful linear relationship between repair time and the two model predictors? Carry out a test of the appropriate hypotheses using a significance level blockquote>b. Given that elapsed time since the last service remains in the model, does type of repair provide useful information about repair time? State and test the appropriate hypotheses using a significance level of .01.

c. Calculate and interpret a 95% CI for β_2.

d. The estimated standard deviation of a prediction for repair time when elapsed time is 6 months and the repair is electrical is .192. Predict repair time under these circumstances by calculating a 99% prediction interval. Does the interval suggest that the estimated model will give an accurate prediction? Why or why not?

47. Efficient design of certain types of municipal waste incinerators requires that information about energy content of the waste be available. The authors of the article “Modeling the Energy Content of Municipal Solid Waste Using Multiple Regression Analysis” (J. of the Air and Waste Mgmt. Assoc, 1996: 650-656) kindly provided us with the accompanying data on y = energy content (kcal/kg), the three physical composition variables x_1 = % plastics by weight, x_2 = % paper by weight, and x_3 = % garbage by weight, and the proximate analysis variable x_4 = % moisture by weight for waste specimens obtained from a certain region.

Using MINITAB to fit a multiple regression model with the four aforementioned variables as predictors of energy content resulted in the following output:

a. Interpret the values of the estimated regression coefficients β_1 and β_4.

b. State and test the appropriate hypotheses to decide whether the model fit to the data specifies a useful linear relationship between energy content and at least one of the four predictors.

c. Given that % plastics, % paper, and % water remain in the model, does % garbage provide useful information about energy content? State and test the appropriate hypotheses using a significance level of .05.

d. Use the fact that S = 7.46 when x_1 = 20, x_2 = 25, x_3 = 40, and x_4 = 45 to calculate a 95% confidence interval for true average energy content under these circumstances. Does the resulting interval suggest that mean energy content hpecified characteristics in a way that conveys information about precision and reliability.

48. An experiment to investigate the effects of a new technique for degumming of silk yarn was described in the article “Some Studies in Degumming of Silk with Organic Acids” (J. Society of Dyers and Colourists, 1992: 79-86). One response variable of interest was y = weight loss (%). The experimenters made observations on weight loss for various values of three independent variables: x_1 = temperature (°C) = 90, 100, 110; x_2 = time of treatment (min) = 30, 75, 120; x_3 = tartaric acid concentration (g/L) = 0, 8, 16. In the regression analyses, the three values of each variable were coded as -1, 0, and 1, respectively, giving the accompanying data (the value y_8 = 19.3 was reported, but our value y_8 = 20.3 results in regression output identical to that appearing in the article).

A multiple regression model with k = 9 predictors—x_1, x_2, x_3, x_4 = , x_5 = , x_6 = , x_7 = x_1x_2, x_8 = x_1x_3, and x_9 = x_2x_3—was fit to the data, resulting in 0 = 21.967, 1 = 2.8125, 2 = 1.2750, 3 = 3.4375, 4 = –2.208, 5 = 1.867, 6 = –4.208, 7 = –.975, 8 = –3.750, 9 = –2.325, SSE = 23.379, and R 2 = .938.

a. Does this model specify a useful relationship? State and test the appropriate hypotheses using a significance level of .01.

b. The estimated standard deviation of μ when x_1 = · · · = xg = 0 (i.e., when temperature = 100, time = 75, and concentration = 8) is 1.248. Calculate a 95% CI for expected weight loss when temperature, time, and concentration have the specified values.

c. Calculate a 95% PI for a single weight-loss value to be observed when temperature, time, and concentration have values 100, 75, and 8, res712; satisfies t 2 = f , where t is the value of the t statistic from part (f).

50. a. When the model is fit to the hops data of Exercise 49, the estimate of β_5 is 5 = .557 with estimated standard deviation S 5 = .94. Test H_0: 5 = 0 versus Ha: β_5 ≠ 0.

b. Each t -ratio i /S_5 (i = 1, 2, 3, 4, 5) for the model of part (a) is less than 2 in absolute value, yet R 2 = .861 for this model. Would it be correct to drop each term from the model because of its small t -ratio? Explain.

c. Using R 2 = .861 for the model of part (a), test H_0: β_3 = β_4 = β_5 = 0 (which says that all second-order terms can be deleted).

51. The article “The Undrained Strength of Some Thawed Permafrost Soils” (Canadian Geotechnical J., 1979: 420–427) contains the following data on undrained shear strength of sandy soil (y , in kPa), depth (x_1 in m), and water content (x_2, in %).

The predicted values and residuals were computed by fitting a full quadratic model, which resulted in the estimated regression function

a. Do plots of e * versus x_1, e * versus x_2, and e * versus suggest that the full quadratic model should be modified? Explain your answer.

b. The value of R 2 for the full quadratic model is .759. Test at level .05 the null hypothesis stating that there is no linear relationship between the dependent variable and any of the five predictors.

c. It can be shown that V (Y ) = σ 2 = V ( ) + V (Y— ). The estimate of σ is img> (i.e., ) when x_1 = 8.0 and x_2 = 33.1. Finally, compute a 95% CI for mean strength. [Hint: What is (y— )/e*?

d. Fitting the first-order model with regression function μY·x_1·x_2 = β_0 + β_1x_1 + β_2x_2 results in SSE = 894.95. Test at level .05 the null hypothesis that states that all quadratic terms can be deleted from the model.

52. Utilization of sucrose as a carbon source for the production of chemicals is uneconomical. Beet molasses is a readily available and low-priced substitute. The article “Optimization of the Production of ¿8-Carotene from Molasses by Blakeslea Trispora (J. of Chemical Technology and Biotechnology, 2002: 933-943) carried out a multiple regression analysis to relate the dependent variable y = amount of ¿8-carotene (g/dm^3) to the three predictors amount of lineolic acid, amount of kerosene, and amount of antioxidant (all g/dm^3).

a. Fitting the complete second-order model in the three predictors resulted in R 2 = .987 and adjusted R 2 = .974, whereas fitting the first-order model gave R 2 = .016. What would you conclude about the two models?

b. For x_1 = x_2 = 30, x_3 = 10, a statistical software package reported that y = .66573, S = .01785 based on the complete second-order model. Predict the amount of β -carotene that would result from a single experimental run with the designated values of the independent variables, and do so in a way that conveys information about precision and reliability.

53. Snowpacks contain a wide spectrum of pollutants that may represent environmental hazards. The article “Atmospheric PAH Deposition: Deposition Velocities and Washout Ratios” (J. of Environmental Engineering, 2002: 186-195) focused on the deposition of polyaromatic hydrocarbons. The authors proposed a multiple regression model for relating deposition over a specified time period (y, in μ g/m^2) to two rather complicated predictors x_1 (μ g-sec/m^2) and x_2 (μ g/m^2) defined in terms of PAH air concentrations for various species, total time, and total amount of precipitation. Here is data on the species fluoranthene and corresponding MINITAB output:

54. The accompanying data resulted from a study of the relationship between brightness of finished paper (y ) and the variables H_2O_2% by weight (x_1), NaOH% by weight (x_2), silicate % by weight (x_3), and process temperature (x_4) (“Advantages of CEHDP Bleaching for High Brightness Kraft Pulp Production,” TAPPI, 1964: 170A–173A). Each independent variable was allowed to assume five different values, and these values were coded for regression analysis as –2, –1, 0, 1, and 2.

a. When a (coded) model involving all linear terms, all quadratic terms, and all cross-product terms was fit, the estimated regression function was

Use this estimated model to predict brightness when H_2O_2 is .4%, NaOH is .4%, silicate is 3.5%, and temperature is 175. What are the values of the residuals for these values of the variables?

b. Express the estimated regression function in uncoded form.

c. SST = 17.2567, and R 2 for the model of part (a) is .885. When a model that includes only the four linear terms is fit, the resulting value of R 2 is .721. State and test at level .05 the null hypothesis that specifies that the coefficients of all quadratic and cross-product terms in the regression function are zero.

d. The estimated (coded) regression function when only linear terms are included is = 84.5548 + .6500x_1—.2583x_2 + .1333x_3 + .1083x_4. When x_1 = x_2 = x_3 = x_4 = 0, the estimated standard deviation of is .0772. Suppose it had been believed that expected brightness for these values of the xis was at least 85.0. Does the given information contradict this belief? State and test the appropriate hypotheses.

13.5 Other Issues in Multiple Regression

In this section, we touch upon a number of issues that may arise when a multiple regression analysis is carried out. Consult the chapter references for a more extensive treatment of any particular topic.

Transformations in Multiple Regression

Often, theoretical considerations suggest a nonlinear relation between ahe model.

Example 13.18

An article in Lubrication Eng .(“Accelerated Testing of Solid Film Lubricants,” 1972: 365-372) reports on an investigation of wear life for solid film lubricant. Three sets of journal bearing tests were run on a Mil-L-8937-type film at each combination of three loads (3000, 6000, and 10,000 psi) and three speeds (20, 60, and 100 rpm), and the wear life (hours) was recorded for each run, as shown in Table 13.7.

Table 13.7 Wear-life Data for Example 13.18

The article contains the comment that a lognormal distribution is appropriate for W, since ln(W ) is known to follow a normal law (recall from Chapter 4 that this is what defines a lognormal distribution). The model that appears is W = (c/salb ) · ∈ , from which ln(W ) = ln(c )—a ln(s )—b ln(l ) + ln(∈ ); so with Y = ln(W ), x_1 = ln(s ), x_2 = ln(l ), β_0 = ln(c ), β_1 =—a , and β_2 =—b , we have a multiple linear regression model. After computing ln(wi ), ln(si ), and ln(li) for the data, a first-order model in the transformed variables yielded the results shown in Table 13.8.

Table 13.8 Estimated Coefficients and t -ratios for Example 13.18

The coefficient of multiple determination (for the transformed fit) has value R 2 = .781. The estimated regression function for the transformed variables is

ln(w ) = 10.87—1.21 ln(s )—1.40 ln(l )

so that the o full second-order model requires solving the system of six normal equations in six unknowns. A computer analysis yielded 0 = 40.2660, 1 = –13.4041, 2 = 10.2553, 3 = 2.3313, 4 = –2.3405, and 5 = 2.5978. The estimated regression equation is then

Thus if x_1 = 3200 and x_2 = 57.0, = .539, = .211, = (.539)2 = .2901, = (–.211)2 = .0447, and = (.539)(–.211) = –.1139, so

= 40.27—(13.40)(.539) + (10.26)(—.211) + (2.33)(.2901)—(2.34)(.0447) + (2.60)(–.1139) = 31.16

Variable Selection

Often an experimenter will have data on a large number of predictors and then wish to build a regression model involving a subset of the predictors. The use of the subset will make the resulting model more manageable, especially if more data is to be subsequently collected, and also result in a model that is easier to interpret and understand than one with many more predictors. Two fundamental questions in connection with variable selection are the following:

1. If we can examine regressions involving all possible subsets of the predictors for which data is available, what criteria should be used to select a model?

2. If the number of predictors is too large to permit all regressions to be examined, is there a way of examining a reduced number of subsets among which a good model (or models) will be found?

To address Question (1) first, if the number of predictors is small (≤ 5, say), then it would not be too tedious to examine all possible regressions using any one of the readily available statistical computer packages. If data on at least six predictors is available, all possible regressions involve at least 64 (= 2^6) different models. Several packages will, for any specified m between 1 and 10, give the m best one-predictor models, the m best two-predictor models, and so on (“best” here means smallest SSE or, equivalently, largest R 2). MINITAB will do this for as many as 20 predictors, whereas BMDP will handle up to 27. The corresponding SSEs (or functions of them) can then be compared according to any of the criteria described next. The reason for specifying an m greater than 1 is to Selection Again we use a subscript k to denote a quantity (say, SSEk ) computed from a model with k predictors (and thus k + 1 /i s, because β_0 will always be included). For a fixed value of k, it is reasonable to identify the best model as the one having minimum SSEk . The more difficult issue concerns comparison of SSEk s for different values of k . Three different criteria, each one a simple function of SSEk , are widely used.

1. , the coefficient of multiple determination for a k -predictor model. Because will virtually always increase as k does (and can never decrease), we are not interested in the k that maximizes . Instead we wish to identify a small k for which is nearly as large as R 2 for all predictors in the model.

2. MSEk = SSE/(n—k—1), the mean squared error for a k -predictor model. This is often used in place of , because although never decreases with increasing k, a small decrease in SSEk obtained with one extra predictor can be more than offset by a decrease of 1 in the denominator of MSEk . The objective is then to find the model having minimum MSEk . Since adjusted = 1—MSEk /MST, where MST = SST/(n—1) is constant in k, examination of adjusted is equivalent to consideration of MSEk .

3. The rationale for the third criterion, Ck , is more difficult to understand, but the criterion is widely used by data analysts. Suppose the true regression model is specified by m predictors—that is,

Y = β_0 + β_1x_1 + · · · + βmxm + ∈ V (∈ ) = σ 2

so that

E (Y ) = β_0 + β_1x_1 + · · · + βmxm

Consider fitting a model by using a subset of k of these m predictors; for simplicity, suppose we use x_1, x_2, …, xk . Then by solving the system of normal equations, estimates 0, 1, …,k are obtained (but not, of course, estimates of any β’ s corresponding toem> (Y ) can then be estimated by = 0 = 1x_1 + · · · + k xk . Now consider the normalized expected total error of estimation

The second equality in (13.21) must be taken on faith because it requires a tricky expected-value argument. A particular subset is then appealing if its Γk value is small. Unfortunately, though, E (SSEk ) and σ 2 are not known. To remedy this, let s 2 denote the estimate of σ 2 based on the model that includes all predictors for which data is available and define

A desirable model is then specified by a subset of predictors for which Ck is small.

Example 13.20

The review article by Ron Hocking listed in the chapter bibliography reports on an analysis of data taken from the 1974 issues of Motor Trend magazine. The dependent variable y was gas mileage, there were n = 32 observations, and the predictors for which data was obtained were x_1 = engine shape (1 = straight and 0 = V), x_2 = number of cylinders, x_3 = transmission type (1 = manual and 0 = auto), x_4 = number

of transmission speeds, x_5 = engine size, x_6 = horsepower, x_7 = number of carburetor barrels, x_8 = final drive ratio, x_9 = weight, and x_10 = quarter-mile time. In Table 13.10, we present summary information from the analysis. The table describes for each k the subset having minimum SSEk ; reading down the variables column indicates which variable is added in going from k to k + 1 (in going from k = 2 to k = 3, both x_3 and x_10 are added, and x_2is deleted). Figure 13.18 contains plots of , adjusted , and Ck against k; these plots are an important visual aid in selecting a subset. The estimate of σ^2 is s^2 = 6.24, which is MSE_10. A simple model that rates highly according to all criteria is the one containing predictors x_3, x_9, and x_10.

Table 13.10 Best Subsets for Gas Mileage Data of Example 13.20

ors then specify the model. The value tin = 2 is often used for the same reason that tout = 2 is used in BE. For the tar-content data, FS resulted in the sequence of models given in Steps 5, 4, …, 1 in Table 13.11 and thus is in agreement with BE. This will not always be the case.

The stepwise procedure most widely used is a combination of FS and BE, denoted by FB. This procedure starts as does forward selection, by adding variables to the model, but after each addition it examines those variables previously entered to see whether any is a candidate for elimination. For example, if there are eight predictors under consideration and the current set consists of x_2, x_3, x_5, and x_6 with x_5having just been added, the t -ratios 2/s_2, 3/s_3, and 6/s_6 are examined. If the smallest absolute ratio is less than tout, then the corresponding variable is eliminated from the model. The idea behind FB is that with forward selection, a single variable may be more strongly related to y than either of two or more other variables individually, but the combination of these variables may make the single variable subsequently redundant. This actually happened with the gas mileage data discussed in Example 13.20, with x_2 entering and subsequently leaving the model.

The FB procedure is part of several standard computer packages. The BMDP package specifies tin = 2 and (most packages actually use f = t 2 rather than t itself).

Although in most situations these automatic selection procedures will identify a good model, there is no guarantee that the best or even a nearly best model will result. Close scrutiny should be given to data sets for which there appear to be strong relationships among some of the potential predictors; we will say more about this shortly.

Identification of Influential Observations

In simple linear regression, it is easy to spot an observation whose x value is much larger or much smaller than other x values in the sample. Such an observation may have a great impact on the estimated regression equation (whether it actually does depends on how consistent the corresponding y value is to the remainder of the data). In multiple regression, it is also desirable to know whether the values of the predictors for a particular observation are such that it has the potential for exerting great influence on the estimated equation. One method for identifying potentially influential observations relies on the fact that because each i is a linear function of y_1, y_2,…, yn , each predicted y value of the standard deviation of the coefficient:

There exist efficient computational formulas that allow all this information to be obtained from the “no-deletion” regression, so that the additional n regressions are unnecessary.

Example 13.24 (Example 13.23 continued)

Consider separately deleting observations 1 and 6, whose residuals are the largest, and observation 4, where hjj is large. Table 13.12 contains the relevant information.

Table 13.12 Changes in Estimated Coefficients for Example 13.24

For deletion of both point 1 and point 4, the change in each estimate is in the range 1-1.5 standard deviations, which is reasonably substantial (this does not tell us what would happen if both points were simultaneously omitted). For point 6, however, the change is roughly .25 standard deviation. Thus points 1 and 4, but not 6, might well be omitted in calculating a regression equation.

Multicollinearity

In many multiple regression data sets, the predictors x_1, x_2,…, xk are highly interdependent. Suppose we consider the usual model

Y = β_0 + β_1x_1 + … + βk xk + ε

with data (x1j,…, xkj , yj) (j = 1, …, n ) available for fitting. If we use the principle of least squares to regress x_1 on the other predictors x_1,…, xi-1, xi + 1,…, xk , obtaining

i = a_0 + a_1x_1 + … + a_1xi-1 + ai+1xi+1 + … + akxk

it can be shown that

When the sample xi values can be predicted very well from the other predictor values, the denominator of (13.22) will be small, so V (i) will be qata is said to exhibit multicollinearity. Multicollinearity is often suggested by a regression computer output in which R^2 is large but some of the t -ratios j /si are small for predictors that, based on prior information and intuition, seem important. Another clue to the presence of multicollinearity lies in a i value that has the opposite sign from that which intuition would suggest, indicating that another predictor or collection of predictors is serving as a “proxy” for xi.

An assessment of the extent of multicollinearity can be obtained by regressing each predictor in turn on the remaining k –1 predictors. Let denote the value of R 2 in the regression with dependent variable xi and predictors x_1, …, xi -1, xi + 1, …, xk. It has been suggested that severe multicollinearity is present if > .9 for any i. MINITAB will refuse to include a predictor in the model when its value is quite close to 1.

There is unfortunately no consensus among statisticians as to what remedies are appropriate when severe multicollinearity is present. One possibility involves continuing to use a model that includes all the predictors but estimating parameters by using something other than least squares. Consult a chapter reference for more details.

EXERCISES Section 13.5 (55–64)

55. The article “Bank Full Discharge of Rivers” (Water Resources J., 1978: 1141-1154) reports data on discharge amount (q, in m^3/sec), flow area (a, in m^2), and slope of the water surface (b, in m/m) obtained at a number of floodplain stations. A subset of the data follows. The article proposed a multiplicative power model Q = ααβb γε

a. Use an appropriate transformation to make the model linear and then estimate the regression parameters for the transformed model. Finally, estimate α, β, and γ (the parameters of the original model). What would be your prediction of discharge amount when flow area is 10 and slope is .01?

b. Without actually doing any analysis, how would you fit a multiplicative exponential model Q = αeβaeγbε ?

c. After the transformation to linearity in part (a), avity data referred to in Exercise 56. The following R^2 values resulted from regressing each predictor on the other four predictors (in the first regression, the dependent variable was x_1 and the predictors were x_2-x_5, etc.): .628, .711, .341, .403, and .403. Does mul-ticollinearity appear to be a substantial problem? Explain.

