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CHAPTER 7 Sampling and Sampling Distributions

CHAPTER 8 Confidence Interval Estimation

CHAPTER 9 Hypothesis Testing

P A R T 3 STATISTICAL INFERENCE

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CHAPTER 7 Sampling and Sampling Distributions

SAMPLE SIZE SELECTION IN A LEGAL CASE This chapter introduces the important problem of estimat- ing an unknown population quantity by randomly sam- pling from the population. Sampling is often expensive and/or time-consuming, so a key step in any sampling plan is to determine the sample size that produces a prescribed level of accuracy. Some of the issues in finding an appro- priate sample size are discussed in Afshartous (2008). The author was involved as an expert statistical witness for the plaintiff in a court case. Over a period of several years, a service company had collected a flat “special service han-

dling fee” from its client during any month in which a special service request was made. The plaintiff claimed that many of these fees had been charged erroneously and sought to recover all the money collected from such erroneous fees. The statistical question con- cerns either the proportion of all monthly billing records that were erroneous or the total number of all erroneous billing records. Both sides had to agree on a sampling method for sampling through the very large population of billing records. They eventually agreed to simple random sampling, as discussed in this chapter. However, there was some contention (and confusion) regarding the appropriate sample size.

Their initial approach was to find a sample size n sufficiently large to accurately esti- mate p, the unknown proportion of all monthly billing records in error. Specifically, if they wanted to be 95% confident that the error in their estimate of p would be no more than 5%, then a standard sample size formula (provided in Chapter 8) requires n to be 385. (This number is surprisingly independent of the total number of billing records.) Then, for example, if the sample discovered 77 errors, or 20% of the sampled items, they would be 95% confident that between 15% and 25% (20% plus or minus 5%) of all billing records were in error.

The author argued that this “plus or minus 5%” does not necessarily provide the desired level of accuracy for the quantity of most interest, the total number of erroneously charged fees. A couple of numerical examples illustrate his point. Let’s suppose that there were 100,000 billing records total and that 20%, or 20,000, were billed erroneously. Then the plus or minus 5% interval translates to an interval from 15,000 to 25,000 bad bill- ings. That is, we are 95% confident that the estimate is not off by more than 5000 billing records on either side. The author defines the relative error in this case to be 0.25: the potential error, 5000, divided by the number to be estimated, 20,000. Now change the example slightly so that 60%, or 60,000, were billed erroneously. Then plus or minus 5% translates to the interval from 55,000 to 65,000, and the relative error is 5000>60,000, or 0.083. The point is that the same plus or minus 5% absolute error for p results in a much smaller relative error in the second example.

Using this reasoning, the author suggested that they should choose the sample size to achieve a prescribed relative error in the number of bad billings. This can change the mag- nitude of the sample size considerably. For example, the author demonstrated by means of a rather complicated sample size formula that if a relative error of 0.10 is desired and the value of p is somewhere around 0.10, a sample size of about 3600 is required. On the

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7-2 Sampling Terminology 2 9 5

other hand, if a relative error of 0.10 is still desired but the value of p is somewhere around 0.5, then the required sample size is only about 400.

Sample size formulas, and statistical arguments that lead to them, are far from intui- tive. In this legal case, by keeping the math to a minimum and using simple terminology like relative error, the author eventually convinced the others to use his approach, even though it led to a considerably larger sample size than the 385 originally proposed.

7-1 Introduction This chapter sets the stage for statistical inference, a topic that is explored in the following two chapters. In a typical statistical inference problem, you want to discover one or more characteristics of a given population. For example, you might want to know the proportion of toothpaste customers who have tried, or intend to try, a particular brand. Or you might want to know the average amount owed on credit card accounts for a population of cus- tomers at a shopping mall. Generally, the population is large and/or spread out, and it is difficult, maybe even impossible, to contact each member. Therefore, you identify a sam- ple of the population and then obtain information from the members of the sample.

There are two main objectives of this chapter. The first is to discuss the sampling schemes that are generally used in real sampling applications. We focus on several types of random samples and see why these are preferable to nonrandom samples. The second objective is to see how the information from a sample of the population—for example, 1% of the population—can be used to infer the properties of the entire popu- lation. The key here is the concept of a sampling distribution. In this chapter we focus on the sampling distribution of the sample mean, and we discuss the role of a famous mathematical result called the central limit theorem. Specifically, we discuss how the central limit theorem is the reason for the importance of the normal distribution in sta- tistical inference.

7-2 Sampling Terminology We begin by introducing some terminology that is used in sampling. In any sampling prob- lem there is a relevant population. The population is the set of all members about which a study intends to make inferences, where an inference is a statement about a numerical characteristic of the population, such as an average income or the proportion of incomes below $50,000. It is important to realize that a population is defined in relationship to any particular study. Any analyst planning a survey should first decide which population the conclusions of the study will concern, so that a sample can be chosen from this population.

For example, if a marketing researcher plans to use a questionnaire to infer consum- ers’ reactions to a new product, she must first decide which population of consumers is of interest—all consumers, consumers over 21 years old, consumers who do most of their shopping online, or others. Once the relevant consumer population has been designated, a sample from this population can then be surveyed. However, it is important to remember that inferences made from the study pertain only to this particular population.

The relevant population contains all members about which a study intends to make inferences.

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In this chapter we assume that the population is finite and consists of N sampling units. We also assume that a frame of these N sampling units is available. Unfortunately, there are many situations where a complete frame is practically impossible to obtain. For example, if the purpose of a study is to survey the attitudes of all unemployed teenagers in Chicago, it is practically impossible to obtain a complete frame of them. In this situation the best alternative is to obtain a partial frame from which the sample can be selected. If the partial frame omits any significant segments of the population that a complete frame would include, then the resulting sample could be biased. For instance, if you use a restau- rant guide to choose a sample of restaurants, you automatically omit all restaurants that do not advertise in the guide. Depending on the purposes of the study, this could be a serious omission.

There are two basic types of samples: probability samples and judgmental samples. A probability sample is a sample in which the sampling units are chosen from the pop- ulation according to a random mechanism. In contrast, no formal random mechanism is used to select a judgmental sample. In this case the sampling units are chosen according to the sampler’s judgment.

It is customary in virtually all statistical literature to let uppercase N be the population size and lowercase n be the sample size. We follow this convention as well.

Before you can choose a sample from a given population, you typically need a list of all members of the population. In sampling terminology, this list is called a frame, and the potential sample members are called sampling units. Depending on the context, sampling units could be individual people, households, companies, cities, or others.

A frame is a list of all members, called sampling units, in the population.

The members of a probability sample are chosen according to a random mecha- nism, whereas the members of a judgmental sample are chosen according to the sampler’s judgment.

We will not discuss judgmental samples. The reason is very simple—there is no way to measure the accuracy of judgmental samples because the rules of probability do not apply to them. In other words, if a population characteristic is estimated from the obser- vations in a judgmental sample, there is no way to measure the accuracy of this estimate. In addition, it is very difficult to choose a representative sample from a population with- out using some random mechanism. Because our judgment is usually not as good as we think, judgmental samples are likely to contain our own built-in biases. Therefore, we focus exclusively on probability samples from here on.

Why random Sampling?

One reason for sampling randomly from a population is to avoid biases (such as choosing mainly stay-at-home mothers because they are easier to contact). An equally important reason is that random sampling allows you to use probability to make inferences about unknown population parameters. If sampling were not random, there would be no basis for using probability to make such inferences.

Fundamental Insight

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7-3 Methods for Selecting random Samples 2 9 7

7-3 Methods for Selecting Random Samples This section discusses the types of random samples that are used in real sampling applications. Different types of sampling schemes have different properties. There is typically a trade-off between cost and accuracy. Some sampling schemes are cheaper and easier to administer, whereas others are more costly but provide more accurate information. Some of these issues are discussed here. However, anyone who intends to make a living in survey sampling needs to learn much more about the topic than we can cover here.

7-3a Simple Random Sampling The simplest type of sampling scheme is appropriately called simple random sampling. Suppose you want to sample n units from a population of size N. Then a simple random sample of size n has the property that every possible sample of size n has the same prob- ability of being chosen. Simple random samples are the easiest to understand, and their statistical properties are the most straightforward. Therefore, we will focus primarily on simple random samples in the rest of this book. However, as we discuss shortly, more complex random samples are often used in real applications.

A simple random sample of size n is one where each possible sample of size n has the same chance of being chosen.

Let’s illustrate the concept with a simple random sample for a small population. Sup- pose the population size is N 5 5, and the five members of the population are labeled a, b, c, d, and e. Also, suppose the sample size is n 5 2. Then the possible samples are (a, b), (a, c), (a, d), (a, e), (b, c), (b, d), (b, e), (c, d), (c, e), and (d, e). That is, there are 10 possible samples—the number of ways two members can be chosen from five members. Then a simple random sample of size n 5 2 has the property that each of these 10 possible samples has the same probability, 1>10, of being chosen.

One other property of simple random samples can be seen from this example. If you focus on any member of the population, such as member b, you will see that b is a mem- ber of 4 of the 10 samples. Therefore, the probability that b is chosen in a simple random sample is 4>10, or 2>5. In general, any member has the same probability n>N of being chosen in a simple random sample. If you are one of 100,000 members of a population, then the probability that you will be selected in a simple random sample of size 100 is 100>100,000, or 1 out of 1000.

There are several ways simple random samples can be chosen, all of which involve random numbers. One approach that works well for the small example with N 5 5 and n 5 2 is to generate a single random number with the RAND function in Excel®. You divide the interval from 0 to 1 into 10 equal subintervals of length 1>10 each and see which of these subintervals the random number falls into. You then choose the corresponding sample. For example, suppose the random number is 0.465. This is in the fifth subinterval, that is, the interval from 0.4 to 0.5, so you choose the fifth sample, (b, c).

This method is clearly consistent with simple random sampling—each of the samples has the same chance of being chosen—but it is prohibitive when n and N are large. In this case there are too many possible samples to list. Fortunately, there is another method that can be used. The idea is simple. You sort the N members of the population randomly, using Excel’s RAND function to generate random numbers for the sort. Then you include the first n members from the sorted sequence in the random sample. This procedure is illustrated in Example 7.1.

The RAND function in Excel generates numbers that are distributed randomly and uniformly between 0 and 1.

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EXAMPLE

7.1 SAMPLING FAMILIES TO ANALYZE ANNUAL INCOMES Consider the frame of 40 families with annual incomes shown in column B of Figure 7.1, with several rows hidden. (See the file Random Sampling.xlsm. The extension is .xlsm because this file contains a macro. When you open it, you will need to enable the macro.) We want to choose a simple random sample of size 10 from this frame. How can this be done? And how do summary statistics of the chosen families compare to the corresponding summary statistics of the population?

Objective To illustrate how Excel’s random number function, RAND, can be used to generate simple random samples.

Solution The idea is very simple. You first generate a column of random numbers next to the data. Then you sort the rows according to the random numbers and choose the first 10 families in the sorted rows. The following procedure produces the results in Figure 7.2. (See the first sheet in the finished version of the file.)

1. Random numbers next to a copy. Copy the original data to columns D and E. Then enter the formula

= RAND()

in cell F10 and copy it down column F. 2. Replace with values. To enable sorting, you must first “freeze” the random numbers—that is, replace their formulas

with values. To do this, copy the range F10:F49 and select Paste Values from the Paste dropdown menu on the Home ribbon.

3. Sort. Sort on column F in ascending order. Then the 10 families with the 10 smallest random numbers are the ones in the sample. These are shaded in the figure. (Note that you could instead have chosen the 10 families with the 10 largest ran- dom numbers. This would be equally valid.)

4. Means. Use the AVERAGE, MEDIAN, and STDEV.S functions in row 6 to calculate summary statistics of the first 10 incomes in column E. Similar summary statistics for the population have already been calculated in row 5. (Cell D5 uses the STDEV.P function because this is the population standard deviation.)

To obtain more random samples of size 10 (for comparison), you would need to go through this process repeatedly. To save you the trouble of doing so, we wrote a macro to automate the process. (See the Automated sheet in the finished version of the file.) This sheet looks essentially the same as the sheet in Figure 7.2, except that there is a button to run the macro, and

Figure 7.1 Population Income Data 1

2 3 4 5 6 7 8 9

10 11 12

47 48 49

A B C D Simple random sampling

Summary statistics Mean Median Stdev

$39,985 $38,500 $7,377 Sample Population

Population Family Income

1 $43,300 2 $44,300 3 $34,600

38 $46,900 39 $37,300 40 $41,000

13 4 $38,000

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7-3 Methods for Selecting random Samples 2 9 9

only the required data remain on the spreadsheet. Try clicking this button. Each time you do so, you will get a different random sample—and different summary measures in row 6. By doing this many times and keeping track of the sample summary data, you can see how the summary measures vary from sample to sample. We will have much more to say about this variation later in the chapter.

Figure 7.2 Selecting a Simple Random Sample 1

2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20

A B C D E F Simple random sampling

Summary statistics Mean Median Stdev

$39,985 $38,500 $7,377 Sampl Population

e $41,490 $42,850 $5,323

Population Random sample Family Income Family Income Random #

1 $43,300 1 $43,300 0.04545 2 $44,300 2 $44,300 0.1496768 3 $34,600 12 $51,500 0.23527 4 $38,000 7 $42,700 0.2746325 5 $44,700 13 $35,900 0.3003506 6 $45,600 15 $43,000 0.3197393 7 $42,700 6 $45,600 0.3610983 8 $36,900 3 $34,600 0.3852641 9 $38,400 9 $38,400 0.4427564

10 $33,700 14 $35,600 0.4447877 11 $44,100 5 $44,700 0.4505899 12 $51,500 $41,000 0.459736121

47 48 49

40 38 $46,900 39 $37,300 0.8644119 39 $37,300 8 $36,900 0.9059098 40 $41,000 10 $33,700 0.9637509

The procedure described in Example 7.1 can be used in Excel to select a simple ran- dom sample of any size from any population. All you need is a frame that lists the pop- ulation values. Then it is just a matter of inserting random numbers, freezing them, and sorting on the random numbers.

Perhaps surprisingly, simple random samples are used infrequently in real applica- tions. There are several reasons for this.

• Because each sampling unit has the same chance of being sampled, simple random sampling can result in samples that are spread over a large geographical region. This can make sampling extremely expensive, especially if personal interviews are used.

• Simple random sampling requires that all sampling units be identified prior to sampling. Sometimes this is infeasible.

• Simple random sampling can result in underrepresentation or overrepresentation of cer- tain segments of the population. For example, if the primary—but not sole—interest is in the graduate student subpopulation of university students, a simple random sam- ple of all university students might not provide enough information about the graduate students.

Despite this, most of the statistical analysis in this book assumes simple random samples. The analysis is considerably more complex for other types of random samples and is best left to more advanced books on sampling.

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EXAMPLE

7.2 SAMPLING ACCOUNTS RECEIVABLE AT SPRING MILLS The file Accounts Receivable Finished.xlsx contains 280 accounts receivable for Spring Mills Company. There is a single variable, the amount of the bill owed. The file contains the bill amounts for 25 random samples of size 15 each. (They were generated by the method in Example 7.1.) Calculate the average amount owed in each random sample, and create a histogram of these 25 averages.

Objective To demonstrate how sample means are distributed.

Solution In most real-world applications, you would generate only a single random sample from a population, so why have we gener- ated 25 random samples in this example? The reason is that we want to introduce the concept of a sampling distribution, in this case the sampling distribution of the sample mean. This is the distribution of all possible sample means you could generate from all possible samples (of a given size) from a population. By generating a fairly large number of random samples from the population of accounts receivable, you can see what the approximate sampling distribution of the sample mean looks like.

The 25 random samples, one per row, are listed in Figure 7.3. We then used the AVERAGE function in column Q to calculate their sample means, and we created a histogram of these 15 sample means. As you can see, the 15 sample means vary quite a lot, from a low of $332.00 to a high of $799.33. You can check that the population mean, the average of all 280 bill amounts, is $464.29. So the 15 sample means vary around this population mean and are spread out roughly as a bell-shaped curve, as shown in the histogram.

Figure 7.3 Random Samples and Sample Means

$610 $410 $200 $250 $260 $260 $450 $310 $260 $280 $620 $500 $240 $310 $510 $210 $240 $260 $240 $430 $250 $240 $580 $310 $220

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28

$460 $570 $470

$1,340 $350 $180 $240 $280 $300 $210

$1,340 $320

$1,580 $240 $220

$1,360 $180 $750 $460 $320

$1,480 $510 $420 $500 $390

$370 $570 $250 $210

$1,380 $190 $460 $220 $930

$1,010 $2,220

$260 $410 $240 $410 $280 $520 $270 $610

$2,030 $210 $460 $410 $290 $490

$370 $280 $570 $200 $260 $000

$000 $460

$1,100 $180

$1,550 $200

$280 $470 $350 $240 $410 $000

$000 $350 $500

$1,590 $1,380

$230

$000 $1,330

$260 $250 $750 $930

$000 $240 $520 $400 $250 $220

$000 $2,030

$380 $310 $580 $250

$000 $560 $560

$1,340 $660 $350

$000 $260 $350 $350 $460 $240

$640 $190 $270 $570 $250 $000

$300 $280 $420

$1,460 $560 $000

$410 $270 $580 $570 $520 $000

$510 $220 $330 $430 $290 $000

$1,340 $270

$2,030 $220 $260 $000

$660 $2,220

$220 $460

$1,460 $650 $580 $240 $370 $290 $650 $290 $560 $220 $240 $260 $250 $190 $210 $190 $200 $300 $560 $350 $320

$1,480 $580 $240 $280 $260 $370 $620 $410 $190 $290 $420 $380 $260 $930 $510 $280

$1,590 $200 $270 $510 $370 $260 $230 $580 $240

$2,030 $540 $210 $270

$1,010 $310 $400 $240 $210 $430 $560

$1,460 $310

$1,340 $540 $250 $540 $350 $230 $560 $250

$1,400 $210 $220 $540

$1,010 $620 $330 $370 $560

$1,330 $350 $290 $220

$1,330 $210 $300 $200 $220

$1,250 $460 $270 $280 $190 $180

$1,520 $220 $560 $270 $540

$1,520 $1,360 $1,520

$330 $270 $180 $330 $510

$1,520 $250 $420 $330 $510 $210 $260 $420 $380 $460 $380 $260 $380 $230 $520 $390 $240

Sample mean $799.33 $590.00 $532.67 $480.00 $540.00 $475.33 $519.33 $415.33 $471.33 $502.00 $581.33 $459.33 $421.33 $500.67 $403.33 $458.67 $395.33 $332.00 $468.00 $458.00 $659.33 $486.00 $527.33 $569.33 $360.00

Sample

Random samples of size 15

Bill Amounts

A B C D E F G H I J K L M N P QO

Histogram of Sample Means

0 [$332.00, $452.00] [$452.00, $572.00] [$572.00, $692.00] [$692.00, $812.00]

This histogram approximates the sampling distribution of the sample mean. It is approximate because it is based on only 15 random samples out of all possible random samples of size 15. (The number of possible random samples of size 15 from a population of size 280 is an astronomically large number.) We will come back to this important idea when we discuss sampling distributions in Section 7-4.

In the next several subsections we describe sampling plans that are often used. These plans differ from simple random sampling both in the way the samples are chosen and in the way the data analysis is performed. However, we will barely touch on this latter issue. The details are quite complicated and are better left to a book devoted entirely to sampling.1

1 See, for example, the excellent book by Levy and Lemeshow (2008).

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7-3b Systematic Sampling Suppose you are asked to select a random sample of 250 names from a large company’s directory of employees. Let’s also say that there are 55,000 names listed in alphabetical order in the directory. A systematic sample provides a convenient way to choose the sam- ple. First, you divide the population size by the sample size: 55,000/250 5 220. Concep- tually, you can think of dividing the directory into 250 “blocks” with 220 names per block. Next, you use a random mechanism to choose a number between 1 and 220. Suppose this number is 131. Then you choose the 131st name and every 220th name thereafter. So you would choose name 131, name 351, name 571, and so on. The result is a systematic sam- ple of size n 5 250.

In general, one of the first k members is selected randomly, and then every kth mem- ber after this one is selected. The value k is called the sampling interval and equals the ratio N/n, where N is the population size and n is the desired sample size.

Systematic sampling is quite different from simple random sampling because not every sample of size 250 has a chance of being chosen. In fact, there are only 220 different samples possible (depending on the first number chosen), and each of these is equally likely. Nevertheless, systematic sampling is generally similar to simple random sampling in its statistical properties. The key is the relationship between the ordering of the sam- pling units in the frame (the employee directory in this case) and the purpose of the study.

If the purpose of the study is to analyze personal incomes, then there is probably no relationship between the alphabetical ordering of names in the directory and personal income. However, there are situations where the ordering of the sampling units is not random, which could make systematic sampling more or less appealing. For example, suppose that a com- pany wants to sample randomly from its customers, and its customer list is in decreasing order of order volumes. That is, the largest customers are at the top of the list and the smallest are at the bottom. Then systematic sampling might be more representative than simple ran- dom sampling because it guarantees a wide range of customers in terms of order volumes.

However, some type of cyclical ordering in the list of sampling units can lead to very unrepresentative samples. As an extreme, suppose a company has a list of daily trans- actions (Monday through Saturday) and it decides to draw a systematic sample with the sampling interval equal to 6. Then if the first sampled day is Monday, all other days in the sample will be Mondays. This could clearly bias the sample. Except for obvious examples like this one, however, systematic sampling can be an attractive alternative to simple ran- dom sampling and is sometimes used because of its convenience.

7-3c Stratified Sampling Suppose various subpopulations within the total population can be identified. These sub- populations are called strata. Then instead of taking a simple random sample from the entire population, it might make more sense to select a simple random sample from each

Systematic random samples are typically chosen because of their convenience.

Types of random Samples

There are actually many methods for choosing random samples, some described only briefly in this book, and they all have their advantages and disadvantages from practical and statistical standpoints. Surprisingly, the simplest of these, where each subset of the population has the same chance of being chosen, is not the most frequently used method in real applications. This is basically because other more complex methods can make more efficient use of a given sample size. Nevertheless, the concepts you learn here remain essentially the same, regardless of the exact sampling method used.

Fundamental Insight

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There are several advantages to stratified sampling. One obvious advantage is that separate estimates can be obtained within each stratum—which would not be obtained with a simple random sample from the entire population. Even if the samples from the individual strata are eventually pooled, it can be useful to have the total sample broken down into separate samples initially.

A more important advantage of stratified sampling is that the accuracy of the resulting population estimates can be increased by using appropriately defined strata. The trick is to define the strata so that there is less variability within the individual strata than in the population as a whole. You want strata such that there is relative homogeneity within the strata, but relative heterogeneity across the strata, with respect to the variable(s) being analyzed. By choosing the strata in this way, you can generally obtain more accuracy for a given sampling cost than you could obtain from a simple random sample at the same cost. Alternatively, you can achieve the same level of accuracy at a lower sampling cost.

The key to using stratified sampling effectively is selecting the appropriate strata. Sup- pose a company that advertises its product on television wants to estimate the reaction of viewers to the advertising. Here the population consists of all viewers who have seen the advertising. But what are the appropriate strata? The answer depends on the company’s objectives and its product. The company could stratify the population by gender, by income, by age, by amount of television watched, by the amount of the product class consumed, and probably others. Without knowing more specific information about the company’s objec- tives, it is impossible to say which of these stratification schemes is most appropriate.

Suppose that you have identified I nonoverlapping strata in a given population. Let N be the total population size, and let Ni be the population size of stratum i, so that

N 5 N1 1 N2 1 c 1 NI

To obtain a stratified random sample, you must first choose a total sample size n, and then choose a sample size ni from each stratum i such that

n 5 n1 1 n2 1 c 1 nI

You can then select a simple random sample of the specified size from each stratum exactly as in Example 7.1.

However, how do you choose the individual sample sizes n1 through nI, given that the total sample size n has been chosen? For example, if you decide to sample 500 custom- ers in total, how many should come from each stratum? There are many ways to choose sample sizes n1 through nI that sum to n, but probably the most popular method is to use proportional sample sizes. The idea is very simple. For example, if one stratum has 15% of the total population, you select 15% of the total sample from this stratum.

Stratified samples are typically chosen because they provide more accurate estimates of population parameters for a given sampling cost.

stratum separately. This sampling method is called stratified sampling. It is a particularly useful approach when there is considerable variation between the various strata but rela- tively little variation within a given stratum.

In stratified sampling, the population is divided into relatively homogeneous sub- sets called strata, and then random samples are taken from each stratum.

With proportional sample sizes, the proportion of a stratum in the sample is the same as the proportion of that stratum in the population.

The advantage of proportional sample sizes is that they are very easy to determine. The disadvantage is that they ignore differences in variability among the strata. If such differences are large, more complex sample size calculations can be used, but they are not discussed here.

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The primary advantage of cluster sampling is sampling convenience (and possibly lower cost). If an agency is sending interviewers to interview heads of household, it is much easier for them to concentrate on particular city blocks than to contact households throughout the city. The downside, however, is that the inferences drawn from a cluster sample can be less accurate for a given sample size than from other sampling plans.

Consider the following scenario. A nationwide company wants to survey its sales- people with regard to management practices. It decides to randomly select several sales districts (the clusters) and then interview all salespeople in the selected districts. It is likely that in any particular sales district the attitudes toward management are somewhat similar. This overlapping information means that the company is probably not getting the maxi- mum amount of information per sampling dollar spent. Instead of sampling 20 salespeo- ple from a given district, all of whom have similar attitudes, it might be better to sample 20 salespeople from different districts who have a wider variety of attitudes. Neverthe- less, the relative convenience of cluster sampling sometimes outweighs these statistical considerations.

Selecting a cluster sample is straightforward. The key is to define the sampling units as the clusters—the city blocks, for example. Then a simple random sample of clusters can be chosen exactly as in Example 7.1. Once the clusters are selected, it is typical to sample all of the population members in each selected cluster, although it is also possible to ran- domly sample from the selected clusters.

7-3e Multistage Sampling The cluster sampling scheme just described, where a sample of clusters is chosen and then all of the sampling units within each chosen cluster are taken, is called a single-stage sam- pling scheme. Real applications are often more complex than this, resulting in multistage sampling. For example, the Gallup organization uses multistage sampling in its nation- wide surveys. A random sample of approximately 300 locations is chosen in the first stage of the sampling process. City blocks or other geographical areas are then randomly sam- pled from the first-stage locations in the second stage of the process. This is followed by a systematic sampling of households from each second-stage area. A total of about 1500 households comprise a typical Gallup poll.2

We will not pursue the topic of multistage sampling in this book. However, you should realize that real-world sampling procedures can be complex.

Cluster analysis is typically more convenient and less costly than other random sampling methods.

7-3d Cluster Sampling Suppose that a company is interested in various characteristics of households in a particular city. The sampling units are households. You could select a random sample of households by one of the sampling methods already discussed. However, it might be more convenient to proceed somewhat differently. You could first divide the city into city blocks and con- sider the city blocks as sampling units. You could then select a simple random sample of city blocks and then sample all of the households in the chosen blocks. In this case the city blocks are called clusters and the sampling scheme is called cluster sampling.

In cluster sampling, the population is separated into clusters, such as cities or city blocks, and then a random sample of the clusters is selected.