62. A study carried out to investigate the relationship between a response variable relating to pressure drops in a screen-plate bubble column and the predictors x_1 = superficial fluid velocity, x_2 = liquid viscosity, and x_3 = opening mesh size resulted in the accompanying data (“A Correlation of

Data for Exercise 62

Two-Phase Pressure Drops in Screen-Plate Bubble Column,'' Canad. J. of Chem. Engr., 1993: 460-463). The standardized residuals and hii values resulted from the model with just x_1, x_2, and x_3 as predictors. Are there any unusual observations?

63. Multiple regression output from MINITAB for the PAH data of Exercise 53 in the previous section included the following information:

What does this suggest about the appropriateness of using the previously given fitted equation as a basis for inferences? The investigators actually eliminated observation #7 and re-regressed. Does this make sense?

64. Refer to the water-discharge data given in Exercise 55 and let y = ln(q ), x_1 = ln(a ), and x_2 = ln(b ). Consider fitting the model Y = β_0 + β_1x_1 + β_2x_2 + ∈.

a. The resulting hii s are .138, .302, .266, .604, .464, .360, .215, .153, .214, and .284. Does any observation appear to be influential?

b. The estimated coefficients are 0 = 1.5652, 1 = .9450, 2 = .1815, and the corresponding estimated standard deviations are s_0 = .7328, s_1 = .1528, and s_2 = .1752. The second standardized residual is = 2.19. When the second observation is omitted from the data set, the man">d. Predict flux when inverse thickness is 45 in a way that conveys information about precision and reliability.

e. Investigate model adequacnd y = palladium content (%) of the coating.

a. Fit the first-order model with five predictors and assess its utility. Do all the predictors appear to be important?

b. Fit the complete second-order model and assess its utility.

c. Does the group of second-order predictors (interaction and quadratic) appear to provide more useful information about y than is contributed by the first-order predictors? Carry out an appropriate test of hypotheses.

d. The authors of the cited article recommended the use of all five first-order predictors plus the additional predictor x_6 = (pH)2. Fit this model. Do all six predictors appear to be important?

72. The article “An Experimental Study of Resistance Spot Welding in 1 σm Thick Sheet of Low Carbon Steel” (J. of Engr. Manufacture, 1996: 341-348) discussed a statistical analysis whose basic aim was to establish a relationship that could explain the variation in weld strength (y) by relating strength to the process characteristics weld current (wc), weld time (wt), and electrode force (ef).

a. SST = 16.18555, and fitting the complete second-order model gave SSE = .80017. Calculate and interpret the coefficient of multiple determination.

b. Assuming that n = 37, carry out a test of model utility (the ANOVA table in the article states that n—(k + 1) = 1, but other information given contradicts this and is consistent with the sample size we suggest).

c. The given F ratio for the current-time interaction was 2.32. If all other predictors are retained in the model, can this interaction predictor be eliminated? [Hint: As in simple linear regression, an F ratio for a coefficient is the square of its t -ratio.]

d. The authors proposed eliminating two interaction predictors and a quadratic predictor and recommended the estimated equation y = 3.352 + .098wc + .222wt + .297ef—.0102(wt)2—.037(ef)2 + .0128(wc)(wt). Consider a weld current of 10 kA, a weld time of 12 ac cycles, and an electrode force of 6 kN. Supposing that the estimated standard deviation of the predicted strength in this situation is .0750, calculate a 95% PI for strength. Does the interval suggest that the value of strength can be accurately predicted?

73. The accompanying data on x = frequency (MHz) and y = output power (W) for a certain laser configuration was read from a graph in the article “Frequency Dependence in RF Dischacindex="00783" alt="Image">1 = 1.002, s_2 = .3073, and SSE = 26.98.

a. Is the quadratic model useful for describing the relationship between x and y ? [Hint: Quadratic regression is a special case of multiple regression with k = 2, x_1 = x and x_2 = x 2. Apply an appropriate procedure.

b. Should the quadratic predictor be eliminated?

c. Estimate expected height for wheat treated with 10 /aM of Mn using a 90% CI. [Hint: The estimated standard deviation of 0 + 1 + 2 is 1.031.]

76. The article “Chemithermomechanical Pulp from Mixed High Density Hardwoods” (TAPPI, July 1988: 145-146) reports on a study in which the accompanying data was obtained to relate y = specific surface area (cm^3/g) to x_1 = % NaOH used as a pretreatment chemical and x_2 = treatment time (min) for a batch of pulp.

The accompanying MINITAB output resulted from a request to fit the model Y = β_0 + β_1x_1 + β_2 x_2 + ∈.

a. What proportion of observed variation in specific surface area can be explained by the model relationship?

b. Does the chosen model appear to specify a useful relationship between the dependent variable and the predictors?

c. Provided that % NaOH remains in the model, would you suggest that the predictor treatment time be eliminated?

d. Calculate a 95% CI for the expected change in specific surface area associated with an increase of 1% in NaOH when treatment time is held fixed.

e. MINITAB reported that the estimated standard deviation of 0 + 1 (9) + 2(60) is .162. Calculate a prediction interval for the value of specific surface area to be observed when % NaOH = 9 and treatment time = 60.

77. A multiple regressiion analysis was carried out to relate y = tensile strength of a synthetic-fiber specimen to the variables x_1 = percent cotton and x_2 = drying time. The data set consisted of n = 12 observations.

a. The estimated coefficients were 0 = 180.00, 1 = 1.000, and 2 = 10.500. Calculate a point estimate of the expected tensile strength when percent cotton = 15 and drying time = 3.5.

b. Sums of squares were SST = 1210.30 and SSE = 117.40. What proportion of observed variation in tensile strength can be attributed to the model relationship?

c. Use the information in part (b) to decide whether the model with variables x_1 and x_2 specifies a useful relationship.

d. The estimated standard deviation of 0 + 1x_1 + 2 x_2 when x_1 = 18 and x_2 = 3.0 was 1.20. Calculate a 95% PI for tensile strength of a fabric specimen for which x1 = 18 and x_2 = 3.0.

78. A study was carried out to relate time to failure (y) of a certain machine component to the variables operating voltage (x_1), motor speed (x_2),and operating temperature (x_3). The resulting data set consisted of n = 20 observations. When the model with the three variables x_1, x_2 and x_3 was fit to the data, the value of error sum of squares was 8212.5. Fitting the second-order interaction model (with quadratic predictors and all products of pairs of variables) gave an error sum of squares of 5027.1. Should at least one of the quadratic or interaction predictors be retained in the model? State and test the relevant hypotheses.

79. The article “A Statistical Analysis of the Notch Toughness of 9% Nickel Steels Obtained from Production Heats” (J. of Testing and Eval, 1987: 355-363) reports on the results of a multiple regression analysis relating Charpy v-notch toughness y (joules) to the following variables: x_1 = plate thickness (mm), x_2 = carbon content (%), x_3 = manganese content (%), x_4 = phosphorus content (%), x_5 = sulphur content (%), x_6 = silicon content (%), x_7 = nickel content (%), x_8 = yield strength (Pa), and x_9 = tensile strength (Pa).

a. The best possible subsets involved ad>. The values of MSEk, and Ck are as follows:

Which model would you recommend? Explain the rationale for your choice.

b. The authors also considered second-order models involving predictors and xixj . Information on the best such models starting with the variables x_2, x_3, x_5, x_6, x_7, and x_8 is as follows (in going from the best four-predictor model to the best five-predictor model, x8 was deleted and both x2x6 and x7x8 were entered; x8 reentered at a later stage):

Which of these models would you recommend, and why? [Note: Models based on eight of the original variables did not yield marked improvement on those under consideration here.]

80. A sample of n = 20 companies was selected, and the values of y = stock price and k = 15 predictor variables (such as quarterly dividend, previous year’s earnings, and debt ratio) were determined. When the multiple regression model using these 15 predictors was fit to the data, R^2 = .90 resulted.

a. Does the model appear to specify a useful relationship between y and the predictor variables? Carry out a test using significance level .05. [Hint: The F critical value for 15 numerator and 4 denominator df is 5.86.]

b. Based on the result of part (a), does a high R^2 value by itself imply that a model is useful? Under what circumstances might you be suspicious of a model with a high R^2 value?

c. With n and k as given previously, how large would R 2 have to be for the model to be judged useful at the .05 level of significance?

81. Does exposure to air pollution result in decreased life expectancy? This question was examined in the article “Does Air Pollution Shorten Lives?” (Statistics and Public Policy, Reading, MA, Addison-Wesley, 1977). Data on

y = total mortality rate (deaths per 10,000)

x_1 = mean suspended particle reading (μg/m^3)

> (2nd ed.), Wiley, New York, 1980. Contains many insights and methods that evolved from the authors’ extensive consulting experience.

Draper, Norman, and Harry Smith, Applied Regression Analysis (3rd ed.), Wiley, New York, 1999. See Chapter 12 bibliography.

Hoaglin, David, and Roy Welsch, “The Hat Matrix in Regression and ANOVA,” American Statistician, 1978: 17-23. Describes methods for detecting influential observations in a regression data set.

Hocking, Ron, “The Analysis and Selection of Variables in Linear Regression,” Biometrics, 1976: 1-49. An excellent survey of this topic.

Neter, John, Michael Kutner, Christopher Nachtsheim, and William Wasserman, Applied Linear Statistical Models (4th ed.), Irwin, Homewood, IL, 1996. See Chapter 12 bibliography.

14 Goodness-of-Fit Tests and Categorical Data Analysis

INTRODUCTION

In the simplest type of situation considered in this chapter, each observation in a sample is classified as belonging to one of a finite number of categories (e.g., blood type could be one of the four categories O, A, B, or AB). With pi denoting the probability that any particular observation belongs in category i (or the proportion of the population belonging to category i ), we wish to test a null hypothesis that completely specifies the values of all the (such as H_0: p_1 = .45, p_2 = .35, p_3 = .15, p_4 = .05, when there are four categories). The test statistic will be a measure of the discrepancy between the observed numbers in the categories and the expected numbers when H_0 is true. Because a decision will be reached by comparing the computed value of the test statistic to a critical value of the chi-squared distribution, the procedure is called a chi-squared goodness-of-fit test.

Sometimes the null hypothesis specifies that the depend on some smaller number of parameters without specifying the values of these parameters. For example, with three categories the null hypothesis might state that p_1 = θ 2, p_2 = 2θ (1 –θ ), and p_3 = (1 –θ )2. For a chi-squared test to be performed, the values of any unspecified parameters must be estimated from the sample data. These problems are discussed in Section 14.2. The methods are then applied to test a null hypothesis that states that the sample comes from a particular family of distributions, such as the Poisson family (with λ estimated from the sample) or the normal family (with μ and σ estimated).

Chi-squared tests f same for several different populations. The second type of situation involves taking a sample from a single population and classifying each individual with respect to two different categorical factors (such as religious preference and political party registration). The null hypothesis in this situation is that the two factors are independent within the population.

14.1 Goodness-of-Fit Tests When Category Probabilities Are Completely Specified

A binomial experiment consists of a sequence of independent trials in which each trial can result in one of two possible outcomes, S (for success) and F (for failure). The probability of success, denoted by p , is assumed to be constant from trial to trial, and the number n of trials is fixed at the outset of the experiment. In Chapter 8, we presented a large-sample z test for testing H_0: p = p_0. Notice that this null hypothesis specifies both P (S ) and P (F ), since if P (S ) = p_0, then P (F ) = 1—p_0. Denoting P (F ) by q and 1—p_0 by q_0, the null hypothesis can alternatively be written as H_0: p = p_0, q = q_0. The z test is two-tailed when the alternative of interest is p ≠ p_0.

A multinomial experiment generalizes a binomial experiment by allowing each trial to result in one of k possible outcomes, where k > 2. For example, suppose a store accepts three different types of credit cards. A multinomial experiment would result from observing the type of credit card used—type 1, type 2, or type 3—by each of the next n customers who pay with a credit card. In general, we will refer to the k possible outcomes on any given trial as categories, and pi will denote the probability that a trial results in category i. If the experiment consists of selecting n individuals or objects from a population and categorizing each one, then pi is the proportion of the population falling in the ith category (such an experiment will be approximately multinomial provided that n is much smaller than the population size).

The null hypothesis of interest will specify the value of each pi . For example, in the case k = 3, we might have H_0: p_1 = .5, p_2 = .3, p_3 = .2. The alternative hypothesis will state that H_0 is not true—that is, that at least one of the pi s has a value different from that asserted by H_0 (in which case at least two must be different, since they sum to 1). The symbol pi 0 will represent the value of pi claimed by the null hypothesis. In the example just given, p_10 = .5, p_20 = .3, and p_30 = .2.

Ni and its observed value by ni . Since each trial results in exactly one of the k categories, Σ Ni = n, and the same is true of the ni s. As an example, an experiment with n = 100 and k = 3 might yield N_1 = 46, N_2 = 35, and N_3 = 19.

The expected number of successes and expected number of failures in a binomial experiment are np and nq , respectively. When H_0: p = p_0, q = q_0 is true, the expected numbers of successes and failures are np_0 and nq_0, respectively. Similarly, in a multinomial experiment the expected number of trials resulting in category i is E (Ni ) = npi (i = l, …, k ). When H_0: p_1 = p_10, …, pk = pk 0 is true, these expected values become E (N_1) = np_10, E (N_2) = np_20, …, E (Nk ) = npk 0. For the case k = 3, H_0: p_1 = .5, p_2 = .3, p_3 = .2, and n = 100, E (N_1) = 100(.5) = 50, E (N_2) = 30, and E (N_3) = 20 when H_0 is true. The ni s are often displayed in a tabular format consisting of a row of k cells, one for each category, as illustrated in Table 14.1. The expected values when H_0 is true are displayed just below the observed values. The Ni s and ni s are usually referred to as observed cell counts (or observed cell frequencies ), and np_10, np_20,…, npk 0 are the corresponding expected cell counts under H_0.

Table 14.1 Observed and Expected Cell Counts

The ni s should all be reasonably close to the corresponding npi 0s when H_0 is true. On the other hand, several of the observed counts should differ substantially from these expected counts when the actual values of the pi s differ markedly from what the null hypothesis asserts. The test procedure involves assessing the discrepancy between the ni s and the npi 0 s, with H_0 being rejected when the discrepancy is sufficiently large. It is natural to base a measure of discrepancy on the squared deviations (n_1—np_10)2, (n_2—np_20)2, em>np_10 = 100 and np_20 = 10. Then if n_1 = 95 and n_2 = 5, the two categories contribute the same squared deviations to the proposed measure. Yet ni is only 5% less than what would be expected when H_0 is true, whereas n_2 is 50% less. To take relative magnitudes of the deviations into account, we will divide each squared deviation by the corresponding expected count and then combine.

Before giving a more detailed description, we must discuss a type of probability distribution called the chi-squared distribution. This distribution was first introduced in Section 4.4 and was used in Chapter 7 to obtain a confidence interval for the variance σ 2 of a normal population. The chi-squared distribution has a single parameter v , called the number of degrees of freedom (df) of the distribution, with possible values 1, 2, 3,…. Analogous to the critical value ta,v for the t distribution, is the value such that a of the area under the χ 2 curve with v df lies to the right of (see Figure 14.1). Selected values of are given in Appendix Table A.7.

Figure 14.1 A critical value for a chi-squared distribution

THEOREM

Provided that npi > 5 for every i (i = 1, 2, …, k ), the variable

has approximately a chi-squared distribution with k –1 df.

The fact that df = k— 1 is a consequence of the restriction Σ Ni = n. Although there are k observed cell counts, once any k— 1 are known, the remaining one is uniquely determined. That is, there are only k— 1 “freely determined” cell counts, and thus k— 1 df.

If npi 0 is substituted for npi in χ 2, the resulting test statistic has a chi-squared distribution when H_0 is true. Rejection of H_0 is appropriate when χ 2 ≥ c (because large discrepancies between observed and expected counts lead to a large value of χ 2), and the choice yields a test with significance level α .

Null hypothesis: H_0: p_1 = p_10, p_2 = p_20, used when k = 2. The null hypothesis in this case can be stated as H_0: p_1 = p_10, since the relations p_2 = 1 –p_1 and p_20 = 1 –p_10 make the inclusion of p_2 = p_20 in H_0 redundant. The alternative hypothesis is Ha : p_1 ≠ p_10. These hypotheses can also be tested using a two-tailed z test with test statistic

Surprisingly, the two test procedures are completely equivalent. This is because it can be shown that Z 2 = χ 2 and , so that if and only if (iff) |Z| ≥ za /2.* If the alternative hypothesis is either Ha: p_1 > p_10 or Ha: p_1 < p_10, the chi-squared test cannot be used. One must then revert to an upper- or lower-tailed z test.

As is the case with all test procedures, one must be careful not to confuse statistical significance with practical significance. A computed χ 2 that exceeds may be a result of a very large sample size rather than any practical differences between the hypothesized pi 0s and true pis. Thus if , but the true pi s have values .330, .340, and .330, a large value of χ 2 is sure to arise with a sufficiently large n . Before rejecting H_0, the should be examined to see whether they suggest a model different from that of H_0 from a practical point of view.

P -Values for Chi-Squared Tests

The chi-squared tests in this chapter are all upper-tailed, so we focus on this case. Just as the P -value for an upper-tailed t test is the area under the tv curve to the right of the calculated t , the P -value for an upper-tailed chi-squared test is the area under the curve to the right of the calculated χ 2. Appendix Table A.7 provides limited P -value information because only five upper-tail critical values are tabulated for each different v . We have therefore included another appendix table, analogous to Table A.8, that facilitates making more precise P -value statements.

The fact that t curves were all centered at zero allowed us to tabulate t -curve tail areas in a relatively compact way, with the left margin giving values ranging from_0.0 to 4.0 on the horizontal t scale and various columns displaying corresponding upper-tail areas for various df’s. The rightward movement of chi-squared curves as df increases necessitates a somewhat different type of tabulation. The left margin of Appendix Table A.11 displays various upper-tail areas: .100, .095, .090, …, .005, and .001. Each column of the table is for a differenft value of df, and the entries are values on the horizontal chi-squared axis that capture these corresponding tail areas. For example, moving down to tail area .085 and across to the 4 df column, we see that the area to the right of 8.18 under the 4 df chi-squared curve is .085 (see Figure 14.2).

* The fact that is a consequence of the relationship between the standard normal distribution and the chi-squared distribution with 1 df; if Z ≈ N (0, 1), then Z 2 has a chi-squared distribution with v = 1.

Figure 14.2 A P -value for an upper-tailed chi-squared test

To capture this same upper-tail area under the 10 df curve, we must go out to 16.54. In the 4 df column, the top row shows that if the calculated value of the chi-squared variable is smaller than 7.77, the captured tail area (the P -value) exceeds .10. Similarly, the bottom row in this column indicates that if the calculated value exceeds 18.46, the tail area is smaller than .001 (P -value < .001).

χ^2 When the Are Functions of Other Parameters

Frequently the pi s are hypothesized to depend on a smaller number of parameters θ_1, …, θm (m < k ). Then a specific hypothesis involving the θis yields specific pi 0s, which are then used in the χ 2 test.

Example 14.2

In a well-known genetics article (“The Progeny in Generations F_12 to F_17 of a Cross Between a Yellow-Wrinkled and a Green-Round Seeded Pea,” J. Genetics, 1923: 255-331), the early statistician G. U. Yule analyzed data resulting from crossing garden peas. The dominant alleles in the experiment were Y = yellow color and R = round shape, resulting in the double dominant YR. Yule examined 269 four-seed pods resulting from a dihybrid cross and counted the number of YR seeds in each pod. Letting X denote the number of YR’s in a randomly selected pod, possible X values are 0, 1, 2, 3, 4, which we identify with cells 1, 2, 3, 4, and 5 of a rectangular table (so, e.g., a pod with X = 4 yields an observed count in cell 5).

The hypothesis that the Mendelian laws are operative and that genotypes of individual seeds within a pod are independent of one another implies that X has a binomial distribution with n = 4 and . We thus wish to test H_0: p_1 = p_10, …, p_5 = p_50, where

Table 14.3 with expected cell counts

Table 14.3 Observed and Expected Cell Counts for Example 14.2

Thus χ 2 = 3.823 + …. + .032 = 4.582. Since , H_0 is not rejected at level .01. Appendix Table A.11 shows that because 4.582 < 7.77, the P -value for the test exceeds .10. H_0 should not be rejected at any reasonable significance level.

χ^2 When the Underlying Distribution Is Continuous

We have so far assumed that the k categories are naturally defined in the context of the experiment under consideration. The χ 2 test can also be used to test whether a sample comes from a specific underlying continuous distribution. Let X denote the variable being sampled and suppose the hypothesized pdf of X is f_0(x ). As in the construction of a frequency distribution in Chapter 1, subdivide the measurement scale of X into k intervals [a_0, a_1), [a_1, a_2), …, [ak -1, ak ), where the interval [ai -1, a ) includes the value ai -1 but not ai . The cell probabilities specified by H_0 are then

The cells should be chosen so that npi 0 ≥ 5 for i = 1, …, k . Often they are selected so that the npi 0s are equal.

Example 14.3

To see whether the time of onset of labor among expectant mothers is uniformly distributed throughout a 24-hour day, we can divide a day into k periods, each of length 24/k . The null hypothesis states that f (x ) is the uniform pdf on the interval [0, 24], so that pi 0 = 1/k . The article “The Hour of Birth” (British J. Preveutive aud Social Medicine, 1953: 43-59) reports on 1186 onset times, which were categorized into k = 24 1-hour intervals beginning at midnight, resulting in cell counts of 52, 73, 89, 88, 68, 47, 58, 47, 48, 53, 47, 34, 21, 31, 40, 24, 37, 31, 47, 34, 36, 44, 78, and 59. Each expected cell count is , and the resulting value of χ 2 is 162.77. Since , the computed value is highly significant, and the null hypothesis is resoundingly ckquote>

7. Criminologists have long debated whether there is a relationship between weather conditions and the incidence of violent crime. The author of the article “Is There a Season for Homicide?” (Criminology, 1988: 287-296) classified 1361 homicides according to season, resulting in the accompanying data. Test the null hypothesis of equal proportions using a = .01 by using the chi-squared table to say as much as possible about the P -value.

8. The article “Psychiatric and Alcoholic Admissions Do Not Occur Disproportionately Close to Patients’ Birthdays” (Psychological Reports, 1992: 944-946) focuses on the existence of any relationship between date of patient admission for treatment of alcoholism and patient’s birthday. Assuming a 365-day year (i.e., excluding leap year), in the absence of any relation, a patient’s admission date is equally likely to be any one of the 365 possible days. The investigators established four different admission categories: (1) within 7 days of birthday, (2) between 8 and 30 days, inclusive, from the birthday, (3) between 31 and 90 days, inclusive, from the birthday, and (4) more than 90 days from the birthday. A sample of 200 patients gave observed frequencies of 11, 24, 69, and 96 for categories 1, 2, 3, and 4, respectively. State and test the relevant hypotheses using a significance level of .01.

9. The response time of a computer system to a request for a certain type of information is hypothesized to have an exponential distribution with parameter λ = 1 sec (so if X = response time, the pdf of X under H_0 is f_0(x ) = e-x for x ≥0).

a. If you had observed X_1, X_2, …, Xn and wanted to use the chi-squared test with five class intervals having equal probability under H_0, what would be the resulting class intervals?

b. Carry out the chi-squared test using the following data resulting from a random sample of 40 response times:

10. a. Show that another expression for the chi-squared statistic is

Why is it more efficient to compute χ 2k for all i ), how does the formula of part (a) simplify? Use the simplified expression to calculate χ 2 for the pigeon/direction data in Exercise 4.

11. a. Having obtained a random sample from a population, you wish to use a chi-squared test to decide whether the population distribution is standard normal. If you base the test on six class intervals having equal probability under H_0, what should be the class intervals?

b. If you wish to use a chi-squared test to test H_0: the population distribution is normal with μ = .5, σ = .002 and the test is to be based on six equiprobable (under H_0) class intervals, what should be these intervals?

c. Use the chi-squared test with the intervals of part (b) to decide, based on the following 45 bolt diameters, whether bolt diameter is a normally distributed variable with μ = .5 in., σ = .002 in.

14.2 Goodness-of-Fit Tests for Composite Hypotheses

In the previous section, we presented a goodness-of-fit test based on a χ 2 statistic for deciding between H_0: p_1 = p_10, …, pk = pk 0 and the alternative Ha stating that H_0 is not true. The null hypothesis was a simple hypothesis in the sense that each pi 0 was a specified number, so that the expected cell counts when H_0 was true were uniquely determined numbers.

In many situations, there are k naturally occurring categories, but H_0 states only that the pis are functions of other parameters θ_1, …, θm without specifying the values of these θ s. For example, a population may be in equilibrium with respect to proportions of the three genotypes AA, Aa, and aa. With p_1, p_2, and p_3 denoting these proportions (probabilities), one may wish to test

where θ represents the proportion of gene A in the population. This hypothesis is composite because knowing that H_0 is true does not uniquely determine the cell probabilities and expected cell counts but only their general form. To carry out a χ 2 test, the unknown θi s must first be estimated.

Similarly, we may be interested in testing to see whether a sample cam, the parameter λ must be estimated. In addition, because there are actually an infinite number of possible values of a Poisson variable, these values must be grouped so that there are a finite number of cells. If H_0 states that the underlying distribution is normal, use of a χ 2 test must be preceded by a choice of cells and estimation of μ and σ .

χ 2 When Parameters Are Estimated

As before, k will denote the number of categories or cells and pi will denote the probability of an observation falling in the ith cell. The null hypothesis now states that each pi is a function of a small number of parameters θ_1, …, θm with the θi s otherwise unspecified:

For example, for H_0 of (14.1), m = 1 (there is only one θ ), π_1(θ ) = θ 2, π_2(θ ) = 2θ (1 –θ ), and π_3(θ ) = (1 –θ )2.

In the case k = 2, there is really only a single rv, N_1 (since N_1 + N_2 = n ), which has a binomial distribution. The joint probability that N_1 = n_1 and N_2 = n_2 is then

where p_1 + p_2 = 1 and n_1 + n_2 = n . For general k , the joint distribution of N_1, …, Nk is the multinomial distribution (Section 5.1) with

When H_0 is true, (14.3) becomes

To apply a chi-squared test, θ = (θ_1, …, θm ) must be estimated.

METHOD OF ESTIMATION

Let n_1, n_2, …, nk denote the observed values of N_1, …, Nk. Then 1,…,m are those values of the b>1, …, θm and the π,(θ)s, if θ_1, …, θm are estimated by the method of maximum likelihood as described previously and n is large,

has approximately a chi-squared distribution with k –1 –m df when H_0 of (14.2) is true. An approximately level a test of H_0 versus Ha is then to reject H_0 if . In practice, the test can be used if nπi () ≥ 5 for every i .

Notice that the number of degrees of freedom is reduced by the number of θ-.s estimated.

Example 14.6 (Example 14.5 continued)

With = .475 and n = 500, the estimated expected cell counts are n π_1() = 500()2 = 112.81, nπ_2() = (500)(2)(.475)(1 –.475) = 249.38, and nπ_3() = 500 –112.81 –249.38 = 137.81. Then

Since and 4.78 ≥ 3.843, H_0 is rejected. Appendix Table A.11 shows that P -value ≈ .029.

Example 14.7

Consider a series of games between two teams, I and II, that terminates as soon as one team has won four games (with no possibility of a tie). A simple probability model for such a series assumes that outcomes of successive games are independent and that the probability of team I winning any particular game is a constant θ. We arbitrarily designate I the better team, so that θ ≥ .5. Any particular series can then terminate after 4, 5, 6, or 7 games. Let π_1(θ), π_2(θ), π_3(θ), π_4(θ) denote the probability of termination in 4, 5, 6, and 7 games, respectively. Then

π_1 (θ ) = P(I wins in 4 games) + P(II wins in 4 games)

= θ 4 + (1 –θ )4

π3(θ) = 10θ^2(1 –θ )2[θ^2 + (1 –θ )2

However, the required for χ 2 is obtained by maximizing Expression (14.4) rather than the likelihood of the full sample. The cell probabilities are

so the right-hand side of (14.4) becomes

There is no nice formula for , the maximizing value of λ, in this latter expression, so it must be obtained numerically.

Because the parameter estimates are usually much more difficult to compute from the grouped data than from the full sample, they are virtually always computed using this latter method. When these “full” estimators are used in the chi-squared statistic, the distribution of the statistic is altered and a level a test is no longer specified by the critical value

THEOREM

Let 1,…, m be the maximum likelihood estimators of θ_1, …, θm based on the full sample x_1,…, xn, and let χ 2 denote the statistic based on these estimators. Then the critical value cα that specifies a level α upper-tailed test satisfies

The test procedure implied by this theorem is the following:

Example 14.9 (Example 14.8 continued)

Using = 2.10, the estimated expected cell counts are computed from nπi ( ), where n = 48. For example,

Similarly, nπ_2() = 12.34, nπ_3() = 12.96, nπ_4() = 9.07, and nπ_5() = 48 –5.88 –dex="00728" alt="Image"> is a method-of-moments estimate and not an mle). Then with replacing λ in f(x ), the estimated expected cell frequencies as displayed previously are computed as

where [ai–1, ai ) is the ith class interval. To obtain expected cell counts of at least 5, the last six cells are combined to yield observed counts of 20, 8, 7, 5 and expected counts of 18.0, 9.9, 5.5, 6.6. The computed value of chi-squared is then χ 2 = 1.34. Because , H_0 is not rejected, so the truncated exponential model provides a good fit.

A Special Test for Normality

Probability plots were introduced in Section 4.6 as an informal method for assessing the plausibility of any specified population distribution as the one from which the given sample was selected. The straighter the probability plot, the more plausible is the distribution on which the plot is based. A normal probability plot is used for checking whether any member of the normal distribution family is plausible. Lets denote the sample x s when ordered from smallest to largest by x(1), x(2),…, x(n). Then the plot suggested for checking normality was a plot of the points (x (i ), yi ), where yi = Φ–1((i –.5)/n ).

A quantitative measure of the extent to which points cluster about a straight line is the sample correlation coefficient r introduced in Chapter 12. Consider calculating r for the n pairs (x(1), y_1), …, (x(n), yn). The yi s here are not observed values in a random sample from a y population, so properties of this r are quite different from those described in Section 12.5. However, it is true that the more r deviates from 1, the less the probability plot resembles a straight line (remember that a probability plot must slope upward). This idea can be extended to yield a formal test procedure: Reject the hypothesis of population normality if r ≤ cα, where cα is a critical value chosen to yield the desired significance level α. That is, the critical value is chosen so that when the population distribution is actually normal, the probability of obtaining an r value that is at most cα (and thus incorrectly rejecting H_0) is the desired a. The developers of the MINITAB statistical computer package give critical values for α = .10, .05, and .01 in combination with different sample sizes. These critical values are based on a slightly different definition of the yi s than that give/em>i s. When there are several tied x(i)s, MINITAB computes r by using the average of the corresponding yi s as the second number in each pair.

Let yi = Φ–[(i –.375)/(n + .25)] and compute the sample correlation coefficient r for the n pairs (x(1), y_1), …, (xw, yn). The Ryan-Joiner test of

H_0: the population distribution is normal

versus

Ha: the population distribution is not normal

consists of rejecting H_0 when r ≤ cα. Critical values cα are given in Appendix Table A.12 for various significance levels α and sample sizes n .

Example 14.12

The following sample of n = 20 observations on dielectric breakdown voltage of a piece of epoxy resin first appeared in Example 4.29.

We asked MINITAB to carry out the Ryan-Joiner test, and the result appears in Figure 14.3. The test statistic value is r = .9881, and Appendix Table A.12 gives .9600 as the critical value that captures lower-tail area .10 under the r sampling distribution curve when n = 20 and the underlying distribution is actually normal. Since .9881 > .9600, the null hypothesis of normality cannot be rejected even for a significance level as large as .10.

Figure 14.3 MINITAB output from the Ryan-Joiner test for the data of Example 14.12

EXERCISES Section 14.2 (12–23)

12. Consider a large population of families in which each family has exactly three children. If the genders of the three children in any family are independent of one another, the number of male children in a randomly selected family will have a binomial distribution based on three trials.

a. Suppose a random sample of 160 families yields the following results. Test the relevant hypotheses by proceeding as align="center">

[Hint: To obtain the mle of θ, write the likelihood (the function to be maximized) as θu(1—θ)ν, where the exponents u and v are linear functions of the cell counts. Then take the natural log, differentiate with respect to θ, equate the result to 0, and solve for .]

16. In a genetics experiment, investigators looked at 300 chromosomes of a particular type and counted the number of sister-chromatid exchanges on each (“On the Nature of Sister-Chromatid Exchanges in 5-Bromodeoxyuridine-Substituted Chromosomes,” Genetics, 1979: 1251-1264). A Poisson model was hypothesized for the distribution of the number of exchanges. Test the fit of a Poisson distribution to the data by first estimating λ and then combining the counts for x = 8 and x = 9 into one cell.

17. An article in Annals of Mathematical Statistics reports the following data on the number of borers in each of 120 groups of borers. Does the Poisson pmf provide a plausible model for the distribution of the number of borers in a group? [Hint: Add the frequencies for 7, 8, …, 12 to establish a single category “ ≥ 7.”]

18. The article “A Probabilistic Analysis of Dissolved Oxygen-Biochemical Oxygen Demand Relationship in Streams” (J. Water Resources Control Fed., 1969: 73-90) reports data on the rate of oxygenation in streams at 20°C in a certain region. The sample mean and standard deviation were computed as = .173 and s = .066, respectively. Based on the accompanying frequency distribution, can it be concluded that oxygenation rate is a normally distributed variable? Use the chi-squared test with α = .05.

19. Each headlight on an automobile undergoing an annual vehicle inspection can be focused either too high (H), too low (L), or properly (N). Checking the two headlights simultaneously (and not distinguishing between left and right) results in the six possible outcomes HH, LL, NN, HL, HN, and LN. If the probabilities (population proportions) for the single headlight focus direction are P (H ) = θ_1, P (L ) = θ_2, and P(N) = 1 –θ_1 –θ_2 and the two headlights are focused independently of one another, the probabilities of the six outcomes for a randomly selected car are the following:

Use the accompanying data to test the null hypothesis

H_0: p 1 = π_1(θ_1, θ_2),…, p_6 = π_6 = (θ_1, θ_2 )

where the π_1 (θ_1, θ_2)s, are given previously.

[Hint: Write the likelihood as a function of θ_1 and θ_2, take the natural log, then compute ∂/∂θ_1 and ∂/∂θ_2, equate them to_0, and solve for 1, 2.]

20. The article “Compatibility of Outer and Fusible Interlining Fabrics in Tailored Garments (Textile Res. J., 1997: 137-142) gave the following observations on bending rigidity (μN • m) for medium-quality fabric specimens, from which the accompanying MINITAB output was obtained:

Would you use a one-sample t confidence interval to estimate true average bending rigidity? Explain your reasoning.

21. The article from which the data in Exercise 20 was obtained also gave the accompanying data on the composite mass/ outer fabric mass ratio for high-quality fabric specimens.

MINITAB gave r = .9852 as the value of the Ryan-Joiner test statistic and reported that P -value > .10. Would you use the one-sample t test to test hypotheses about the value of the true average ratio? Why or why not?

22. The following data set consists of 25 observations on fracture toughness of base plate of 18% nickel maraging steel (from “Fracture Testing of Weldments,” ASTM Special Publ. No. 381, 1965: 328-356). Use the normal probability plot correlation coefficient test to decide whether a normal distribution provides a plausible model for fracture toughness.

23. The article “Nonbloated Burned Clay Aggregate Concrete” (Jg>Testing for Homogeneity

We assume that each individual in every one of the I populations belongs in exactly one of J categories. A sample of ui individuals is taken from the ith population; let n = Σ ui and

nij = the number of individuals in the ith sample who fall into category j

The nijs are recorded in a two-way contingency table with I rows and J columns. The sum of the uijs in the ith row is ui , whereas the sum of entries in the jth column is nj.

Let

Thus, for population 1, the J proportions are p_11, p_12,…, p1J (which sum to 1) and similarly for the other populations. The null hypothesis of homogeneity states that the proportion of individuals in category j is the same for each population and that this is true for every category; that is, for every j, p1j = p2j- = ••• = pIj •

When H_0 is true, we can use p_1, p_2,…, pJ to denote the population proportions in the J different categories; these proportions are common to all I populations. The expected number of individuals in the ith sample who fall in the jth category when H_0 is true is then E (Nj = ni · pi . To estimate E(Ni j j), we must first estimate pj, the proportion in category j. Among the total sample of n individuals, Nj fall into category j, so we use j = N.j/n as the estimator (this can be shown to be the maximum likelihood estimator of pj ). Substitution of the estimate j for pj in ni p pj yields a simple formula for estimated expected counts under H_0:

The test statistic also has the same form as in previous problem situations. The number of degrees of freedom comes from the general rule of thumb. In each row of Table 14.9 there are J – 1 freely determined cell counts (each sample size ni is fixed), so there are a total of 1(J – 1)m>I(J –1) –(J –1) = (J –1)(I –1).

Null hypothesis: H_0: p1j = p2j = … = pIj j = 1, 2, …, J

Alternative hypothesis: H a: H_0 is not true

Test statistic value:

Rejourn region:

P -value information can be obtained as described in Section 14.1. The test can safely be applied as long as ij ≥ 5 for all cells.

Example 14.13

A company packages a particular product in cans of three different sizes, each one using a different production line. Most cans conform to specifications, but a quality control engineer has identified the following reasons for nonconformance:

1. Blemish on can

2. Crack in can

3. Improper pull tab location

4. Pull tab missing

5. Other

A sample of nonconforming units is selected from each of the three lines, and each unit is categorized according to reason for nonconformity, resulting in the following contingency table data:

Does the data suggest that the proportions falling in the various nonconformance categories are not the same for the three lines? The parameters of interest are the various proportions, and the relevant hypotheses are

H_0: the production lines are homogeneous with respect to the five nonconformance categories; that is, p1j = p2j = p3j for j = 1, …, 5

H a: the production lines are not homogeneous with respect to the categories

The estimated expected frequencies (assuming homogeneity) must now be calculated. Consider the first nonconformance categoryiv>
Again, P -value information can be obtained as described in Section 14.1. The test can safely be applied as long as ij > 5 for all cells.