2 If you watch discussions of politics on the news, you are aware that the results of new polls are available on an almost daily basis from a variety of polling organizations. Surprisingly, the organization probably best known for polling, Gallup, announced in October 2015 that it was getting out of the political polling business. Its traditional polling method, evening calls to landline phones, evidently doesn’t work in an environment dom- inated by cell phones and social media. Gallup will continue polling in other areas, just not political races.

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Problems Solutions for problems whose numbers appear within a colored box can be found in the Student Solution Files.

Level A 1. The file P02_07.xlsx includes data on 204 employees

at the (fictional) company Beta Technologies. For this problem, consider this data set as the population frame. a. Using the method in this section, generate 10 simple

random samples of size 20 from this population. b. Calculate the population mean, median, and standard

deviation of Annual Salary. Then calculate the sam- ple mean, median, and standard deviation of Annual Salary for each of the samples in part a. Comment briefly on how they compare to each other and the population measures.

2. The file P07_02.xlsx contains data on the 1995 students who have gone through the MBA program at State Uni- versity. You can consider this the population of State University’s MBA students. a. Find the mean and standard deviation for each of the

numerical variables in this population. Also, find the following proportions: the proportion of students who are male, the proportion of students who are interna- tional (not from the USA), the proportion of students under 30 years of age, and the proportion of students with an engineering undergrad major.

b. Using the method in this section, generate a simple random sample of 100 students from this population, and find the mean and standard deviation of each numerical variable in the sample. Is there any way to know (without the information in part a) whether your summary measures for the sample are lower or higher than the (supposedly unknown) population summary measures?

c. Use the method in this section to generate 10 simple random samples of size 100 of School Debt. For each, find the mean of School Debt and its deviation from the population mean in part a (negative if it is below the population mean, positive if it is above the popu- lation mean). What is the average of these 10 devia- tions? What would you expect it to be?

d. We want random samples to be representative of the population in terms of various demographics. For each of the samples in part c, find each of the pro- portions requested in part a. Do these samples appear to be representative of the population in terms of age, gender, nationality, and undergrad major? Why or why not? If they are not representative, is it because there is something wrong with the sampling procedure?

3. The file P02_35.xlsx contains data from a survey of 500 randomly selected households. a. Suppose you decide to generate a systematic random

sample of size 25 from this population of data. How many such samples are there? What is the mean of

Debt for each of the first three such samples, using the data in the order given?

b. If you wanted to estimate the (supposedly unknown) population mean of Debt from a systematic random sample as in part a, why might it be a good idea to sort first on Debt? If you do so, what is the mean of Debt for each of the first three such samples?

4. Recall from Chapter 2 that the file Supermarket Trans- actions.xlsx contains over 14,000 transactions made by supermarket customers over a period of approximately two years. For this problem, consider this data set the population of transactions. a. If you were interested in estimating the mean of

Revenue for the population, why might it make sense to use a stratified sample, stratified by product family, to estimate this mean?

b. Suppose you want to generate a stratified random sample, stratified by product family, and have the total sample size be 250. If you use proportional sample sizes, how many transactions should you sample from each of the three product families?

c. Using the sample sizes from part b, generate a corre- sponding stratified random sample. What are the indi- vidual sample means from the three product families? What are the sample standard deviations?

Level B 5. This problem illustrates an interesting variation of sim-

ple random sampling. a. Open a blank spreadsheet and use the RAND() func-

tion to create a column of 1000 random numbers. Don’t freeze them. This is actually a simple random sample from the uniform distribution between 0 and 1. Use the COUNTIF function to count the number of values between 0 and 0.1, between 0.1 and 0.2, and so on. Each such interval should contain about 1/10 of all values. Do they? (Keep pressing the F9 key to see how the results change.)

b. Repeat part a, generating a second column of ran- dom numbers, but now generate the first 100 as uni- form between 0 and 0.1, the next 100 as uniform between 0.1 and 0.2, and so on, up to 0.9 to 1. (Hint: For example, to create a random number uniformly distributed between 0.5 and 0.6, use the formula 50.510.1*RAND(). Do you see why?) Again, use COUNTIF to find the number of the 1000 values in each of the intervals, although there shouldn’t be any surprises this time. Why might this type of random sampling be preferable to the random sampling in part a? (Note: The sampling in part a is called Monte Carlo sampling, whereas the sampling in part b is basically Latin Hypercube sampling.)

6. Another type of random sample is called a bootstrap sample. (It comes from the expression “pulling yourself up by your own bootstraps.”) Given a data set with n

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observations, a bootstrap sample, also of size n, is when you randomly sample from the data set with replace- ment. To do so, you keep choosing a random integer from 1 to n and include that item in the sample. The “with replacement” part means that you can sample the same item more than once. For example, if n 5 4, the sampled items might be 1, 2, 2, and 4. Using the data in the file Accounts Receivable.xlsx, illustrate a simple

method for choosing bootstrap samples with the RAND- BETWEEN and VLOOKUP functions. For each boot- strap sample, find the mean and standard deviation of Bill Amount. How do these compare to the similar measures for the original data set? (For more on boot- strap sampling, do a Web search. Wikipedia has a nice overview.)

7-4 Introduction to Estimation The purpose of any random sample, simple or otherwise, is to estimate properties of a population from the data observed in the sample. The following is a good example to keep in mind. Suppose a government agency wants to know the average household income over the population of all households in Indiana. Then this unknown average is the population parameter of interest, and the government is likely to estimate it by sampling several rep- resentative households in Indiana and reporting the average of their incomes.

The mathematical procedures appropriate for performing this estimation depend on which properties of the population are of interest and which type of random sampling scheme is used. Because the details are considerably more complex for more complex sampling schemes such as multistage sampling, we will focus on simple random samples, where the mathematical details are relatively straightforward. Details for other sampling schemes such as stratified sampling can be found in Levy and Lemeshow (2008). How- ever, even for more complex sampling schemes, the concepts are the same as those we discuss here; only the details change.

Throughout most of this section, we focus on the population mean of some variable such as household income. Our goal is to estimate this population mean by using the data in a randomly selected sample. We first discuss the types of errors that can occur.

7-4a Sources of Estimation Error There are two basic sources of errors that can occur when sampling randomly from a popu- lation: sampling error and all other sources, usually lumped together as nonsampling error. Sampling error results from “unlucky” samples. As such, the term error is somewhat misleading. Suppose, for example, that the mean household income in Indiana is $58,225. (We can only assume that this is the true value. It wouldn’t actually be known without taking a census.) A government agency wants to estimate this mean, so it randomly sam- ples 500 Indiana households and finds that their average household income is $60,495. If the agency then infers that the mean of all Indiana household incomes is $60,495, the resulting sampling error is the difference between the reported value and the true value: $60,495 2 $58,225 5 $2270. In this case, the agency hasn’t done anything wrong. This sampling error is essentially due to bad luck.

Sampling error is the inevitable result of basing an inference on a random sample rather than on the entire population.

We will soon discuss how to measure the potential sampling error involved. The point here is that the resulting estimation error is not caused by anything the government agency is doing wrong—it might just get unlucky.

Nonsampling error is quite different and can occur for a variety of reasons. We discuss a few of them.

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• Perhaps the most serious type of nonsampling error is nonresponse bias. This occurs when a portion of the sample fails to respond to the survey. Anyone who has ever conducted a questionnaire, whether by mail, by phone, or any other method, knows that the percentage of nonrespondents can be quite large. The question is whether this introduces estimation error. If the nonrespondents would have responded similarly to the respondents, not much is lost by not hearing from them. However, because the nonrespondents don’t respond, there is typically no way of knowing whether they differ in some important respect from the respondents.

• Another source of nonsampling error is nontruthful responses. This is particularly a problem when there are sensitive questions in a questionnaire. For example, if the questions “Have you ever had an abortion?” or “Do you regularly use cocaine?” are asked, most people will answer “no,” regardless of whether the true answer is “yes” or “no.”

• Another type of nonsampling error is measurement error. This occurs when the responses to the questions do not reflect what the investigator had in mind. It might result from poorly worded questions, questions the respondents don’t fully understand, questions that require the respondents to supply information they don’t have, and so on. Undoubtedly, there have been times when you were filling out a questionnaire and said to yourself, “OK, I’ll answer this as well as I can, but I know it’s not what they want to know.”

• One final type of nonsampling error is voluntary response bias. This occurs when the subset of people who respond to a survey differ in some important respect from all potential respondents. For example, suppose a population of students is surveyed to see how many hours they study per night. If the students who respond are predominantly those who get the best grades, the resulting sample mean number of hours could be biased on the high side.

From this discussion and your own experience with questionnaires, you should realize that the potential for nonsampling error is enormous. However, unlike sampling error, it cannot be measured with probability theory. It can be controlled only by using appropriate sampling procedures and designing good survey instruments. We will not pursue this topic any further here. If you are interested, however, you can learn about methods for controlling nonsampling error, such as proper questionnaire design, from books on survey sampling.

7-4b Key Terms in Sampling We now set the stage for the rest of this chapter, as well as for the next two chapters. Suppose there is some numerical population parameter you want to know. This parameter could be a population mean, a population proportion, the difference between two popu- lation means, the difference between two population proportions, or others. Unless you measure each member of the population—that is, unless you take a census—you cannot learn the exact value of this population parameter. Therefore, you instead take a random sample of some type and estimate the population parameter from the data in the sample.

You typically begin by calculating a point estimate (or, simply, an estimate) from the sample data. This is a “best guess” of the population parameter. The difference between the point estimate and the true value of the population parameter is called the sampling error (or estimation error). You then use probability theory to estimate the magnitude of the sampling error. The key to this is the sampling distribution of the point estimate, which is defined as the distribution of the point estimates you would see from all possible samples (of a given sample size) from the population. Often you report the accuracy of the point estimate with an accompanying confidence interval. A confidence interval is an interval around the point estimate, calculated from the sample data, that is very likely to contain the true value of the population parameter. (We will say much more about confi- dence intervals in the next chapter.)

A point estimate is a single numeric value, a “best guess” of a population param- eter, based on the data in a random sample.

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Unbiased estimates are desirable because they average out to the correct value. How- ever, this isn’t enough. Point estimates from different samples should vary as little as pos- sible from sample to sample. If they vary wildly, a point estimate from a single random sample isn’t very reliable. Therefore, it is common to measure the standard deviation of the sampling distribution of the estimate. This indicates how much point estimates from different samples vary. In the context of sampling, this standard deviation is called the standard error of the estimate. Ideally, estimates should have small standard errors.

Additionally, there are two other key terms you should know. First, consider the mean of the sampling distribution of a point estimate. It is the average value of the point esti- mates you would see from all possible samples. When this mean is equal to the true value of the population parameter, the point estimate is unbiased. Otherwise, it is biased. Nat- urally, unbiased estimates are preferred. Even if they sometimes miss on the low side and sometimes miss on the high side, they tend to be on target on average.

The sampling error (or estimation error) is the difference between the point esti- mate and the true value of the population parameter being estimated.

The sampling distribution of any point estimate is the distribution of the point estimates from all possible samples (of a given sample size) from the population.

A confidence interval is an interval around the point estimate, calculated from the sample data, that is very likely to contain the true value of the population parameter.

An unbiased estimate is a point estimate such that the mean of its sampling dis- tribution is equal to the true value of the population parameter being estimated.

The standard error of an estimate is the standard deviation of the sampling dis- tribution of the estimate. It measures how much estimates vary from sample to sample.

The terms in this subsection are relevant for practically any population parameter you might want to estimate. In the following subsection we discuss them in the context of esti- mating a population mean.

7-4c Sampling Distribution of the Sample Mean In this section we discuss the estimation of the population mean from some population. For example, you might be interested in the mean household income for all families in a particular city, the mean diameter of all parts from a manufacturing process, the mean

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amount of underreported taxes by all U.S. taxpayers, and so on. We label the unknown population mean by m. (It is common to label population parameters with Greek letters.)

The point estimate of m typically used, based on a sample from the population, is the sample mean X, the average of the observations in the sample. There are other pos- sible point estimates for a population mean besides the sample mean, such as the sample median, the trimmed mean (where all but the few most extreme observations are aver- aged), and others. However, it turns out that this “natural” estimate, the sample mean, has very good theoretical properties, so it is the point estimate used most often.

How accurate is X in estimating m? That is, how large does the estimation error X 2 m tend to be? The sampling distribution of the sample mean X provides the key. Before describing this sampling distribution in some generality, we provide some insight into it by revisiting the population of 40 incomes in Example 7.1. There we showed how to generate a single random sample of size 10. For the particular sample we generated (see Figure 7.2), the sample mean was $41,490. Because the population mean of all 40 incomes is $39,985, the estimation error based on this particular sample is the difference $41,490 2 $39,985, or $1505 on the high side.

However, this is only one of many possible samples. To see other possibilities, you could generate many random samples of size 10 from the population of 40 incomes. You could then calculate the sample mean for each random sample and create a histogram of these sample means. We did this, generating 100 random samples of size 10, with the result shown in Figure 7.4. Although this is not exactly the sampling distribution of the sample mean (because there are many more than 100 possible samples of size 10 from a population of size 40), it indicates approximately how the possible sample means are distributed. They are most likely to be near the population mean ($39,985), very unlikely to be more than about $3000 from the population mean, and have an approximately bell-shaped distribution.

Figure 7.4 Approximate Sampling Distribution of Sample Mean

30 Histogram of Sample Means

(34820,…

(36086.25,…

(37352.5,…

(38618.75,…

(41151.25,…

(42417.5,…

(39885,…

(43683.75,…

The insights in the previous paragraph can be generalized. It turns out that the sam- pling distribution of the sample mean has the following properties, regardless of the under- lying population. First, it is an unbiased estimate of the population mean, as indicated in Equation (7.1). The sample means from some samples will be too low, and those from other samples will be too high, but on the average, they will be on target.

Unbiased Property of Sample Mean

E(X) 5 m (7.1)

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The second property involves the variability of the X estimate. Recall that the stan- dard deviation of an estimate, called the standard error, indicates how much the estimate varies from sample to sample. The standard error of X is given in Equation (7.2). Here, SE(X) is an abbreviation for the standard error of X, s is the standard deviation of the population, and n is the sample size. You can see that the standard error is large when the observations in the population are spread out (large s), but that the standard error can be reduced by taking a larger sample.3

Standard Error of Sample Mean

SE(X) 5 s/!n (7.2)

Approximate Standard Error of Sample Mean

SE(X) 5 s>!n (7.3)

(Approximate) Confidence Interval for Population Mean

X { 2s>!n (7.4)

There is one problem with the standard error in Equation (7.2). Its value depends on another unknown population parameter, s. Therefore, it is customary to approximate the standard error by substituting the sample standard deviation, s, for s. This leads to Equation (7.3).

As we discuss in the next subsection, the shape of the sampling distribution of X is approximately normal. Therefore, you can use the standard error exactly as you have used standard deviations in previous chapters to obtain confidence intervals for the population mean. Specifically, if you go out two standard errors on either side of the sample mean, as shown in Expression (7.4), you can be approximately 95% confident of capturing the population mean. Equivalently, you can be 95% confident that the estimation error will be no greater than two standard errors in magnitude.

3 This formula for SE(X ) assumes that the sample size n is small relative to the population size N. As a rule of thumb, we assume that n is no more than 5% of N. Later we provide a “correction” to this formula when n is a larger percentage of N.

Sampling Distributions and Standard errors

Any point estimate, such as the sample mean, is random because it depends on the random sample that happens to be chosen. The sampling distribution of the point estimate is the probability distribution of point estimates from all possible random samples. This distribution describes how the sample means vary from one sample to another. The corresponding standard error is the standard deviation of the sam- pling distribution. These two concepts, sampling distribution and standard error, are the keys to statistical inference, as discussed here and in the next two chapters.

Fundamental Insight

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The following example illustrates a typical use of sample information.

EXAMPLE

7.3 ESTIMATING THE MEAN OF ACCOUNTS RECEIVABLE An internal auditor for a furniture retailer wants to estimate the average of all accounts receivable, where this average is taken over the population of all customer accounts. Because the company has approximately 10,000 accounts, an exhaustive enu- meration of all accounts receivable is impractical. Therefore, the auditor randomly samples 100 of the accounts. The data from the sample appear in Figure 7.5, with several hidden rows. (See the file Auditing Receivables.xlsx.) What can the auditor conclude from this sample?

Figure 7.5 Sampling in Auditing Example 1

2 3 4 5 6 7 8 9

10 11 12 13

105 106 107

A B C D E Random sample of accounts receivable

10000 Sample size 100

Sample of receivables Summary measures from sample Account Amount Sample mean $278.92

1 $85 Sample stdev $419.21 2 $1,061 Std Error of mean $41.92 3 $0 4 $1,260 With fpc $41.71 5 $924 6 $129

98 $657 99 $86

100 $0

Popula�on size

Objective To illustrate the meaning of standard error of the mean in a sample of accounts receivable.

Solution The receivables for the 100 sampled accounts appear in column B. This is the only information available to the auditor, so he must base all conclusions on these sample data. Begin by calculating the sample mean and sample standard deviation in cells E7 and E8 with the AVERAGE and STDEV.S functions. Then use Equation (7.3) to calculate the (approximate) standard error of the mean in cell E9 with the formula

=E8/SQRT(B4)

The auditor should interpret these values as follows. First, the sample mean $279 is a point estimate of the unknown pop- ulation mean. It provides a best guess for the average of the receivables from all 10,000 accounts. In fact, because the sample mean is an unbiased estimate of the population mean, there is no reason to suspect that $279 either underestimates or over- estimates the population mean. Second, the standard error $42 provides a measure of accuracy of the $279 estimate. Specifi- cally, there is about a 95% chance that the estimate differs by no more than two standard errors (about $84) from the true but unknown population mean. Therefore, the auditor can be approximately 95% confident that the mean from all 10,000 accounts is within the interval $279 { $84, that is, between $195 and $363.

It is important to distinguish between the sample standard deviation s and the standard error of the mean, approximated by s>!n. The sample standard deviation in the auditing example, $419, measures the variability across individual receivables in the sample (or in the population). By scrolling down column B, you can see some very low amounts

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(many zeros) and some fairly large amounts. This variability is indicated by the rather large sample standard deviation s. However, this value does not measure the accuracy of the sample mean as an estimate of the population mean. To judge its accuracy, you need to divide s by the square root of the sample size n. The resulting standard error, about $42, is much smaller than the sample standard deviation. It indicates that you can be about 95% confident that the sampling error is no greater than $84. In short, sample means vary much less than individual observations from a given population.

The Finite Population Correction We mentioned that Equation (7.2) [or Equation (7.3)] for the standard error of X is appro- priate when the sample size n is small relative to the population size N. Generally, “small” means that n is no more than 5% of N . In most realistic samples this is certainly true. For example, political polls are typically based on samples of approximately 1000 people from the entire U.S. population.

There are situations, however, when the sample size is greater than 5% of the popu- lation. In this case the formula for the standard error of the mean should be modified with a finite population correction, or fpc, factor. The modified standard error of the mean appears in Equation (7.5), where the fpc is given by Equation (7.6). Note that this factor is always less than 1 (when n 7 1) and it decreases as n increases. Therefore, the standard error of the mean decreases—and the accuracy increases—as n increases.

Standard Error of Mean with Finite Population Correction Factor

SE(X) 5 fpc 3 (s>!n) (7.5)

Finite Population Correction Factor

fpc 5 Å N 2 n

N 2 1 (7.6)

To see how the fpc varies with n and N, consider the values in Table 7.1. Rather than listing n, we have listed the percentage of the population sampled, that is, n>N 3 100%. It is clear that when 5% or less of the population is sampled, the fpc is very close to 1 and can safely be ignored. In this case you can use s>!n as the standard error of the mean. Otherwise, you should use the modified formula in Equation (7.5).

Table 7.1 Finite Population Correction Factors

N % Sampled fpc

100 5 0.980

100 10 0.953

10,000 1 0.995

10,000 5 0.975

10,000 10 0.949

1,000,000 1 0.995

1,000,000 5 0.975

1,000,000 10 0.949

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In the auditing example, n>N 5 100>100,000 5 0.1%. This suggests that the fpc can safely be omitted. We illustrate this in cell E11 of Figure 7.5, which uses the formula from Equation (7.5):

=SQRT((B3-B4)/(B3-1))*E9

Clearly, it makes no practical difference in this example whether you use the fpc or not. The standard error, rounded to the nearest dollar, is $42 in either case.

Virtually all standard error formulas used in sampling include an fpc factor. However, because it is rarely necessary—the sample size is usually very small relative to the popula- tion size—we omit it from here on.

7-4d The Central Limit Theorem Our discussion to this point has concentrated primarily on the mean and standard deviation of the sampling distribution of the sample mean. In this section we discuss this sampling distribution in more detail. Because of an important theoretical result called the central limit theorem, this sampling distribution is approximately normal with mean m and stan- dard deviation s>!n. This theorem is the primary reason why the normal distribution appears in so many statistical results. The theorem can be stated as follows.

If less than 5% of the population is sampled, as is often the case, the fpc can safely be ignored.

For any population distribution with mean m and standard deviation s, the sam- pling distribution of the sample mean X is approximately normal with mean m and standard deviation s>!n, and the approximation improves as n increases.

The important part of this result is the normality of the sampling distribution. We know, without any conditions placed upon the sample size n, that the mean and standard deviation are m and s>!n. However, the central limit theorem also implies normality, provided that n is reasonably large.

The central limit theorem is not a simple concept to grasp. To help explain it, we use simulation in the following example.

The Central Limit Theorem

This important result states that when you sum or average n randomly selected values from any distribution, normal or otherwise, the distribution of the sum or average is approximately normal, provided that n is sufficiently large. This is the primary reason why the normal distribution is relevant in so many real applications.

How large must n be for the approximation to be valid? Many textbooks suggest n $ 30 as a rule of thumb. However, this depends heavily on the population distribution. If the population distribution is very nonnormal— extremely skewed or bimodal, for example—the normal approximation might not be accurate unless n is considerably greater than 30. On the other hand, if the population distribution is already approximately symmetric, the normal approximation is quite good for n considerably less than 30. In fact, in the special case where the population distribution itself is normal, the sampling distribution of X is exactly normal for any value of n.

Fundamental Insight

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EXAMPLE

7.4 AVERAGE WINNINGS FROM A WHEEL OF FORTUNE Suppose you have the opportunity to play a game with a “wheel of fortune” (similar to the one in a popular television game show). When you spin a large wheel, it is equally likely to stop in any position. Depending on where it stops, you win any- where from $0 to $1000. Let’s suppose your winnings are actually based on not one spin, but on the average of n spins of the wheel. For example, if n 5 2, your winnings are based on the average of two spins. If the first spin results in $580 and the second spin results in $320, you win the average, $450. How does the distribution of your winnings depend on n?

Objective To illustrate the central limit theorem by a simulation of winnings in a game of chance.

Solution First, what does this experiment have to do with random sampling? Here, the population is the set of all outcomes you could obtain from a single spin of the wheel—that is, all dollar values from $0 to $1000. Each spin results in one randomly sam- pled dollar value from this population. Furthermore, because we have assumed that the wheel is equally likely to land in any position, all possible values in the continuum from $0 to $1000 have the same chance of occurring. The resulting population distribution is called the uniform distribution on the interval from $0 to $1000. (See Figure 7.6, where the 1 on the horizontal axis corresponds to $1000.) It can be shown that the mean and standard deviation of this uniform distribution are m 5 $500 and s 5 $289.4

4 In general, if a distribution is uniform on the interval from a to b, its mean is the midpoint (a 1 b)>2 and its standard deviation is (b 2 a)>!12.

Figure 7.6 Uniform Distribution

Before we go any further, take a moment to test your own intuition. If you play this game once and your winnings are based on the average of n spins, how likely is that you will win at least $600 if n 5 1? if n 5 3? if n 5 10? (The answers are 0.4, 0.27, and 0.14, respectively, where the last two answers are approximate and are based on the central limit theorem or the simulation. So you are much less likely to win big if your winnings are based on the average of many spins.)

Now we analyze the distribution of winnings based on the average of n spins. We do so by means of a sequence of simu- lations in Excel. (See the file Wheel of Fortune Simulation.xlsx, which is set up to work for any number of spins up to 10.) For each simulation, consider 1000 replications of an experiment. Each replication of the experiment simulates n spins of the wheel and calculates the average—that is, the winnings—from these n spins. Based on these 1000 replications, the average and standard deviation of winnings can be calculated, and a histogram of winnings can be formed, for any value of n. These will show clearly how the distribution of winnings depends on n.

The values in Figure 7.7 and the histogram in Figure 7.8 show the results for n 5 1. Here there is no averaging—you spin the wheel once and win the amount shown. To replicate this experiment 1000 times and collect statistics, proceed as follows.

Calculating the Distribution of Winnings by Simulation 1. Random outcomes. To generate outcomes uniformly distributed between $0 and $1000, enter the formula

=IF(B$9<=$B$6,$B$3+($B$4−$B$3)*RAND(), “ ”)

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in cell B11 and copy it to the entire range B11:K1010. The effect of this formula, given the values in cells B3 and B4, is to generate a random number between 0 and 1 and multiply it by $1000. The effect of the IF part is to fill as many Outcome columns as there are spins in cell B6 and to leave the rest blank.

2. Winnings. Calculate the winnings in each row in column L as the average of the outcomes of the spins in that row. (Note that the AVERAGE function ignores blanks.)

3. Summary measures. Calculate the average and standard deviation of the 1000 winnings in column L with the AVERAGE and STDEV.S functions. These values appear in cells L4 and L5.

4. Histogram. Create a histogram of the values in column L.

Note the following from Figures 7.7 and 7.8:

• The sample mean of the winnings (cell L4) is very close to the population mean, $500.

• The standard deviation of the winnings (cell L5) is very close to the population standard deviation, $289. • The histogram is nearly flat.

These properties should come as no surprise. When n 5 1, the sample mean is a single observation—that is, no averaging takes place. Therefore, the sampling distribution of the sample mean is equivalent to the flat population distribution in Figure 7.6.

But what happens when n 7 1? Figure 7.9 shows the results for n 5 2. All you need to do is change the number of spins in cell B6, and everything updates automatically. The average winnings are again very close to $500, but the standard deviation of winnings is much lower. In fact, you should find that it is close to s>!2 5 289>!2 5 $204, exactly as the theory pre- dicts. In addition, the histogram of winnings is no longer flat. It is triangularly shaped—symmetric, but not really bell-shaped.