Example 14.14

A study of the relationship between facility conditions at gasoline stations and aggressiveness in the pricing of gasoline (“An Analysis of Price Aggressiveness in Gasoline Marketing,” J. Marketing Research, 1970: 36-42) reports the accompanying data based on a sample of n = 441 stations. At level .01, does the data suggest that facility conditions and pricing policy are independent of one another? Observed and estimated expected counts are given in Table 14.10.

Table 14.10 Observed and Estimated Expected Counts for Example 14.14

Thus

and because , the hypothesis of independence is rejected.

We conclude that knowledge of a station’s pricing policy does give information about the condition of facilities at the station. In particular, stations with an aggressive pricing policy appear more likely to have substandard facilities than stations with a neutral or nonaggressive policy.

Models and methods for analyzing data in which each individual is categorized with respect to three or more factors (multidimensional contingency tables) are discussed in several of the chapter references.

EXERCISES Section 14.3 (24–36)

24. The accompanying two-way table was constructed using data in the article “Television Viewing and Physical Fitness in Adults” (Research Quarterly for Exerci se and Sport, 1990: 315-320). The author hoped to determine whether time spent watching television is associated with cardiovascular fitness. Subjects were asked about their television-viewing habits and were classified as physically fit if they scored in the excellent or very good category on a step test. We include MINITAB output from a chi-squared analysis. The four TV groups corresponded to different amounts of time per day spent watching TV (0, 1 -2, 3-4, or 5 or more hours). The 168 individuals represented in the first column were those judged physically fit. Expected counts appear below observed counts, and MINITAB displays the contribution to χ 2 from each cell. State and test the appropriate hypotheses using α = .05.

25. The accompanying data refers to leaf marks found on white clover samples selected from both long-grass areas and short-grass areas (“The Biology of the Leaf Mark Polymorphism in Trifolium repens L.,” Heredity, 1976: 306-325). Use a χ 2 test to decide whether the true proportions of different marks are identical for the two types of regions.

26. The following data resulted from an experiment to study the effects of leaf removal on the ability of fruit of a certain type to mature (“Fruit Set, Herbivory, Fruit Reproduction, and the Fruiting Strategy of Catalpa speciosa,” Ecology, 1980: 57-64):

Does the data suggest that the chance of a fruit maturing is affected by the number of leaves removed? State and test the appropriate hypotheses at level .01.

27. The article “Human Lateralization from Head to Foot: Sex-Related Factors” (Sci ence, 1978: 1291 -1292) reports for both a sample of right-handed men and a sample of right-handed women the number of individuals whose feet were the same size, had a bigger left than right foot (a difference of half a shoe size or more), or had a bigger right than left foot.

Does the data indicate that gender has a strong effect on the development of foot asymmetry? State the appropriate null and alternative hypotheses, compute the value of χ 2, and obtain information about the P -value.

28. The article “Susceptibility of Mice to Audiogenic Seizure Is Increased by Handling Their Dams During Gestation” (Science, 1976: 427-428) reports on research into the effect of different injection treatments on the frequencies of audio-genic seizures.

Does the data suggest that the true percentages in the different response categories depend on the nature of the injection treatment? State and test the appropriate hypotheses using a = .005.

29. The accompanying data on sex combinations of two recombinants resulting from six different male genotypes appears in the article “A New Method for Distinguishing Between Meiotic and Premeiotic Recombinational Events in Drosophi la melanogastef (Genetics, 1979: 543-554). Does the data support the hypothesis that the frequency distribution among the three sex combinations is homogeneous with respect to the different genotypesm>Ha, and perform the analysis.

30. Three different design configurations are being considered for a particular component. There are four possible failure modes for the component. An engineer obtained the following data on number of failures in each mode for each of the three configurations. Does the configuration appear to have an effect on type of failure?

31. A random sample of individuals who drive alone to work in a large metropolitan area was obtained, and each individual was categorized with respect to both size of car and commuting distance. Does the accompanying data suggest that commuting distance and size of car are related in the population sampled? State the appropriate hypotheses and use a level .05 chi-squared test.

32. Each individual in a random sample of high school and college students was cross-classified with respect to both political views and marijuana usage, resulting in the data displayed in the accompanying two-way table (“Attitudes About Marijuana and Political Views,” Psychological Reports, 1973: 1051-1054). Does the data support the hypothesis that political views and marijuana usage level are independent within the population? Test the appropriate hypotheses using level of significance .01.

33. Show that the chi-squared statistic for the test of independence can be written in the form

Why is this formula more efficient computationally than the defining formula for χ 2?

34. Suppose that in Exercise 32 each student had been categorized with respect to political views, marijuana usage, and religious preference, with the categories of this latter factor being Protestant, Catholic, and other. The data could be displayed in three different two-way tables, one corresponding to each category of the third factor. With pijk = P(political categoryi , marijuana category j, and religious category k), the null hypothesis of independence of all three factors states that pijk = pi . pj. pk. Let ni jk denote the observed frequency in cell (i , j, k). Show how to estimate the expected cell counts assuming that H_0 is true (o determine the number of degrees of freedom for the chi-squared statistic.

35. Suppose that in a particular state consisting of four distinct regions, a random sample of nk voters is obtained from the kth region for k = 1, 2, 3, 4. Each voter is then classified according to which candidate (1, 2, or 3) he or she prefers and according to voter registration (1 = Dem., 2 = Rep., 3 = Indep.). Let pjk denote the proportion of voters in region k who belong in candidate category and registration category j. The null hypothesis of homogeneous regions is H_0: pj1 = pj2 = pj3 = pj4 for all, j (i.e., the proportion within each candidate/registration combination is the same for all four regions). Assuming that H_0 is true, determine jjk and jjk as functions of the observed njks s, and use the general rule of thumb to obtain the number of degrees of freedom for the chi-squared test.

36. Consider the accompanying 2 X 3 table displaying the sample proportions that fell in the various combinations of categories (e.g., 13% of those in the sample were in the first category of both factors).

a. Suppose the sample consisted of n = 100 people. Use the chi-squared test for independence with significance level .10.

b. Repeat part (a) assuming that the sample size was n = 1000.

c. What is the smallest sample size n for which these observed proportions would result in rejection of the independence hypothesis?

SUPPLEMENTARY EXERCISES(37-49)

37. The article “Birth Order and Political Success” (Psych. Reports, 1971: 1239-1242) reports that among 31 randomly selected candidates for political office who came from families with four children, 12 were firstborn, 11 were middle-born, and 8 were lastborn. Use this data to test the null hypothesis that a political candidate from such a family is equally likely to be in any one of the four ordinal positions.

38. The results of an experiment to assess the effect of crude oil on fish parasites are described in the article “Effects of Crude Oils on the Gastrointestinal Parasites of Two Species of Marine Fish” (J. Wi ldli fe Di seases, 1983: 253-258). Three treatments (corresponding to populations in the procedure described) were compared: (1) no contamination, (2) contamination by 1-year-old weathered oil, and (3) contamination by new oil. For each treatment condition. Data compatible with that in the article is given. Does the data indicate that the three treatments differ with respect to the true proportion of parasitized and nonparasitized fish? Test using a = .01.

39. Qualifications of male and female head and assistant college athletic coaches were compared in the article “Sex Bias and the Validity of Believed Differences Between Male and Female Interscholastic Athletic Coaches” (Research Quarterlyfor Exerci se and Sport, 1990: 259-267). Each person in random samples of 2225 male coaches and 1141 female coaches was classified according to number of years of coaching experience to obtain the accompanying two-way table. Is there enough evidence to conclude that the proportions falling into the experience categories are different for men and women? Use α = .01.

40. The authors of the article “Predicting Professional Sports Game Outcomes from Intermediate Game Scores” (Chance, 1992: 18-22) used a chi-squared test to determine whether there was any merit to the idea that basketball games are not settled until the last quarter, whereas baseball games are over by the seventh inning. They also considered football and hockey. Data was collected for 189 basketball games, 92 baseball games, 80 hockey games, and 93 football games. The games analyzed were sampled randomly from all games played during the 1990 season for baseball and football and for the 1990-1991 season for basketball and hockey. For each game, the late-game leader was determined, and then it was noted whether the late-game leader aem">

42. In a study of 2989 cancer deaths, the location of death (home, acute-care hospital, or chronic-care facility) and age at death were recorded, resulting in the given two-way frequency table (“Where Cancer Patients Die,” Public Health Reports, 1983: 173). Using a .01 significance level, test the null hypothesis that age at death and location of death are independent.

43. In a study to investigate the extent to which individuals are aware of industrial odors in a certain region (“Annoyance and Health Reactions to Odor from Refineries and Other Industries in Carson, California,” Envi ronmental Research, 1978: 119-132), a sample of individuals was obtained from each of three different areas near industrial facilities. Each individual was asked whether he or she noticed odors (1) every day, (2) at least once/week, (3) at least once/month, (4) less often than once/month, or (5) not at all, resulting in the data and SPSS output at the bottom of this page. State and test the appropriate hypotheses.

44. Many shoppers have expressed unhappiness because grocery stores have stopped putting prices on individual grocery items. The article “The Impact of Item Price Removal on Grocery Shopping Behavior” (J. Marketing, 1980: 73-93) reports on a study in which each shopper in a sample was classified by age and by whether he or she felt the need for item pricing. Based on the accompanying data, does the need for item pricing appear to be independent of age?

45. Let p_1 denote the proportion of successes in a particular population. The test statistic value in Chapter 8 for testing

H_0: p_1 = p_10 was , where p_20 = 1 –p_10. Show that for the case k = 2, the chi-squared test statistic value of Section 14.1 satisfies χ 2 = z^2. [Hi nt: First show that (n_1 –np_10)2 = (n_2 –np_20)2.]

46. The NCAA basketball tournament begins with 64 teams that are apportioned into four regional tournaments, each involving 16 teams. The 16 teams in each region are then ranked (seeded) from 1 to 16. During the 12-year period from 1991 to 2002, the toP -ranked team won its regional tournament 22 times, the second-ranked team won 10 times, the third-ranked team won 5 times, and the remaining 11 regional tournaments were won by teams ranked lower than 3. Let Pi j denote the probability that the team ranked in its region is victoing in positively signed ranks and a large value of s+. On the other hand, if the median is_0, magnitudes of positively signed observations should be intermingled with those of negatively signed observations, in which case s+ will not be very large. Thus we should reject H_0: =0 when s+ is “quite large”–the rejection region should have the form s+ ≥ c.

The critical value c should be chosen so that the test has a desired significance level (type I error probability), such as .05 or .01. This necessitates finding the distribution of the test statistic S+ when the null hypothesis is true. Let’s consider n = 5, in which case there are 2^5 = 32 ways of applying signs to the five ranks 1, 2, 3, 4, and 5 (each rank could have a –sign or a + sign). The key point is that when H_0 is true, any collection of five signed ranks has the same chance as does any other collection. That is, the smallest observation in absolute magnitude is equally likely to be positive or negative, the same is true of the second smallest observation in absolute magnitude, and so on. Thus the collection –1, 2, 3, –4, 5 of signed ranks is just as likely as the collection 1, 2, 3, 4, –5, and just as likely as any one of the other 30 possibilities.

Table 15.1 Possible Signed-Rank Sequences for n = 5

Table 15.2 Null Distribution of S+ When n = 5

Table 15.1 lists the 32 possible signed-rank sequences when n = 5 along with the value s+ for each sequence. This immediately gives the “null distribution” of S+ displayed in Table 15.2. For example, Table 15.1 shows that three of the 32 possible sequences have s+ = 8, so P(S+ = 8 when H_0 is true) = . Notice that the null distribution is symmetric about 7.5 [more generally, symmetrically distributed over the possible values_0, 1, 2,… ,n(n + 1)/2]. This symmetry is important in relating the rejection region of lower-tailed and two-tailed tests to that of an upper-tailed test.

For n = 10 there are 2^10 = 1024 possible signed-rank sequences, so a listing would involve much effort. Each sequence, though, would have probability when H_0 is true, from which the distribution of S+ when H_0 is true can be easily obtained.

We are now in a position to determine a rejection region for testing H_0: = 0 versus Ha: & {s+: s+ ≥ 13} = {13, 14, 15}. Then

α = P(reject H_0 when H_0 is true)

= P(S+ = 13, 14, or 15 when H_0 is true)

=

= .094

so that R = {13, 14, 15} specifies a test with approximate level .1. For the rejection region {14, 15}, α = 2/32 = .063. For the sample x_1 = .58, x_2 = 2.50, x_3 = –.21, x_4 = 1.23, x_5 = .97, the signed rank sequence is –1, +2, +3, +4, +5, so s+ = 14 and at level .063 H_0 would be rejected.

A General Description of the Wilcoxon Signed-Rank Test

Because the underlying distribution is assumed symmetric, μ = , so we will state the hypotheses of interest in terms of μ rather than .[*]

* If the tails of the distribution are “too heavy,” as was the case with the Cauchy distribution mentioned in Chapter 6, then μ will not exist. In such cases, the Wilcoxon test will still be valid for tests concerning .

ASSUMPTION

X_1, x_2,…, xn is a random sample from a continuous and symmetric probability distribution with mean (and median) μ.

When the hypothesized value of μ is μ_0, the absolute differences |x_1 –μ_0|, …, |xn –μ_0| must be ranked from smallest to largest.

Null hypothesis: H_0: μ = μ_0

Test statistic value: s+ = the sum of the ranks associated with positive (xi –μ_0)s

where the critical values c1
Example 15.1

A manufacturer of electric irons, wishing to test the accuracy of the thermostat control at the 500°F setting, instructs a test engineer to obtain actual temperatures at that setting for 15 irons using a thermocouple. The resulting measurements are as follows:

The engineer believes it is reasonable to assume that a temperature deviation from 500° of any particular magnitude is just as likely to be positive as negative (the assumption of symmetry) but wants to protect against possible nonnormality of the actual temperature distribution, so she decides to use the Wilcoxon signed-rank test to see whether the data strongly suggests incorrect calibration of the iron.

The hypotheses are H_0: μ = 500 versus Ha: μ ≠ 500, where μ = the true average actual temperature at the 500°F setting. Subtracting 500 from each xi gives

The ranks are obtained by ordering these from smallest to largest without regard to sign.

Thus s+ = 2 + 4 + 7 + 9 + 13 = 35. From Appendix Table A.13, P(S+ ≥ 95) = P(S+ ≤ 25) = .024 when H_0 is true, so the two-tailed test with approximate level .05 rejects H_0 when either s+ ≥ 95 or ≤ 25 [the exact α is 2(.024) = .048]. Since s+ = 35 is not in the rejection region, it cannot be concluded at level .05 that μ is anything other than 500. Even at level .094 (approximately .1), H_0 is not rejected, since P(S+ ≤ 30) = .047 implies that s+ values between 30 and 90 are not significant at that level. The P -value of the data is thus greater than .1.

Although a theoretical implication of the continuity of the underlying distribution is that ties will not occur, in practice they often do because of the discreteness of measuring instruments. If there are several data values with the same absolute magnitude, then they would be assigned the average of the ranks they would receive if they differed very slightly from one another. For example, if in Example 15.1 x_8 = 498.2 is changed to 498.4, then two different values of (xi –500) would have absolute magnitude 1.6. The ranks to be averaged would be 2 and 3, so each would be assigned rank 2.5.

Pa = (Ri –i )2. Then small values of D give support to Ha (e.g., the smallest value is_0 for R_1 = 1, R_2 = 2, …, Rn = n), so H_0 should be rejected in favor of Ha if d ≤ c. When H_0 is true, any sequence of ranks has probability 1ln!. Use this to find c for which the test has a level as close to .10 as possible in the case n = 4. [Hint: List the 4! rank sequences, compute d for each one, and then obtain the null distribution of D. See the Lehmann book (in the chapter bibliography), p. 290, for more information.]

15.2 The Wilcoxon Rank-Sum Test

When at least one of the sample sizes in a two-sample problem is small, the t test requires the assumption of normality (at least approximately). There are situations, though, in which an investigator would want to use a test that is valid even if the underlying distributions are quite nonnormal. We now describe such a test, called the Wilcoxon rank-sum test. An alternative name for the procedure is the Mann-Whitney test, though the Mann-Whitney test statistic is sometimes expressed in a slightly different form from that of the Wilcoxon test. The Wilcoxon test procedure is distribution-free because it will have the desired level of significance for a very large class of underlying distributions.

ASSUMPTIONS

X_1,…, xm and Y_1,…, Yn are two independent random samples from continuous distributions with means μ_1 and μ_2, respectively. The X and Y distributions have the same shape and spread, the only possible difference between the two being in the values of μ_1 and μ_2.

When H_0: μ_1—μ_2 = Δ_0 is true, the X distribution is shifted by the amount Δ_0 to the right of the Y distribution; whereas when H_0 is false, the shift is by an amount other than Δ_0.

Development of the Test When m = 3, n = 4

Consider first testing H_0: μ_1 –μ_2 =0. If μ_1 is actually much larger than μ_2, then most of the observed jt’s will fall to the right of the observed y’s. However, if H_0 is true, then the observed values from the two samples should be intermingled. The test statistic will provide a quantification of how much intermingling there is in the two samples.

Consider the case m = 3, n = 4. Then if all three observed jt’s were to the right of all four observed y’s, this would provide strong evidence for rejecting H_0 in favor of H< G`sub>a: μ_1 –μ_2 ≠ 0, with a similar conclusion being appropriate if all three x’s fall below all four of the y’s. Suppose we pool the X’s and y’s into a combined sample of size m + n = 7 and rank these observations from smallest to largest, with the smallest receiving rank 1 and the largest, rank 7. If either most of the largest ranks or most of the smallest ranks were associated with X observations, we would begin to doubt H_0. This suggests the test statistic

For the values of m and n under consideration, the smallest possible value of W is w = 1 + 2 + 3 = 6 (if all three x’s are smaller than all four y’s), and the largest possible value is w = 5 + 6 + 7 = 18 (if all three x’s are larger than all four y’s).

As an example, suppose x_1 = –3.10, x_2 = 1.67, x_3 = 2.01, y_1 = 5.27, y_2 = 1.89, y_3 = 3.86, and y_4 = .19. Then the pooled ordered sample is -3.10, .19, 1.67, 1.89, 2.01, 3.86, and 5.27. The X ranks for this sample are 1 (for -3.10), 3 (for 1.67), and 5 (for 2.01), so the computed value of W is w = 1 + 3 + 5 = 9.

The test procedure based on the statistic (15.3) is to reject H_0 if the computed value w is “too extreme”—that is, ≥ c for an upper-tailed test, ≤ c for a lower-tailed test, and either ≥ c_1 or < c_2 for a two-tailed test. The critical constant(s) c (c_1, c_2) should be chosen so that the test has the desired level of significance a. To see how this should be done, recall that when H_0 is true, all seven observations come from the same population. This means that under H_0, any possible triple of ranks associated with the three x’s—such as (1, 4, 5), (3, 5, 6), or (5, 6, 7)—has the same probability as any other possible rank triple. Since there are = 35 possible rank triples, under H_0 each rank triple has probability . From a list of all 35 rank triples and the w value associated with each, the probability distribution of W can immediately be determined. For example, there are four rank triples that have w value 11—(1, 3, 7), (1,4, 6), (2, 3, 6), and (2,4, 5)—so P (W = 11) = The summary of the listing and computations appears in Table 15.4.