1 2 3 4 5 6 7 8

A B C D E F G H I J K L Wheel of fortune simulation

Minimum winnings Summary measures of winnings

Maximum winnings Mean $497 $290

0.406 Stdev

Number of spins P(>600)

Simulation of spins 9

10 11 12 13 14 15 16

1007 1008

Spin 1 2 3 4 5 6 7 8 9 10 Outcome Outcome Outcome Outcome Outcome Outcome Outcome Outcome Outcome Outcome Winnings

1 $404 $893 $111 $764 $960 $363 $569 $869

$24 $255

$404 $893 $111 $764 $960 $363 $569 $869

$24 $255

2 3 4 5 6

997 998

1009 1010

999 1000

Replication

$0 $1,000

Figure 7.7 Simulation of Winnings from a Single Spin

Figure 7.8 Histogram of Simulated Winnings from a Single Spin

Histogram of Winnings

100

120

($1, $ 101)

($101, $ 200)

($200, $ 300)

($300, $ 400)

($400, $ 500)

($500, $ 600)

($600, $ 700)

($700, $ 799)

($799, $ 899)

($899, $ 999)

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7-4 Introduction to estimation 3 1 5

To develop similar simulations for n 5 3, n 5 6, n 5 10, or any other n, simply change the number of spins in cell B6. The resulting histograms appear in Figures 7.10 through 7.12. They clearly show two effects of increasing n: (1) the histogram becomes more bell-shaped, and (2) there is less variability. However, the mean stays right at $500. This behavior is exactly what the central limit theorem predicts. In fact, because the population distribution is symmetric in this example—it is flat— you can see the effect of the central limit theorem for n much less than 30; it is already evident for n as low as 6.

Figure 7.9 Histogram of Simulated Winnings from Two Spins

Histogram of Winnings

100

120

140

200

160

180

($35, $ 126)

($126, $ 217)

($217, $ 308)

($308, $ 399)

($399, $ 490)

($490, $ 581)

($581, $ 672)

($672, $ 763)

($763, $ 854)

($854, $ 945)

Figure 7.10 Histogram of Simulated Winnings from Three Spins

Histogram of Winnings

250

200

150

100

($50, $ 143)

($143, $ 236)

($236, $ 329)

($329, $ 422)

($422, $ 515)

($515, $ 608)

($608, $ 701)

($701, $ 794)

($794, $ 887)

($887, $ 980)

Figure 7.11 Histogram of Simulated Winnings from Six Spins

Histogram of Winnings

250

200

150

100

($142, $ 217)

($217, $ 292)

($292, $ 366)

($366, $ 441)

($441, $ 516)

($516, $ 591)

($591, $ 666)

($666, $ 741)

($741, $ 816)

($816, $ 891)

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Finally, it is easy to answer the question we posed previously: How does the probability of winning at least $600 depend on n? For any specific value of n, you can find the fraction of the 1000 replications where the average of n spins is greater than $600 with a COUNTIF formula in cell L6. (The value shown in Figure 7.7, 0.406, is only a point estimate of the true probabil- ity, which turns out to be very close to 0.4.)

Figure 7.12 Histogram of Simulated Winnings from Ten Spins

Histogram of Winnings

300

250

200

150

100

($197, $ 257)

($257, $ 318)

($318, $ 378)

($378, $ 438)

($438, $ 498)

($498, $ 558)

($558, $ 619)

($619, $ 679)

($679, $ 739)

($739, $ 799)

What are the main lessons from this example? For one, you can see that the sampling distribution of the sample mean (winnings) is bell-shaped when n is rea- sonably large. This is in spite of the fact that the population distribution is flat— far from bell-shaped. Actually, the population distribution could have any shape, not just uniform, and the bell-shaped property would still hold (although n might have to be larger than in the example). This bell-shaped normality property allows you to perform probability calculations with the NORM.DIST and NORM.INV functions, as discussed in Chapter 5.

Equally important, this example demonstrates the decreased variability in the sample means as n increases. Why should an increased sample size lead to decreased variability? This is due to averaging. Think about winning $750 based on the average of two spins. All you need is two lucky spins. In fact, one really lucky spin and an average spin will do. But think about winning $750 based on the average of 10 spins. Now you need a lot of really lucky spins—and virtually no unlucky ones. The point is that you are much less likely to obtain a really large (or really small) sample mean when n is large than when n is small. This is exactly what we mean when we say that the variability of the sample means decreases with larger sample sizes.

This decreasing variability is predicted by the formula for the standard error of the mean, s>!n. As n increases, the standard error decreases. This is what drives the behavior in Figures 7.9 through 7.12. In fact, using s 5 $289, the (theoretical) standard errors for n 5 2, n 5 3, n 5 6, and n 5 10 are $204, $167, $118, and $91, respectively.

Finally, what does this decreasing variability have to do with estimating a population mean with a sample mean? Very simply, it means that the sample mean tends to be a more accurate estimate when the sample size is large. Because of the approximate normality from the central limit theorem, you know from Chapter 5 that there is about a 95% chance that the sample mean will be within two standard errors of the population mean. In other words, there is about a 95% chance that the sampling error will be no greater than two standard errors in magnitude. Therefore, because the standard error decreases as the sam- ple size increases, the sampling error is likely to decrease as well.

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7-4 Introduction to estimation 3 1 7

7-4e Sample Size Selection The problem of selecting the appropriate sample size in any sampling context is not an easy one (as illustrated in the chapter opener), but it must be faced in the planning stages, before any sampling is done. We focus here on the relationship between sampling error and sample size. As we discussed previously, the sampling error tends to decrease as the sample size increases, so the desire to minimize sampling error encourages us to select larger sample sizes. We should note, however, that several other factors encourage us to select smaller sample sizes. The ultimate sample size selection must achieve a trade-off between these opposing forces.

What are these other factors? First, there is the obvious cost of sampling. Larger sam- ples cost more. Sometimes, a company or agency might have a budget for a given sam- pling project. If the sample size required to achieve an acceptable sampling error is 500, but the budget allows for a sample size of only 300, budget considerations will probably prevail.

Another problem caused by large sample sizes is timely collection of the data. Sup- pose a retailer wants to collect sample data from its customers to decide whether to run an advertising blitz in the coming week. Obviously, the retailer needs to collect the data quickly if they are to be of any use, and a large sample could require too much time to collect.

Finally, a more subtle problem caused by large sample sizes is the increased chance of nonsampling error, such as nonresponse bias. As we discussed previously in this chapter, there are many potential sources of nonsampling error, and they are usually very difficult to quantify. However, they are likely to increase as the sample size increases. Arguably, the potential increase in sampling error from a smaller sample could be more than offset by a decrease in nonsampling error, especially if the cost saved by the smaller sample size is used to reduce the sources of nonsampling error—conducting more follow-up of nonre- spondents, for example.

Nevertheless, the determination of sample size is usually driven by sampling error considerations. If you want to estimate a population mean with a sample mean, then the key is the standard error of the mean, given by

SE(X) 5 s>!n

The central limit theorem says that if n is reasonably large, there is about a 95% chance that the magnitude of the sampling error will be no more than two standard errors. Because s is fixed in the formula for SE(X), n can be chosen to make 2SE(X) acceptably small.

The averaging effect

As you average more and more observations from a given distribution, the vari- ance of the average decreases. This has a very intuitive explanation. For example, suppose you average only two observations. Then it is easy to get an abnormally large (or small) average. All it takes are two abnormally large (or small) observa- tions. But if you average a much larger number of observations, you aren’t likely to get an abnormally large (or small) average. The reason is that a few abnor- mally large observations will typically be cancelled by a few abnormally small observations. This cancellation produces the averaging effect. It also explains why a larger sample size tends to produce a more accurate estimate of a popula- tion mean.

Fundamental Insight

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We postpone further discussion of sample size selection until the next chapter, where we will discuss in detail how it can be used to control confidence interval length.

7-4f Summary of Key Ideas in Simple Random Sampling To this point, we have covered some very important concepts. Because we build on these concepts in later chapters, we summarize them here.

• To estimate a population mean with a simple random sample, the sample mean is typically used as a “best guess.” This estimate is called a point estimate. That is, X is a point estimate of m.

• The accuracy of the point estimate is measured by its standard error. It is the standard deviation of the sampling distribution of the point estimate. The standard error of X is approximately s>!n, where s is the sample standard deviation.

• A confidence interval (with 95% confidence) for the population mean extends to approximately two standard errors on either side of the sample mean.

• From the central limit theorem, the sampling distribution of X is approximately normal when n is reasonably large.

• There is approximately a 95% chance that any particular X will be within two standard errors of the population mean m.

• The sampling error can be reduced by increasing the sample size n. Appropriate sample size formulas for controlling confidence interval length are given in the next chapter.

effect of Larger Sample Sizes

Accurate estimates of population parameters require small standard errors, and small standard errors require large sample sizes. However, standard errors are typically inversely proportional to the square root of the sample size (or sample sizes). The implication is that if you want to decrease the standard error by a given factor, you must increase the sample size by a much larger factor. For example, to decrease the standard error by a factor of 2, you must increase the sample size by a factor of 4. Accurate estimates are not cheap.

Fundamental Insight

Problems Solutions for problems whose numbers appear within a colored box can be found in the Student Solution Files.

Level A 7. A manufacturing company’s quality control personnel

have recorded the proportion of defective items for each of 500 monthly shipments of one of the computer com- ponents that the company produces. The data are in the file P07_07.xlsx. The quality control department man- ager does not have sufficient time to review all of these data. Rather, she would like to examine the proportions of defective items for a sample of these shipments. For

this problem, you can assume that the population is the data from the 500 shipments. a. Generate a simple random sample of size 25 from the

data. b. Calculate a point estimate of the population mean

from the sample selected in part a. What is the sam- pling error, that is, by how much does the sample mean differ from the population mean?

c. Calculate a good approximation for the standard error of the mean.

d. Repeat parts b and c after generating a simple random sample of size 50 from the population. Is this estimate bound to be more accurate than the one in part b? Is its standard error bound to be smaller than the one in part c?

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7-4 Introduction to estimation 3 1 9

8. The manager of a local fast-food restaurant is interested in improving the service provided to customers who use the restaurant’s drive-up window. As a first step in this process, the manager asks his assistant to record the time it takes to serve a large number of customers at the final window in the facility’s drive-up system. The results are in the file P07_08.xlsx, which consists of nearly 1200 service times. For this problem, you can assume that the population is the data in this file. a. Generate a simple random sample of size 30 from the

data. b. Calculate a point estimate of the population mean

from the sample selected in part a. What is the sam- pling error, that is, by how much does the sample mean differ from the population mean?

c. Calculate a good approximation for the standard error of the mean.

d. If you wanted to halve the standard error from part c, what approximate sample size would you need? Why is this only approximate?

9. The file P02_16.xlsx contains traffic data from 256 weekdays on four variables. Each variable lists the num- ber of arrivals during a specific 5-minute period of the day. For this problem, consider this data set a simple random sample from all possible weekdays. a. For each of the four variables, find the sample mean.

If each of these is used as an estimate from the cor- responding (unknown) population mean, is there any reason to believe that they either underestimate or overestimate the population means? Why or why not?

b. What are the (approximate) standard errors of the esti- mates in part a? How can you interpret these standard errors? Be as specific as possible.

c. Is it likely that the estimates in part a are accurate to within 0.4 arrival? Why or why not? (Answer for each variable separately.)

10. The file P02_35.xlsx contains data from a survey of 500 randomly selected households. For this problem, con- sider this data set a simple random sample from all pos- sible households, where the number of households in the population is well over 1,000,000. a. Create a new variable, Total Income, that is the sum of

First Income and Second Income. b. For each of the four variables Total Income, Monthly

Payment, Utilities, and Debt, find the sample mean. If each of these is used as an estimate from the cor- responding (unknown) population mean, is there any reason to believe that they either underestimate or overestimate the corresponding population means? Why or why not?

c. What are the (approximate) standard errors of the esti- mates in part b? How can you interpret these standard errors? Be as specific as possible. Is the finite popula- tion correction required? Why or why not?

d. Is it likely that the estimate of Total Income in part b is accurate to within $1500? Why or why not?

11. The file P02_10.xlsx contains midterm and final exam scores for 96 students in a corporate finance course. For this problem, assume that these 96 students represent a sample of the 175 students taking the course, and that these 175 students represent the relevant population. a. Assuming the same instructor is teaching all four sec-

tions of this course and that the 96 students are the students in two of these sections, is it fair to say that the 96 students represent a random sample from the population? Does it matter?

b. Find the sample mean and the standard error of the sample mean, based on the 96 students in the file. Should the finite population correction be used? What is the standard error without it? What is the standard error with it?

Level B 12. Create a simulation similar to the one in the Wheel of

Fortune Similation.xlsx file. However, suppose that the outcome of each spin is no longer uniformly distributed between $0 and $1000. Instead, it is the number of 7’s you get in 20 rolls of two dice. In other words, each spin results in a binomially distributed random number with parameters n 5 20 and p 5 1>6 (because the chance of rolling a 7 is 1 out of 6). The simulation should still allow you to vary the number of “spins” from 1 to 10, and the “winnings” is still the average of the outcomes of the spins. What is fundamentally different from the simulation in the text? Does the central limit theorem still work? Explain from the results you obtain.

13. Suppose you plan to take a simple random sample from a population with N members. Specifically, you plan to sample a percentage p of the population. If p is 1%, is the finite population correction really necessary? Does the answer depend on N? Explain. Then answer the same questions when p is 5%, 10%, 25%, and 50%, respec- tively. In general, explain what goes wrong if the finite population correction is really necessary but isn’t used.

14. The file P07_14.xlsx contains a very small population of only five members. For each member, the height of the person is listed. The purpose of this problem is to let you see exactly what a sampling distribution is. Find the exact sampling distribution of the sample mean with sample size 3. Verify that Equation (7.1) holds, that is, the mean of this sampling distribution is equal to the population mean. Also, verify that Equation (7.2) holds, that is, the standard deviation of this sampling distribution is equal to the population standard deviation divided by the square root of 3. (Hint: You will have to do this by brute force. There are 125 different samples of size 3 that could be drawn from this population. These include samples with duplicate members, and order counts. For example, they include (1,1,2), (1,2,1), (2,1,1), and (1,1,1). You will need to find the sample mean of each and then find the mean and standard deviation of these sample means.)

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7-5 Conclusion This chapter has provided the fundamental concepts behind statistical inference. We discussed ways to obtain random samples from a population; how to calculate a point estimate of a particular population parameter, the population mean; and how to measure the accuracy of this point estimate. The key idea is the sampling distribution of the estimate and specifically its stan- dard deviation, called the standard error of the estimate. Due to the central limit theorem, the sampling distribution of the sam- ple mean is approximately normal, which implies that the sample mean will be within two standard errors of the population mean in approximately 95% of all random samples. In the next two chapters we build on these important concepts.

Summary of Key Terms TERM SYMBOL EXPLANATION PAGE EQUATION population Contains all members about which a study intends to make

inferences 281

Frame A list of all members of the population 281

Sampling units Potential members of a sample from a population 281

probability sample Any sample that is chosen by using a random mechanism 281

Judgmental sample Any sample that is chosen according to a sampler’s judgment rather than a random mechanism

281

Simple random sample

A sample where each member of the population has the same chance of being chosen

282

Systematic sample A sample where one of the first k members is selected randomly, and then every kth member after this one is selected

287

Stratified sampling Sampling in which the population is divided into relatively homogeneous subsets called strata, and then random samples are taken from each of the strata

288

proportional sam- ple sizes (in strati- fied sampling)

The property of each stratum selected having the same proportion from stratum to stratum

289

Cluster sampling A sample where the population is separated into clusters, such as cities or city blocks, and then a random sample of the clusters is selected

289

Sampling error The inevitable result of basing an inference on a sample rather than on the entire population

292

Nonsampling error Any type of estimation error that is not sampling error, including nonresponse bias, nontruthful responses, measurement error, and voluntary response bias

293

point estimate A single numeric value, a “best guess” of a population parameter, based on the data in a sample

294

Sampling error (or estimation error)

Difference between the estimate of a population parameter and the true value of the parameter

294

Sampling distribution

The distribution of the point estimates from all possible samples (of a given sample size) from the population

294

Confidence interval An interval around the point estimate, calculated from the sample data, where the true value of the population parameter is very likely to be

294

Unbiased estimate An estimate where the mean of its sampling distribution equals the value of the parameter being estimated

294

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7-5 Conclusion 3 2 1

TERM SYMBOL EXPLANATION PAGE EQUATION

Standard error of an estimate

The standard deviation of the sampling distribution of the estimate

294

Mean of sample mean

E(X) Indicates property of unbiasedness of sample mean 296 7.1

Standard error of sample mean

SE(X) Indicates how sample means from different samples vary 296 7.2, 7.3

Confidence inter- val for population mean

An interval that is very likely to contain the population mean mean

296 7.4

Finite population correction

fpc A correction for the standard error when the sample size is fairly large relative to the population size

298 7.5, 7.6

Central limit theorem

States that the distribution of the sample mean is approximately normal for sufficiently large sample sizes

299

Problems Note: Because the material in this chapter is more conceptual than calculation-based, we have included only conceptual questions here. You will get plenty of practice with calculations in the next two chapters, which build upon the concepts in this chapter.

Conceptual Questions C.1. Suppose that you want to know the opinions of American

secondary school teachers about establishing a national test for high school graduation. You obtain a list of the members of the National Education Association (the largest teachers’ union) and mail a questionnaire to 3000 teachers chosen at random from this list. In all, 823 teachers return the questionnaire. Identify the relevant population. Do you believe there is a good possibility of nonsampling error? Why or why not?

C.2. A sportswriter wants to know how strongly the res- idents of Indianapolis, Indiana, support the local minor league baseball team, the Indianapolis Indi- ans. He stands outside the stadium before a game and interviews the first 30 people who enter the stadium. Suppose that the newspaper asks you to comment on the approach taken by this sportswriter in performing the survey. How would you respond?

C.3. A large corporation has 4520 male and 567 female employees. The organization’s equal employment opportunity officer wants to poll the opinions of a random sample of employees. To give adequate atten- tion to the opinions of female employees, exactly how should the EEO officer sample from the given popula- tion? Be specific.

C.4. Suppose that you want to estimate the mean monthly gross income of all households in your local commu- nity. You decide to estimate this population parameter

by calling 150 randomly selected residents and ask- ing each individual to report the household’s monthly income. Assume that you use the local phone directory as the frame in selecting the households to be included in your sample. What are some possible sources of error that might arise in your effort to estimate the pop- ulation mean?

C.5. Provide an example of when you might want to take a stratified random sample instead of a simple random sample, and explain what the advantages of a stratified sample might be.

C.6. Provide an example of when you might want to take a cluster random sample instead of a simple random sample, and explain what the advantages of a cluster sample might be. Also, explain how you would choose the cluster sample.

C.7. Do you agree with the statement that nonresponse error can be overcome with larger samples? If you

agree, explain why. If you disagree, provide an exam- ple that backs up your opinion.

C.8. When pollsters take a random sample of about 1000 people to estimate the mean of some quantity over a population of millions of people, how is it possible for them to estimate the accuracy of the sample mean?

C.9. Suppose you want to estimate the population mean of some quantity when the population consists of mil- lions of members (such as the population of all U.S. households). How is it possible that you can obtain a fairly accurate estimate, using the sample mean of only about 1000 randomly selected members?

C.10. What is the difference between a standard deviation and a standard error? Be precise.

C.11. Explain as precisely as possible what it means that the sample mean is an unbiased estimate of the population mean [as indicated in Equation (7.1)].

Key Terms (continued)

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C.12. Explain the difference between the standard error for- mulas in equations (7.2) and (7.3). Why is Equation (7.3) the one necessarily used in real situations?

C.13. Explain as precisely as possible what Equation (7.4) means, and the reason for the 2 in the formula.

C.14. Explain as precisely as possible the role of the finite population correction. In which types of situations is it necessary? Is it necessarily used in the typical polls you see in the news?

C.15. In the wheel of fortune simulation with, say, three spins, many people mistakenly believe that the distri- bution of the average is the flat graph in Figure 7.7, that is, they believe the average of three spins is uni- formly distributed between $0 and $1000. Explain intuitively why they are wrong.

C.16. Explain the difference between a point estimate for the mean and a confidence interval for the mean. Which provides more information?

C.17. Explain as precisely as possible what the central limit theorem says about averages.

C.18. Many people seem to believe that the central limit the- orem “kicks in” only when n is at least 30. Why is this not necessarily true? When is such a large n necessary?

C.19. Suppose you are a pollster and are planning to take a sample that is very small relative to the population. In terms of estimating a population mean, can you say that a sample of size 9n is about 3 times as accurate as a sample of size n? Why or why not? Does the answer depend on the population size? For example, would it matter if the population size were 50 million instead of 10 million?

C.20. You saw in Equation (7.1) that the sample mean is an unbiased estimate of the population mean. How- ever, some estimates of population parameters are biased. In such cases, there are two sources of error in estimating the population parameter: the bias and the standard error. To understand these, imagine a rifleman shooting at a bull’s-eye. The rifleman could be aiming wrong and/or his shots could vary wildly from shot to shot. If he is aiming wrong but his shots are very consistent, what can you say about his bias and standard error? Answer the same question if he is correctly aiming at the bull’s-eye but is very incon- sistent. Can you say which of these two situations is worse?

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CHAPTER 8 Confidence Interval Estimation

ESTIMATING A COMPANY’S TOTAL TAXABLE INCOME In Example 7.3 in the previous chapter, we illustrated how sampling can be used in auditing. Another illustra- tion of sampling in auditing appears in Example 8.5 of this chapter. In both examples, the point of the sampling is to discover some property (such as a mean or a proportion) from a large population of a company’s accounts by exam- ining a small fraction of these accounts and projecting the results to the population. An article by Press (1995) offers an interesting variation on this problem. He poses the ques- tion of how a government revenue agency should assess a

business taxpayer’s income for tax purposes on the basis of a sample audit of the compa- ny’s business transactions. A sample of the company’s transactions will indicate a taxable income for each sampled transaction. The methods of this chapter will be applied to the sample information to obtain a confidence interval for the total taxable income owed by the company.

Suppose for the sake of illustration that this confidence interval extends from $1,000,000 to $2,200,000 and is centered at $1,600,000. In other words, the government’s best guess of the company’s taxable income is $1,600,000, and the government is fairly confident that the true taxable income is between $1,000,000 and $2,200,000. How much tax should it assess the company? Press argues that the agency would like to maximize its revenue while minimizing the risk that the company will be assessed more than it really owes. This last assumption, that the government does not want to overassess the company, is crucial. By making several reasonable assumptions, he argues that the agency should base the tax on the lower limit of the confidence interval, in this case, $1,000,000.

On the other hand, if the agency were indifferent between overcharging and under- charging, then it would base the tax on the midpoint, $1,600,000, of the confidence interval. Using this strategy, the agency would overcharge in about half the cases and undercharge in the other half. This would certainly be upsetting to companies—it would appear that the agency is flipping a coin to decide whether to overcharge or undercharge.

If the government agency does indeed decide to base the tax on the lower limit of the confidence interval, Press argues that it can still increase its tax revenue—by increas- ing the sample size of the audit. When the sample size increases, the confidence interval shrinks in width, and the lower limit, which governs the agency’s tax revenue, almost surely increases. But there is some point at which larger samples are not warranted, for the simple reason that larger samples cost more money to obtain. Therefore, there is an opti- mal size that will balance the cost of sampling with the desire to obtain more tax revenue.

8-1 Introduction This chapter expands on the ideas from the previous chapter. Given an observed data set, the goal is to make inferences to some larger population. Two typical examples follow:

• A mail-order company has accounts with thousands of customers. The company would like to infer the average time its customers take to pay their bills, so it randomly

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samples a relatively small number of its customers, sees how long these customers take to pay their bills, and draws inferences about the entire population of customers.

• A manufacturing company is considering two compensation schemes to implement for its workers. It believes that these two different compensation schemes might provide different incentives and hence result in different worker productivity. To see whether this is true, the company randomly assigns groups of workers to the two different com- pensation schemes for a period of three months and observes their productivity. Then it attempts to infer whether any differences observed in the experiment can be general- ized to the overall worker population.

In each of these examples, there is an unknown population parameter a company would like to estimate. In the mail-order example, the unknown parameter is the mean length of time customers take to pay their bills. Its true value could be discovered only by learning how long every customer in the entire population takes to pay its bills. This is not really possible, given the large number of customers. In the manufacturing exam- ple, the unknown parameter is a mean difference, the difference between the mean pro- ductivities with the two different compensation schemes. This mean difference could be discovered only by subjecting each worker to each compensation scheme and measuring their resulting productivities. This procedure would almost certainly be impossible from a practical standpoint. Therefore, the companies in these examples are likely to select random samples and base their estimates of the unknown population parameters on the sample data.

The inferences discussed in this chapter are always based on an underlying proba- bility model, which means that some type of random mechanism must generate the data. Two random mechanisms are generally used. The first involves sampling randomly from a larger population, as we discussed in the previous chapter. This is the mechanism respon- sible for generating the sample of customers in the mail-order example. Regardless of whether the sample is a simple random sample or a more complex random sample, such as a stratified sample, the fact that it is random allows us to use the rules of probability to make inferences about the population as a whole.

The second commonly used random mechanism is called a randomized experiment. The compensation scheme example just described is a typical randomized experiment. Here the company selects a set of subjects (employees), randomly assigns them to two different treatment groups (compensation schemes), and then compares some quantitative measure (productivity) across the groups. The fact that the subjects are randomly assigned to the two treatment groups is useful for two reasons. First, it allows us to rule out a num- ber of factors that might have led to differences across groups. For example, assuming that males and females are randomly spread across the two groups, we can rule out gender as the cause of any observed group differences. Second, the random selection allows us to use the rules of probability to infer whether observed differences can be generalized to all employees.

Generally, statistical inferences are of two types: confidence interval estimation and hypothesis testing. The first of these is the subject of the current chapter; hypothesis testing is discussed in the next chapter. They differ primarily in their point of view. For example, the mail-order company might sample 100 customers and find that they average 15.5 days before paying their bills. In confidence interval estimation, the data are used to obtain a point estimate and a confidence interval around this point estimate. In this exam- ple the point estimate is 15.5 days. It is a best guess for the mean bill-paying time in the entire customer population. Then, using the methods in this chapter, the company might find that a 95% confidence interval for the mean bill-paying time in the population is from 13.2 days to 17.8 days. The company is now 95% confident that the true mean bill-paying time in the population is within this interval.

Hypothesis testing takes a different point of view. There, the goal is to check whether the observed data provide support for a particular hypothesis. In the compensation scheme example, suppose the manager believes that workers will have higher productiv- ity if they are paid by salary than by an hourly wage. He runs the three-month randomized

We actually introduced 95% confidence intervals for the mean in the previous chapter. We generalize this method in the current chapter.

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experiment described previously and finds that the salaried workers produce on average eight more parts per day than the hourly workers. Now he must make one of two conclu- sions. Either salaried workers are in general no more productive than hourly workers and the ones in the experiment just got lucky, or salaried workers really are more productive. The next chapter explains how to decide which of these conclusions is more reasonable.

There are only a few key ideas in this chapter, and the most important of these, sampling distributions, was introduced in the previous chapter. It is important to concen- trate on these key ideas and not get bogged down in formulas or numerical calculations. Software such as Excel is generally available to take care of these calculations. The job of a businessperson is much more dependent on knowing which methods to use in which situations and how to interpret computer outputs than on memorizing and plugging into formulas.

8-2 Sampling Distributions Most confidence intervals are of the form in Equation (8.1). For example, when estimat- ing a population mean, the point estimate is the sample mean, the standard error is the sample standard deviation divided by the square root of the sample size, and the multiple is approximately 2. To learn why it works this way, you must first understand sampling distributions. This knowledge will then be put to use in the next section.

In the previous chapter, we introduced the sampling distribution of the sample mean X and saw how it was related to the central limit theorem. In general, whenever you make inferences about one or more population parameters, such as a mean or the difference between two means, you always base this inference on the sampling distribution of a point estimate, such as the sample mean. Although the concepts of point estimates and sampling distributions are no different from those in the previous chapter, there are some new details to learn.

We again begin with the sample mean X. The central limit theorem states that if the sample size n is reasonably large, then for any population distribution, the sampling dis- tribution of X is approximately normal with mean m and standard deviation s/!n, where m and s are the population mean and standard deviation. An equivalent statement is that the standardized quantity Z defined in Equation (8.2) is approximately normal with mean 0 and standard deviation 1.

Typical Form of Confidence Interval

Point Estimate { Multiple 3 Standard Error (8.1)

Standardized Z-Value

Z 5 X 2 m

s/!n (8.2)

Typically, this fact is used to make inferences about an unknown population mean m. There is one problem, however—the population standard deviation s is almost never known. This parameter, s, is called a nuisance parameter. Although it is typically not

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the parameter of primary interest, its value is needed for making inferences about the mean m. The solution appears to be straightforward: Replace the nuisance parameter s by the sample standard deviation s in the formula for Z and proceed from there. However, when s is replaced by s, this introduces a new source of variability, and the sampling dis- tribution is no longer normal. It is instead called the t distribution, a close relative of the normal distribution that appears in a variety of statistical applications.

8-2a The t Distribution We first set the stage for this new sampling distribution. We are interested in estimating a population mean m with a sample of size n. We assume the population distribution is nor- mal with unknown standard deviation s. We intend to base inferences on the standardized value of X from Equation (8.2), where s is replaced by the sample standard deviation s, as shown in Equation (8.3). Then the standardized value in Equation (8.3) has a t distribution with n − 1 degrees of freedom.

Standardized Value

t 5 X 2 m

s>!n (8.3)

The degrees of freedom is a numerical parameter of the t distribution that defines the precise shape of the distribution. Each time we encounter a t distribution, we will specify its degrees of freedom. In this particular sampling context, where we are basing inferences about m on the sampling distribution of X, the degrees of freedom turns out to be 1 less than the sample size n.

The t distribution looks very much like the standard normal distribution. It is bell-shaped and centered at 0. The only difference is that it is slightly more spread out, and this increase in spread is greater for small degrees of freedom. In fact, when n is large, so that the degrees of freedom is large, the t distribution and the standard normal distribu- tion are practically indistinguishable. This is illustrated in Figure 8.1. With 5 degrees of freedom, it is possible to see the increased spread in the t distribution. With 30 degrees of freedom, the t and standard normal curves are practically the same curve.

The t distribution and the standard normal distribution are practically the same when the degrees of freedom parameter is large.

Figure 8.1 The t and Standard Normal Distributions

The t-value in Equation (8.3) is very much like a typical Z-value such as in Equation (8.2). That is, the t-value represents the number of standard errors by which the sam- ple mean differs from the population mean. For example, if a t-value is 2.5, the sample

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mean is 2.5 standard errors above the population mean. Or if a t-value is 22.5, the sample mean is 2.5 standard errors below the population mean. Also, t-values greater in magnitude than 3 are quite unusual because of the same property of the normal distribu- tion: It is very unlikely for a random value to be more than three standard deviations from its mean.

Because of this interpretation, t-values are perfect candidates for the multiple term in Equation (8.1), as you will soon see. First, however, we briefly examine some Excel® functions that are useful for working with the t distribution in Excel.

Chapter 5 explained how to use Excel’s NORM.S.DIST and NORM.S.INV func- tions to calculate probabilities or percentiles from the standard normal distribution. There are similar Excel functions for the t distribution. Unfortunately, these functions are some- what more difficult to master than their normal counterparts. The file t Calculations.xlsx spells out the possibilities (see Figure 8.2). The top three examples show how to find the probability to the left or right of a given value. The bottom three examples show how to find the value with a given probability beyond it in one or both tails. You can refer to this figure as often as necessary when you work with the t distribution.

A t-value indicates the number of standard errors by which a sample mean differs from a population mean.

Figure 8.2 Excel Functions for the t Distribution 1

2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

Calculations for the t distribution

Sample size Degrees of freedom

One-tailed probabilities Value Probability in left tail

Value Probability in right tail

Two-tailed probability Value Probability in both tails

Inverse calculations Probability in left tail Value

Probability in right tail Value

Probability in both tails Value

30 29

–2 0.9725

2 0.0275

2 0.0549

0.05 –1.699

0.05 1.699

0.05 2.045

A B C D E F G

Formulas in Excel 2010 (or later)

=T.DIST(–B7,B4,TRUE)

=T.DIST.RT(B10,B4)

=T.DIST.2T(B14,B4) Half of this probability is in each tail

=T.INV(B18,B4)

=T.INV(1–B21,B4)

Half of this probability is in each tail =T.INV.2T(B24,B4)

8-2b Other Sampling Distributions The t distribution, a close relative of the normal distribution, is used to make inferences about a population mean when the population standard deviation is unknown. Through- out this chapter (and later chapters) you will see other contexts where the t distribution appears. The theme is always the same—one or more means are of interest, and one or more standard deviations are unknown.

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The t (and normal) distributions are not the only sampling distributions you will encounter. Two other close relatives of the normal distribution that appear in various con- texts are the chi-square and F distributions. These are used primarily to make inferences about variances (or standard deviations), as opposed to means. We omit the details of these distributions for now, but you will see them in later sections.

Problems Solutions for problems whose numbers appear within a colored box can be found in the Student Solution Files.

Level A 1. Calculate the following probabilities using Excel.

a. P(t10 $ 1.75), where t10 has a t distribution with 10 degrees of freedom.

b. P(t100 $ 1.75), where t100 has a t distribution with 100 degrees of freedom. How do you explain the dif- ference between this result and the one obtained in part a?

c. P(Z $ 1.75), where Z is a standard normal random variable. Compare this result to the results obtained in parts a and b. How do you explain the differences in these probabilities?

d. P(t20 # 20.80), where t20 has a t distribution with 20 degrees of freedom.

e. P(t3 # 20.80), where t3 has a t distribution with 3 degrees of freedom. How do you explain the difference between this result and the result obtained in part d?

2. Calculate the following quantities using Excel. a. P(22.00 # t10 # 1.00), where t10 has a t distribution

with 10 degrees of freedom.

b. P(22.00 # t100 # 1.00), where t100 has a t distri- bution with 100 degrees of freedom. How do you explain the difference between this result and the one obtained in part a?

c. P(22.00 # Z # 1.00), where Z is a standard normal random variable. Compare this result to the results obtained in parts a and b. How do you explain the dif- ferences in these probabilities?

d. Find the 68th percentile of the t distribution with 20 degrees of freedom.

e. Find the 68th percentile of the t distribution with 3 degrees of freedom. How do you explain the difference between this result and the result obtained in part d?

3. Calculate the following quantities using Excel. a. Find the value of x such that P(t10 7 x) 5 0.75, where

t10 has a t distribution with 10 degrees of freedom. b. Find the value of y such that P(t100 7 y) 5 0.75, where

t100 has a t distribution with 100 degrees of freedom. How do you explain the difference between this result and the result obtained in part a?

c. Find the value of z such that P(Z 7 z) 5 0.75, where Z is a standard normal random variable. Compare this result to the results obtained in parts a and b. How do you explain the differences in the values of x, y, and z?

8-3 Confidence Interval for a Mean We now come to the main topic of this chapter: using properties of sampling distribu- tions to construct confidence intervals. We assume that data have been generated by some random mechanism, either by observing a random sample from some population or by performing a randomized experiment. The goal is to infer the values of one or more pop- ulation parameters such as the mean, the standard deviation, or a proportion from sample data. For each such parameter, you use the data to calculate a point estimate, which can be considered a best guess for the unknown parameter. You then calculate a confidence inter- val around the point estimate to measure its accuracy.

We begin by deriving a confidence interval for a population mean m, and we discuss its interpretation. Although the particular details pertain to a specific parameter, the mean, the same ideas carry over to other parameters as well, as will be described in later sections. As usual, the sample mean X is used as the point estimate of the unknown population mean m.

To obtain a confidence interval for m, you first specify a confidence level, usually 90%, 95%, or 99%. You then use the sampling distribution of the point estimate to deter- mine the multiple of the standard error (SE) to go out on either side of the point esti- mate to achieve the given confidence level. If the confidence level is 95%, the value used

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8-3 Confidence Interval for a Mean 3 2 9

most frequently in applications, the multiple is approximately 2. More precisely, it is a t-value. That is, a typical confidence interval for m is of the form in Equation (8.4), where SE(X) 5 s>!n.

Confidence Interval for Population Mean

X { t-multiple 3 SE (X) (8.4)

To obtain the correct t-multiple, let a be one minus the confidence level (expressed as a decimal). For example, if the confidence level is 90%, then a 5 0.10. Then the appropri- ate t-multiple is the value that cuts off probability a>2 in each tail of the t distribution with n 2 1 degrees of freedom. For example, if n 5 30 and the confidence level is 95%, cell B25 of Figure 8.2 indicates that the correct t-value is 2.045. Then the corresponding 95% confidence interval for m is

X { 2.045(s>!n)

If the confidence level is instead 90%, the appropriate t-value is 1.699 (change the proba- bility in cell B24 to 0.10 to see this), and the resulting 90% confidence interval is

X { 1.699(s>!n)

If the confidence level is 99%, the appropriate t-value is 2.756 (change the probability in cell B24 to 0.01 to see this), and the resulting 99% confidence interval is

X { 2.756(s>!n)

Note that as the confidence level increases, the length of the confidence interval also increases. Because narrow confidence intervals are desirable, this presents a trade-off. You can either have less confidence and a narrow interval, or you can have more confi- dence and a wide interval. However, you can also take a larger sample. As n increases, the standard error s>!n tends to decrease, so the length of the confidence interval tends to decrease for any confidence level. (Why doesn’t it decrease for sure? The larger sample might result in a larger value of s that could offset the increase in n.)

Example 8.1 illustrates confidence interval estimation for a population mean.

EXAMPLE

8.1 CUSTOMER RESPONSE TO A NEW SANDWICH A fast-food restaurant recently added a new sandwich to its menu. To estimate the popularity of this sandwich, a random sample of 40 customers who ordered the sandwich were surveyed. Each of these customers was asked to rate the sand- wich on a scale of 1 to 10, 10 being the best. (See the file Satisfaction Ratings.xlsx.) The manager wants to estimate the mean satisfaction rating over the entire population of customers by finding a 95% confidence interval. How should she proceed?

Objective To obtain a 95% confidence interval for the mean satisfaction rating of the new sandwich.

Solution The method is spelled out in Figure 8.3 by the formulas shown in column D. The calculations follow directly from Equation (8.4). As in Figure 8.2, the T.INV.2T function is used to find the correct multiple. Its arguments are one minus the confidence level and the degrees of freedom. The result is that the 95% confidence interval extends from 5.739 to 6.761.

Confidence interval lengths increase when you ask for higher confidence levels, but they tend to decrease when you use larger sample sizes.

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Because the Excel calculations for a confidence interval are always the same (for a mean or for other population parame- ters), we have created a template file, Confidence Interval Template.xlsx, you can use. It has a sheet for each of the population parameters discussed in this chapter. For example, the sheet CI Mean shown in Figure 8.4 explains how to use the template for a population mean. The best way to proceed is to copy the appropriate template sheet to your data file and then follow the directions in the text box. If you do this for the sandwich rating data, you get the results in Figure 8.3, where we have shaded the only cells you need to change in orange and the confidence interval in yellow. The template sheets not only provide a quick way to get the correct results, but they also provide formulas you can examine to see how the confidence intervals are calculated.

Figure 8.3 Analysis of New Sandwich Data 1

2 3 4 5 6 7 8 9

10 11

Confidence level

Sample size Sample mean Sample standard deviation Standard error of mean Degrees of freedom Multiple Lower limit Upper limit

A B C D E F Confidence interval for population mean

=COUNT(Data!B2:B41)

=B5+B9*B7 =B5–B9*B7 =T.INV.2T(1–$B$2,B8) =B4–1 =B6/SQRT(B4) =STDEV.S(Data!B2:B41) =AVERAGE(Data!B2:B41)

95%

40 6.250 1.597 0.253

39 2.02

5.739 6.761

Figure 8.4 Template Sheet for a Confidence Interval for a Mean

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16

Confidence level

Sample size Sample mean Sample standard deviation Standard error of mean Degrees of freedom Multiple Lower limit Upper limit

Confidence interval for population mean A B C D E F HG JI K L

This template calculates a confidence interval for a population mean. Proceed as follows.

1. If you like, change the confidence level in cell B2.

2. Enter the sample size, sample mean, and sample standard deviation in cells B4 to B6. You can use the current formulas in these cells, replacing “Data” with the range of your numeric data variable, but feel free to fill these cells in any appropriate way.

3. Optionally, change the labels to fit the context of your problem.

The confidence interval will appear automatically in the yellow cells.

95%

0 #REF! #REF! #REF!

–1 #NUM!

#REF! #REF!

In any case, the principal results are that (1) the best guess for the population mean rating is 6.250, and (2) a 95% confi- dence interval for the population mean rating extends from 5.739 to 6.761. The manager can be 95% confident that the true mean rating over all customers who might try the sandwich is within this confidence interval.

We stated previously that as the confidence level increases, the length of the confidence interval increases. You can con- vince yourself of this by entering different confidence levels such as 90% or 99%. The lower and upper limits of the confi- dence interval will change automatically, getting closer together for the 90% level and farther apart for the 99% level. Just remember that you, the analyst, can choose the confidence level, but 95% is the level most commonly chosen.

Before leaving this example, we discuss the assumptions that lead to the confidence interval. First, you might question whether the sample is really a random sample—or whether it matters. Perhaps the manager used some random mechanism to select the customers to be surveyed. More likely, however, she simply surveyed 40 consecutive customers who tried the sand- wich on a given day. This is called a convenience sample and is not really a random sample. However, unless there is some reason to believe that these 40 customers differ in some relevant aspect from the entire population of customers, it is probably safe to treat them as a random sample.

A second assumption is that the population distribution is normal. We made this assumption when we introduced the t distri- bution. Obviously, the population distribution cannot be exactly normal because it is concentrated on the 10 possible satisfaction ratings, and the normal distribution describes a continuum. However, this is probably not a problem for two reasons. First, confi- dence intervals based on the t distribution are robust to violations of normality. This means that the resulting confidence intervals are approximately valid for any populations that are approximately normal. Second, the normal population assumption is less crucial for larger sample sizes because of the central limit theorem. A sample size of 40 should be large enough.

Finally, it is important to recognize what this confidence interval implies and what it doesn’t imply. In the entire popula- tion of customers who ordered this sandwich, there is a distribution of satisfaction ratings. Some fraction rate it as 1, some rate it as 2, and so on. All we are trying to determine here is the average of all these ratings. Based on the analysis, the manager can be 95% confident that this (still unknown) average is between 5.739 and 6.761. However, this confidence interval doesn’t tell her other characteristics of the population of ratings that might be of interest, such as the proportion of customers who rate the sandwich 6 or higher. It only provides information about the mean rating. Later in this chapter, you will see how to find a confidence interval for a proportion, which allows you to analyze another important characteristic of a population distribution.

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In the sandwich example, we said that the manager can be 95% confident that the true mean rating is between 5.739 and 6.761. What does this statement really mean? Contrary to what you might expect, it does not mean that the true mean lies between 5.739 and 6.761 with probability 0.95. Either the true mean is inside this interval or it is not. The true meaning of a 95% confidence interval is based on the procedure used to obtain it. Specif- ically, if you use this procedure on a large number of random samples, all from the same population, then approximately 95% of the resulting confidence intervals will be “good” ones that include the true mean, and the other 5% will be “bad” ones that do not include the true mean. Unfortunately, when you have only a single sample, as in the sandwich example, you have no way of knowing whether your confidence interval is one of the good ones or one of the bad ones, but you can be 95% confident that you obtained one of the good intervals.

Because this is such an important concept, we illustrate it in Figure 8.5 with simu- lation. (See the file Confidence Interval Simulation Finished.xlsx. There is no “unfin- ished” version of this file.) The data in column B are generated randomly from a normal distribution with the known values of m and s in cells B3 and B4. (Examine cell A7 to see how a random value from a normal distribution can be generated.) Then the confidence interval procedure is used to calculate a 95% confidence interval for the true value of m, exactly as in the sandwich example. However, because the true value of m is known, it is possible to record a 1 in cell H6 if the true mean is inside the interval and a 0 otherwise. The appropriate formula is

=IF(AND(B3>=D14,B3<=D15),1,0)

This simulation is performed only to illustrate the true meaning of a “95% confidence interval.”

Figure 8.5 Simulation Demonstration of Confidence Intervals

80.00 90.00 100.00 110.00 120.00

Confidence limits

Mean

5 6 7

Confidence level Sample size Sample mean Sample Std Dev Std Error of mean Degrees of freedom t multiple Lower limit Upper limit

This simulation uses a normal population for illustration. But you could generate the random sample from another distribution (e.g., triangular) to see if the confidence intervals are still valid, i.e, if the % in cell H7 is about 95%.

HGFEDCBA

Interpretation of a "95% confidence interval"

Population Population

Random sample Mean captured? 1 94.8%120.45

102.64 90.53 96.62 79.30

104.31 105.24 113.68

91.96 77.90

102.31 65.95

146.16 137.09

85.01 60.69

132.82 148.33 107.77 108.08 111.81

80.89 107.94 120.05

72.57 100.84 115.28 106.82

84.95 89.48

Confidence interval for mean

Graphical representation

% of Cl's capturing mean

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

95% 30

102.25 21.900

3.998 29

2.045 94.071

110.426

Limit 94.07

110.43

Mean 100

Height 1 1

Height 1

Data table to replicate confidence interval Replication Mean captured?

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 1 1 1

mean 100 stdev 20

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Finally, a data table can be used to replicate the simulated results 1000 times.1 Specifically, the formula in H11 is

5G6

Then to build the data table in the range G11:H1011, leave the row input cell box empty and specify any blank cell as the column input cell. Finally, the AVERAGE function can be used in cell H7 to find the fraction of 1’s in the range H12:H1011.

You can see that 948 of the simulated confidence intervals (each based on a different random sample of size 30) contain the true mean 100. In theory, 950 of the 1000 inter- vals should cover the true mean, and this is almost exactly what occurred. Of course, in a particular application you might unluckily obtain the 20th sample (in row 31). However, without knowing that the true mean is 100, you would have no way of knowing that you obtained a “bad” interval.

True Meaning of a 95% Confidence Interval

Given the data in a particular sample, a 95% confidence interval for the mean will either include the (unknown) population mean or it won’t. The true meaning of a 95% confidence interval is that if the same procedure is used on many different random samples, about 95% of the resulting confidence intervals will include the population mean, and only about 5% won’t. Therefore, you can be 95% confident that any particular confidence interval you happen to get is a “good” one.

Fundamental Insight

1 Depending on the speed of your computer, it can take a few seconds to simulate 1000 samples of size 30 in this data table. Therefore, it is a good idea to set the recalculation mode to “automatic except tables.” (You can find this option under the Calculation Options dropdown menu on the Formulas ribbon.) That way, the data table recalculates only if you explicitly tell it to (by pressing the F9 key).

We also show this graphically in the file. (See Figure 8.5.) The small square in this graph is positioned at the known mean and never changes. The blue line represents a par- ticular confidence interval. Press the F9 key to force a recalculation. The position of the blue line will change. About 95% of the time, the blue line will straddle the small square— the confidence interval will include the true mean—but about 1 time out of 20, it will not. This also illustrates the meaning of a “95% confidence interval.”

Problems Solutions for problems whose numbers appear within a colored box can be found in the Student Solution Files.

Level A 4. A manufacturing company’s quality control personnel

b. Using the sample generated in part a, calculate a 95% confidence interval for the mean proportion of defec- tive items over all monthly shipments. Assume that the population consists of the proportion of defective items for each of the given 500 monthly shipments.

c. Interpret the 95% confidence interval constructed in part b.

d. Does the 95% confidence interval contain the actual population mean in this case? If not, explain why not. What proportion of many similarly constructed confi- dence intervals should include the true population mean?

5. The file P08_05.xlsx contains salary data on all NFL players in each of the years 2002 to 2009. Because this file contains all players for each of these years, you can calculate the population mean for each year if population

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is defined as all NFL players that year. However, pro- ceed as in the previous chapter to select a random sam- ple of size 50 from the 2009 population. Based on this random sample, calculate a 95% confidence interval for the mean NFL total salary in 2009. Does it contain the population mean? Repeat this procedure several times until you find a random sample where the population mean is not included in the confidence interval.

6. The file P08_06.xlsx contains data on repetitive task times for each of two workers. John has been doing this task for months, whereas Fred has just started. Each time listed is the time (in seconds) to perform a routine task on an assembly line. The times shown are in chronologi- cal order. a. Calculate a 95% confidence interval for the mean

time it takes John to perform the task. Do the same for Fred.

b. Do you believe both of the confidence intervals in part a are valid and/or useful? Why or why not? Which of the two workers would you rather have, assuming that task time is the only issue?

7. The manager of a local fast-food restaurant is interested in improving the service provided to customers who use the restaurant’s drive-up window. As a first step in this process, the manager asks an assistant to record the time (in seconds) it takes to serve a large number of custom- ers at the final window in the facility’s drive-up system. The file P08_07.xlsx contains a random sample of 200 service times during the busiest hour of the day. a. Identify the relevant population. b. Calculate and interpret a 95% confidence interval for

the mean service time of all customers arriving during the busiest hour of the day at this fast-food operation.

c. If the manager wants to improve service, at least during the busiest time of day, does this confidence

interval provide useful information? What useful information does it not provide?

Level B 8. Continuing Problem 5, generate a random sample of 50

players for each of the eight years in the file P08_05.xlsx. For each of these samples, calculate a 95% confidence interval for the mean total salary for that year. What is the confidence level that any particular one of these con- fidence intervals includes the population mean for that year? Is this the same confidence level that all eight of these confidence intervals include the respective popula- tion means? Why or why not?

9. The file Confidence Interval Simulation.xlsx generates observations randomly from a normal population. Sup- pose instead that each observation in column A is expo- nentially distributed with mean 10. (Refer to Section 5-6 for a brief explanation of the exponential distribution.) Unlike a normal distribution, an exponential distribution is very skewed to the right. A value from this distribution can be generated with the formula =−10*LN(RAND()). Rerun the simulation, still with sample size 30, with this exponential distribution. Are 95% confidence intervals still valid? That is, is the percentage in cell H7 approxi- mately equal to 95%, as it should be? Press the F9 key a few times to check this.

10. Answer the questions in the previous problem when the population is a mixture of two normal distributions. Spe- cifically, suppose each observation has a 65% chance of coming from a normal distribution with mean 100 and standard deviation 20, and a 35% chance of coming from a normal distribution with mean 200 and standard deviation 40. What is the mean of this mixture distribu- tion? (Hint: Use an IF function to generate each random value in column A.)

2 This section can be omitted without any loss of continuity.

8-4 Confidence Interval for a Total2 There are situations where a population mean is not the population parameter of most interest. A good example is the auditing example discussed in the previous chapter (Example 7.3). Rather than estimating the mean amount of receivables per account, the auditor might be more interested in the total amount of all receivables, summed over all accounts. In this section we provide a point estimate and a confidence interval for a population total.

First, we introduce some notation. Let T be a population total we want to estimate, such as the total of all receivables, and let T̂ be a point estimate of T based on a simple ran- dom sample of size n from a population of size N. We first need a point estimate of T . For the population total T , it is reasonable to sum all of the values in the sample, denoted Ts, and then “project” this total to the population with Equation (8.5), where the second equal- ity follows because the sample total Ts divided by the sample size n is the sample mean X.

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Equation (8.5) is quite intuitive. For example, suppose there are 1000 accounts in the population, you sample 50 of them, and you observe a sample total of $5000. Then, because only 1/20 of the population was sampled, a natural estimate of the population total is 20 3 $5000 5 $100,000.

Like the sample mean X, the estimate T̂ has a sampling distribution. The mean and standard deviation of this sampling distribution are given in Equations (8.6) and (8.7), where s is again the population standard deviation.

Point Estimate for Population Total

T̂ 5 N n

Ts 5 NX (8.5)

Mean and Standard Error of Point Estimate for Population Total

E(T ̂) 5 T (8.6)

SE(T̂ ) 5 Ns/!n (8.7)

Approximate Standard Error of Point Estimate for Population Total

SE(T̂ ) 5 Ns/!n 5 N 3 SE(X) (8.8)

Because s is usually unknown, s is used instead of s to obtain the approximate stan- dard error of T̂ given in Equation (8.8). The second equality follows because s/!n is the standard error of X.

Note from Equation (8.6) that T̂ is an unbiased estimate of the population total T . Therefore, it has no tendency to either overestimate or underestimate T .

From Equations (8.5) and (8.8), the point estimate of T is the point estimate of the mean multiplied by N, and the standard error of this point estimate is the standard error of the sample mean multiplied by N. We illustrate this procedure in the following example.

EXAMPLE

8.2 ESTIMATING TOTAL TAX REFUNDS The Internal Revenue Service would like to estimate the total net amount of refund due to a particular set of 1,000,000 tax- payers. Each taxpayer will either receive a refund, in which case the net refund is positive, or will have to pay an amount due, in which case the net refund is negative. Therefore the total net amount of refund is a natural quantity of interest; it is the net amount the IRS will have to pay out (or receive, if negative). Find a 95% confidence interval for this total using the refunds from a random sample of 500 taxpayers in the file IRS Refunds.xlsx.

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8-4 Confidence Interval for a Total2 3 3 58-4 Confidence Interval for a Total 3 3 5

Objective To find a 95% confidence interval for the total (net) amount the IRS must pay out to these 1,000,000 taxpayers.

Solution The confidence interval appears in Figure 8.6. We used the CI Total sheet from the template file to perform the calculations. Note in particular how the sample mean is multiplied by the population size in cell E8 and how the standard error of the mean is also multiplied by the population size in cell E9. The effect is to scale the usual confidence interval for the mean by the population size.

Problems Solutions for problems whose numbers appear within a colored box can be found in the Student Solution Files.

Level A 11. The file P02_16.xlsx contains the number of arrivals at

a turnpike tollbooth for each of four 5-minute intervals for each of 256 days. For this problem, assume that each column, such as arrivals from 8:00 am to 8:05 am, is a random sample of all arrivals from the corresponding hour of the day, such as 8:00 am to 9:00 am. Calculate a 95% confidence interval for the mean number of arrivals during each corresponding hour of the day, that is, one for 8:00 am to 9:00 am, one for 9:00 am to 10:00 am, and so on.

12. A lightbulb manufacturer wants to estimate the total number of defective bulbs contained in all of the boxes shipped by the company during the past week. Produc- tion personnel at this company have recorded the num- ber of defective bulbs found in each of 50 randomly selected boxes shipped during the past week. These data are provided in the file P08_12.xlsx. Calculate a 95% confidence interval for the total number of defective bulbs contained in the 1000 boxes shipped by this com- pany during the past week.