Table 15.4 Probability Distribution of W (m = 3, n = 4) When H_0 Is True

The distribution of Table 15.4 is symmetric about the value w = (6 + 18)/2 = 12, which is the mid triples (r, s, t ) (with r < s < t ) and (8 –t,8 –s,8 –r) have values of w symmetric about 12, so for each triple with w value below 12, there is a triple with w value above 12 by the same amount.

If the alternative hypothesis is Ha: μ_1 –μ_2 >0, then H_0 should be rejected in favor of Ha for large W values. Choosing as the rejection region the set of W values {17, 18}, a = P(type I error) = P(reject H_0 when H_0 is true) = P(W = 17 or 18 when H_0 is true) = = .057; the region {17, 18} therefore specifies a test with level of significance approximately .05. Similarly, the region {6, 7}, which is appropriate for Ha: μ_1 –μ_2 <0, has α = .057 ≈ .05. The region {6, 7, 17, 18}, which is appropriate for the two-sided alternative, has α = = .114. The W value for the data given several paragraphs previously was w = 9, which is rather close to the middle value 12, so H_0 would not be rejected at any reasonable level a for any one of the three Has.

General Description of the Wilcoxon Rank-Sum Test

The null hypothesis H_0: μ_1 –μ_2 = Δ_0 is handled by subtracting Δ_0 from each Xi and using the (X¿ –Δ_0)s as the Xi s were previously used. Recalling that for any positive integer K, the sum of the first K integers is K(K + 1)/2, the smallest possible value of the statistic W is m(m + 1)/2, which occurs when the (Xi –Δ_0s are all to the left of the Y sample. The largest possible value of W occurs when the (Xi –Δ_0)s lie entirely to the right of the Y ’s; in this case, W = (n + 1) +••• + (m + n )= (sum of first m + n integers) –(sum of first n integers), which gives m (m + 2n + 1)/2. As with the special case m = 3, n = 4, the distribution of W is symmetric about the value that is halfway between the smallest and largest values; this middle value is m(m + n + 1)/2. Because of this symmetry, probabilities involving lower-tail critical values can be obtained from corresponding upper-tail values.

Null hypothesis: H_0: μ_1 –μ_2 = Δ_0

where P (W ≤ c_1 when H_0 is true) ≈ α, P (W ≥ c when H_0 is true) ≈ α/2.e probability distribution, there will not always exist a critical value corresponding exactly to one of the usual levels of significance. Appendix Table A.14 gives upper-tail critical values for probabilities closest to .05, .025, .01, and .005, from which level .05 or .01 one- and two-tailed tests can be obtained. The table gives information only for m = 3, 4, …,8 and n = m, m + 1, …,8 (i.e., 3 ≤ m ≤ n < 8). For values of m and n that exceed 8, a normal approximation can be used. To use the table for small m and n, though, the X and Y samples should be labeled so that m ≤ n.

Example 15.4

The urinary fluoride concentration (parts per million) was measured both for a sample of livestock grazing in an area previously exposed to fluoride pollution and for a similar sample grazing in an unpolluted region:

Does the data indicate strongly that the true average fluoride concentration for livestock grazing in the polluted region is larger than for the unpolluted region? Use the Wilcoxon rank-sum test at level a = .01.

The sample sizes here are 7 and 5. To obtain m ≤ n, label the unpolluted observations as the x’s (x_1 = 14.2, …, x_5 = 20.0) and the polluted observations as the y s. Thus μ_1 is the true average fluoride concentration without pollution, and μ_2 i is the true average concentration with pollution. The alternative hypothesis is Ha: μ_1—μ_2 <0 (pollution causes an increase in concentration), so a lower-tailed test is appropriate. From Appendix Table A.14 with m = 5 and n = 7, P (W ≥ 47 when H_0 is true) ≈ .01. The critical value for the lower-tailed test is therefore m (m + n + 1)—47 = 5(13)—47 = 18; H_0 will now be rejected if w ≤ 18. The pooled ordered sample follows; the computed W is w = r_1 + r_2 +•••+ r_5 (where ri is the rank of xi ) = 1 + 5 + 4 + 6 + 9 = 25. Since 25 is not < 18, H_0 is not rejected at (approximately) level .01.

Theoretically, the assumption of continuity of the two distributions ensures that all m + n observed x’s and y s will have different values. In practice, though, there will often be ties in the observed values. As with the Wilcoxon signed-rank test, the common practice in dealing with ties is to assign each of the tied observations in a particular set of ties sub> versus Ha: μ ≠μ_0, the t test at level .10 specifies that H_0 should be rejected if t is either ≥ 1.711 or ≤ –1.711, where

Consider now the null value μ_0 = 95. Then t = 1.25 so H_0 is not rejected. Similarly, if μ_0 = 104, then t = –1 so again H_0 is not rejected. However, if μ_0 = 90, then t = 2.5 so H_0 is rejected, and if μ_0 = 108, then t = – 2 so H_0 is again rejected. By considering other values of μ_0 and the decision resulting from each one, the following general fact emerges: Every number inside the interval (15.7) specifies a value of μ_0 for which t of (15.8) leads to nonrejection of H_0, whereas every number outside interval (15.7) corresponds to a t for which H_0 is rejected. That is, for the fixed values of n, , ands, the set of all μ_0 values for which testing H_0: μ = μ_0 versus Ha: μ ≠ μ_0 results in nonrejection of H_0 is precisely the interval (15.7).

PROPOSITION

Suppose we have a level a test procedure for testing H_0: θ = θ_0 versus Ha: θ ≠ θ_0. For fixed sample values, let A denote the set of all values θ_0 for which H_0 is not rejected. Then A is a 100(1—a)% CI for θ.

There are actually pathological examples in which the set A defined in the proposition is not an interval of θ values, but instead the complement of an interval or something even stranger. To be more precise, we should really replace the notion of a CI with that of a confidence set. In the cases of interest here, the set A does turn out to be an interval.

The Wilcoxon Signed-Rank Interval

To test H_0: μ = μ_0 versus Ha: μ ≠ μ_0 using the Wilcoxon signed-rank test, where μ is the mean of a continuous symmetric distribution, the absolute values |x_1 –μ_0|,…, |xn –μ_0| are ordered from smallest to largest, with the smallest receiving rank 1 and the largest, rank n. Each rank is then given the sign of its associated xi – μ_0, and the test statistic is the sum of the positively signed ranks. The two-tailed test rejects H_0 if s+ is either ≥ c ors n(n + l)/2—c, where c is o>, …, xn, the 100(1 –a)% signed-rank interval will consist of all μ_0 for which H_0: μ = μ_0 is not rejected at level a. To identify this interval, it is convenient to express the test statistic S+ in another form.

That is, if we average each xj in the list with each xi to its left, including (xj + xj )/2 (which is just xj), and count the number of these averages that are ≥ μ_0, s+ results. In moving from left to right in the list of sample values, we are simply averaging every pair of observations in the sample [again including (xj+ xj)/2] exactly once, so the order in which the observations are listed before averaging is not important. The equivalence of the two methods for computing s+ is not difficult to verify. The number of pairwise averages is + n (the first term due to averaging of different observations and the second due to averaging each xi with itself), which equals n(n + 1)/2. If either too many or too few of these pairwise averages are ≥ μ_0, H_0 is rejected.

Example 15.6

The following observations are values of cerebral metabolic rate for rhesus monkeys: x_1 = 4.51, x_2 = 4.59, x_3 = 4.90, x_4 = 4.93, x_5 = 6.80, x_6 = 5.08, x_7 = 5.67. The 28 pairwise averages are, in increasing order,

The first few and the last few of these are pictured on a measurement axis in Figure 15.2.

Figure 15.2 Plot of the data for Example 15.6

Because of the discreteness of the distribution of S+, α = .05 cannot be obtained exactly. The rejection region {0, 1, 2, 26, 27, 28} has α = .046, which is as close as possible to .05, so the level is approximately .05. Thus if the number of pair-wise averages ≥ μ_0 is between 3 and 25, inclusive, H_0 is not rejected. From Figure 15.2 the (approximate) 95% CI for μ is (4.59, 5.94).

In general, once the pairwise averages are ordered from smallest to largest, the endpoints of the Wilcoxon intervalallest by (2),…, and the largest by (n(n+1)/2).

PROPOSITION

If the level α Wilcoxon signed-rank test for H_0: μ = μ_0 versus Ha: μ ≠ μ_0 is to reject H_0 if either S+ ≥ c or S+ ≤ n(n + 1)/2 –c , then a 100(1 –α)% CI for μ is

In words, the interval extends from the dth smallest pairwise average to the dth largest average, where d = n(n + 1)/2 –c + 1. Appendix Table A.15 gives the values of c that correspond to the usual confidence levels for n = 5, 6, …, 25.

Example 15.7 (Example 15.6 continued)

For n = 7, an 89.1% interval (approximately 90%) is obtained by using c = 24 (since the rejection region {0,1, 2, 3,4, 24, 25, 26, 27, 28} has a = .109). The interval is ( (28–24+1), (24)) = ( (5), (24)) = (4.72, 5.85), which extends from the fifth smallest to the fifth largest pairwise average.

The derivation of the interval depended on having a single sample from a continuous symmetric distribution with mean (median) μ. When the data is paired, the interval constructed from the differences d_1, d_2,…, dn is a CI for the mean (median) difference μD . In this case, the symmetry of X and Y distributions need not be assumed; as long as the X and Y distributions have the same shape, the X –Y distribution will be symmetric, so only continuity is required.

For n > 20, the large-sample approximation to the Wilcoxon test based on standardizing S+ gives an approximation to c in (15.10). The result [for a 100(1—a)% interval] is

The efficiency of the Wilcoxon interval relative to the t interval is roughly the same as that for the Wilcoxon test relative to the t test. In particular, for large samples when the underlying population is normal, the Wilcoxon interval will tend to be slightly longer than the t interval, but if the population is quite nonnormal (symmetric but with heavy tails), then the Wilcoxon interval will tend to be much shwhereas others will consist mostly of observations having large ranks. More specifically, if Rij denotes the rank of xij among the N observations, and Ri . and . denote, respectively, the total and average of the ranks in the ith sample, then when H_0 is true

The Kruskal-Wallis test statistic is a measure of the extent to which the .s deviate from their common expected value (N – 1)/2, and H_0 is rejected if the computed value of the statistic indicates too great a discrepancy between observed and expected rank averages

TEST STATISTIC

The second expression for K is the computational formula; it involves the rank totals (Ri. s) rather than the averages and requires only one subtraction.

If H_0 is rejected when k ≥ c, then c should be chosen so that the test has level a. That is, c should be the upper-tail critical value of the distribution of K when H_0 is true. Under H_0, each possible assignment of the ranks to the I samples is equally likely, so in theory all such assignments can be enumerated, the value of K determined for each one, and the null distribution obtained by counting the number of times each value of K occurs. Clearly, this computation is tedious, so even though there are tables of the exact null distribution and critical values for small values of the Ji s, we will use the following “large-sample” approximation.

PROPOSITION

When H_0 is true and either

I = 3 Ji ≥ 6 (i = 1,2,3)

or

I > 3 Ji ≥ 5(i = 1, …, I )

then K has approximately a chi-squared distribution with I –1 df. This implies that a test with approximate significance level α rejects H_0 if k ≥ l-1.

Example 15.9

The accompanying observations (Table 15.6) on axial stiffness index resulted from a study of metal-plate connected trusses in which five different plate lengths—4 in.,cise 33, and consider a confidence interval associated with the sign test, the sign interval. The relevant hypotheses are now H_0: = 0 versus Ha: ≠ 0. Let’s use the following rejection region: either Y ≥ 15 or Y ≤ 5.

a. What is the significance level for this test?

b. The confidence interval will consist of all values 0 for which H_0 is not rejected. Determine the CI for the given data, and state the confidence level.

35. Suppose we wish to test

H_0: the X and Y distributions are identical

versus

Ha: the X distribution is less spread out than the Y distribution

The accompanying figure pictures X and Y distributions for which Ha is true. The Wilcoxon rank-sum test is not appropriate in this situation because when Ha is true as pictured, the Y’s will tend to be at the extreme ends of the combined sample (resulting in small and large Y ranks), so the sum of X ranks will result in a W value that is neither large nor small.

Consider modifying the procedure for assigning ranks as follows: After the combined sample of m + n observations is ordered, the smallest observation is given rank 1, the largest observation is given rank 2, the second smallest is given rank 3, the second largest is given rank 4, and so on. Then if Ha is true as pictured, the X values will tend to be in the middle of the sample and thus receive large ranks. Let W denote the sum of the X ranks and consider rejecting H_0 in favor of Ha when w’> c. When H_0 is true, every possible set of X ranks has the same probability, so W has the same distribution as does W when H_0 is true. Thus c can be chosen from Appendix Table A.14 to yield a level a test. The accompanying data refers to medial muscle thickness for arterioles from the lungs of children who died from sudden infant death syndrome (x ’s) and a control group of children (y’s). Carry out the test of H_0 versus Ha at level .05.

Consult the Lehmann book (in the chapter bibliography) for more informs, the height of the center line is estimated from the data. If the points on the chart all lie between the two control limits, the process is deemed to be in control. That is, the process is believed to be operating in a stable fashion reflecting only natural random variation. An out-of-control “signal” occurs whenever a plotted point falls outside the limits. This is assumed to be attributable to some assignable cause, and a search for such causes commences. The limits are designed so that an in-control process generates very few false alarms, whereas a process not in control quickly gives rise to a point outside the limits.

There is a strong analogy between the logic of control charting and our previous work in hypothesis testing. The null hypothesis here is that the process is in control. When an in-control process yields a point outside the control limits (an out-of-control signal), a type I error has occurred. On the other hand, a type II error results when an out-of-control process produces a point inside the control limits. Appropriate choice of sample size and control limits (the latter corresponding to specifying a rejection region in hypothesis testing) will make the associated error probabilities suitably small.

We emphasize that “in control” is not synonymous with “meets design specifications or tolerance.” The extent of natural variation may be such that the percentage of items not conforming to specification is much higher than can be tolerated. In such cases, a major restructuring of the process will be necessary to improve process capability. An in-control process is simply one whose behavior with respect to variation is stable over time, showing no indications of unusual extraneous causes.

As a final introductory comment, software for control charting is now widely available. The journal Quality Progress contains many advertisements for statistical quality control computer packages. In addition, SAS and the newest version of MINITAB, among other general-purpose packages, have attractive quality control capabilities.

EXERCISES Section 16.1 (1–3)

1. A control chart for thickness of rolled-steel sheets is based on an upper control limit of .0520 in. and a lower limit of .0475 in. The first ten values of the quality statistic (in this case the sample mean thickness of n = 5 sample sheets) are .0506, .0493, .0502, .0501, .0512, .0498, .0485, .0500, .0505, and .0483. Construct the initial part of the quality control chart, and comment on its appearance.

2. Refer to Exercise 1 and suppose the ten most recent values of the quality statistic are .0493, .0485, .0490, .0503, .0492, .0486, .0495, .0494, .0493, and .0488. Construct the relevant portion of the corresponding control chart, and comment on its appearance.

3. Suppose a control chart is constructed so that the probability of a point falling outside the control limits when the process is actually in control is .002. What is the probability that ten successive points (based on independently s (1, 1), (2, 2), (3, 3), and so on—and then drawing horizontal lines across the plot at

Such a plot is often called a 3-sigma chart. Any point outside the control limits suggests that the process may have been out of control at that time, so a search for assignable causes should be initiated.

Example 16.1

Once each day, three specimens of motor oil are randomly selected from the production process, and each is analyzed to determine viscosity. The accompanying data (Table 16.1) is for a 25-day period. Extensive experience with this process suggests that when the process is in control, viscosity of a specimen is normally distributed with mean 10.5 and standard deviation .18. Thus σ = = = .104, so the 3 SD control limits are

Table 16.1 Viscosity Data for Example 16.1

All points on the control chart shown in Figure 16.2 are between the control limits, indicating stable behavior of the process mean over this time period (the standard deviation and range for each sample will be used in the next subsection).

Figure 16.2 chart for the viscosity data of Example 16.1

Charts Based on Estimated Parameters

In practice it frequently happens that values of μ and σ are unknown, so they must be estimated from sample data prior to determining the control limits. This is especially true when a process is first subjected to a quality control analysis. Again denote the number of observations in each sample by n and let k represent the number of samples available. Typical values of n are 3, 4, 5, or 6, and it is recommended that k be at least 20. We assume that the k samples were gathered during a period when the process was believed to be in control. More will be said about this assumption shortly.

We present two different commonly used methods for estimating σ, one based on the k sample standard deviations and the other on the k sample ranges (recall that the sample range is the difference between the largest and smallest sample observations). Prior to the wide availability of good calculators and statistical computer software, ease of hand calculation was of paramount consideration, so the range method predominated. However, in the case of a normal population distribution, the unbiased estimator of σ based on S is known to have smaller variance than that based on the sample range. Statisticians say that the former estimator is more efficient than the latter. The loss in efficiency for the estimator is slight when n is very small but becomes important for n > 4.

Recall that the sample standard deviation is not an unbiased estimator for σ. When X_1, …, Xn is a random sample from a normal distribution, it can be shown (cf. Exercise 6.37) that

E(S) = an · σ

where

and Γ(·) denotes the gamma function (see Section 4.4). A tabulation of an for selected n follows:

where S_1, S_2, …, Sk are the sample standard deviations for the k samples. Then

Thus

so is an unbiased estimator of σ.

Control Limits Based on the Sample Standard Deviations

where

comes from an unbiased estimator for σ. Selected values of bn appear in the accompanying table [their computation is based on using statistical theory and numerical integration to determine E (min(Z_1, …, Zn)) and E (max(Z_1, …, Zn))].

Control Limits Based on the Sample Ranges

where are the k individual sample ranges.

Example 16.3 (Example 16.2 continued)

It is easily verified from Table 16.1 that = .292, so = .292/b_3 = .292/1.693 = .172 and

These limits are identical to those based on , and again every i lies between the limits.

Recomputing Control Limits

We have assumed that the sample data used for estimating μ and σ was obtained from an in-control process. Suppose, though, that one of the points on the resulting control chart falls outside the control limits. Then if an assignable cause for this out-of-control situation can be found and verified, it is recommended that new control limits be calculated after deleting the corresponding sample from the data set. Similarly, if more than one point falls outside the original limits, new limits should be determined after eliminating any such point for which an assignable cause can be identified and dealt with. It may even happen that one or more points fall outside the new limits, in which case the deletion/recomputation process must be repeated.

Performance Characteristics of Control Charts

Generally speaking, a control chart will be effective if it gives very few out-of-control signals when the process is in control, but shows a point outside the control limits almost as soon as the process goes out of control. One assessment of a chart's effectiveness is based on the notion of ”error probabilities.” Suppose the variable of interest is normally distributed with known σ n the process is in control. One error probability is

α = P (a single sample gives a point outside the control limits when μ = μ_0)

The standardized variable Z = ( − μ_0)/(σ/) has a standard normal distribution when μ = μ_0 , so

α = P(Z > 3 or Z <-3) = Φ(-3.00) + 1 -Φ(3.00) = .0026

If 3.09 rather than 3 had been used to determine the control limits (this is customary in Great Britain), then

α = P(Z > 3.09 or Z <-3.09) = .0020

The use of 3-sigma limits makes it highly unlikely that an out-of-control signal will result from an in-control process.