13. Auditors of a particular bank are interested in comparing the reported value of all 2265 customer savings account balances with their own findings regarding the actual value of such assets. Rather than reviewing the records of each savings account at the bank, the auditors decide to examine a representative sample of savings account

balances. The population from which they will sample is given in the file P08_13.xlsx. a. Select 10 simple random samples, each consisting of

100 savings account balances from this population. b. For each sample generated in part a, calculate a 95%

confidence interval for the total value of all 2265 sav- ings account balances within this bank. How many of them include the (known) population total?

Level B 14. Suppose you are gambling on a roulette wheel. Each

time the wheel is spun, the result is one of the out- comes 0, 1, and so on through 36. Of these outcomes, 16 are red, 16 are black, and 1 is green. On each spin you bet $5 that a red outcome will occur and $1 that the green outcome will occur. If red occurs, you win a net $4. (You win $10 from red and nothing from green.) If green occurs, you win a net $24. (You win $30 from green and nothing from red.) If black occurs, you lose everything you bet for a loss of $6. a. Use random numbers to generate 20 plays from this

strategy. Each play should indicate the net amount won or lost. Then, based on these 20 outcomes, calcu- late a 95% confidence interval for the total net amount won or lost from 1000 plays of the game. Would you conclude that this strategy is a winning one for you?

b. Repeat part a, but with slightly changed rules. Now your betting strategy is the same, but if red occurs, your net gain is $5 (you win $11 from red, nothing from green). Comment on whether this slight change makes much of a difference in the mean total from 1000 bets.

Figure 8.6 Analysis of IRS Refund Data 1

2 3 4 5 6 7 8 9

10 11 12 13

Confidence level

Population size Sample size Sample mean Sample standard deviation Estimate of total Standard error of total Degrees of freedom Multiple Lower limit Upper limit

A B C D E F

95%

1,000,000 500

294.980 581.312

$294,980,000 $25,997,048

499 1.96

$243,902,836 $346,057,164

Confidence interval for population total

=COUNT(Data!B2:B501)

=T.INV.2T(1–$B$2,B10)

=B8+B11*B9 =B8–B11*B9

=B5–1 =B4*B7/SQRT(B5) =B4*B6 =STDEV.S(Data!B2:B501) =AVERAGE(Data!B2:B501)

Based on these calculations, the IRS can be 95% confident that it will need to pay out somewhere between about 244 and 346 million dollars to these 1,000,000 taxpayers.

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8-5 Confidence Interval for a Proportion How often have you heard on the evening news a survey finding such as, “52% of the public agree with the president’s handling of the economy, with a sampling error of plus or minus 3%”? Surveys are often used to estimate proportions, such as the proportion of the public who agree with the president’s handling of the economy. We now discuss how to form a confidence interval for any population proportion p.

The basic procedure is very similar to the procedure for a population mean. It requires a point estimate, the standard error of this point estimate, and a multiple that depends on the confidence level. Then the confidence level has the same form as in Equation (8.1):

point estimate { multiple 3 standard error

In the news example the point estimate is 52% and the “multiple 3 standard error” is 3%. Therefore, the confidence interval extends from 49% to 55%. Although the news show doesn’t state the confidence level explicitly, it is 95% by convention. In words, they are 95% confident that the percentage of the public who agree with the president’s handling of the economy is somewhere between 49% and 55%.

The theory that leads to this result is fairly straightforward. Consider any property that each member of a population either has or does not have. As examples, the property might be that

• a person agrees with the president’s handling of the economy • a person has purchased a company’s product at least once in the past three months • the diameter of a part is within specification limits • a customer’s account is at least two months overdue • a customer’s rating of a new sandwich is at least 6 on a 10-point scale.

In each of these examples, let p be the proportion of the population with the property. From a random sample of size n, let p̂ be the sample proportion of members with the property. For example, if 10 out of 50 sampled members have the property, then p̂ 5 10>50 5 0.2. Then p̂ is used as a point estimate of p.

It can be shown that for sufficiently large n, the sampling distribution of p̂ is approx- imately normal with mean p and standard error !p(1 2 p)>n. Because p is the unknown parameter, p̂ is substituted for p in this standard error to obtain the following approximate standard error of p̂:

Finally, the multiple used to obtain a confidence interval for p is a Z-value. (It is not a t-value.) It is the standard normal value that cuts off an appropriate probability in each tail. For example, the z-multiple for a 95% confidence interval is 1.96 because this value cuts off probability 0.025 in each tail of the standard normal distribution. In general, the confidence interval has the form in Equation (8.10):

Standard Error of Sample Proportion

SE( p̂) 5 Å p̂(1 2 p̂)

n (8.9)

Confidence Interval for a Proportion

p̂ { z-multiple 3 Å p̂(1 2 p̂)

n (8.10)

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This confidence interval is based on the assumption of a large sample size. A rule of thumb for checking the validity of this assumption is the following. Let pL and pU be the lower and upper limits of the confidence interval. Then the sample size is suffi- ciently large—and the confidence interval is valid—if npL 7 5, n(1 2 pL) 7 5, npU 7 5, and n(1 2 pU) 7 5. Essentially, these mean that n should be reasonably large and the two limiting values of p should not be too close to 0 or 1. We illustrate the procedure in the following example.

EXAMPLE

8.3 ESTIMATING THE RESPONSE TO A NEW SANDWICH The fast-food manager from Example 8.1 has already sampled 40 customers to estimate the population mean rating of the restaurant’s new sandwich. Recall that each rating is on a 1-to-10 scale, 10 being the best. The manager would now like to use the same sample to estimate the proportion of customers who rate the sandwich at least 6. Her thinking is that these are the customers who are likely to purchase the sandwich on subsequent visits.

Objective To illustrate the procedure for finding a confidence interval for the proportion of customers who rate the new sandwich at least 6 on a 10-point scale.

Solution The calculations appear in Figure 8.7. (See the file Satisfaction Ratings Finished.xlsx.) We used the CI Proportion sheet from the template file to perform the calculations. These calculations correspond exactly to Equation (8.10). Note in particular how the sample proportion is calculated directly from the sample data with a COUNTIF function. Also, note the formula for the z multiple in row 6. The argument for the NORM.S.INV function is 95% 1 5%>2, or 97.5%. The effect is to find the value that cuts off 2.5% in the right tail. You will see this type of calculation quite often.

The output is fairly good news for the manager. Based on this sample of size 40, she can be 95% confident that the per- centage of all customers who would rate the sandwich 6 or higher is somewhere between 47.5% and 77.5%. Of course, she realizes that this is a very wide interval, so there is still a lot of uncertainty about the true population proportion. To reduce the length of this interval, she would need to sample more customers—quite a few more customers. Typically, confidence intervals for proportions are fairly wide unless n is quite large.

1 2 3 4 5 6 7 8 9

Confidence level

Sample size Number with property of interest Sample proportion Standard error of proportion Multiple Lower limit Upper limit

A B C D E F

95%

40 25

0.625 0.077 1.960 0.475 0.775

Confidence interval for proportion with rating at least 6

=COUNT(Data!B2:B41)

=NORM.S.INV($B$2+(1–$B$2)/2)

=B6+B8*B7 =B6–B8*B7

=SQRT(B6*(1–B6)/B4) =B5/B4 =COUNTIF(Data!B2:B41,“>=6”)

Figure 8.7 Confidence Interval for Proportion

The data for this example are in “long” form, where there is an observation for each customer. In this case, formulas are required in cells B4 and B5 for the counts. However, the required counts for proportions examples are often given in tables. In this case, you can enter the given counts directly into the template.

In news shows, you have probably noticed that they almost always quote a sampling error of plus or minus 3%. In words, the “plus or minus” part of their 95% confidence interval is 3%, or 0.03. How large a sample size must they use to achieve this? The “plus

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or minus” part of the confidence interval is 1.96 times the standard error of p̂, so we must have

1.963!p̂(1 2 p̂)>n 5 0.03

Now, the quantity p̂(1 2 p̂) is fairly constant for values of p̂ between 0 and 1, provided that p̂ isn’t too close to 0 or 1. To get a reasonable estimate of the required n, we use p̂ 5 0.5. Then we have

1.96 3 !(0.5)(0.5)>n 5 0.03

Solving for n, we obtain n 5 3(1.96)(0.5)>0.0342 . 1067. This is a rather remarkable result. To obtain a 95% confidence interval of this length

for a population proportion, where the population consists of millions of people, only about 1000 people need to be sampled. The remarkable fact is that this small a sample can provide such accurate information about such a large population.

One of many business applications of confidence intervals for proportions is in auditing. Auditors typically use attribute sampling to check whether certain procedures are being fol- lowed correctly. The term “attribute” means that each item checked is done either correctly or incorrectly—there is no in-between. Examples of items not done correctly might include (1) an invoice copy that is not initialed by an accounting clerk, (2) an invoice quantity that does not agree with the quantity on the shipping document, (3) an invoice price that does not agree with the price on an authorized price list, and (4) an invoice with a clerical inaccuracy. Typically, an auditor focuses on one of these types of errors and then estimates the propor- tion of items with this type of error.

Because auditors are concerned primarily with how large the proportion of errors might be, they usually calculate one-sided confidence intervals for proportions. Instead of using sample data to find lower and upper limits pL and pU of a confidence interval, they automatically use pL 5 0 and then determine an upper limit pU such that the 95% confidence interval is from 0 to pU. A simple modification of the confidence interval in Equation (8.10) provides the result in Equation (8.11), where the z-multiple is chosen so that the entire probability (0.05 for a 95% interval) is in the right tail. For a 95% confi- dence level, the relevant z-multiple is 1.645.

Sample Size for Estimating a Proportion

To obtain an estimate of a proportion that is accurate to within 3 percentage points with 95% confidence, it is sufficient to sample approximately 1000 mem- bers of the population, regardless of the population size. This remarkable fact allows news broadcasters to make such statements about various proportions on a nightly basis. By sampling only about 1000 people from the entire country, they can estimate quite accurately what the entire population believes.

Fundamental Insight

Upper Limit of a One-Sided Confidence Interval for a Proportion

pU 5 p̂ 1 z-multiple 3 !p̂(1 2 p̂)>n (8.11)

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One further complication occurs, however. This formula for pU relies on the large-sample approximation of the normal distribution to the binomial distribution. Audi- tors typically use an exact procedure to find pU that is based directly on the binomial dis- tribution. We illustrate how this is done in the following example.

EXAMPLE

8.4 AUDITING FOR PRICE ERRORS An auditor wants to check the proportion of invoices that contain price errors—that is, prices that do not agree with those on an authorized price list. He checks 93 randomly sampled invoices and finds that two of them include price errors. What can he conclude, in terms of a one-sided 95% confidence interval, about the proportion of all invoices with price errors?

Objective To find the upper limit of a one-sided 95% confidence interval for the proportion of errors in the context of attribute sampling in auditing.

Solution The results appear in Figure 8.8. (See the file One-Sided Confidence Interval Finished.xlsx. It is not based on the template file because the template file only calculates two-sided confidence intervals.) The sample proportion is p 5 2>93 5 0.0215 and the upper confidence limit based on the large-sample approximation is 0.046. This latter value is calculated in cell B14 with the formula

=B7+B13*SQRT(B7*(1-B7)/B5)

However, note that npU 5 93(0.046) 5 4.278, which is less than 5. This indicates that the large-sample approximation might not be valid.

Figure 8.8 Analysis of Auditing Example

An exact one-sided confidence interval in auditing

Confidence level Number of errors

A 1 2 3 4 5 6 7 8 9

10 11 12 13 14

B C D E F G H I J

Sample size

Sample proportion 0.0215

0.066 0.050 Goal seek condition

0.05=

93 2

95%

Exact upper confidence limit for p

Large-sample upper confidence limit for p

Upper limit

Upper limit 0.046 1.645z-multiple

A more valid procedure, based on the binomial distribution, appears in row 10. It turns out that if pU is the appropriate upper confidence limit, then pU satisfies the following equation.

Formula for Upper Confidence Limit

P(X # k) 5 a (8.12)

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Here, X is binomially distributed with parameters n and pU, k is the observed number of errors, and a is one minus the confi- dence level. There is no way to find pU directly (by means of a formula) from Equation (8.12). However, you can use Excel’s Goal Seek tool, as illustrated in the figure. First, enter any trial value of pU in cell B10 and the binomial formula

=BINOM.DIST(B4,B5,B10,TRUE)

in cell D10. [This formula calculates P(X # k) from the trial value in cell B10.] Then use Goal Seek from the What-If Analysis dropdown menu on the Data ribbon, with cell D10 as the Set cell, 0.05 as the target value, and cell B10 as the changing cell.

The resulting value of pU is 0.066. This is considerably different (from the auditor’s point of view) from the 0.046 value found from the large-sample approximation. It allows the auditor to state with 95% confidence that the percentage of invoices with price errors is no greater than 6.6%, based on the two errors out of 93 observed in the sample.

Problems

Level A 15. A drugstore manager needs to purchase adequate

supplies of various brands of toothpaste to meet the ongoing demands of its customers. In particular, the company is interested in estimating the propor- tion of its customers who favor the country’s leading brand of toothpaste, Crest. The Data sheet of the file P08_15 .xlsx contains the toothpaste brand preferences of 200 randomly selected customers, obtained recently through a customer survey. Calculate a 95% confidence interval for the proportion of all of the company’s cus- tomers who prefer Crest toothpaste. How might the manager use this confidence interval for purchasing decisions?

16. The employee benefits manager of a large public uni- versity would like to estimate the proportion of full- time employees who prefer adopting the first (plan A) of three available health care plans in the next annual enrollment period. A random sample of the university’s employees and their tentative health care preferences are given in the file P08_16.xlsx.

a. Calculate a 90% confidence interval for the proportion of all the university’s employees who favor plan A.

b. The file also includes the classification of each employee (administrative staff, support staff, or faculty). Calculate a separate 90% confidence interval for each of these groups for the proportion who favor plan A. How do these confidence intervals compare to one another? How do their lengths compare to the confidence interval in part a? Is this what you would expect? Explain.

17. A market research consultant hired by a leading soft-drink company wants to determine the proportion of consum- ers who favor its low-calorie brand over the leading low- calorie competitor in a particular geographic region. A random sample of 250 consumers from the market under investigation is provided in the file P08_17.xlsx. a. Calculate a 90% confidence interval for the propor-

tion of all consumers in this market who prefer the company’s brand.

b. The file contains the gender and age group for each customer in the sample. Calculate a separate 90% confidence for each gender for the proportion who prefer the company’s brand. Then do the same for each age group. Explain briefly how these confidence intervals compare to each other and to the confidence interval in part a.

8-6 Confidence Interval for a Standard Deviation3

In Section 8-3 we focused primarily on estimation of a population mean. We had to deal with the population standard deviation s in its role as a nuisance parameter. That is, we needed an estimate of s to estimate the standard error of the sample mean. However, there are cases where the variability in the population, measured by s, is of primary interest in its own right. We describe a procedure for obtaining a confidence interval for s in this section.

The theory is somewhat more complex than for the case of the mean. As you might expect, the sample standard deviation s is used as a point estimate of s. However, the sampling distribution of s is not symmetric—in particular, it is not the normal distribu- tion or the t distribution. Rather, the appropriate sampling distribution is a right-skewed

3 This section can be omitted without any loss of continuity.

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distribution called the chi-square distribution. Like the t distribution, the chi-square dis- tribution has a degrees of freedom parameter, which (for this procedure) is again n 2 1.

Tables of the chi-square distribution, for selected degrees of freedom, appear in many statistics books, but the necessary information can be obtained more easily with Excel’s chi-square functions, as illustrated in Figure 8.9. (See the file Chi-Square Calculations Finished.xlsx.)

Figure 8.9 Excel Functions for the Chi-Square Distribution 1

2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17

Calculations for the chi-square distribution A B C D E F

Degrees of freedom

One-tailed probabilities Value Probability in left tail

Value Probability in right tail

Inverse calculations Probability in left tail Value

Probability in right tail Value

10 0.180

0.050

0.05 7.261

0.05 24.996

=CHISQ.DIST(B6,B3,TRUE)

=CHISQ.DIST.RT(B9,B3)

=CHISQ.INV(B13,B3)

=CHISQ.INV.RT(B16,B3)

Because of the skewness of the sampling distribution of s, a confidence interval for s is not centered at s. That is, the confidence interval is not the point estimate plus or minus a multiple of a standard error. Instead, s is always closer to the left endpoint of the confi- dence interval than to the right endpoint. Specifically, the endpoints of a confidence for the population standard deviation extends from s!(n 2 1)>c1 to s!(n 2 1)c2. Here, c1 is the value that cuts off probability (1 2 a)>2 in the right tail of the chi-square distribution with n 2 1 degrees of freedom, and c2 similarly cuts off probability (1 2 a)>2 in the left tail. For example, if the confidence level is 95%, these values cut off probability 0.025 in each tail. Note that c1 is larger than c2, so the left endpoint of the confidence interval is indeed lower than the right endpoint. The procedure is illustrated in the following example.

EXAMPLE

8.5 ANALYZING VARIABILITY IN DIAMETERS OF MACHINE PARTS

A machine produces parts that are supposed to have diameter 10 centimeters. However, due to inherent variability, some diam- eters are greater than 10 and some are less. The production supervisor is concerned about two things. First, he is concerned that the mean diameter is not what it should be, 10 centimeters. Second, he is worried about the extent of variability in the diameters. Even if the mean is on target, excessive variability implies that many of the parts will fail to meet specifications. To analyze the process, he randomly samples 50 parts during the course of a day and measures the diameter of each part to the nearest millimeter. (See the file Part Diameters.xlsx.) Should the supervisor be concerned about the results from this sample?

Objective To find a confidence interval for the standard deviation of part diameters, and to see how variability affects the proportion of unusable parts produced.

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Solution Because the manager is concerned about the mean and the standard deviation of diameters, it is useful to obtain 95% confi- dence intervals for both. We calculated each with the template file, using the CI Mean and CI Std Dev sheets. The confidence interval for the mean, not shown here, extends from 9.986 to 10.005. The confidence interval for the standard deviation, shown in Figure 8.10, extends from 0.043 to 0.131. Two chi-square values that cut off probability 0.025 in each tail are calculated first, and they are then used to calculate the endpoints of the confidence interval.

This interval extends from 0.029 cm to 0.043 cm. Is this good news or bad news? It depends. Let’s say that a part is unus- able if its diameter is more than 0.065 cm from the target. Let’s also assume that the true mean is right on target and that the standard deviation is at the upper end of the confidence interval, that is, s 5 0.043 cm. Finally, assume that the population distribution of diameters is normal. Then the calculation in cell E23 shows that 13.1% of the parts will be unusable. It adds the normal probabilities of being below or above the usable range.

To pursue this analysis one step further, the two-way data table in Figure 8.10 is useful. The means used in column B are the lower confidence limit, the sample mean, and the upper confidence limit. Similarly, the assumed standard deviations used in row 20 are the lower confidence limit, the sample standard deviation, and the upper confidence limit. To form the table, enter the formula =B16 in cell B20, highlight the range B20:E23, and create a data table with cells B15 and B14 as the row and column input cells.

Each value in the body of the data table is the resulting proportion of unusable parts. Obviously, a mean close to the target and a small standard deviation are best, but even this best-case scenario results in 2.5% unusable parts (see cell C22). However, a mean off target and a large standard deviation can lead to as many as 14.9% unusable parts (see cell E21). In any case, the message for the supervisor is clear—he must work to reduce the underlying variability in the process. This variability is hurt- ing him much more than an off-target mean.

Figure 8.10 Analysis of Parts Data

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22 23

Sample size Sample standard deviation Degrees of freedom Lower chi-sq value Upper chi-sq value Lower limit Upper limit

Confidence interval for population standard deviation A B C D E F HG JI

=COUNT(Data!B2:B51) =STDEV.S(Data!B2:B51) =B4–1 =CHISQ.INV((1–$B$2)/2,B6) =CHISQ.INV.RT((1–$B$2)/2,B6) =SQRT(B6)*B5/SQRT(B8) =SQRT(B6)*B5/SQRT(B7)

Proportion of unusable parts Maximum deviation for usability Assumed mean Assumed standard deviation Proportion unusuable

0.065 10

0.043 0.131

0.131 9.986 9.996

10.005

0.029 0.041 0.025 0.026

0.034 0.080 0.060 0.061

0.043 0.149 0.130 0.131

Assumed mean

Assumed standard deviation

=NORM.DIST(10–B13,B14,B15,TRUE)+(1–NORM.DIST(10+B13,B14,B15,TRUE))

Twoway data table for finding proportion unusable as a function of mean and stdev

95%Confidence level

50 0.034

49 31.555 70.222

0.029 0.043

Problems Solutions for problems whose numbers appear within a colored box can be found in the Student Solution Files.

Level A 18. Senior management of a large consulting firm is con-

cerned about a growing decline in the organization’s weekly number of billable hours. Ideally, the organi- zation expects each professional employee to spend at

least 40 hours per week on work. The file P08_18.xlsx contains the work hours reported by a random sample of employees in a typical week. a. Calculate a 95% confidence interval for the mean

number of hours worked by the company’s employees in a typical week.

b. Calculate a 95% confidence interval for the standard deviation of the number of hours worked by the com- pany’s employees in a typical week.

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8-7 Confidence Interval for the Difference Between Means 3 4 3

c. Given the target range of 40 to 60 hours of work per week, should senior management be concerned about the number of hours their employees are currently devoting to work? Explain how the answers to both parts a and b help to answer this question.

Level B 19. The file P08_06.xlsx contains data on repetitive task

times for each of two workers. John has been doing this task for months, whereas Fred has just started. Each time

listed is the time (in seconds) to perform a routine task on an assembly line. The times shown are in chronolog- ical order. a. Calculate a 95% confidence interval for the standard

deviation of times for John. Do the same for Fred. What do these indicate?

b. Given that these times are listed chronologically, how useful are the confidence intervals in part a? Specifi- cally, is there any evidence that the variation in times is changing over time for either of the two workers?

8-7 Confidence Interval for the Difference Between Means One of the most important applications of statistical inference is the comparison of two population means. There are many applications to business, including the following.

Applications of Comparisons of Means in Business

• Men and women shop at a retail clothing store. The manager would like to know how much more (or less), on average, a woman spends on a typical purchase occasion than a man.

• Two airline companies fly similar routes. A consumer organization would like to check how much the average delay differs between the two airlines, where delay is defined as the actual arrival time at the destination minus the scheduled arrival time.

• A supermarket chain mails coupons for various products to a randomly selected subset of its customers in a particular city. Its other customers in this city receive no such coupons. The chain would like to check how much the average amount spent on these products differs between the two sets of customers over the next couple of months.

• A computer company has a customer service center that responds to customers’ questions and complaints. The center employs two types of people: those who have had a recent course in dealing with customers (but little actual experience) and those with a lot of experience dealing with customers (but no formal course). The company would like to know how these two types of employees differ with respect to the average number of customer complaints of poor service in the last six months.

• A consulting company hires business students directly out of undergraduate school. The new hires all take a problem-solving test. They then go through an intensive three-month training program, after which they take another similar problem-solving test. The company wants to know how much the average test score improves after the training program.

• A car dealership often deals with husband–wife pairs shopping for cars. To check whether husbands react differently than their wives to the sales presentation, husbands and wives are asked (separately) to rate the quality of the sales presentation. The deal- ership wants to know how much husbands differ from their wives in terms of average ratings.

Each of these examples deals with a difference between means from two populations. However, the first four examples differ in one important respect from the last two. In the last two examples, there is a natural pairing across the two samples. In the first of these, each employee takes a test before a course and then a test after the course, so that each employee is naturally paired with himself or herself. In the final example, husbands and wives are naturally paired with one another. There is no such pairing in the first four exam- ples. Instead, we assume that the samples in these first four examples are chosen inde- pendently of one another. We need to distinguish these two cases, independent samples and paired samples, in the discussion that follows.

Statisticians call these general types of problems “comparison problems.” They are among the most important types of prob- lems tackled with statistical methods.

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8-7a Independent Samples The framework for this situation is the following. We are interested in some quantity, such as dollars spent or airplane delay, for each of two populations. The population means are m1 and m2, and the population standard deviations are s1 and s2. We take random samples of sizes n1 and n2 (not necessarily equal) from the populations to estimate the difference between means, m1 2 m2. A point estimate of this difference is the natural one, the dif- ference between sample means, X1 2 X2. Starting with this estimate, we want to form a confidence interval for the unknown population mean difference, m1 2 m2.

The appropriate sampling distribution of the difference between sample means is again the t distribution, now with n1 1 n2 2 2 degrees of freedom.4 Therefore, a confi- dence interval for m1 2 m2 is given by Expression (8.13). The t-multiple is the value that cuts off the appropriate probability (depending on the confidence level) in each tail of the t distribution with n1 1 n2 2 2 degrees of freedom. For example, if the confidence level is 95% and n1 5 n2 5 30, the appropriate t-multiple is 2.002, which can be found in Excel with the function T.INV(0.025,58).

Confidence Interval for Difference Between Means

X1 2 X2 { t-multiple 3 SE(X1 2 X2) (8.13)

Pooled Estimate of Common Standard Deviation

sp 5 Å (n1 2 1)s2

1 1 (n2 2 1)s2 2

n1 1 n2 2 2 (8.14)

Standard Error of Difference Between Sample Means

SE(X1 2 X2) 5 spÅ 1 n1

1 1 n2

(8.15)

4 This assumes that either the population distributions are normal or that the sample sizes are reasonably large, conditions that are at least approximately met in a wide variety of applications.

The standard error, SE(X1 2 X2), is more involved. We must first make the assump- tion that the population standard deviations are equal, that is, s1 5 s2. (We shortly pres- ent an alternative procedure for the situation where the population standard deviations are not assumed to be equal.) Then an estimate of this common standard deviation is provided by the “pooled” estimate from both samples, labeled sp.

Here, s1 and s2 are the sample standard deviations from the two samples. This pooled estimate is somewhere between s1 and s2, with the relative sample sizes determining its exact value. Then the standard error of X1 2 X2 is given by Equation (8.15):

This procedure, usually referred to as “two-sample t,” is illustrated in the following example.

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8-7 Confidence Interval for the Difference Between Means 3 4 5

EXAMPLE

8.6 RELIABILITY OF TREADMILL MOTORS AT SURESTEP SureStep Company manufactures high-quality treadmills for use in exercise clubs. SureStep currently purchases its motors for these treadmills from supplier A. However, it is considering a change to supplier B, which offers a slightly lower cost. The only question is whether supplier B’s motors are as reliable as supplier A’s. To check this, SureStep installs motors from supplier A on 30 of its treadmills and motors from supplier B on another 30 of its treadmills. It then runs these treadmills under typical conditions and, for each treadmill, records the number of hours until the motor fails. (See the file Treadmill Motors.xlsx. It has two data columns, one for the supplier and one for the lifetime.) What can SureStep conclude?

Objective To find a confidence interval for the difference between mean lifetimes of motors, and to see how this confidence interval can help SureStep choose the better supplier.

Solution In any comparison problem it is a good idea to look initially at side-by-side box plots of the two samples. These appear in Figure 8.11. These show that (1) the distributions of times until failure are skewed to the right for each supplier, (2) the mean for supplier A is somewhat greater than the mean for supplier B, and (3) there are several outliers. There seems to be little doubt that supplier A’s motors will last longer on average than supplier B’s—or is there? A confidence interval for the mean difference allows you to see whether the differences apparent in the box plots can be generalized to all motors from the two suppliers.