Now suppose the process goes out of control because μ has shifted to μ +Δσ (Δ might be positive or negative); Δ is the number of standard deviations by which μ has changed. A second error probability is

We now standardize by first subtracting μ_0 + Δσ from each term inside the parentheses and then dividing by σ/. This gives

This error probability depends on Δ, which determines the size of the shift, and on the sample size n. In particular, for fixed Δ, β will decrease as n increases (the larger the sample size, the more likely it is that an out-of-control signal will result), and for fixed n, β decreases as | Δ | increases (the larger the magnitude of a shift, the more likely it is that an out-of-control signal will result). The accompanying table gives β for selected values of Δ when n = 4.

It is clear that a small shift is quite likely to go undetected in a single sample.

If 3 is replaced by 3.09 in the control limits, then α decreases from .0026 to .002, but for any fixed n and σ, β will increase. This is just a manifestation of the inverse relationship between the two types of error probabilities in hypothesis testing. For example, changing 3 to 2.5 will increase are plotted. Simulations reported in the article indicated that the performance of the chart with these limits is superior to that of the traditional chart.

EXERCISES Section 16.2 (4–13)

4. In the case of known μ and σ, what control limits are necessary for the probability of a single point being outside the limits for an in-control process to be .005?

5. Consider a 3-sigma control chart with center line at μ_0 and based on n = 5. Assuming normality, calculate the probability that a single point will fall outside the control limits when the actual process mean is

a. μ_0 + .5σ

b. μ_0—σ

c. μ_0 + 2σ

6. The table on page 636 gives data on moisture content for specimens of a certain type of fabric. Determine control limits for a chart with center line at height 13.00 based on σ = .600, construct the control chart, and comment on its appearance.

Data for Exercise 6

7. Refer to the data given in Exercise 6, and construct a control chart with an estimated center line and limits based on using the sample standard deviations to estimate σ. Is there any evidence that the process is out of control?

8. Refer to Exercises 6 and 7, and now employ control limits based on using the sample ranges to estimate σ. Does the process appear to be in control?

9. The accompanying table gives sample means and standard deviations, each based on n = 6 observations of the refractive index of fiber-optic cable. Construct a control chart, and comment on its appearance. [Hint: = 2317.07 and = 30.34.]

10. Refer to Exercise 9. An assignable cause was found for the unusually high sample average refractive index on day 22. Recompute control limits after deleting the data from this day. What do you conclude?

a. What is the ARL when the process is in control?

b. What is the ARL when n = 4 and the process mean has shifted to μ = μ_0 + σ?

c. How do the values of parts (a) and (b) compare to the corresponding values for a 3-sigma chart?

12. Apply the supplemental rules suggested in the text to the data of Exercise 6. Are there any out-of-control signals?

13. Calculate control limits for the data of Exercise 6, using the robust procedure presented in this section.

16.3 Control Charts for Process Variation

The control charts discussed in the previous section were designed to control the location (equivalently, central tendency) of a process, with particular attention to the mean as a measure of location. It is equally important to ensure that a process is under control with respect to variation. In fact, most practitioners recommend that control be established on variation prior to constructing an chart or any other chart for controlling location. In this section, we consider charts for variation based on the sample standard deviation S and also charts based on the sample range R. The former are generally preferred because the standard deviation gives a more efficient assessment of variation than does the range, but R charts were used first and tradition dies hard.

The S Chart

We again suppose that k independently selected samples are available, each one consisting of n observations on a normally distributed variable. Denote the sample standard deviations by s_1, s_2, …, sk, with . The values s_1, s_2, s_3, … are plotted in sequence on an S chart. The center line of the chart will be at height , and the 3-sigma limits necessitate determining 3σS (just as 3-sigma limits of an chart required with σ then estimated from the data).

Recall that for any rv Y, V(Y) = E(Y^2) – [E(Y)]2, and that a sample variance S 2 is an unbiased estimator r an S chart from the refractive index data of Exercise 9. Does the process appear to be in control with respect to variability? Why or why not?

18. When S 2 is the sample variance of a normal random sample, (n –1)S^2/σ^2 has a chi-squared distribution with n – 1 df, so

from which

This suggests that an alternative chart for controlling process variation involves plotting the sample variances and using the control limits

Construct the corresponding chart for the data of Exercise 9. [Hint: The lower- and upper-tailed chi-squared critical values for 5 df are .210 and 20.515, respectively.]

16.4 Control Charts for Attributes

The term attribute data is used in the quality control literature to describe two situations:

1. Each item produced is either defective or nondefective (conforms to specifications or does not).

2. A single item may have one or more defects, and the number of defects is determined.

In the former case, a control chart is based on the binomial distribution; in the latter case, the Poisson distribution is the basis for a chart.

The p Chart for Fraction Defective

Suppose that when a process is in control, the probability that any particular item is defective is p (equivalently, p is the long-run proportion of defective items for an in-control process) and that different items are independent of one another with respect to their conditions. Consider a sample of n items obtained at a particular time, and let X be the number of defectives and = X/n. Because X has a binomial distribution, E(X) = np and V(X) = np(l—p), so

Also, if np > 10 and n(1—p) > 10, has approximately a normal distribution.

In the case of known p

The control chart for number of defectives is based on the Poisson probability distribution. Recall that if Y is a Poisson random variable with parameter θ, then

E(Y) = θ V(Y) = θ σy =

Also, Y has approximately a normal distribution when θ is large (θ ≥ 10 will suffice for most purposes). Furthermore, if Y_1, Y_2, …, Yn are independent Poisson variables with parameters θ_1, θ_2, …, θn, it can be shown that Y_1 + … + Yn has a Poisson distribution with parameter θ_1 + … + θn. In particular, if θ_1 = … = θn = θ (the distribution of the number of defects per item is the same for each item), then the Poisson parameter is λ = n θ.

Let λ denote the Poisson parameter for the number of defects in a unit (it is the expected number of defects per unit). In the case of known λ (or a chart based on a target value),

LCL = λ – UCL = λ +

With xi denoting the total number of defects in the ith unit (i = 1, 2, 3, …), then points at heights x_1, x_2, x_3, … are plotted on the chart. Usually the value of λ must be estimated from the data. Since E(Xi) = λ, it is natural to use the estimate (based on x_1, x_2, …, xk ).

The c chart for the number of defectives in a unit has center line at height and

If LCL is negative, it is replaced by 0.

Example 16.7

A company manufactures metal panels that are baked after first being coated with a slurry of powdered ceramic. Flaws sometimes appear in the finish of these panels, and the company wishes to establish a control chart for the number of flaws. The numbers of flaws in each of the 24 panels sampled at regular time intervals are as follows:

with = 235 and = 235/24 = 9.79. The control limits are

LCL = 9.79 –18

The control chart is in Figure 16.6 (page 644). The point corresponding to the fifteenth panel lies above the UCL. Upon investigation, the slurry used on that panel was discovered to be of unusually low viscosity (an assignable cause). Eliminating that observation from the data set gives = 214/23 = 9.30 and new control limits

LCL = 9.30 – = .15 UCL = 9.30 + = 18.45

The remaining 23 observations all lie between these limits, indicating an in-control process.

Figure 16.6 Control chart for number of flaws data of Example 16.7

Control Charts Based on Transformed Data

The use of 3-sigma control limits is presumed to result in p (statistic < LCL) ≈ P (statistic > UCL) ≈ .0013 when the process is in control. However, when p is small, the normal approximation to the distribution of = X/n will often not be very accurate in the extreme tails. Table 16.3 gives evidence of this behavior for selected values of p and n (the value of p is used to calculate the control limits). In many cases, the probability that a single point falls outside the control limits is very different from the nominal probability of .0026.

Table 16.3 In-Control Probabilities for a p Chart

This problem can be remedied by applying a transformation to the data. Let h(X) denote a function applied to transform the binomial variable X. Then h(¸) should be chosen so that h(X) has approximately a normal distribution and this approximation is accurate in the tails. A recommended transformation is based on the arcsin (i.e., sin-1) function:

Then Y is approximately normal with mean value sin-1 and variance 1/(4n); note that the variance is independent of p . Let yi = sin-1. Then points on the control chart are plotted at heights y1
where ”size” might refer to area, length, volume, or simply the number of items inspected. Then a u chart plots u_1, u_2, …, has center line , and the control limits for the ith observations are ± .

Painted panels were examined in time sequence, and for each one, the number of blemishes in a specified sampling region was determined. The surface area (ft^2) of the region examined varied from panel to panel. Results are given below. Construct a u chart.

26. Construct a control chart for the data of Exercise 23 by using the transformation suggested in the text.

16.5 CUSUM Procedures

A defect of the traditional X chart is its inability to detect a relatively small change in a process mean. This is largely a consequence of the fact that whether a process is judged out of control at a particular time depends only on the sample at that time, and not on the past history of the process. Cumulative sum (CUSUM) control charts and procedures have been designed to remedy this defect.

There are two equivalent versions of a CUSUM procedure for a process mean, one graphical and the other computational. The computational version is used almost exclusively in practice, but the logic behind the procedure is most easily grasped by first considering the graphical form.

The V-Mask

Let μ_0 denote a target value or goal for the process mean, and define cumulative sums by

(in the absence of a target value, is used in place of μ_0). These cumulative sums are plotted over time. That is, at time l, we plot a point at height Sl . At the current time point r , the plotted points are (1, S_1), (2, S_2), (3, S_3), …, (r , Sr ).

Now a V-shaped ”mask” is superimposed on the plot, as shown in Figure 16.7. The point 0, which lies a distance d behind the point at which the two arms of the mask intersect, is positioned at the current CUSUM point (r , Sr ). At time r, the process is judged out of control if any of the plotted points lies outside the V-mask—either above the upper arm or below the lower arm. When the process is in control, the i s will vary around the target value μ

μ_0 + k = 40.15 μ_0 –k = 39.85

so

dl = max[0, dl–1 + (l –40.15)]

el = max[0, el–1 –( –39.85)]

Calculations of the first few dl s proceeds as follows:

d_0 = 0

d_1 = max[0, d_0 + (1 –40.15)]

= max[0, 0 + (40.20 –40.15)]

= .05

d_2 = max[0, d_1 + (2 –40.15)]

= max[0, .05 + (39.72 –40.15)]

= 0

d_3 = max[0, d_2 + (3 –40.15)]

= max[0, 0 + (40.42 –40.15)]

= .27

The remaining calculations are summarized in Table 16.5.

Table 16.5 CUSUM Calculations for Example 16.9

The value h = .95 gives a CUSUM procedure with desirable properties—false alarms (incorrect out-of-control signals) rarely occur, yet a shift of Δ= .3 will usually be detected rather quickly. With this value of h, the first out-of-control signal comes after the 13th sample is available. Since d_13 = 1.17 > .95, it appears that the mean has shifted to a value larger than the target. This is the same message as the one given by the V-mask in Figure 16.9(b).

Equivalence of the V-Mask and Computational Form

Again let r denote the current time point, so that 2, …, r are available. Figure 16.10 displays a V-mask with the point labeled 0 at (r , Sr ). The slope of

Figure 16.10 A V-mask with slope of lower arm = k

the lower arm, which we denote by k, is h/d. Thus the points on the lower arm above r, r − 1, r − 2,… are at heights Sr − h, Sr − h − k , Sr − h − 2k , and so on.

The process is in control if all points are on or between the arms of the mask. We wish to describe this condition algebraically. To do so, let

The conditions under which all points are on or above the lower arm are

Now subtract rk from both sides of each inequality to obtain

Thus all plotted points lie on or above the lower arm if and only if (iff) Tr –Tr ≤ h , Tr –Tr—1 ≤ h , Tr –Tr—2 ≤ h, and so on. This is equivalent to

Tr –min(T_1, T_2, …, Tr ) ≤ h

In a similar manner, if we let

it can be shown that all points lie on or below the upper arm iff

max(V_1, …, Vr )—Vr ≤ h

If we now let

dr = Tr –min(T_1,…, Tr )

er = max(V_1, …, Vr ) –Vr

it is easily seen that d_1, d_2, … and e_1, e_2, … can be calculated recursively as illpreviously. For example, the expression for dr follows from consideration of two cases:

1. min(T_1, …, Tr ) = Tr , whence dr = 0

2. min(T_1, …, Tr ) = min(T_1, …, Tr-j ), so that

dr = Tr –min(T_1, …, Tr-1 )

= r –(μ_0 + k) + Tr-j –min(T_1, …,Tr-1 )

= r –(μ_0 + k) + dr-1

Since dr cannot be negative, it is the larger of these two quantities.

Designing a CUSUM Procedure

Let Δ denote the size of a shift in μ that is to be quickly detected using a CUSUM procedure.* It is common practice to let k = Δ/2. Now suppose a quality control practitioner specifies desired values of two average run lengths:

1. ARL when the process is in control (μ = μ 0)

2. ARL when the process is out of control because the mean has shifted by Δ (μ = μ_0 + Δ or μ = μ_0 –Δ)

A chart developed by Kenneth Kemp (“The Use of Cumulative Sums for Sampling Inspection Schemes,” Applied Statistics, 1962: 23), called a nomogram, can then be used to determine values of h and n that achieve the specified ARLs.[†] This chart is shown as Figure 16.11. The method for using the chart is described in the accompanying box. Either the value of σ must be known or an estimate is used in its place.

Using the Kemp Nomogram

1. Locate the desired ARLs on the in-control and out-of-control scales. Connect these two points with a line.

2. Note where the line crosses the k′ scale, and solve for n using the equation

Then up to the nearest integer.

3. Connect the point on the k′ scale with the point on the in-control ARL scale using a second line, and note where this line crosses the h′ scale. Then h = ( • h′.

* This contrasts with previous notation, where Δ represented the number of standard deviations by which μ changed.

† The word nomogram is not specific to this chart; nomograms are used for many other purposes.

Figure 16.11 The Kemp nomogram*

The value h = .95 was used in Example 16.9. In that situation, it follows that the in-control ARL is 500 and the out-ences should be consulted for information on these procedures.

EXERCISES Section 16.5 (27–30)

27. Containers of a certain treatment for septic tanks are supposed to contain 16 oz of liquid. A sample of five containers is selected from the production line once each hour and the sample average content is determined. Consider the following results: 15.992, 16.051, 16.066, 15.912, 16.030, 16.060, 15.982, 15.899, 16.038, 16.074, 16.029, 15.935, 16.032, 15.960, 16.055. Using Δ = .10 and h = .20, employ the computational form of the CUSUM procedure to investigate the behavior of this process.

28. The target value for the diameter of a certain type of drive-shaft is .75 in. The size of the shift in the average diameter considered important to detect is .002 in. Sample average diameters for successive groups of n = 4 shafts are as follows: .7507, .7504, .7492, .7501, .7503, .7510, .7490, .7497, .7488, .7504, .7516, .7472, .7489, .7483, .7471, .7498, .7460, .7482, .7470, .7493, .7462, .7481. Use the computational form of the CUSUM procedure with h = .003 to see whether the process mean remained on target throughout the time of observation.

29. The standard deviation of a certain dimension on an aircraft part is .005 cm. What CUSUM procedure will give an in-control ARL of 600 and an out-of-control ARL of 4 when the mean value of the dimension shifts by .004 cm?

30. When the out-of-control ARL corresponds to a shift of 1 standard deviation in the process mean, what are the characteristics of the CUSUM procedure that has ARLs of 250 and 4.8 for the in-control and out-of-control conditions, respectively?

16.6 Acceptance Sampling

Items coming from a production process are often sent in groups to another company or commercial establishment. A group might consist of all units from a particular production run or shift, in a shipping container of some sort, sent in response to a particular order, and so on. The group of items is usually called a lot, the sender is referred to as a producer, and the recipient of the lot is the consumer. Our focus will be on situations in which each item is either defective or nondefective, with p denoting the proportion of defective units in the lot. The consumer would naturally want to accept the lot only if the value of p is suitably small. Acceptance sampling is that part of applied statistics dealing with methods for deciding whether the consumer should accept or reject a lot.

Until quite recently, control chart procedures and acceptance sampling techniques were regarded by practitioners as equally important parts of quality control methodology. This is no longer the case. The reason is that the use of control charts and other recently developed strategies offers the opportunity to design quality into a product, whereas acceptance sampling deals with what has already been produced and thus does not provide for any direct control over process quality. This led the late American quality control expert n a double-sampling plan, the number of defective items x_1 in an initial sample of size n_1 is determined. There are then three possible courses of action: immediately accept the lot, immediately reject the lot, or take a second sample of n_2 items and reject or accept the lot depending on the total number x_1 + x_2 of defective items in the two samples. Besides the two sample sizes, a specific plan is characterized by three further numbers—c_1, r_1, and c_2—as follows:

1. Reject the lot if x_1 ≥ r_1.

2. Accept the lot if x_1 ≤ c_1.

3. If c_1 < x_1 < r_1, take a second sample; then accept the lot if x_1 + x_2 ≤ c_2 and reject it otherwise.

Example 16.13

Consider the double-sampling plan with n_1 = 80, n_2 = 80, c_1 = 2, r_1 = 5, and c_2 = 6. Thus the lot will be accepted if (1) x_1 = 0, 1, or 2; (2) x_1 = 3 and x_2 = 0, 1, 2, or 3; or (3) x_1 = 4 and x_2 = 0, 1, or 2.

Assuming that the lot size is large enough for the binomial approximation to apply, the probability P(A) of accepting the lot is

Again the graph of P(A) versus p is the plan's OC curve. The OC curve for this plan appears in Figure 16.13.

Figure 16.13 OC curve for the double-sampling plan of Example 16.13

One standard method for designing a double-sampling plan involves proceeding as suggested earlier for single-sample plans. Specify values p_1 and p_2 along with corresponding acceptance probabilities 1—α and β. Then find a plan that satisfies these conditions. The book by Montgomery provides tables similar to Table!6.6 for this purpose in the cases n_2 = n_1 and n_2 = 2n_1 with 1—α = .95, β = .10. Much more extensive tabulations of plans are available in other sources.

Analogous to standard practice with single-sample plans, it is recommended that alrior to inspection of the n_1th item. However, it is customary to terminate inspection of the second sample if the number of defectives is sufficient to justify rejection before all items have been examined. This is referred to as curtailment in the second sample. Under curtailment, it can be shown that the expected number of items inspected in a double-sampling plan is smaller than the number of items examined in a single-sampling plan when the OC curves of the two plans are close to being identical. This is the major virtue of double-sampling plans. For more on these matters as well as a discussion of multiple and sequential sampling plans (which involve selecting items for inspection one-by-one rather than in groups), a book on quality control should be consulted.

Rectifying Inspection and Other Design Criteria

In some situations, sampling inspection is carried out using rectification. For single-sample plans, this means that each defective item in the sample is replaced with a satisfactory one, and if the number of defectives in the sample exceeds the acceptance cutoff c, all items in the lot are examined and good items are substituted for any defectives. Let N denote the lot size. One important characteristic of a sampling plan with rectifying inspection is average outgoing quality, denoted by AOQ. This is the long-run proportion of defective items among those sent on after the sampling plan is employed. Now defectives will occur only among the N—n items not inspected in a lot judged acceptable on the basis of a sample. Suppose, for example, that P(A) = P(X < c) = .985 when p = .01. Then in the long run, 98.5% of the N—n items not in the sample will not be inspected, of which we expect 1% to be defective. This implies that the expected number of defectives in a randomly selected batch is (N—n ) · P (A ) · p = .00985(N—n ). Dividing this by the number of items in a lot gives average outgoing quality:

Because AOQ = 0 when either p = 0 or p = 1 [P (A ) = 0 in the latter case], it follows that there is a value of p between 0 and 1 for which AOQ is a maximum. The maximum value of AOQ is called the average outgoing quality limit, AOQL. For example, for the plan with n = 137 and c = 3 discussed previously, AOQL = .0142, the value of AOQ at p ≈ .02.