The calculations appear in Figure 8.12. They were performed by the CI Mean Diff sheet from the template file. As you can see, the sample means differ by approximately 93 hours, the sample standard deviations are of roughly the same magnitude, and the confidence interval for the difference between means extends from 247.549 to 233.815.

These calculations are fairly complex, but remember that you have to fill in only the orange cells; the template does the rest. Still, the following comments are useful:

1. The sample means are calculated with AVERAGEIF functions. The formula for sample mean 1 averages only over the A’s, and the formula for sample mean 2 averages only over the B’s.

2. There is no corresponding “STDEVIF” function, so the array formulas in cells B11 and B12 are necessary. The template only requires you to fill in the data ranges, but you have to complete the formulas by pressing Ctrl1Shift1Enter (all three keys at once), not just Enter. This creates the curly brackets; you don’t type them.

3. We stated earlier that there is a version of the procedure that does not require the equal-standard deviation assumption. The template takes care of this automatically. It performs a test for equal variances (discussed in the next chapter) in

Box Plots for Two Suppliers

1400

1600

1200

1000

800

600

400

200

0 A B

Figure 8.11 Box Plots for Treadmill Motors Data

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3 4 6 C H A P T E R 8 C o n f i d e n c e I n t e r v a l E s t i m a t i o n

rows 20–22. A large (7 0.1) p-value for this test is evidence of unequal variances or equivalently of unequal standard deviations. The formulas in cells B14 and B15 check for this and enter the appropriate formulas in either case. For this example, the p-value is large, 0.639, so the equal-standard deviation assumption is warranted.

The confidence interval extends from a negative value to a positive value, but it’s mostly positive. If SureStep had to make a guess, it would say that supplier A’s motors last longer on average. Should SureStep continue with supplier A? This depends on the trade-off between the cost of the motors and warranty costs (and any other relevant costs). Because the warranty prob- ably depends on whether a motor lasts a certain amount of time, warranty costs probably depend on a proportion (the propor- tion that fail before 500 hours, say) rather than a mean. Confidence intervals for differences between proportions are discussed in Section 8-8.

JG I K

Confidence level

Category 1

Category 2

Sample size 1

Sample size 2

Sample mean 1

Sample mean 2

Sample mean difference (1 minus 2)

Sample std dev 1

Sample std dev 2

Pooled std dev

Standard error of diff

Degrees of freedo

Multiple

Lower limit

Upper limit

Test for equal variances

Ratio of sample variances

p-value

BA C D E F H

95%

748.800

655.667

93.133

283.881

259.986

272.196

70.281

2.00

–47.549

233.815

1.192

0.639

Confidence interval for difference between means

=(B11/B12)^2

=2*(0.5 – ABS(0.5 – F.DIST(B21,B6–1,B7–1,TRUE)))

=COUNTIF(Data!$A$2:$A$61,B4)

=COUNTIF(Data!$A$2:$A$61,B5)

=AVERAGEIF(Data!$A$2:$A$61,B4,Data!$B$2:$B$61)

=AVERAGEIF(Data!$A$2:$A$61,B5,Data!$B$2:$B$61)

=B8–B9

{=STDEV.S(IF(Data!$A$2:$A$61=B4,Data!$B$2:$B$61))}

{=STDEV.S(IF(Data!$A$2:$A$61=B5,Data!$B$2:$B$61))}

=SQRT(((B6–1)*B11^2+(B7–1)*B12^2)/B15)

=IF(B22>0.1,B13*SQRT(1/B6+1/B7),SQRT(B11^2/B6+B12^2/B7))

=IF(B22>0.1,B6+B7–2,ROUND(B14^4/((B11^2/B6)^2/(B6–1)+(B12^2/B7)^2/(B7–1)),0))

=T.INV.2T(1–$B$2,B15)

=B10–B16*B14

=B10+B16*B14

Figure 8.12 Analysis of Treadmill Motors Data

8-7b Paired Samples When the samples to be compared are paired in some natural way, such as a pre-test and post-test for each person, or husband–wife pairs, there is a more appropriate form of analy- sis than the two-sample procedure. Consider the example where each new employee takes a test, then receives a three-month training course, and afterward takes another similar test. There is likely to be a fairly strong correlation between the pre-test and post-test scores. Employees who score relatively low on the first test are likely to score relatively low on the second test, and employees who score relatively high on the first test are likely to score relatively high on the second test. The two-sample procedure in the previous section for independent samples does not take this correlation into account and therefore ignores

Role of Variances in Estimating the Difference Between Means

It might be surprising that variances (or standard deviations) play such an import- ant role in estimating the difference between means, but this is actually quite intuitive. If there is a lot of variability in the populations, it is more difficult to get accurate estimates of the population means, and hence the difference between the means. But if there is very little variability, it is much easier to estimate the means accurately.

Fundamental Insight

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8-7 Confidence Interval for the Difference Between Means 3 4 7

important information. The paired procedure described in this section, on the other hand, uses this information to advantage.

The procedure itself is straightforward. It does not directly analyze two separate vari- ables (pre-test scores and post-test scores, for example); it analyzes their differences. For each pair in the sample, the difference between the two scores for the pair is calculated. Then a one-sample analysis, as in Section 8-3, is performed on these differences. This is illustrated in the following example.

EXAMPLE

8.7 HUSBAND AND WIFE REACTIONS TO SALES PRESENTATIONS

Stevens Honda-Buick automobile dealership often sells to husband-wife pairs. The manager would like to check whether the sales presentation is viewed any more or less favorably by the husbands than by the wives. If it is, then some new training might be recommended for its salespeople. To check for differences, a random sample of husbands and wives are asked (sep- arately) to rate the sales presentation on a scale of 1 to 10, 10 being the most favorable rating. (See the Sales Presentation Ratings.xlsx file, where there are two columns, one for husbands and one for their wives.) What can the manager conclude from these data?

Objective To use a paired-sample procedure to find a confidence interval for the mean difference between husbands’ and wives’ ratings of sales presentations.

Solution The procedure is straightforward, as illustrated in Figure 8.13. It uses the CI Paired Mean Diff sheet from the template file. You are first asked to calculate the differences in column A, where each difference is a husband rating minus a wife rating. Then the template uses the same one-sample mean formulas on the differences as in Example 8.1. The sample mean Husband minus Wife difference is 1.629 and a 95% confidence interval for this difference extends from 1.057 to 2.200.

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15

A B C D E F HG Difference Confidence interval for difference (husbands - wives)

Confidence level

Sample size Sample mean diff Sample std dev of diff Standard error Degrees of freedom Multiple Lower limit Upper limit

=COUNT(Difference) =AVERAGE(Difference) =STDEV.S(Difference) =D6/SQRT(D4) =D4–1 =T.INV.2T(1–D2,D8) =D5–D9*D7 =D5+D9*D7

95%3 –1

3 2 3 1 3

–1 –1

2 3 1 3

–1

35 1.629

1.664256 0.281

34 2.032 1.057 2.200

Figure 8.13 Sales Presentation Analysis

Before leaving this example, we again note that the finished version of the file shows what would have happened if the two-sample procedure had been used on the Husband and Wife variables. The resulting confidence interval for the mean differ- ence extends from 0.895 to 2.362, which is somewhat longer than the confidence interval from the paired-sample procedure. This is typical. When the two-sample procedure is used in a situation where the paired-sample procedure is more appropriate, the data are not used as efficiently. The effect is that the standard error of the difference tends to be larger, and the resulting confidence interval tends to be longer.

Why is the paired-sample procedure appropriate here? It is not just because husbands and wives naturally come in pairs. It is because they tend to react similarly to one another. You can check that the correlation between the husbands’ scores and their wives’ scores is 0.442. (This can be found with Excel’s CORREL function on the Husband and Wife variables.) This is far from a perfect correlation, but it is large enough to warrant using the paired-sample procedure.

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In general, the paired-sample procedure is appropriate when the samples are naturally paired in some way and there is a reasonably large positive correlation between the pairs. In this case the paired-sample procedure makes more efficient use of the data and gener- ally results in narrower confidence intervals.

Problems Solutions for problems whose numbers appear within a colored box can be found in the Student Solution Files.

Level A 20. The director of a university’s career development cen-

ter is interested in comparing the starting annual sala- ries of male and female students who recently graduated from the university and commenced full-time employ- ment. The director has formed pairs of male and female graduates with the same major and similar grade-point averages. Specifically, she has collected a random sample of 50 such pairs and has recorded the starting annual salary of each person. These data are provided in the file P08_20.xlsx. Calculate a 95% confidence inter- val for the mean difference between similar male and female graduates of this university. Interpret your result.

21. A real estate agent has collected a random sample of 75 houses that were recently sold in a suburban com- munity. She is particularly interested in comparing the appraised value and recent selling price of the houses in this particular market. The data are provided in the file P08_21.xlsx. Using this sample data, calculate a 95% confidence interval for the mean difference between the appraised values and selling prices of the houses sold in this suburban community. Interpret the confidence inter- val for the real estate agent.

22. The Wall Street Journal CEO Compensation Study ana- lyzed CEO pay from many U.S. companies with fiscal year 2008 revenue of at least $5 billion that filed their proxy statements between October 2008 and March 2009. The data are in the file P02_30.xlsx. a. Create a new column, Total, that is the sum of col-

umns D and E. b. After combining Telecommunications and Technology

into a single company type, there are nine company types. For each of these, calculate a 95% confidence interval for the difference between the mean of Total for

that company type and mean of Total for all other com- pany types. Comment on what these nine confidence intervals indicate about CEO pay in different industries.

Level B 23. The file P02_35.xlsx contains data from a survey of 500

randomly selected households. a. Separate the households in the sample by the loca-

tion of their residence within the given community. For each of the four locations, use the sample infor- mation to calculate a 90% confidence interval for the mean annual income (sum of first income and second income) of all relevant households. Compare these four interval estimates. You might also consider gen- erating box plots of the total income for households in each of the four locations.

b. Calculate a 90% confidence interval for the difference between the mean annual income of all households in the first (i.e., SW) and second (i.e., NW) sectors of this community. Calculate similar 90% confidence intervals for the differences between the mean annual income levels of all households from all other pairs of locations (i.e., first and third, first and fourth, second and third, second and fourth, and third and fourth). Summarize your findings.

24. A company employs two shifts of workers. Each shift pro- duces a type of gasket where the thickness is the critical dimension. The average thickness and the standard devia- tion of thickness for shift 1, based on a random sample of 30 gaskets, are 10.53 mm and 0.14 mm. The similar fig- ures for shift 2, based on a random sample of 25 gaskets, are 10.55 mm and 0.17 mm. Let m1 2 m2 be the mean difference in thickness between shifts 1 and 2. a. Calculate a 95% confidence interval for m1 2 m2. b. Based on your answer to part a, are you convinced

that the gaskets from shift 2 are, on average, wider than those from shift 1? Why or why not?

c. How would your answers to parts a and b change if the sample sizes were instead 300 and 250?

8-8 Confidence Interval for the Difference Between Proportions

The final confidence interval we examine is a confidence interval for the difference between two population proportions. As in the previous section, this comparison proce- dure finds many real applications. Several potential business applications follow.

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8-8 Confidence Interval for the Difference Between Proportions 3 4 9

Applications of Comparisons of Proportions in Business • When an appliance store is about to have a sale, it sometimes sends selected

customers a mailing to notify them of the sale. On other occasions it includes a coupon for 5% off the sale price in these mailings. The store’s manager would like to know whether the inclusion of coupons affects the proportion of customers who respond.

• A manufacturing company has two plants that produce identical products. The company wants to know how much the proportion of out-of-spec products differs across the two plants.

• A pharmaceutical company has developed a new over-the-counter sleeping pill. To judge its effectiveness, the company runs an experiment where one set of randomly chosen people takes the new pill and another set takes a placebo. (Neither set knows which type of pill they are taking.) The company judges the effectiveness of the new pill by comparing the proportions of people who fall asleep within a certain amount of time with the new pill and with the placebo.

• An advertising agency would like to check whether men are more likely than women to switch TV channels when a commercial comes on. The agency runs an experiment where the channel-switching behavior of randomly chosen men and women can be monitored, and it collects data on the proportion of viewers who switch channels on at least half of the commercial times. The agency then compares these proportions across gender.

The basic form of analysis in each of these examples is the same as in the two-sample analysis for differences between means. However, instead of comparing two means, we now compare two proportions.

Formally, let p1 and p2 represent the two unknown population proportions, and let p̂1 and p̂2 be the two sample proportions, based on samples of sizes n1 and n2. Then the point estimate of the difference between proportions, p1 2 p2, is the difference between sample proportions, p̂1 2 p̂2. If the sample sizes are reasonably large, the sampling distribution of p̂1 2 p̂2 is approximately normal.5

Therefore, a confidence interval for p1 2 p2 is given by Equation (8.16). Here, the z-multiple is the usual value from the standard normal distribution that cuts off the appro- priate probability in each tail (1.96 for a 95% confidence interval, for example). Also, the standard error of p̂1 2 p̂2 is given by Equation (8.17).

The following example illustrates this procedure.

Confidence Interval for Difference Between Proportions

p̂1 2 p̂2 { z-multiple 3 SE( p̂1 2 p̂2) (8.16)

Standard Error of Difference Between Sample Proportions

SE( p̂1 2 p̂2) 5 Å p̂1(1 2 p̂1)

n1 1

p̂2(1 2 p̂2) n2

(8.17)

5 This large-sample assumption is valid as long as nip̂i 7 5 and ni(1 2 p̂i) 7 5 for i 5 1 and i 5 2.

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EXAMPLE

8.8 SALES RESPONSE TO COUPONS FOR DISCOUNTS ON APPLIANCES

An appliance store is about to have a big sale. It selects 300 of its best customers and randomly divides them into two sets of 150 customers each. It then mails a notice of the sale to all 300 customers but includes a coupon for an extra 5% off the sale price to the second set of customers only. As the sale progresses, the store keeps track of which of these customers purchase appliances. Some of the resulting data appear in Figure 8.14. (See the file Coupon Effectiveness.xlsx.) What can the store’s manager conclude about the effectiveness of the coupons?

Figure 8.14 Coupon Data 1 2 3 4 5 6 7 8 9

10 11

1 2 3 4 5 6 7 8 9

No No No No Yes Yes No Yes Yes No

No No No Yes No Yes Yes No No No

Customer Received coupon Purchased A B C

Figure 8.15 Analysis of Coupon Data

Confidence level

Sample size 2

Lower limit

=COUNTIFS(Data!$B$2:$B$301,B4,Data!$C$2:$C$301,“Yes”)

=COUNTIFS(Data!$B$2:$B$301,B5,Data!$C$2:$C$301,“Yes”)

C IHGFEDBA

8 9

12 13

15 16

=NORM.S.INV($B$2+(1–$B$2)/2)

=SQRT(B10*(1–B10)/B6+B11*(1–B11)/B7)

=B10–B11

=B9/B7

=B8/B6

=COUNTIF(Data!$B$2:$B$301,B5)

=COUNTIF(Data!$B$2:$B$301,B4)

0.236

0.031

1.960

0.052

0.133

0.233

0.367

150

Yes

95%

Upper limit

Multiple

Standard error

Sample proportion diff

Sample proportion 2

Sample proportion 1

Number 2 purchased

Number 1 purchased

Sample size 1

Received coupon 2

Received coupon 1

=B12+B14*B13

=B12–B14*B13

Confidence interval for difference between proportions purchased

Objective To find a confidence interval for the difference between proportions of customers purchasing appliances with and without 5% discount coupons.

Solution The solution, using the CI Proportion Diff sheet from the template file, is shown in Figure 8.15. (See the file Coupon Effectiveness Finished.xlsx.) You can see that 36.67% of customers who received a coupon purchased something, as

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8-9 Sample Size Selection 3 5 1

The data for this last example are in “long” form, where there is a Yes/No value for each customer. This means that formulas are required for the counts in cells B6 to B9. However, the data are often in tabular form, where the required counts are given. In this case, you can enter these counts directly into the template.

opposed to only 23.33% of those who didn’t receive a coupon. The difference, 36.67% 2 23.33% 5 13.33% (or 0.1333), is the quantity of interest. Specifically, the sample difference is 13.33%, and the objective is to find a confidence interval for this difference for the entire population. Using Equations (8.16) and (8.17), the relevant confidence interval formulas are spelled out in column D. They show that the confidence interval for the difference between proportions extends from 0.031 to 0.236 (all positive). So there is good reason to conclude that the proportion who purchase is larger for those who receive a coupon.

8-9 Sample Size Selection In this section we discuss the most widely used methods for achieving a confidence inter- val of a specified length. Confidence intervals are a function of three things: (1) the data in the sample, (2) the confidence level, and (3) the sample size(s). We briefly discuss the role of the first two in terms of their effect on confidence interval length and then discuss the effect of sample size in more depth.

Problems Solutions for problems whose numbers appear within a colored box can be found in the Student Solution Files.

Level A 25. A company that advertises on the Web wants to know

which search engine its customers prefer as their pri- mary search engine: Google or Bing. Specifically, the company wants to know whether the preference depends on the browser being used. The file P08_25.xlsx con- tains counts of 800 customers’ favorite search engine, broken down by the browser used. a. Calculate a 95% confidence interval for the difference

between two proportions: the proportion of Internet Explorer users whose favorite search engine is Google and the similar proportion of Firefox users.

b. Repeat part a, replacing Google with Bing. c. Interpret the results in parts a and b. Do the search

engine preferences seem to depend on the browser used? 26. A market research consultant hired by a leading soft-drink

company is interested in estimating the difference between the proportions of female and male consumers who favor the company’s low-calorie brand over the leading com- petitor’s low-calorie brand in a particular geographical region. A random sample of 250 consumers from the mar- ket under investigation is provided in the file P08_17.xlsx. After separating the 250 randomly selected consumers by gender, calculate a 95% confidence interval for the differ- ence between these two proportions. Of what value might this interval estimate be to marketing managers at the company?

Level B 27. The file P02_35.xlsx contains data from a survey of 500

randomly selected households. Researchers would like to use the available sample information to see whether home ownership rates vary by household location. For example, is there a nonzero difference between the proportions of individuals who own their homes (as opposed to those who rent their homes) in households located in the first (i.e., SW) and second (i.e., NW) sec- tors of this community? Use the given sample to cal- culate a 95% confidence interval that estimates this difference between proportions in home ownership rates for each pair of locations. Interpret and summarize your results. (The solution should include six confidence intervals.)

28. Continuing from problem 26, marketing managers at the soft-drink company have asked their market research consultant to explore further the difference between the proportions of women and men who prefer drinking their brand over the leading competitor. Specifically, the company’s managers would like to know whether the difference between the proportions of female and male consumers who favor their brand varies by the age of the consumers. Use the same data as in problem 29 to assess whether estimates of this difference vary across the four given age categories: under 20, between 20 and 40, between 40 and 60, and over 60. Use a 95% confi- dence level for each of the four required confidence inter- vals. Summarize your findings. What recommendations would you make to the marketing managers in light of your findings?

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The data in the sample directly affect the length of a confidence interval through their sample standard deviation(s). It might appear that because of random sampling, you have no control over the sample data, but this is not entirely true. In the case of surveys from a population, there are random sampling plans that can reduce the amount of variability in the sample and hence reduce confidence interval length. Indeed, this is the primary reason for using the stratified sampling procedure discussed in the previous chapter.

Variance reduction is also possible in randomized experiments. There is a whole area of statistics called experimental design that suggests how to perform experiments to obtain the most information from a given amount of sample data. Although this is often aimed at scientific and medical research, it is also appropriate in business contexts. For exam- ple, the automobile dealership in Example 8.7 was wise to use paired husband–wife data rather than two independent samples of men and women. The pairing leads to a potential reduction in variability and hence a narrower confidence interval.

The confidence level has a clear effect on confidence interval length. As the confi- dence level increases, the length of the confidence interval increases as well. For example, a 99% confidence interval is always longer than a 95% confidence interval, assuming that they are both based on the same data. However, the confidence level is rarely used to con- trol the length of the confidence interval. Instead, the confidence level choice is usually based on convention, and 95% is by far the most commonly used value. In fact, it is the default level built into most software packages.

The most obvious way to control confidence interval length is to choose the sam- ple size(s) appropriately. The rest of this section explains to do this. For each parameter we discuss, the goal is to make the length of a confidence interval sufficiently narrow. Because each confidence interval discussed so far (with the exception of the confidence interval for a standard deviation) is a point estimate plus or minus some quantity, we focus on the “plus or minus” part, called the half-length of the interval. The usual approach is to specify the half-length B you would like to obtain. Then you find the sample size(s) neces- sary to achieve this half-length.

Confidence Interval Length

The length of any confidence interval is influenced by three things: the sample size, the confidence level, and the variability in the population. The confidence level is typically set at 95%, and you have no control over the variability in the population (except possibly by choosing an appropriate experimental design). Therefore, the best way to control confidence interval length is through the choice of the sample size.

Fundamental Insight

We begin with a confidence interval for the mean. From Section 8-3, the relevant formula is

X { t-multiple 3 s>!n

The goal is to make the half-length of this interval equal to some prescribed value B. For example, if you want the confidence interval to be of the form X { 5, you use B 5 5. Actually, it is not possible to achieve this half-length B exactly, but you can come close.

By setting

t-multiple 3 s>!n 5 B

and solving for n, the appropriate sample size is

n 5 a t-multiple 3 s

B b

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8-9 Sample Size Selection 3 5 3

Unfortunately, sample size selection must be done before a sample is observed. There- fore, no value of s is yet available. Also, because the t-multiple depends on n (through the degrees of freedom parameter), it is not clear which t-multiple to use.

The usual solution is to replace s by some reasonable estimate sest of the pop- ulation standard deviation s, and to replace the t-multiple with the corresponding z - multiple from the standard normal distribution. The latter replacement is justified because z-values and t-values are practically equal unless n is very small. The result- ing sample size formula is given in Equation (8.18). This formula generally results in a noninteger value of n, in which case the practice is to round n up to the next larger integer.

Keep in mind that the sample size must be determined before the data are observed.

The following example, an extension of Example 8.1, shows how to implement Equation (8.18).

EXAMPLE

8.9 SAMPLE SIZE SELECTION FOR ESTIMATING REACTION TO A NEW SANDWICH

The fast-food manager in Example 8.1 surveyed 40 customers, each of whom rated a new sandwich on a scale of 1 to 10. Based on the data, a 95% confidence interval for the mean rating of all potential customers extended from 5.739 to 6.761, with a half-length of (6.761 2 5.739)>2 5 0.511. How large a sample would be needed to reduce this half-length to approx- imately 0.3?

Objective To find the sample size of customers required to achieve a sufficiently narrow confidence interval for the mean rating of the new sandwich.

Solution Equation (8.18) for n uses three inputs: the z-multiple, which is 1.96 for a 95% confidence level; the prescribed confidence interval half-length B, which is 0.3 for this example; and an estimate sest of the standard deviation. This final quantity must be guessed, but based on the given sample of size 40, the observed sample standard deviation, 1.597, from Example 8.1 can be used. Therefore, Equation (8.18) yields

n 5 a 1.96(1.597)

0.3 b

5 108.86

which is rounded up to n 5 109. The claim, then, is that if the manager surveys 109 customers, a 95% confidence interval will have approximate half-length 0.3. Its exact half-length will differ slightly from 0.3 because the standard deviation from the sample will almost surely not be exactly 1.597.

Figure 8.16 shows how this formula can be implemented in Excel. It is part of the CI Sample Sizes sheet from the tem- plate file. Here, the CEILING function is used to round up to the nearest integer. You can see that a sample size of 109 is required.

Sample Size Formula for Estimating a Mean

n 5 a z-multiple 3 sest

B b

(8.18)

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What if the manager is at the planning stage and doesn’t have a “preliminary” sample of size 40? What standard devia- tion estimate should she use for sest (because the value 1.597 is no longer available)? This is not an easy question to answer, but because of the role of sest in Equation (8.18), it is crucial for the determination of n. The manager basically has three choices: (1) she can base her estimate of the standard deviation on historical data, assuming relevant historical data are avail- able; (2) she can take a small preliminary sample to get an estimate of the standard deviation; or (3) she can simply guess a value for the standard deviation. We do not recommend the third option, but there are cases in which it is the only feasible option available.

We have demonstrated the use of Equation (8.18) for a sample mean. In the same way, it can also be used in the paired-sample procedure. In this case the resulting value of n refers to the number of pairs that should be included in the sample, and sest refers to an estimate of the standard deviation of the differences (Husband scores minus Wife scores, for example).

The sample-size analysis for the mean carries over with very few changes to other parameters. We discuss three other parameters in this section: a proportion, the difference between two means, and the difference between two proportions. In each case the required confidence interval can be obtained by setting the half-length equal to a prescribed value B and solving for n.

There are two points worth mentioning. First, the confidence interval for the differ- ence between means uses a t-multiple. As was done for the mean, this can be replaced by a z-multiple, which is perfectly acceptable in most situations. Second, the confidence inter- vals for differences between means or proportions require two sample sizes, one for each sample. The formulas below assume that each sample uses the same sample size, denoted by n. Each of these formulas is implemented in the CI Sample Sizes sheet of the template file.

The sample size formula for a proportion p is given by Equation (8.19). Here, pest is an estimate of the population proportion p. A conservative value of n can be obtained by using pest 5 0.5. It is conservative in the sense that the sample size obtained by using pest 5 0.5 guarantees a confidence interval half-length no greater than B, regardless of the true value of p.

Figure 8.16 Sample Size for Mean

Confidence level

Half-length of interval

Std dev (estimate)

z multiple

Sample size

95%

0.30

1.597

1.960

109

=NORM.S.INV(B2+(1–B2)/2)

=CEILING((B5*B4/B3)^2,1)

Confidence interval for a mean A B C D E F

Sample Size Formula for Estimating a Proportion

n 5 a z-multiple

B b

pest(1 2 pest) (8.19)

The sample size formula for the difference between means is given by Equation (8.20). Here, sest is an estimate of the standard deviation of each population, where we assume (as in Section 8-7a) that the two populations have a common standard deviation s.

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8-9 Sample Size Selection 3 5 5

Finally, the sample size formula for the difference between proportions is given by Equation (8.21). Here, p1est and p2est are estimates of the two unknown population propor- tions p1 and p2. As in the case of a single proportion, a conservative value of n is obtained by using the estimates p1est 5 p2est 5 0.5.

Sample Size Formula for Estimating the Difference Between Means

n 5 2a z-multiple 3 sest

B b

(8.20)

Sample Size Formula for Estimating the Difference Between Proportions

n 5 a z-multiple

B b

3p1est(1 2 p1est) 1 p2est(1 2 p2est)4 (8.21)

EXAMPLE

8.10 SAMPLE SIZE SELECTION FOR ESTIMATING THE PROPORTION WHO HAVE TRIED A NEW SANDWICH

Suppose that the fast-food manager from the previous example wants to estimate the proportion of customers who have tried its new sandwich. She wants a 90% confidence interval for this proportion to have half-length 0.05. For example, if the sample proportion turns out to be 0.42, a 90% confidence interval should be (approximately) 0.42 { 0.05. How many customers need to be surveyed?

Objective To find the sample size of customers required to achieve a sufficiently narrow confidence interval for the proportion of custom- ers who have tried the new sandwich.

Solution If the manager has no idea what the proportion is, she can use pest 5 0.5 in Equation (8.19) to obtain a conservative value of n. The appropriate z-multiple is now 1.645 because this value cuts off probability 0.05 in each tail of the standard normal dis- tribution. (Remember that we are asking for a 90% confidence level, not the usual 95% level.) Therefore, the required value of n is

n 5 a 1.645

0.05 b

(0.5)(1 2 0.5) . 271

On the other hand, if the manager is fairly sure that the proportion who have tried the new sandwich is around 0.3, she can use pest 5 0.3 instead. The resulting output is shown in Figure 8.17.

These calculations indicate that if you have more specific information about the unknown proportion, you can use a smaller sample size—in this case 228 rather than 271. Also, note

Again, remember that lower confidence levels result in narrower confidence intervals.

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that we selected a 90% confidence level rather than the usual 95% level. There is a trade-off here. Using 90% rather than 95% obviously provides less confidence in the result, but it requires a smaller sample size. You can check that the required sample sizes for a 95% confidence level increase from the current values, 271 and 228, to 385 and 323, respectively.

Confidence level

Half-length of interval

Proportion (estimate)

z multiple

Sample size

90%

0.05

0.30

1.645

228

=NORM.S.INV(B9+(1–B9)/2)

=CEILING((B12/B10)^2*B11*(1–B11),1)

Confidence interval for a proportion

A B C D E F G

Figure 8.17 Sample Size for Proportion

EXAMPLE

8.11 SAMPLE SIZE SELECTION FOR ANALYZING CUSTOMER COMPLAINTS

A computer company has a customer service center that responds to customers’ questions and complaints. The center employs two types of people: those who have had a recent course in dealing with customers (but little actual experience) and those with a lot of experience dealing with customers (but no formal course). The company wants to estimate the difference between these two types of employees in terms of the average number of customer complaints regarding poor service in the last six months. The company plans to obtain information on a randomly selected sample of each type of employee, using equal sample sizes. How many employees should be in each sample to achieve a 95% confidence interval with approximate half-length 2?

Objective To see how many employees in each experimental group must be sampled to achieve a sufficiently narrow confidence interval for the difference between the mean numbers of complaints.

Solution Equation (8.20) should be used with z-multiple 1.96 and B 5 2. However, this formula also requires a value for sest, an esti- mate of the (assumed) common standard deviation for each group of employees, and there is no obvious estimate available. The manager might use the following argument. Based on a brief look at complaint data, he believes that some employees receive as few as 6 complaints over a six-month period, whereas others receive as many as 36 (about six per month). Now he can estimate sest by arguing that all observations are likely to be within three standard deviations of the mean, so that the range of data—minimum to maximum—is about six standard deviations. Therefore, he sets

6sest 5 36 2 6 5 30

and obtains sest 5 5. Using this value in Equation (8.20), the required sample size is

n 5 2a 1.96(5)

2 b

. 49

The output in Figure 18.18 confirms this value. Some analysts prefer the estimate

4sest 5 36 2 6 5 30

that is, sest 5 7.5, arguing that the quoted range (6 to 36) might not include “extreme” values and hence might extend to only two standard deviations on either side of the mean. By using this estimate of the standard deviation, you can check that the required sample size increases from 49 to 109. The important point here is that the estimate of the standard deviation can have a dramatic effect on the required sample size. (And don’t forget that this size sample must be taken from each group of employees.)

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8-9 Sample Size Selection 3 5 7

The following example illustrates what can happen when you ask for extremely accu- rate confidence intervals.

EXAMPLE

8.12 SAMPLE SIZE SELECTION FOR ANALYZING PROPORTIONS OF OUT-OF-SPEC PRODUCTS

A manufacturing company has two plants that produce identical products. The production supervisor wants to know how much the proportion of out-of-spec products differs across the two plants. He suspects that the proportion of out-of-spec products in each plant is in the range of 3% to 5%, and he wants a 99% confidence interval to have approximate half-length 0.005 (or 0.5%). How many items should he sample from each plant?

Objective To see how many products in each plant must be sampled to achieve a sufficiently narrow confidence interval for the difference between the proportions of out-of-spec products.

Solution Equation (8.21) should be used with z-multiple 2.576 (the value that cuts off probability 0.005 in each tail of the standard normal distribution), B 5 0.005, and p1est 5 p2est 5 0.05. The reasoning for the latter is that the supervisor believes each pro- portion is around 3% to 5%, and the most conservative (largest) sample size corresponds to using the larger 5% value. Then the required sample size is

n 5 a 2.576

0.005 b

30.05(0.95) 1 0.05(0.95)4 . 25,213

This sample size (from each sample) is almost certainly prohibitive, so the supervisor realizes he must lower his goals. One way is to decrease the confidence level, say, from 99% to 95%. Another way is to increase the desired half-length from 0.005 to, say, 0.025. We implemented both of these changes in Figure 8.19. Even now each required sample size is 584. Obvi- ously, narrow confidence intervals for differences between proportions can require large sample sizes.

Confidence level

Half-length of interval

Common std dev (estimate)

z multiple

Sample size (each sample)

95%

2.00

5.00

1.960

=NORM.S.INV(B16+(1–B16)/2)

=CEILING(2*(B19*B18/B17)^2,1)

Confidence interval for the difference between two means A B C D E FFigure 8.18 Sample

Size for Difference Between Means

Figure 8.19 Sample Size for Difference Between Proportions

Confidence level

Half-length of interval

Proportion 1 (estimate)

Proportion 2 (estimate)

z multiple

Sample size (each sample)

95%

0.025

0.050

1.960

584

=NORM.S.INV(B23+(1–B23)/2)

=CEILING((B27/B24)^2*(B25*(1–B25)+B26*(1–B26)),1)

Confidence interval for the difference between two proportions A B C D E F G H

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Problems Solutions for problems whose numbers appear within a colored box can be found in the Student Solution Files.

Level A 29. Elected officials in a California city are preparing the

annual budget for their community. They would like to estimate how much their constituents living in this city are typically paying each year in real estate taxes. Given that there are over 100,000 homeowners in this city, the officials have decided to sample a representative subset of taxpayers and study their tax payments. a. What sample size is required to generate a 95% con-

fidence interval for the mean annual real estate tax payment with a half-length of $100? Assume that the best estimate of the population standard deviation s is $535.

b. If a random sample of the size from part a is selected and a 95% confidence interval for the mean is calculated from this sample, will the half-length of the confidence interval be equal to $100? Explain why or why not.

c. Now suppose that the officials want to construct a 95% confidence interval with a half-length of $75. What sample size is required to achieve this objective? Again, assume that the best estimate of the population standard deviation s is $535. Explain the difference between this result and the result from part a.

30. You have been assigned to determine whether more peo- ple prefer Coke or Pepsi. Assume that roughly half the population prefers Coke and half prefers Pepsi. How large a sample do you need to take to ensure that you can estimate, with 95% confidence, the proportion of people preferring Coke within 2% of the actual value?

31. You are trying to estimate the average amount a fam- ily spends on food during a year. In the past the stan- dard deviation of the amount a family has spent on food during a year has been approximately $1000. If you want to be 99% sure that you have estimated average family food expenditures within $50, how many fami- lies do you need to survey?

32. In past years, approximately 20% of all U.S. families purchased potato chips at least once a month. You are

interested in determining the fraction of all U.S. fami- lies that currently purchase potato chips at least once a month. How many families must you survey if you want to be 99% sure that your estimate of the relevant propor- tion is accurate within 2%?

33. Continuing Problem 29, suppose that the officials in this city want to estimate the proportion of taxpayers whose annual real estate tax payments exceed $2000. a. What sample size is required to generate a 99% con-

fidence interval for this proportion with a half-length of 0.10? Assume for now that the relevant population proportion p is close to 0.50.

b. Assume now that officials discover another source that suggests that approximately 30% of all property owners in this community pay more than $2000 annually in real estate taxes. What sample size is now required to gener- ate the 99% confidence interval requested in part a?

c. Why is there a difference between your answers to parts a and b?

d. If a random sample of the size from part a is selected and a 99% confidence for the proportion is calculated from this sample, will the half-length of the confidence interval be equal to 0.10? Explain why or why not.

Level B 34. Continuing the previous problem, suppose that the offi-

cials in this city want to estimate the difference between the proportions, labeled p1 and p2, of taxpayers living in neighborhood 1 whose annual real estate tax payments exceed $2000 and the similar proportion for taxpayers living in neighborhood 2. a. What sample size (randomly selected from all taxpayers

residing in each of neighborhoods 1 and 2) is required to generate a 90% confidence interval for this difference between proportions with a half-length of 0.10? Assume for now that p1 and p2 are both close to 0.5.

b. We assumed that the two population proportions in part a are both close to 0.5. Use a two-way data table to find the required (common) sample size when each of the population proportions is allowed to vary from 0.1 to 0.9 in increments of 0.1. Comment on the sensi- tivity of the required sample size to the magnitudes of the population proportions.

8-10 Conclusion When you want to estimate a population parameter from sample data, one of the most useful ways to do so is to report a point estimate and a corresponding confidence interval. This confidence interval provides a quick sense of where the true param- eter lies. It essentially quantifies the amount of uncertainty in the point estimate. Obviously, narrow confidence intervals are desired. You have seen that the length of a confidence interval is determined by the variability in the data, the confidence level, usually set at 95%, and the sample size(s). Finally, you have also seen how sample size formulas can be used at the planning stage to achieve confidence intervals that are sufficiently narrow.

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8-10 Conclusion 3 5 9

Summary of Key Terms TERM EXPLANATION EXCEL PAGES EQUATION

Confidence interval An interval that, with a stated level of confidence, captures a population parameter

313 8.1

t distribution The sampling distribution of the stan- dardized sample mean when the sample standard deviation is used in place of the population standard deviation

T.DIST, T.INV, and other Excel T functions

314 8.2

Confidence level Percentage (usually 90%, 95%, or 99%) that indicates how confident you are that the confidence interval captures the true population parameter

318

Confidence interval for a mean

Interval that is likely to capture a popula- tion mean

Excel formulas in Confidence Interval Template.xlsx file

318 8.4

Confidence interval for a total

Interval that is likely to capture the total of all observations in a population

Excel formulas in Confidence Interval Template.xlsx file

324

Confidence interval for a proportion

Interval that is likely to capture the pro- portion of all population members that satisfy a specified property

Excel formulas in Confidence Interval Template.xlsx file

327 8.10

Confidence interval for a standard deviation

Interval that is likely to capture a popula- tion standard deviation

Excel formulas in Confidence Interval Template.xlsx file

331

Chi-square distribution Skewed distribution useful for estimating standard deviations

CHI.DIST, CHI.INV, and other Excel CHI functions

332

Confidence interval for difference between means with independent samples

Interval that is likely to capture the difference between two popula- tion means when the samples are independent

Excel formulas in Confidence Interval Template.xlsx file

335–336 8.13–8.15

Confidence interval for difference between means with paired samples

Interval that is likely to capture the difference between two population means when the samples are paired in a natural way

Excel formulas in Confidence Interval Template.xlsx file

339

Confidence interval for difference between proportions

Interval that is likely to capture the differ- ence between similarly defined propor- tions from two populations

Excel formulas in Confidence Interval Template.xlsx file

342 8.16, 8.17

Sample size formulas Formulas that specify the sample size(s) required to obtain sufficiently narrow confidence intervals

Excel formulas in Confidence Interval Template.xlsx file

345–348 8.18–8.21

Problems Solutions for problems whose numbers appear within a colored box can be found in the Student Solution Files.

Conceptual Questions C.1. Under what conditions, if any, is it not correct to

assume that the sampling distribution of the sample mean is approximately normally distributed?

C.2. When, if ever, is it appropriate to use the standard nor- mal distribution as a substitute for the t distribution with n 2 1 degrees of freedom in estimating a popula- tion mean?

C.3. “Assuming that all else remains constant, the length of a confidence interval for a population mean increases

whenever the confidence level and sample size increase simultaneously.” Is this statement true or false? Explain your choice.

C.4. Assuming all else remains constant, what happens to the length of a 95% confidence interval for a popu- lation parameter when the sample size is reduced by half? You can assume that the resulting sample size is still quite large. Justify your answer.

C.5. “The probability is 0.99 that a 99% confidence inter- val contains the true value of the relevant population parameter.” Is this statement true or false? Explain your choice.

C.6. Suppose you have a list of salaries of all professional athletes in a given sport in a given year. For exam- ple, you might have the salaries of all Major League

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Baseball players in 2018. Does it make sense to find a 95% confidence interval for the mean salary? If so, what is the relevant population?

C.7. Suppose that someone proposes a new way to calcu- late a 95% confidence interval for a mean. This could involve any arithmetic on the given data. For example, it could say to go out 1.75 interquartile ranges (IQRs) on either side of the median. What would it mean to say that this procedure produces valid 95% confidence intervals? How could you use simulation to check whether the procedure produces valid 95% confidence intervals?

C.8. The sample size formula for a confidence interval for the population mean requires an estimate of the pop- ulation standard deviation. Intuitively, why is this the case? Specifically, why is the required sample size larger if the population standard deviation is larger?

C.9. Suppose a 95% confidence interval for a population mean has been calculated, and it extends from 123.7 to 155.2. Some people would then state, “The probability that the population mean is between 123.7 and 155.2 is 0.95.” Why is this, strictly speaking, an invalid state- ment? How would you rephrase it to make it a valid statement?

C.10. Researchers often create multiple 95% confidence intervals based on a given data set. For example, if the variable of interest is home price and there are five neighborhoods in the population, they might create 10 confidence intervals, one for each difference between mean home prices for a given pair of neighborhoods. (There are 10 pairs.) Can they then conclude that there is 95% confidence that all 10 of their confidence inter- vals will include the corresponding population mean differences? Why or why not?

C.11. Based on a given random sample, suppose you calcu- late a 95% confidence interval for the following differ- ence: the mean test score for students under 25 years old minus the mean test score for students at least 25 years old, and the confidence interval extends from 914.3 to 1.2. How would you interpret these results? Would you claim that older students, on average, score higher on this test? Would you claim that, on average, it is possible that the younger students score higher on this test?

Level A 35. A sample of 15 quality control managers with more than

20 years experience have an average salary of $68,000 and a sample standard deviation of $19,000. a. You can be 95% confident that the mean salary for all

quality managers with at least 20 years of experience is between what two numbers? What assumption are you making about the distribution of salaries?

b. What size sample is needed to ensure that you can estimate the population mean salary of all quality

managers with more than 20 years of experience and have only one chance in 100 of being off by more than $500?

36. Political polls typically sample randomly from the U.S. population to investigate the percentage of voters who favor some candidate or issue. The number of people polled is usually on the order of 1000. Suppose that one such poll asks voters how they feel about the President’s handling of environmental issues. The results show that 575 out of the 1280 people polled say they either approve or strongly approve of the President’s handling. Find a 95% confidence interval for the proportion of the entire voter population who approve or strongly approve of the President’s handling. If the same sample propor- tion were found in a sample twice as large—that is 1150 out of 2560—how would this affect the confidence inter- val? How would the confidence interval change if the confidence level were 90% instead of 95%?

37. Referring to the previous problem, you often hear the results of such a poll in the news. In fact, the newscast- ers usually report something such as, “44.9% of the pop- ulation approve or strongly approve of the President’s handling of the environment. The margin of error in this result is plus or minus 3%.” Where does this 3% comes from? If the pollsters want the margin of error to be plus or minus 3%, how does this lead to a sample size of approximately 1000?

38. The widths of 100 elevator rails have been measured. The sample mean and standard deviation of the elevator rails are 2.05 inches and 0.01 inch. a. Calculate a 95% confidence interval for the average

width of an elevator rail. Do you need to assume that the widths of elevator rails are normally distributed?

b. How large a sample of elevator rails would you have to measure to ensure that you could estimate, with 95% confidence, the average diameter of an elevator rail within 0.01 inch?

39. You want to determine the percentage of Fortune 500 CEOs who think Indiana University (IU) deserves its current Business Week rating. You mail a questionnaire to all 500 CEOs and 100 respond. Exactly half of the respondents believe IU does deserve its ranking. a. Calculate a 95% confidence interval for the fraction

of Fortune 500 CEOs who believe IU deserves its ranking.

b. Suppose again that you want to estimate the frac- tion of Fortune 500 CEOs who believe IU deserves its ranking. Your goal is to have only a 5% chance of having your estimate be in error by more than 0.02. What size sample would you need to take? Is it possi- ble to implement this result?

c. Is the finite population correct ( fpc) from the previous chapter relevant here? Why or why not?

40. The SEC requires companies to file annual reports concerning their financial status. It is impossible to audit every account receivable. Suppose an auditor audits a

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random sample of 49 accounts receivable invoices and finds a sample average of $128 and a sample standard deviation of $53. a. Calculate a 99% confidence interval for the mean size

of an accounts receivable invoice. Does your answer require the sizes of the accounts receivable invoices to be normally distributed?

b. How large a sample is required for you to be 99% sure that the estimate of the mean invoice size is accurate within $5?

41. An opinion poll surveyed 900 people and reported that 36% believe a certain governor broke campaign financ- ing laws in his election campaign. a. Calculate a 95% confidence interval for the popula-

tion proportion of people who believe the governor broke campaign financing laws. Does the result of the poll convince you that fewer than 38% of all U.S. citi- zens favor that viewpoint?

b. Suppose 10,000 (not 900) people are surveyed and 36% believe that the governor broke campaign financ- ing laws. Would you now be convinced that fewer than 38% of all U.S. citizens favor that viewpoint? Why is your answer different than in part a?

c. How many people would you have to survey to be 99% confident that you can estimate to within 1% the fraction of people who believe the governor broke campaign financing laws?

42. The file P08_07.xlsx contains a random sample of 200 service times during the busiest hour of the day at a par- ticular fast-food restaurant. Calculate a 95% confidence interval for each of the following population parame- ters. Then explain how each result might be useful to the manager of the restaurant in terms of improving service. a. The mean service time b. The standard deviation of service times c. The proportion of service times longer than 90

seconds d. The proportion of service times shorter than 60

seconds 43. We know that IQs are normally distributed with a mean

of 100 and standard deviation of 15. Suppose you want to verify this, so you take 100 random samples of size four each and, for each sample, calculate a 95% confi- dence interval for the mean IQ. You expect that approx- imately 95 of these intervals will contain the true mean IQ (100) and approximately five of these intervals will not contain the true mean. Use simulation in Excel to check whether this is the case.

44. In Section 8-9, we gave a sample size formula for con- fidence interval estimation of a mean. If the confidence level is 95%, then because the z-multiple is about 2, this formula is essentially

n 5 4s2

However, this formula is based on the assumption that the sample size n will be small relative to the population size N . If this is not the case, the appropriate formula turns out to be

n 5 Ns2

s2 1 (N 2 1)B2/4

(This is based on the same idea as the finite popula- tion correction from the previous chapter.) Now sup- pose you want to find a 95% confidence interval for a population mean. Based on preliminary (or historical) data, you believe that the population standard deviation is approximately 15. You want the confidence inter- val to have length 4. That is, you want the confidence interval to be of the form X { 2. What sample size is required if N 5 400? if N 5 800? if N 5 10,000? if N 5 100,000,000? How would you summarize these findings in words?

45. Ritter Manufacturing Company has kept track of machine hours and overhead costs at its main manufacturing plant for the past 52 weeks. The data appear in the file P08_45.xlsx. Ritter has studied these data to understand the relationship between machine hours and overhead costs. Although the relationship is far from perfect, Ritter believes a fairly accurate prediction of overhead costs can be obtained from machine hours through the equation

Estimated Overhead 5 746.5078 1 3.3175* Machine Hours

By substituting any observed value of Machine Hours into this equation, Ritter obtains an estimated value of Overhead, which is always somewhat different from the true value of Overhead. The difference is called the pre- diction error. a. Calculate a 95% confidence interval for the mean

prediction error. Do the same for the absolute pre- diction error. (For example, the prediction error in week 1, actual overhead minus predicted overhead, is 994.5303. The absolute prediction error is the abso- lute value, 94.5303.)

b. A close examination of the data suggests that week 45 is a possible outlier. Illustrate this by creating a box plot of the prediction errors. In what sense is week 45 an outlier? See whether week 45 has much effect on the confidence intervals from part a by recalculating these confidence intervals, this time with week 45 deleted. Discuss your findings briefly.

Problems 46 through 55 are related to the data in the file P08_46.xlsx. This file contains data on 400 customers’ orders from ElecMart, a company that sells electronic appliances by mail order. (This same data set was used in Example 3.4 of Chapter 3.) You can consider the data as a random sample from all of ElecMart’s orders.

46. Calculate a 95% confidence interval for the mean total cost of all customer orders. Then do this separately for each of the four regions. Create side-by-side box plots of

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total cost for the four regions. Does the positive skew- ness in these box plots invalidate the confidence interval procedure used?

47. Calculate a 95% confidence interval for the proportion of all customers whose order is for more than $100. Then do this separately for each of the three times of day.

48. Calculate a 95% confidence interval for the proportion of all customers whose orders contain at least three items and cost at least $100 total.

49. Calculate a 95% confidence interval for the difference between the mean amount of the highest cost item pur- chased for the High customer category and the similar mean for the Medium customer category. Do the same for the difference between the Medium and Low cus- tomer categories. Because of the way these customer categories are defined, you would probably expect these mean differences to be positive. Is this what the data indicate?

50. Of the subpopulation of customers who order in the eve- ning, consider the proportion who are female. Similarly, of the subpopulation of customers who order in the morn- ing, consider the proportion who are female. Calculate a 95% confidence interval for the difference between these two proportions.

51. Calculate a 95% confidence interval for the difference between the following two proportions: the proportion of female customers who order during the evening and the proportion of male customers who order during the evening.

52. Calculate a 95% confidence interval for the difference between the following means: the mean total order cost for West customers and the mean total order cost for Northeast customers. Do the same for the other com- binations: West versus Midwest, West versus South, Northeast versus South, Northeast versus Midwest, and South versus Midwest.

53. Calculate a 95% confidence interval for the difference between the mean cost per item for female orders and the similar mean for males.

Level B 54. Let pE,F be the proportion of female orders that are paid

for with the ElecMart credit card, and let pE,M be the similar proportion for male orders. a. Calculate a 95% confidence interval for pE,F; for pE,M;

and for the difference pE,F 2 pE,M. b. Let pE,F,Wd be the proportion of female orders on

weekdays that are paid for with the ElecMart credit card, and let pE,F,We be the similar proportion for weekends. Define pE,M,Wd and pE,M,We similarly for males. Calculate a 95% confidence interval for the difference (pE,F,Wd 2 pE,M,Wd) 2 (pE,F,We 2 pE,M,We). Interpret this difference in words. Why might it be of interest to ElecMart?

55. Suppose these 400 orders are a sample of the 4295 orders made during this time period, and suppose 2531 of these orders were placed by females. Calculate a 95% confidence interval for the total paid for all 4295 orders. Do the same for all 2531 orders placed by females. Do the same for all 1764 orders placed by males.

Problems 56 through 61 are related to the data in the file P08_56.xlsx. This file contains data on 91 billings from Rebco, a company that sells plumbing supplies to retailers. You can consider the data as a random sample from all of Rebco’s billings.

56. Calculate a 95% confidence interval for the mean amount of all Rebco’s bills. Do the same for each cus- tomer size separately.

57. Calculate a 95% confidence interval for the mean num- ber of days it takes Rebco’s customers (as a combined group) to pay their bills. Do the same for each customer size separately. Create a box plot for the variable Days, based on all 91 billings. Also, create side-by-side box plots for Days for the three separate customer sizes. Do any of these suggest problems with the validity of the confidence intervals?

58. Calculate a 95% confidence interval for the proportion of all large customers who pay bills of at least $1000 at least 15 days after they are billed.

59. Calculate a 95% confidence interval for the proportion of all bills paid within 15 days. Calculate a 95% confi- dence interval for the difference between the proportion of large customers who pay within 15 days and the sim- ilar proportion of medium-size customers. Calculate a 95% confidence interval for the difference between the proportion of medium-size customers who pay within 15 days and the similar proportion of small customers.

60. Suppose a bill is considered late if it is paid after 20 days. In this case its “lateness” is the number of days over 20 . For example, a bill paid 23 days after billing has a late- ness of 3, whereas a bill paid 18 days after billing has a lateness of 0. Calculate a 95% confidence interval for the mean amount of lateness for all customers. Calculate similar confidence intervals for each customer size sep- arately. Why is the distribution of lateness certainly not normal? Do you think this matters for the validity of the confidence interval?

61. Suppose Rebco can earn interest at the rate of 0.011% daily on excess cash. The company realizes that it could earn extra interest if its customers paid their bills more promptly. a. Calculate a 95% confidence interval for the mean

amount of interest it could gain if each of its custom- ers paid exactly one day more promptly. Calculate similar confidence intervals for each customer class separately.

b. Suppose these 91 billings represent a random sam- ple of the 2792 billings Rebco generates during the year. Calculate a 95% confidence interval for the

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total amount of extra interest it could gain by getting each of these 2792 billings to be paid two days more promptly.

62. The file P08_62.xlsx contains data on the first 100 cus- tomers who entered a two-teller bank on Friday. All vari- ables in this file are times, measured in minutes. a. Calculate a 95% confidence interval for the mean

amount of time a customer spends in service with a teller.

b. The bank is most interested in mean waiting times because customers get upset when they have to spend a lot of time waiting in line. Use the usual procedure to calculate a 95% confidence interval for the mean waiting time per customer.

c. Your answer in part b is not valid! (It is much too narrow. It makes you believe you have a much more accurate estimate of the mean waiting time than you really have.) We made two implicit assumptions when we stated the confidence interval procedure for a mean: (1) The individual observations come from the same distribution, and (2) the individual observa- tions are probabilistically independent. Why are both of these, particularly (2), violated for the customer waiting times? [Hint: For (1), how do the first few customers differ from “typical” customers? For (2), if you are behind someone in line who has to wait a long time, what do you suspect about your own wait- ing time?]

d. Following up on assumption (2) of part c, you might expect waiting times of successive customers to be autocorrelated, that is, correlated with each other. Large waiting times tend to be followed by large waiting times, and small by small. Check this with another template file we have provided for Chapter 12, Autocorrelation Template.xlsx. An autocorrelation of a certain lag, say, lag 2, is the correlation in waiting times between a cus- tomer and the second customer behind her. Do these successive waiting times appear to be autocorrelated? (A valid confidence interval for the mean waiting time takes autocorrelations into account—but it is consider- ably more difficult to calculate.)

Problems 63 through 65 are related to the data in the file P08_63.xlsx. SoftBus Company sells PC equipment and customized software to small companies to help them manage their day-to-day business activities. Although SoftBus spends time with all customers to understand their needs, the customers are eventually on their own to use the equipment and software intelligently. To understand its customers better, SoftBus recently sent questionnaires to a large number of prospective cus- tomers. Key personnel—those who would be using the software—were asked to fill out the questionnaire. Soft- Bus received 82 usable responses, as shown in the file. You can assume that these employees represent a ran- dom sample of all of SoftBus’s prospective customers.