Proper choices of n and c will yield a sampling plan for which AOQL is a specified small number. Such a plan is not, however, unique, so another condition can be imposed. Frequently this second condition will involve the average (i.e., expected) total number inspected, denoted by ATI. The number of items inspected in a randomly chosen lot is a random variable that takes on the value n with probability P(A) and N with probability 1—P(A). Thus the expected number of items inspected in a randomly selected lot is

ATI = n · P(A) + N · (1—P (A ))

27. a. The observation 50 falls on a class boundary.

b.

A representative (central) value is either a bit below or a bit above 100, depending on how one measures center. There is a great deal of variability in lifetimes, especially in values at the upper end of the data. There are several candidates for outliers.

c.

There is much more symmetry in the distribution of the ln (x) values than in the x values themselves, and less variability. There are no longer gaps or obvious outliers.

d..38, .14

29.

31.

33. a. = 192.57, = 189.0

b. New = 189.71; unchanged

c. 191.0, 7.14%

d. 122.6

35. a. = 12.55, = 12.50, tr(12.5) = 12.40. Deletion of the largest observation (18.0) causes and tr to be a bit smaller than .

b. By at most 4.0

c. No; multiply the values of and by the conversion factor 1/2.2.

37. tr(10) = 11.46

39. a. = 1.0297, = 1.009

41. a..7

b. Also .7

c. 13

43. = 68.0, tr(20) = 66.2,tr(30) = 67.5

45. a. = 115.58; the deviations are .82, .32, -.98, -.38, .22

b..482, .694

c..482

d..482

47. = 116.2, s = 25.75. The magnitude of s indicates a substantial amount of variation about the center (a “representative” deviation of roughly 25).

49. a. 56.80, 197.8040

b..5016, .708

51. a. 1264.766, 35.564

b..351, .593

53. a. 2.74, 3.88

b. 1.14

c. Unchanged

d. At most .40

e. 1.19

55. a. 33

b. No

c. Slight positive skewness in the middle half, but rather symmetric overall. The extent of variability appears substantial.

d. At most 32

57. a. Yes. 125.8 is an extreme outlier and 25ng> The mean, median, and trimmed mean are virtually identical, suggesting a substantial amount of symmetry in the data; the fact that the quartiles are roughly the same distance from the median and that the smallest and largest observations are roughly equidistant from the center provides additional support for symmetry. The standard deviation is quite small relative to the mean and median.

b. See the comments of (a). In addition, using 1.5(Q3 − Q1) as a yardstick, the two largest and three smallest observations are mild outliers.

73. = .9255, s = .0809, = .93, small amount of variability, slight bit of skewness

75. a. The “five-number summaries” (, the two fourths, and the smallest and largest observations) are identical and there are no outliers, so the three individual boxplots are identical.

b. Differences in variability, nature of gaps, and existence of clusters for the three samples.

c. No. Detail is lost.

77. a.

b.

79. a. n+1 = (n+xn+1/(n+1)

c. 12.53, .532

81. A substantial positive skew (assuming unimodality)

83. a. All points fall on a 45° line. Points fall below a 45° line.

b. Points fall well below a 45° line, indicating a substantial positive skew.

Chapter 2

1. a. S = {1324, 3124, 1342, 3142, 1423, 1432, 4123, 4132, 2314, 2341, 3214, 3241, 2413, 2431, 4213, 4231}

< 3241, 2413, 2431, 4213, 4231},

A ∩ B contains no outcomes (A and B are disjoint),

A′ = {3124, 3142, 4123, 4132, 2314, 2341, 3214, 3241, 2413, 2431, 4213, 4231}

3. a. A = {SSF, SFS, FSS}

b. B = {SSF, SFS, FSS, SSS}

c. C = {SFS, SSF, SSS}

d. C′ = {FFF, FSF, FFS, FSS, SFF},

A ∪ C = {SSF, SFS, FSS, SSS},

A ∩ C = {SSF, SFS},

B ∪ C = {SSF, SFS, FSS, SSS} = B,

B ∩ C = {SSF, SFS, SSS} = C

5. a. S = {(1, 1, 1), (1, 1, 2), (1, 1, 3), (1, 2, 1), (1, 2, 2), (1, 2, 3), (1, 3, 1), (1, 3, 2), (1, 3, 3), (2, 1, 1), (2, 1, 2), (2, 1, 3), (2, 2, 1), (2, 2, 2), (2, 2, 3), (2, 3, 1), (2, 3, 2), (2, 3, 3), (3, 1, 1), (3, 1, 2), (3, 1, 3), (3, 2, 1), (3, 2, 2), (3, 2, 3), (3, 3, 1), (3, 3, 2), (3, 3, 3)}

b. {(1, 1, 1), (2, 2, 2), (3, 3, 3)}

c. {(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1)}

d. {(1, 1, 1), (1, 1, 3), (1, 3, 1), (1, 3, 3), (3, 1, 1), (3, 1, 3), (3, 3, 1), (3, 3, 3)}

7. a. There are 35 outcomes in S .

b. {AABABAB, AABAABB, AAABBAB, AAABABB, AAAABBB}

11. a..07

b..30

>

d..47

e..17

f..75

15. a..572

b..879

17. a. There are statistical software packages other than SPSS and SAS.

b..70

c..80

d..20

19. a..8841

b..0435

21. a..10

b..18, .19

c..41

d..59

e..31

f..69

23. a..067

b..400

c..933

d..533

25. a..98

b..02

c..03

d..24

29. a. 676; 1296

b. 17,576; 46,656

c. 456,976; 1,679,616

d..942

31. a. 243

b. 3645 days (right="0em" width="-3em">
77. a..06235

b..00421

79..0059

81. a..95

83. a..10, .20

b. 0

85. a. p (2 -p )

b. 1 − (1 − p )n

c. (1 − p )3

d..9 + (1 − p )3(.1)

e..1(1 − p )3/[.9 + .1(1 − p )3] = .0137 for p = .5

87..8588, .9897

89. [2π77(1 − π)]/(1 − π^2)

91. a..333, .444

b..150

c..291

93..45, .32

95. a..0083

b..2

c..2

97..905

99. a..956

b..994

101.m" width="-1em">
b..9364

c..9677

d..9364

31..74, .8602, .85

33. a. p

b. p (1 − p )

c. p

35. E[h_3(X)] = 2.4667, E[h_4 ((X)] = 2.667, so 4 copies is better.

37. E(X) = (n + 1)/2, E(X^2) = (n + 1)(2 n + 1)/6, V(X) = (n^2 − 1)/12

39. 2.3, .81, 88.5, 20.25

43. E(X − c ) = E(X ) − c , E (X − μ) = 0

47. a..515

b..218

c..011

d..480

e..965

f..000

g..595

49. a..354

b..115

c..918

51. a. 6.25

b. 2.17

c..030

53lockquote>

c..774

55..1478

57..407, independence

59. a..017

b..811, .425

c..006, .902, .586

61. When p = .9, the probability is .99 for A and .9963 for B. If p = .5, these probabilities are .75 and .6875, respectively.

63. The tabulation for p > .5 is unnecessary.

65. a. 20, 16

b. 70, 21

67. P (|X − μ | ≤ 2σ) = .042 when p = .5 and = .065 when p = .75, compared to the upper bound of .25. Using k = 3 in place of k = 2, these probabilities are .002 and .004, respectively, whereas the upper bound is .11.

69. a..114

b..879

c..121

d. Use the binomial distribution with n = 15, p = .10

71. a. h(x; 15, 10, 20) for x = 5,…,10

b..0325

c..697

73. a. h(x; 10, 10, 20)

b..033

77. nb(x; 6, .5), 6

79. a..932

b..065

c..068

d..492

e..251

81. a..011

b..441

c..554, .459

d..945

83. Poisson(5)

a..492

b..133

85. a..122, .809, .283

b. 12, 3.464

c..530, .011

87. a..099

b..135

c. 2

89. a. 4

b..215

c. At least -ln(.1)/2 ≈ 1.1513 years

91. a..221

b. 6,800,000

c. p(x; 20.106)

95. b. 3.114, .405, .636

9face="Times New Roman">Chapter 4

1. a. 25

b..50

c..4375

3. b..5

c. 6875

d..6328

5. a..375

b..125

c..297

d..578

7. a. f(x) = .1 for 25 ≤ x ≤ 35 and 0 otherwise

b..20

c..40

d..20

9. a..562

b..438, .438

c..071

11. a..25

b..1875

c..9375

d. 1.4142

e. f(x) = x/2 for 0 > x > 2

f. 1.33

g..222, .471

h. 2

c..125, .088

d. 1.5, .866

e..924

15. a. F(x) = 0 for x ≤ 0, = 90 for 0 < x < 1, = 1 for x≥1

b..0107

c..0107, .0107

d..9036

e..818, .111

f..3137

17. a. A + (B − A)p

b. E(X) = (A + B)/2, σX = (B − A)/

c. [Bn + 1 − An + 1]/[(n + 1)(B − A )]

19. a..597

b..369

c. f (x) = .3466 − .25 ln(x) for 0 > x > 4

21. 314.79

23. 248, 3.60

25. b. 1.8(90th percentile for X) + 32

c. a(X percentile) + b

27. 0, 1.814

29. a. 2.14

b..81

c. 1.17

d..97

31. a. 2.54

b. 1.34

c. -.42

33. a..9918

b..0082

c..8664

35. a..3336

b. Approximately 0

c..5795

d. 6.524

e..8028

37. a. 0, .5793, .5793

b..3174, no

c. > 87.6 or < 120.4

39. a. 36.7

b. 22.225

c. 3.179

41..002

43. 10, .2

45. 7.3%

47. 21.155

49. a..1190, .6969

b..0021

c..7054

d. >5020 or <1844 (using z.0005 = 3.295)

e. Normal, = = 7.576, > = 1.064, .7054

69. a. ∩ Ai

b. Exponential with λ = .05

c. Exponential with parameter n λ

73. a..826, .826, .0636

b..664

c. 172.727

77. a. 123.97, 117.373

b..5517

c..1587

79. a. 68.0, 122.1

b..3204

c..7257, skewness

81. a. 149.157, 223.595

b..9573

c..0414

d. 148.41

e. 9.57

f. 125.90

83. α = β

85. b. [Γ(α + β ) • Γ(m + β )]/[Γ(α + β +m ) • Γ(β )], β /(α + β )

87. Yes, since the pattern in the plot is quite linear.

89. Yes

91. Yes

93. Plot ln(x) vs. z percentile. The pattern is straight, so a log-normal p5. The pattern in the plot is quite linear; it is very plausible that strength is normally distributed.

97. There is substantial curvature in the plot. λ isa scale parameter (as is σ for the normal family).

99. a. F(y) = (y^2 − y^3 /18) for 0 ≤ y ≤ 12

b..259, .5, .241

c. 6, 43.2, 7.2

d..518

e. 3.75

101. a. (f(x) = x^2 for 0 ≤ < 1 and = for 1 ≤ x ≤

b..917

c. 1.213

103. a..9162

b..9549

c. 1.3374

105. a..3859

b..0663

c. (72.97, 119.03)

107. b. F(x) = 0 for x <- 1, = (4x − x 3/3)/9 + for -1 ≤ x ≤ 2, and = 1 for x > 2

c. No. F (0) < .5 ⇒ > 0

d. Y ~ Bin(10, )

109. a..368, .828, .460

b. 352.53

c. 1/^face="Times New Roman">Chapter 5

1. a..20

b..42

c. At least one hose is in use at each pump; .70.

d. pX(x ) = .16, .34, .50 for x = 0, 1, 2, respectively; pY (y ) = .24, .38, .38 for y = 0, 1, 2, respectively; .50

e. No; p (0, 0) ≠ pX(0) · pY(0)

3. a..15

b..40

c..22

d..17, .46

5. a..054

b..00018

7. a..030

b..120

c..300

d..380

e. Yes

9. a. 3/380,000

b..3024

c..3593

d. 10Kx 2+ .05 for 20 ≤ x ≤ 30

e. No

11. a. e -λ-μ • λx • μy /x!y!

b. e -λ-μ • [1 + λ + μ]

c. e-(λ+μ) • (λ + μ)m/m!; Poisson (λ + μ)
< gE2="1/p>

13. a. e-x-y for x ≥ 0, y ≥ 0

b..400

c..594

d..330

15. a. F(y) = 1 e-λy + (1 − e-λy )2 − (1 − e-λy )3for y ≥ 0

b. 2/3λ

17. a..25

b..318

c..637

d.fx(x) = /πR 2 for -R ≤ x ≤ R ; no

19. a. K(x^2 + y^2)/(10Kx^2 + .05); K(x^2 + y^2)/(10Ky^2 + .05)

b..556, .549

c. 25.37, 2.87

21. a. f(x_1, x_2, x_3 /fX_1,x_2(x_1, x_2)

b. f (x_1, x_2, x_3)/fx_1(x_1)

23..15

25. L 2

27..25 hr

29.

31. a. -.108t="Image">) = μ = 44.5

b,E (S 2) = 212.25 = σ^2

39.

41. a.

b. b..85

c.

47. a..6826

b..1056

49. a..6026

b..2981

51..7720

53. a..0062

b.0

55. a..9838

b..8926

57..9616

59. a..9986, .9986

b..9015, .3970

c..8357

d..9525, .0003

61. a. 3.5, 2.27, 1.51

b. 15.4, 75.94, 8.71

63. a..695

b. 4.0675 > 2.6775

65. a..9232

b..9660

67..158rong> 97

85..9973

89. b, c. Chi-squared with v = n.

91. a.

b..9999

93. 26, 1.64

95. a..6

b. U = ρX +

Chapter 6

1. a. 8.14,

b..77,

c. 1.66, S

d..148

e..204, S

3. a. 1.348,

b. 1.348,

c. 1.781, + 1.28S

d..6736

e..0905

5. N = 1,703,000; T − N ; T • (/ 1,601,438.281

7. a. 120.6

b. 1,206,000

c..80

d.120.09. a. 2.11

b..119

11. b.

c. Use i = xi /ni and i = 1 -i in place of pi and qi in part (b) for i = 1,2.

d. -.245

e..041

15. a.

b. 74.505

17. b..444

19. a. = 2 − .30 = .20

b. = (100 − 9)/70

21. b. = 5, = 28.0/Γ(1.2)

23. 1 = , 2 = , estimate of (λ_1 − λ_2) is − .

25. a. 384.4, 18.86

b. 415.42

29. a. = min(Xi), = n/Σ[Xi − min(Xi )]

b..64, .202

33. With xi time between birth i − 1 a

37. 1.0132

Chapter 7

1. a. 99.5%

b. 85%

c. 2.96

d. 1.15

3. a. Narrower

b. No

c. No

d. No

5. a. (4.52, 5.18)

b. (4.12, 5.00)

c..55

d. 94

7. By a factor of 4; the width is decreased by a factor of 5.

9. a. ( − 1.645σ/); (4.57, ∞)

b. ( − z∞ • )

c. (-∞, + z∞ • σ/); (-∞, 59.7)

11. 950, .8714

13. a. (608.58, 699.74)

b. 189

15. a. 80%

19. (.513, .615)

21..218

23. a. (.438, .814)

b. 659

25. a. 381

b. 339

29. a. 2.228

b. 2.086

c. 2.845

d. 2.680

e. 2.485

f. 2.571

31. a. 1.812

b. 1.753

c. 2.602

d. 3.747

e. 2.1716 (from MINITAB)

f. Roughly 2.43

33. a. Reasonable amount of symmetry, no outliers

b. Yes (based on a normal probability plot)

c. (430.5, 446.1), yes, no

35. a. 95% CI: (23.1, 26.9)

b. 95% PI: (17.2, 32.8), roughly 4 times as wide

37. a. (.888, .964)

b. (.752, 1.100)

c. (.634, 1.218)

39. a. Yes

b. (6.45, 98.01)

c. (18.63, 85.83)

41. All 70%; (c), because it is shortest

43. a. 18.307

b. 3.940

c..95

d..10

45. (3.6, 8.1); no

47. a. 95% CI: (6.702, 9.456)

b. (.166, .410)

49. a. There appears to be a slight positive skew in the middle half of the sample, but the lower whisker is much longer than the upper whisker. The extent of variability is rather substantial, although there are no outliers.

b. Yes. The pattern of points in a normal probability plot is reasonably linear.

c. (33.53, 43.79)

51. a. (.624, .732)

b. 1080

c. No

53. (- .84, -.16)

55. 246

57. = (65.3, 232.5)

59. a. (max(xi) /(1 − α/2)1/n, max(xi )/(α/2)1/n)

b. (max(xi ), max(xi )/α1/n )

c. (b); (4.2, 7.65)

61. (73.6, 78.8) versus (75.1, 79.6)

1. a. Yes

b. No

c. No

d. Yes

e. No

f. Yes

5. H_0: σ = .05 versus Ha: σ < .05. I: conclude variability in thickness is satisfactory when it isn’t. II: conclude variability in thickness isn’t satisfactory when in fact it is.

7. I: concluding that the plant isn’t in compliance when it is; II: concluding that the plant is in compliance when it isn’t.

9. a. R_1

b. I: judging that one of the two companies is favored over the other when that is not the case; II: judging that neither company is favored over the other when in fact one of the two really is preferred.

c..044

d. β(.3) = β(.7) = .488, β(.4) = β(.6) = .845

e. Reject H_0 in favor of Ha.

11. a. H_0: μ = 10 versus Ha: μ ≠ 10

b..01

c..5319, .0078

d. 2.58

e. 10.1032 is replaced by 10.124, and 9.8968 is replaced by 9.876.

f. = 10.020, so H_0 should not be rejected.

g.z ≥ 2.58 or ≤ -2.58

13. b..0004, 0, lnt>

b..003

c..004

17. a. z = 2.56 ≥ 2.33, so reject H_0.

b..8413

c. 143

d..0052

19. a. z =-2.27, so don’t reject H_0.

b..2266

c. 22

21. a. t.025,12 = 2.179 > 1.6, so don’t reject H_0: μ = .5.

b. -1.6 >-2.179, so don’t reject H_0.

c. Don’t reject H_0.

d. Reject H_0 in favor of Ha: μ ≠ .5.

23.t = 2.24 ≥ 1.708, so H_0 should be rejected. The data does suggest a contradiction of prior belief.

25. a. z =-3.33 ≤ -2.58, so reject H_0.

b..1056

c. 217

27. a. = .750, = .640, s = .3025, fs = .480. A boxplot shows substantial positive skew; there are no outliers. b. No. A normal probability plot shows substantial curvature. No, since n is large.

c. z = -5.79; reject H_0 at any reasonable significance level; yes.

d..821

d..653

e. <.005

f. ≈ .000

51. P -value > α, so don’t reject H_0; no apparent difference.

53. P -value < .0004 < .01, so H_0: μ = 5 should be rejected in favor of Ha: μ ≠ 5.

55. No; P -value ≈ .2

57. t ≈ 1.9, so P -value ≈ .041. Since P -value − ≤ α, H_0: μ = 25 should be rejected in favor of Ha: μ > 25.

59. t ≈ 1.9, so P -value ≈ .116. H_0 should therefore not be rejected.

61. a..8980, .1049, .0014

b. P -value ≈ 0. Yes.

c. No

63. z = -3.12 ≤ -1.96, so H_0 should be rejected.

b..10. Yes.

75. Yes. z =-3.32 ≤ -3.08, so H_0 should be rejected.

77. No, since z = 1.33 < 2.05.

79. P -value ≈ 0, so reject H_0; it appears that μ > 15.

81. a..01 < P -value < .025, so do not reject H_0; no extradiction

83. a. For H_2: μ < μ_0, reject H_0 if

b. Test statistic value = 19.65 > 8.260, so do not reject H_0.

85. a. Yes, α = .002

Chapter 9

1. a. -.4 hr; it doesn’t

b..0724, .2691

c. No

3. z = 1.76 < 2.33, so don’t reject H_0.

5. a. z =-2.90, so reject H_0.

b..0019

c..8212

d. 66

7. Yes, since z = 1.83 ≥ 1.645.

9. a. 6.2; yes

b. z = 1.14, P -value ≈ .25, no

se weight gain.

63. (); (.023, 1.99)

65. No. t = 3.2, df = 15, P -value = .006, so reject H_0: μ_1 − μ_2 = 0 using either α = .05 or .01.

67. z > 0 d P -value > .5, so H_0: p_1 − p_2 = 0 cannot be rejected.

93. (- 1.29, -.59)

Chapter 10

1. a. f = 1.85 < 3.06 = F.05,4,15’ so don’t reject H_0. yn

b. P -value > .10

3. f= 1.30 < 2.57 = F.10,2,21’ so P -value > .10. H_0 cannot be rejected at any reasonable significance level.

5. f= 1.73 < 5.49 = F.01,2,27’ so the three grades don’t appear to differ.

7. f= 1.70 < 2.46 = F.10,3,16’ so P -value > .10. H_0 cannot be rejected at any reasonable significance level.

9. f= 3.96 and F05,3,20 = 3.10 < 3.96 < 4.94 = F.01,3,20’ so .01 < P -value < .05. Thus H_0 can be rejected at significance level .05; there appear to be differences among the grains.

11. w = 36.09

Brands 2 and 5 don’t appear to differ, nor does there appear to be any difference between brands 1, 3, and 4, but each brand in the first group appears to differ significantly from all brands in the second group.

13.

15. w = 5.94 The only significant differences are between 4 and both

17. (- .029, .379)

19. Any value of SSE between 422.16 and 431.88 will work.

21. a. f = 22.6 and F.01,5,78 ≈ 3.3, so reject H_0.

b. (- 99.16, –35.64), (29.34, 94.16)

23.

25. a. Normal, equal variances

b. SSTr = 8.33, SSE = 77.79, f = 1.7, H_0 should not be rejected (P -value > .10)

27. a. f = 3.75 ≥ 3.10 = F.05,3,20, so brands appear to differ.

b. Normality is quite plausible (a normal probability plot of the residuals − xij -. shows a linear pattern).

c. Only brands 1 and 4 appear to differ significantly.

31. Approximately .62

33. arcsin()

35. a. 3.68 < 4.94, so H_0 is not rejected.

b..029 > .01, so again H_0 is not rejected.

37. f = 8.44 > 6.49 = F.001, so P -value < .001 and H_0 should be rejected.

This underscoring pattern is a bit awkward to interpret.

39. The CI is (- .144, .474), which does include 0.

41. f = 3.96 < 4.07, so H_0: = 0 cannot be rejected.

43. (-3.70, 1.04), (-4.83, -.33), (-3.77, 1.27), (-3.99, .15). Only μ_1 − μ_3 among these four contrasts appears to differ significantly from zero.

45. They are identical.

Chapter 11

1. a. fA = 1.55, so don’t reject H0A.

b. fB = 2.98, so don’t reject H0B.

3. a. fA = 12.987 ≥ F.01,3.9, so conclude font face="Times New Roman">Since F.01,8.30 = 3.17, F. 01,2.8 = 8.65, and F. 01,4.30 = 4.02,

23.

H0G is not rejected but both H0A and H0B are rejected.

25. (- 1.39, -1.05)

27. a.

d. Q05,3.27 = 3.51, w = 8.90, and all three of the levels differ significantly from one another.

29.

a. No interaction effects are significant.

b. Factor B and factor C main effects are significant.

c. w = 1.89; only machines 2 and 4 do not differ significantly from one another.

31. The P -value column shows that several interaction effects are significant at level .01.

33.

F .05,6.30 = 2.42, fC = .61, so H0C is not rejected.

35.

Since F.05,4.12 = 3.26, both A and B are significant.

37.

*Denotes a significant F ratio.

39. a.

b.

41.

F.05,1.32 ≈"1em" width="-1em">
c. 130

d. –130

9. a..095

b. -.475

c..830, 1.305

d..4207, .3446

e..0036

11. a. -.01, -.10

b. 3.00, 2.50

c..3627

d..4641

13. a. Yes, because r 2 = .972

15. a.

Typical value in low 40’s, reasonable amount of variability, positive skewness, two potential outliers

b. No

c. y =3.2925 + .10748x = 7.59. No; danger of extrapolation

d. 18.736, 71.605, .738, yes

17. a. We estimate that .144 is the expected change in calcium content associated with a 1 mgfcm2 increase in the amount of dissolved material. Roughly 86% of the observed variation in calcium content can be attributed to the simple linear regression model relationship between content and the amount of dissolved material.

b. 10.88

c. 1.46

19. a. y = –45.5519 + 1.7114x

b. 339.51

c. –85.57

<7;s are 125.6, 168.4, 168.4, 211.1, 211.1, 296.7, 296.7, 382.3, 382.3, 467.9, 467.9, 553.4, 639.0, 639.0; a 45° line through (0, 0).

21. a. Yes; r 2 = .985

b. 368.89

c. 368.89

23. a. 16,213.64; 16,205.45

b. 414,235.71; yes, since r 2 = .961.

27.

29.

31. a..001017

b. (- .00956, -.00516)

33. a. (.081, .133)

b. Ha: β_1 > .1, P -value = .277, no

35. a. (.63, 2.44) is a 95% CI.

b. Yes. t ≈3.6, P -value ≈.004

c. No; extrapolation

d. (.54, 2.82), no

37. a. Yes. t =7.99, P -value ≈ 0. Note: There is one mild outlier, so the resulting normal probability plot is not entirely satisfactory.

b. Yes. t =-5.8, P -value ≈ 0, so reject H_0: β_1 = 1 in favor of Ha: β_1 < 1

39. f =71.97, s = .004837, t =8.48, P -value = .000

43. d =1.20, df = 13, and β ≈ .1.

45. a. (77.80, 78.38v height="0em">

c. wider, since 115 is further from

d. t = –11, P -value = 0

47. a. 95% PI is (20.21, 43.69), no

b. (28.53, 51.92), at least 90%

49. (431.3, 628.5)

51. a..40 is closer to = .7495

b. (.745, .875)

c. (.059, .523)

53. (a) narrower than (b), (c) narrower than (d), (a) narrower than (c), (b) narrower than (d)

57. If, for example, 18 is the minimum age of eligibility, then for most people y ≈ x -18.

59. a..966

b. The percent dry fiber weight for the first specimen tends to be larger than for the second.

c. No change

d. 93.3%

e. t =14.9, P -value ≈ 0, so there does appear to be such a relationship.

61. a. r = .748, t =3.9, P -value = .001. Using either α = .05 or .01, yes.

b..560 (56%), same

63. r = .773, yet t =2.44 < 2.776; so H_0: ρ = 0 cannot be rejected.

65. a. The x plot is a bit curved but not disturbingly so in light of the small sample size. The y plot is quite straight.

b. t =6.3 ≥ 3.355, so there does appear to be a linear relationship.

67. a. Reject H_0

b. No. P -value = .00032 ⇒ z ≈ 3.6 ⇒ r ≈.16, which indicates only a weak relationship.

c. Yes, but very large n ⇒ ρ ≈ .022, so no practical significance.

69. a. 95% CI: (.888, 1.086)

b..95% CI: (47.730, 49.172)

c. 95% PI: (45.378, 51.524)

d. Narrower for x =25, since 25 is closer to

e..981

71. a. t =-1.24 > -2.201, so don’t reject H_0

b..970

73. a..507

b..712

c. P -value = .0013 < .01 = α, so reject H_0: β_1 = 0 and conclude that there is a useful linear relationship.

d. A 95% CI is (1.056, 1.275).

e. 1.0143, .2143

75. a. y =14.1904 –.14892 x

b. t =-1.43, so don’t reject H_0: β_1 = -.10.

c. No; Σ(x − )2 = 143 here and 182 for the given data.

d. A 95% CI for μ_7 Y .28 is (9.599, 10.443).

77. a. A substantial linear relationship

15. a. A curved pattern

b. A linear pattern

c. Y = αx β • ∈

d. A 95% PI is (3.06, 6.50).

e. One standardized residual, corresponding to the third observation, is a bit large. There are only two positive standardized residuals, but two others are essentially 0. The patterns in a standardized residual plot and normal probability plot are marginally acceptable.

17. a. ∑ =15.501, ∑ = 13.352, ∑() = 20.228, ∑ = 18.109, ∑ () = 16.572, 1 = 1.254, 0 = − .468, = .626, =1.254

c. t =-1.07, so don’t reject H_0.

d. H_0: β =1, t =-4.30, so reject H_0.

19. a. No

b. Y′ = β_0 + β_1 • (1/t) + ∈′, where Y′= ln(Y), so Y = αeβ/t • ∈.

c. = 1 = 3735.45, 0 = –10.2045, =(3.70034) • (10-5), = 6.7748, = 875.5

d. SSE = 1.39587, SSPE = 1.36594 (using transformed values), f = .33 < 8.68 = F.01,1,15’ so don’t reject H_0.

21. a. Y.x =18.14 − 1485/x

b. =15.17

23. For the exponential model, V(Y|x ) = α^2e2βxx . A similar result holds for the power model.

25. a. The point estimate of β_1 is = .1772, and the estimated odds ratio is 1.194. H_0: β_1 = 0 is rejected in favor of the conclusion that experience does appear to affect the likelihood of successful task performance.

27. b. 52.88, .12

c..895

d. No

e. (48.54, 57.22)

f. (42.85, 62.91)

29. a. SSE = 16.8, s =2.048

b. R 2 = .995

c. Yes. t =-6.55, P -value = .003 (from MINITAB)

d. 98% individual confidence levels ⇒ joint confidence level ≥ 96%: (.671, 3.706), (- .00498, − .00135)

e. (69.531, 76.186), (66.271, 79.446), using software

31. a. 13.636 + 11.406x − 1.7155x 2

b. Yes. Yes, (6, 20)

c. 2.040, .947. The model utility F test via MINITAB gives f =35.9, P -value = .003, suggesting a useful model.

d. Yes, yes

e. (28.35, 35.28)

33. a..9671, .9407

b..0000492x 3 − .000446058x 2 + .007290688x + .96034944

c. t = 2 < 3.182 = t.025,3’ so the cubic term should be deleted.

d. Identical

e..9994, yes

35 =7 6883e 1799x -.0022x 2

37. a. 4.9

b. When number of deliveries is held fixed, the average change in travel time associated with a 1-mile increase in distance traveled is .060 hr. When distance traveled is held fixed, the average change in travel time associated with one extra delivery is .900 hr.

c..9861

39. a. 77.3

b. 40.4

41. f =24.4 > 5.12 = F001,6.30’ so P -value << .001. The chosen model appears to be useful.

43. a. 48.31, 3.69

b. No. If x_1 increases, either x_3 or x_2 must change.

c. Yes, since f =18.924, P -value = .001.

d. Yes, using α = .01, since t =3.496 and P -value = .003.

45. a. f =87.6, P -value = 0, there does appear to be a useful linear relationship between y and at least one of the predictors.

b..935

c. (9.095, 11.087)

47. b. P -value = .000, so conclude that the model is useful.

c. P -value = .034 ≤ .05 = α, so reject H_0: β_3 = 0; % garbage does appear to provide additional useful information.

d. (1479.8, 1531.1), reasonable precision

e. A 95% PI is (1435.7, 1575.2).

c. No

d. (21.07, 21.65)

e. (20.67, 22.05)

75. a. f = 30.8 ≥ 9.55 = F.01,2.7’, so the model appears useful.

b. t =-7.69 and P -value < .001, so retain the quadratic predictor.

c. (44.01, 47.91)

77. a. 231.75

b..903

c. f =41.9, indicating a useful relationship.

d. (220.9, 238.1)

79. There are several reasonable choices in each case.

81. a. f = 106, P -value ≈ 0

b. (.014, .068)

c. t = 5.9, reject H_0: β_4 = 0, percent nonwhite appears to be important.

d. 99.514, y − = 3.486

Chapter 14

1. a. Reject H_0.

b. Don’t reject H_0.

c. Don’t reject H_0

d. Don’t reject H_0.

3. No. x 2 = 1.57 and P -value > .10, so H_0 cannot be rejected.

5. X 2 =ckquote>

5. Reject H_0 if either s+ ≥ 64 or s+ ≤ 14. Because s+ = 72, H_0is rejected.

7. s+ = 442.5, z = 2.89 ≥ 1.645, so reject H_0.

9.

11. w = 37 and 29 < 37 < 61, so H_0 cannot be rejected.

13. z = 2.27 < 2.58, so H_0 cannot be rejected. P -value ≈ .023

15. w = 39 ≤ 41, so H_0 is rejected.

17. ((5), (32)) = (11.15, 23.80)

19. (–13.0, –6.0)

21. (dij(5), dij(21)) = (16, 87)

23. k = 14.06 ≥ 6.251, so reject H_0.

25. k = 9.23 ≥ 5.992, so dont’t reject H_0.

27. fr = 2.60 < 5.992, so don’t reject H_0.

29. fr = 9.62 > 7.815 = , so reject H_0.

31. (–5.9, –3.8)

33. a..021

b. c = 14 gives α = .058; y = 12, so H_0 cannot be rejected.

35. w′= 26 < 27, so don’t reject H_0.

3..9802, .9512, 53

5. a..0301

b..2236

c..6808

7. LCL = 12.20, UCL = 13.70. No.

9. LCL = 94.91, UCL = 98.17. There appears to be a problem on the 22nd day.

11. a. 200

b. 4.78

c. 384.62 (larger), 6.30 (smaller)

13. LCL = 12.37, UCL = 13.53

15. a. LCL = 0, UCL = 6.48

b. LCL = .48, UCL = 6.60

17. LCL = .045, UCL = 2.484. Yes, since all points are inside the control limits.

19. a. LCL = .105, UCL = .357

b. Yes, since .39 > UCL.

21. > 3/53

23. LCL = 0, UCL = 10.1

25. When area = .6, LCL = 0 and UCL = 14.6; when area = .8, LCL = 0 and UCL = 13.4; when area = 1.0, LCL = 0 and UCL = 12.6.

27.

There are no out–of–control signals.

29. n = 5, h = .00626

31. Hypergeometric probabilities (calculated on an HP21S calculator) are .9919, .9317, .8182, .6775, .5343, .4047, .2964, .2110, .1464, and .0994, whereas/font>

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Normal equations

Normalized expected total error of estimation

Normal probability plot

Normal random variables

Null event

Null hypothesis

Null hypothesis of homogeneity

Null value

Number-defective data

Objective interpretation

Observational studies

Observations

influential

retrospective

Observed cell counts

Observed significance level

Population correlation coefficient

Population mean

Population median

Population moment

Population regression coefficients

Population regression function

Population regression line

Population variance

Positively skewed histogram

Posterior probability

Power

Power models

Practical significance

Predicted values

Prediction intervals

Prediction levels

Predictor variables

Price, Bertram

Principal block

Principle of least squares

Prior probability

Probability

axioms of

conditional

determining systematically

equally likely outcomes and

inferential statistics and

interpretation of

law of total

multiplication rule for

posterior

prior

Probability density function

marginal

Probability distributions

Bernoulli

beta

binomial

bivariate normal

chi-squared

conditional

continuous variable

discrete random variable

exponential

F

family of

gamma

geometric

hypergeometric

joint

linear combination

lognormal

multinomial

negative binomial

normal

parameter of

Poisson

sample mean

sampling

standard normal

statistics and

Studentized range

symmetric

t

uniform

Weibull

See also specific distributions

Probability mass function

conditional

joint

marginal

Probability plots

half-normal

normal

parameters and

sample percentiles and

Product rule

k -tuples

ordered pairs

Proportions

confidence interval

differences between

hypothesis test

sample

Pure birth process

P -values

chi-squared test

definitions of

F test

t test

z test

Quadratic regression

Qualitative data

Quality control methods

acceptance sampling

control charts

CUSUM procedures

Quartiles

Random deviation

Random effects model

multifactor ANOVA

Rice, John

Robust control charts

Robust estimator

Rocke, David M.

Ryan, Thomas

Ryan–Joiner test

Sample

convenience

simple random

stratified

Sample coefficient of variation

Sample correlation coefficient

Sample mean

Sample median

Sample moment

Sample percentiles

Sample proportions

Sample size

confidence intervals and

hypothesis tests and

single-factor ANOVA and

small-sample inferences and

type II errors and

Sample space

Sample standard deviation

Sample variance

computing formula

motivation for

Sampling distributions

examples of deriving

sample mean and

simulation experiments and

Sampling frame

Scale, denm">
LOWESS method

residual plots

S control chart

Second-order multiple regression models

Shape parameter

Siegel–Tukey test

Significance

observed level of

statistical vs. practical

Significance level

Sign interval

Sign test

Simple event

Simple hypothesis

Simple linear regression

linear probabilistic model

model parameter estimates

slope parameter inferences

Simple random sample

Simulation experiments

Simultaneous confidence level

Single-factor ANOVA

explanation of

fixed effects model

F test

model equation

noncentrality parameter

notation and assumptions

random effects model

sample sizes

sums of squares

test statistic

transformations

See also Multifactor ANOVA

Single-sampling plans

Skewed distribution

Slopen

confidence intervals

hypothesis-testing procedures

parameter inferences

Small-sample hypothesis tests

Small-sample inferences

Standard beta distribution

Standard deviation

confidence interval

continuous random variable

discrete random variable

population

sample

Standard distribution

Standard error

Standard gamma distribution

Standardized independent variable

Standardized residuals

Standardized variables

Standard normal curve

Standard normal distribution

definition of

percentiles of

Standard normal random variable

Standard order

Standard sampling plans

Statistic

defined

test

Statistical hypothesis

Statistical significance

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Latin square designs

See also Multifactor ANOVA

Tibshirani, Robert

Time series

T method

Tolerance intervals

Total probability law

Total sum of squares

Transformations

ANOVA

control chart

regression

t ratio

Treatments

mean square for

Treatment sum of squares

Tree diagram

Trials

Trimmed mean

True regression coefficients

True regression function

True regression line

t tests

F tests and

paired

P -values for

two-sample

Tukey’s procedure

Two-factor ANOVA

expected mean squares

fixed effects model

multiple comght="0em" width="0em">Variable selection

backward elimination

criteria for

forward selection

Variance

confidence interval

continuous random variable

discrete random variable

expected value and

hypothesis test

linear combination

pooled estimator of

population

rules of

sample

shortcut formula

two-population

See also ANOVA

Variation

coefficient of

control charts for

Venn diagrams

V-mask

Weibull, Waloddi

Weibull distribution

point estimation

probability plot

Weighted least squares

Wilcoxon rank-sum interval

Wilcoxon rank-sum test

critical values for

efficiency of

general description of

normilcoxon signed-rank interval

Wilcoxon signed-rank test

critical values for

efficiency of

general description of

large-sample approximation and

paired observations and

Winkler, Robert

Wolfe, Douglas

X charts

estimated parameters and

known parameter values and

probability limits and

supplemental rules for

Yates’s method

z critical values

z curves

z tests

large-sample