63. Create a histogram of the PC Knowledge variable. [Because there are only five possible responses (195), this histogram should have only five bars.] Repeat this separately for those who own a PC and those who do not. Then calculate a 95% confidence interval for the mean value of PC Knowledge for all of SoftBus’s pro- spective customers; and of all its prospective customers who own PCs; and of all its prospective customers who do not own PCs. The PC Knowledge variable obviously can’t be exactly normally distributed because it has only five possible values. Do you think this invalidates the confidence intervals? Explain your choice.

64. SoftBus believes it can afford to spend much less time with customers who own PCs and score at least 4 on PC Knowledge. Let’s call these the “PC-savvy” customers. On the other hand, SoftBus believes it will have to spend a lot of time with customers who do not own a PC and score 2 or less on PC Knowledge. Let’s call these the “PC-illiterate” customers. a. Calculate a 95% confidence interval for the propor-

tion of all prospective customers who are PC-savvy. Calculate a similar confidence interval for the propor- tion who are PC-illiterate.

b. Repeat part a twice, once for the subpopulation of customers who have at least 12 years of experience and once for the subpopulation who have less than 12 years of experience.

c. Again repeat part a twice, once for the subpopulation of customers who have no more than a high school diploma and once for the subpopulation who have more than a high school diploma.

d. Calculate a 95% confidence interval for the differ- ence between two proportions: the proportion of all customers with some college education who are PC-savvy and the similar proportion of all customers with no college education. Repeat this, substituting “PC-savvy” with “PC-illiterate.”

e. Discuss any insights you gain from parts a through d that might be of interest to SoftBus.

65. Following up on the previous problem, SoftBus believes its profit from each prospective customer depends on the cus- tomer’s level of PC knowledge. It divides the customers into three classes: PC-savvy, PC-illiterate, and all others (where the first two classes are as defined in the previous problem). As a rough guide, SoftBus figures it can gain profit P1 from each PC-savvy customer, profit P3 from each PC-illiterate customer, and profit P2 from each of the others. a. What values of P1, P2, and P3 seem reasonable? For

example, would you expect P1 6 P2 6 P3 or the opposite?

b. Using any reasonable values for P1, P2, and P3, calcu- late a 95% confidence interval for the mean profit per customer that SoftBus can expect to obtain.

Problems 66 through 69 are related to the data in the file P08_66.xlsx. Comfy Company sells medium-priced

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patio furniture through a mail-order catalog. It has operated primarily in the East but is now expanding to the Southwest. To get off to a good start, it plans to send potential customers a catalog with a discount cou- pon. However, Comfy is not sure how large a discount is needed to entice customers to buy. It experiments by sending catalogs to selected residents in six cities. Tuc- son and San Diego receive coupons for 5% off any fur- niture within the next two months, Phoenix and Santa Fe receive coupons for 10% off, and Riverside and Albuquerque receive coupons for 15% off.

66. Calculate a 95% confidence interval for the proportion of customers who will purchase at least one item if they receive a coupon for 5% off. Repeat for 10% off and for 15% off.

67. Calculate a 95% confidence interval for the proportion of customers who will purchase at least one item and pay at least $500 total if they receive a coupon for 5% off. Repeat for 10% off and for 15% off.

68. Comfy wonders whether the customers who receive larger discounts are buying more expensive items. Recalling that the value in the Total Paid column is after the discount, calculate a 95% confidence interval for the difference between the mean original price per item for customers who purchase something with the 5% coupon and the similar mean for customers who purchase something with the 10% coupon. Repeat with 5% and 10% replaced by 10% and 15%. What can you conclude?

69. Comfy wonders whether there are differences across pairs of cities that receive the same discount. a. Calculate a 95% confidence interval for the difference

between the mean amount spent in Tucson and the simi- lar mean in San Diego. (These means should include the “0 purchases.”) Repeat this for the difference between Phoenix and Santa Fe and then between Riverside and Albuquerque. Does city appear to make a difference?

b. Repeat part a, but instead of analyzing differences between means, analyze differences between propor- tions of customers who purchase something. Does city appear to make a difference?

Problems 70 through 73 are related to the data in the file P08_70.xlsx. Niyaki Company sells Blu-ray disc players through a number of retail stores. On one popular model, there is a standard warranty that cov- ers parts for the first six months and labor for the first year. Customers are always asked whether they wish to purchase an extended service plan for $50 that extends the original warranty two more years—that is, to 30 months on parts and 36 months on service. To get a bet- ter understanding of warranty costs, the company has gathered data on 70 Blu-ray units purchased. This data is listed in the Data1 sheet of the file P08_70.xlsx. The

two costs in this sheet (columns D and E) are tracked only for repairs covered by warranty. [Otherwise, the customer bears the cost(s).]

70. Create a histogram of the time until first failure for this type of disc player. Then calculate a 95% confidence interval for the mean time until failure for this type of disc player. Does the shape of the histogram invalidate the confidence interval? Why or why not?

71. Calculate a 95% confidence interval for the proportion of customers who purchase the extended service plan. Calculate a 95% confidence interval for the proportion of all customers who would benefit by purchasing the extended service plan.

72. Calculate a 95% confidence interval for Niyaki’s mean net warranty cost per unit sold (net of the $50 paid for the plan for those who purchase it). You can assume that this mean is for the first failure only; subsequent failures of the same units are ignored here.

73. This problem follows up on the previous two prob- lems with the data in the Data2 sheet of the file. Here Niyaki did more investigation on the same 70 custom- ers. It tracked subsequent failures and costs (if any) that occurred within the warranty period. (Note: Only two customers had three failures within the warranty period, and parts weren’t covered for either on the third failure. Also, no one had more than three failures within the warranty period.) a. With these data, calculate the confidence intervals

requested in the previous two problems. b. Suppose that Niyaki sold this Blu-ray model to 12,450

customers during the year. Calculate a 95% confi- dence interval for its total net cost due to warranties from all of these sales.

74. The file P08_74.xlsx contains data on 856 customers who have either tried or not tried a company’s new fro- zen lasagna dinner. (This data set was used in Example 3.5 in Chapter 3.) The manager of the company would like to compare the proportion of customers who have tried the lasagna across various subpopulations. For each of the following, calculate a 95% confidence inter- val for the difference between the proportions who have tried the lasagna for the two specified subpopulations. Explain briefly how the results help the manager to understand his customers. (Hint: One approach is to use pivot tables to get the count data you need.) a. Those with weight under 190 versus those with weight

at least 190 b. Females versus males c. Those who live alone versus those who do not live

alone d. Those who live in a home or condo versus those who

live in an apartment e. Those who live in the South or West versus those who

live in the East

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f. Those who average five or more trips to the mall per month versus those who average fewer than five trips to the mall per month.

75. The formula for a 95% confidence interval for a mean (sample mean plus or minus approximately two stan- dard errors) is so well-rooted in statistical theory and practice that you might not even consider other pos- sibilities. However, many researchers and even practi- tioners favor a totally different method of calculating a 95% confidence interval for the mean. It is called the bootstrap method. Starting with a sample of size n, they generate many “bootstrap samples,” calculate the sample mean of each, and report the 2.5 and 97.5

percentiles of these sample means as the endpoints of the confidence interval. Each bootstrap sample is a random sample of size n, with replacement, from the given data. That is, each member of a bootstrap sample is equally likely to be any of the original n data points. Implement this in Excel, starting with the sample of 50 salaries in the file P08_75.xlsx. Create at least 100 bootstrap samples. Compare the resulting bootstrap confidence interval with the traditional one. (Hint: The bootstrap samples can be generated quickly with a combination of the RANDBETWEEN and VLOOKUP functions.)

CASE 8.1 Harrigan University Admissions Harrigan University is a liberal arts university in the Mid- west that attempts to attract the highest-quality students, especially from its region of the country. It has gathered data on 178 applicants who were accepted by Harrigan (a random sample from all acceptable applicants over the past several years). The data are in the file C08_01.xlsx. The variables are as follows:

• Accepted: whether the applicant accepts Harrigan’s offer to enroll

• MainRival: whether the applicant enrolls at Harrigan’s main rival university

• HSClubs: number of high school clubs applicant served as an officer

• HSSports: number of varsity letters applicant earned • HSGPA: applicant’s high school GPA • HSPctile: applicant’s percentile (in terms of GPA) in his

or her graduating class • HSSize: number of students in applicant’s graduating

class • SAT: applicant’s combined SAT score • Combined Score: a combined score for the applicant

used by Harrigan to rank applicants

The derivation of the combined score is a closely kept secret by Harrigan, but it is basically a weighted average of the various components of high school performance and SAT. Harrigan is concerned that it is not getting enough of the best students, and worse yet, that many of these best stu- dents are going to Harrigan’s main rival. Solve the follow- ing problems and then, based on your analysis, comment on whether Harrigan appears to have a legitimate concern.

1. Calculate a 95% confidence interval for the propor- tion of all acceptable applicants who accept Harrigan’s invitation to enroll. Do the same for all acceptable

applicants with a combined score less than 330, with a combined score between 330 and 375, and then with a combined score greater than 375. (Note that 330 and 375 are approximately the first and third quartiles of the Combined Score variable.)

2. Calculate a 95% confidence interval for the proportion of all acceptable students with a combined score less than the median (356) who choose Harrigan’s rival over Harrigan. Do the same for those with a combined score greater than the median.

3. Calculate 95% confidence intervals for the mean combined score, the mean high school GPA, and the mean SAT score of all acceptable students who accept Harrigan’s invitation to enroll. Do the same for all ac- ceptable students who choose to enroll elsewhere. Then calculate 95% confidence intervals for the differences between these means, where each difference is a mean for students enrolling at Harrigan minus the similar mean for students enrolling elsewhere.

4. Harrigan is interested (as are most schools) in getting students who are involved in extracurricular activities (clubs and sports). Does it appear to be doing so? Cal- culate a 95% confidence interval for the proportion of all students who decide to enroll at Harrigan who have been officers of at least two clubs. Calculate a similar confidence interval for those who have earned at least four varsity letters in sports.

5. The combined score Harrigan calculates for each student gives some advantage to students who rank highly in a large high school relative to those who rank highly in a small high school. Therefore, Harrigan wonders wheth- er it is relatively more successful in attracting students from large high schools than from small high schools. Calculate one or more confidence intervals for relevant parameters to shed some light on this issue.

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CASE 8.2 Employee Retention at D&Y Demand for systems analysts in the consulting industry is greater than ever. Graduates with a combination of business and computer knowledge—some even from liberal arts pro- grams—are getting great offers from consulting companies. Once these people are hired, they frequently switch from one company to another as competing companies lure them away with even better offers. One consulting company, D&Y, has collected data on a sample of systems analysts with under- graduate degrees they hired several years ago. The data are in the file C08_02.xlsx. The variables are as follows:

• Starting Salary: employee’s starting salary at D&Y • On Road Pct: percentage of time employee has spent on

the road with clients • State Univ: whether the employee graduated from State

University (D&Y’s principal source of recruits) • CIS Degree: whether the employee majored in Computer

Information Systems (CIS) or a similar computer- related area

• Stayed 3 Years: whether the employee stayed at least three years with D&Y

• Tenure: tenure of employee at D&Y (months) if he or she moved before three years

D&Y is trying to learn everything it can about retention of these valuable employees. You can help by solving the fol- lowing problems and then, based on your analysis, present- ing a report to D&Y.

1. Although starting salaries are in a fairly narrow band, D&Y wonders whether they have anything to do with retention.

a. Calculate a 95% confidence interval for the mean starting salary of all employees who stay at least three years with D&Y. Do the same for those who leave before three years. Then calculate a 95% confidence interval for the difference between these means.

b. Among all employees whose starting salary is below the median ($37,750), calculate a 95% confidence in- terval for the proportion who stay with D&Y for at least three years. Do the same for the employees with starting salaries above the median. Then calculate a 95% confidence interval for the difference between these proportions.

2. D&Y wonders whether the percentage of time on the road might influence who stays and who leaves. Repeat the previous problem, but now do the analysis in terms of percentage of time on the road rather than starting salary. (The median percentage of time on the road is 54%.)

3. Find a 95% confidence interval for the mean tenure (in months) of all employees who leave D&Y within three years of being hired. Why is it not possible with the giv- en data to calculate a confidence interval for the mean tenure at D&Y among all systems analysts hired by D&Y?

4. State University’s students, particularly those in its na- tionally acclaimed CIS area, have traditionally been among the best of D&Y’s recruits. But are they relatively hard to retain? Calculate one or more relevant confidence intervals to help you make an argument one way or the other.

CASE 8.3 Delivery Times at SnowPea Restaurant The SnowPea Restaurant is a Chinese carryout/delivery restaurant. Most of SnowPea’s deliveries are within a 10-mile radius, but it occasionally delivers to customers more than 10 miles away. SnowPea employs a number of delivery people, four of whom are relatively new hires. The restaurant has recently been receiving customer complaints about exces- sively long delivery times. Therefore, SnowPea has collected data on a random sample of deliveries by its four new deliv- ery people during the peak dinner time. The data are in the file C08_03.xlsx. The variables are as follows:

• Deliverer: which person made the delivery • Prep Time: time from when order was placed until deliv-

ery person started driving it to the customer • Travel Time: time to drive from SnowPea to customer • Distance: distance (miles) from SnowPea to customer

Solve the following problems and then, based on your analysis, write a report that makes reasonable recommenda- tions to SnowPea management.

1. SnowPea is concerned that one or more of the new delivery people might be slower than others. a. Let mDi and mTi be the mean delivery time and mean

total time for delivery person i, where the total time is the sum of the delivery and prep times. Calculate 95% confidence intervals for each of these means for each delivery person. Although these might be interesting, give two reasons why they are not really fair measures for comparing the efficiency of the delivery people.

b. Responding to the criticisms in part a, calculate a 95% confidence interval for the mean speed of delivery for each delivery person, where speed is measured as

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miles per hour during the trip from SnowPea to the customer. Then calculate 95% confidence intervals for the mean difference in speed between each pair of de- livery people.

2. SnowPea would like to advertise that it can achieve a to- tal delivery time of no more than M minutes for all cus- tomers within a 10-mile radius. On all orders that take more than M minutes, SnowPea will give the customers a $10 certificate on their next purchase. a. Assuming for now that the delivery people in the sam-

ple are representative of all of SnowPea’s delivery people, calculate a 95% confidence interval for the proportion of deliveries (within the 10-mile limit) that will be on time if M 5 25 minutes; if M 5 30 min- utes; if M 5 35 minutes.

b. Suppose SnowPea makes 1000 deliveries within the 10-mile limit. For each of the values of M in part a, calculate a 95% confidence interval for the total dollar amount of certificates it will have to pay for being late.

3. The policy in the previous problem is simple to state and simple to administer. However, it is somewhat unfair to customers who live close to SnowPea—they will never get $10 certificates. A fairer, but more complex, policy

is the following. SnowPea first analyzes the data and finds that total delivery times can be predicted fairly well with the equation

Predicted Delivery Time 5 14.8 1 2.06*Distance

(This is based on regression analysis, the topic of Chapters 10 and 11.) Also, most of these predictions are within 5 minutes of the actual delivery times. Therefore, whenever SnowPea receives an order over the phone, it looks up the customer’s address in its computerized ge- ographical database to find distance, calculates the pre- dicted delivery time based on this equation, rounds this to the nearest minute, adds 5 minutes, and guarantees this delivery time or else a $10 certificate. It does this for all customers, even those beyond the 10-mile limit. a. Assuming again that the delivery people in the sample

are representative of all of SnowPea’s delivery people, calculate a 95% confidence interval for the proportion of all deliveries that will be within the guaranteed total delivery time.

b. Suppose SnowPea makes 1000 deliveries. Calculate a 95% confidence interval for the total dollar amount of certificates it will have to pay for being late.

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CHAPTER 9 Hypothesis Testing

OFFICIAL SPONSORS OF THE OLYMPICS Hypothesis testing is one of the most frequently used tools in academic research, including research in the area of business. Many studies pose interesting questions, stated as hypotheses, and then test these with appropriate sta- tistical analysis of experimental data. One such study is reported in McDaniel and Kinney (1996). They inves- tigate the effectiveness of “ambush marketing” in prom- inent sports events such as the Olympic Games. Many companies pay significant amounts of money, perhaps $10 million, to become official sponsors of the Olympics. Ambushers are their competitors, who pay no such fees

but nevertheless advertise heavily during the Olympics, with the intention of linking their own brand image to the event in the minds of consumers. The question McDaniel and Kinney investigate is whether consumers are confused into thinking that the ambushers are the official sponsors.

At the time of the 1994 Winter Olympics in Lillehammer, Norway, the researchers ran a controlled experiment using 215 subjects ranging in age from 19 to 49 years old. Approximately half of the subjects—the “control group”—viewed a 20-minute tape of a women’s skiing event in which several actual commercials for official sponsors in four product categories were interspersed. (The categories were fast food, automobile, credit card, and insurance; the official sponsors were McDonald’s, Chrysler, VISA, and John Hancock.) The other half—the “treatment group”—watched the same tape but with commercials for competing ambushers. (The ambushers were Wendy’s, Ford, American Express, and Northwestern Mutual, all of which advertised during the 1994 Olympics.) After watching the tape, each subject was asked to fill out a questionnaire. This question- naire asked subjects to recall the official Olympics sponsors in each product category, to rate their attitudes toward the products, and to state their intentions to purchase the products.

McDaniel and Kinney tested several hypotheses. First, they tested the hypothesis that there would be no difference between the control and treatment groups in terms of which products they would recall as official Olympics sponsors. The experimental evidence allowed them to reject this hypothesis decisively. For example, the vast majority of the control group, who watched the McDonald’s commercial, recalled McDonald’s as being the official sponsor in the fast-food category. But a clear majority of the treatment group, who watched the Wendy’s commercial, recalled Wendy’s as being the official sponsor in this category. Evidently, Wendy’s commercial was compelling.

Because the ultimate objective of commercials is to increase purchases of a com- pany’s brand, the researchers also tested the hypothesis that viewers of official spon- sor commercials would rate their intent to purchase that brand higher than viewers of ambusher commercials would rate their intent to purchase the ambusher brand. After all, isn’t this why the official sponsors were paying large fees? However, except for the credit card category, the data did not support this hypothesis. VISA viewers did indeed rate their intent to use VISA higher than American Express viewers rated their intent to use Amer- ican Express. But in the other three product categories, the ambusher brand came out

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9-1 Introduction 3 6 9

ahead of the official brand in terms of intent to purchase (although the differences were not statistically significant).

There are at least two important messages this research should convey to business. First, if a company is going to spend a lot of money to become an official sponsor of an event such as the Olympic Games, it must create a more vivid link in the mind of con- sumers between its product and the event. Otherwise, the company might be wasting its money. Second, ambush marketing is very possibly a wise strategy. By seeing enough of the ambushers’ commercials during the event, consumers get confused into thinking that the ambusher is an official sponsor. In addition, previous research in the area suggests that consumers do not view ambushers negatively for using an ambush strategy.

9-1 Introduction When you make inferences about a population on the basis of sample data, you can per- form the analysis in either of two ways. You can proceed as in the previous chapter, where you calculate a point estimate of a population parameter and then form a confidence inter- val around this point estimate. In this way you bring no preconceived ideas to the analysis but instead let the data speak for themselves in estimating the parameter’s true value.

In contrast, an analyst often has a particular theory, or hypothesis, that he or she would like to test. This hypothesis might be that a new packaging design will produce more sales than the current design, that a new drug will have a higher cure rate for a given disease than any drug currently on the market, that people who smoke cigarettes are more susceptible to heart disease than nonsmokers, and so on. In this case the analyst typically collects sample data and checks whether the data provide enough evidence to support the hypothesis.

The hypothesis that the analyst is attempting to prove is called the alternative hypothesis. It is also frequently called the research hypothesis. The opposite of the alter- native hypothesis is called the null hypothesis. It usually represents the current thinking or status quo. That is, the null hypothesis is usually the accepted theory that the analyst is trying to disprove. In the previous examples the null hypotheses are:

• The new packaging design is no better than the current design. • The new drug has a cure rate no higher than other drugs on the market. • Smokers are no more susceptible to heart disease than nonsmokers.

The burden of proof is traditionally on the alternative hypothesis. It is up to the analyst to provide enough evidence in support of the alternative; otherwise, the null hypothesis will continue to be accepted. A slight amount of evidence in favor of the alternative is usually not enough. For example, if a slightly higher percentage of people are cured with a new drug in a sequence of clinical tests, this still might not be enough evidence to warrant introducing the new drug to the market. In general, we reject the null hypothesis—and accept the alternative—only if the results of the hypothesis test are statistically significant, a concept explained in this chapter.

Hypothesis testing is a form of decision making under uncertainty, where you decide which of two competing hypotheses to accept, based on sample data. However, it is performed in a very specific way, as described in this chapter.

The null hypothesis is usually the current thinking, or status quo. The alternative, or research, hypothesis is usually the hypothesis a researcher wants to prove. The burden of proof is on the alternative hypothesis.

Confidence interval estimation and hypothesis testing use data in much the same way and they often report basically the same results, only from different points of view. There continues to be a debate (largely among academic researchers) over which of these two procedures is more useful. We believe that in a business context, confidence interval

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estimation is more useful and enlightening than hypothesis testing. However, hypothesis testing continues to be a key aspect of statistical analysis. Indeed, statistical software pack- ages routinely include the elements of standard hypothesis tests in their outputs. You will see this, for example, when you study regression analysis in Chapters 10 and 11. There- fore, it is important to understand the fundamentals of hypothesis testing so that you can interpret this output correctly.

9-2 Concepts in Hypothesis Testing Before we plunge into the details of specific hypothesis tests, it is useful to discuss the concepts behind hypothesis testing. There are a number of concepts and statistical terms involved, all of which lead eventually to the key concept of statistical significance. The following example provides context for this discussion.

EXAMPLE

9.1 A NEW PIZZA STYLE AT PEPPERONI PIZZA RESTAURANT

The manager of Pepperoni Pizza Restaurant has recently begun experimenting with a new method of baking pepperoni pizzas. He personally believes that the new method produces a better-tasting pizza, but he would like to base the decision whether to switch from the old method to the new method on customer reactions. Therefore, he performs an experiment. For 100 ran- domly selected customers who order a pepperoni pizza for home delivery, he includes both an old-style and a free new-style pizza in the order. All he asks is that these customers rate the difference between pizzas on a 210 to 110 scale, where 210 means that they strongly favor the old style, 110 means they strongly favor the new style, and 0 means they are indifferent between the two styles. Once he gets the ratings from the customers, how should he proceed?

We begin by stating that Example 9.1 is used primarily to explain hypothesis-testing concepts. We do not mean to imply that the manager would necessarily use a hypothesis-testing procedure to decide whether to switch from the old method to the new method. First, hypothesis testing does not take costs into account. If the new method of making pizzas uses more expensive cheese, for example, then hypothesis testing would ignore this important aspect of the decision problem. Second, even if the costs of the two pizza-making methods are equivalent, the manager might base his decision on a simple point estimate and possibly a confidence interval. For example, if the sample mean rating is 1.8 and a 95% confidence interval for the mean rating extends from 0.3 to 3.3, this in itself should probably be enough evidence to make the manager switch to the new method.

We come back to these ideas—basically, that hypothesis testing is not necessarily the best procedure to use in a business decision-making context—throughout this chapter. However, with these caveats in mind, we discuss how the manager might proceed by using hypothesis testing.

9-2a Null and Alternative Hypotheses Remember that the hypothesis the manager is trying to prove is called the alternative, or research, hypothesis, whereas the null hypothesis represents the status quo. In this example the manager would like to prove that the new method provides better-tasting pizza, so this becomes the alternative hypothesis. The opposite, that the old-style pizzas are at least as good as the new-style pizzas, becomes the null hypothesis. We assume he judges which of these is true on the basis of the mean rating over the entire customer population, labeled m. If m # 0, the null hypothesis is true. Otherwise, if m 7 0, the alternative hypothesis is true.

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9-2 concepts in hypothesis testing 3 7 1

Usually, the null hypothesis is labeled H0 and the alternative hypothesis is labeled Ha. Therefore, in our example they can be specified as H0:m # 0 and Ha:m 7 0. This is typ- ical. The null and alternative hypotheses divide all possibilities into two nonoverlapping sets, exactly one of which must be true. In our case the mean rating is either less than or equal to 0, or it is positive. Exactly one of these possibilities must be true, and the manag- er’s goal is to use sample data to decide which is true.

Traditionally, hypothesis testing has been phrased as a decision-making problem, where an analyst decides either to accept the null hypothesis or reject it, based on the sample evidence. In our example, accepting the null hypothesis means deciding that the new-style pizza is not really better than the old-style pizza and presumably discontinuing the new style. In contrast, rejecting the null hypothesis means deciding that the new-style pizza is indeed better than the old-style pizza and presumably switching to the new style.

9-2b One-Tailed Versus Two-Tailed Tests The form of the alternative hypothesis can be either one-tailed or two-tailed, depending on what the analyst is trying to prove. The pizza manager’s alternative hypothesis is one- tailed because he is hoping to prove that the customers’ ratings are, on average, greater than 0. The only sample results that will lead to rejection of the null hypothesis are those in a particular direction, namely, those where the sample mean rating is positive. On the other hand, if the manager sets up his rating scale in the reverse order, so that negative ratings favor the new-style pizza, the test is still one-tailed, but now only negative sample means lead to rejection of the null hypothesis.

In contrast, a two-tailed test is one where results in either of two directions can lead to rejection of the null hypothesis. A slight modification of the pizza example where a two-tailed alternative might be appropriate is the following. Suppose the manager cur- rently uses two methods for producing pepperoni pizzas. He is thinking of discontinuing one of these methods if it appears that customers, on average, favor one method over the other. Therefore, he runs the same experiment as before, but now the hypotheses he tests are H0:m 5 0 versus Ha:m ? 0, where m is again the mean rating across the customer pop- ulation. In this case either a large positive sample mean or a large negative sample mean will lead to rejection of the null hypothesis—and presumably to discontinuing one of the production methods.

Hypotheses for Pizza Example

Null hypothesis: m # 0 Alternative hypothesis: m 7 0 where m is the mean population rating

A one-tailed alternative is one that is supported only by evidence in a single direction.

A two-tailed alternative is one that is supported by evidence in either of two directions.

Once the hypotheses are set up, it is easy to detect whether the test is one-tailed or two-tailed. One-tailed alternatives are phrased in terms of “ 7 ” or “6” whereas two-tailed alternatives are phrased in terms of “?”. The question is whether to set up hypotheses

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for a particular problem as one-tailed or two-tailed. There is no statistical answer to this question. It depends entirely on what an analyst is trying to prove. If the pizza manager is trying to prove that the new-style pizza is better than the old-style pizza—only results on one side will lead to a switch—a one-tailed alternative is appropriate. However, if he is trying to decide whether to discontinue either of two existing production methods—where results on either side will lead to a switch—then a two-tailed alternative is appropriate.

9-2c Types of Errors Regardless of whether the manager decides to accept or reject the null hypothesis, it might be the wrong decision. He might incorrectly reject the null hypothesis when it is true 1m # 02, and he might incorrectly accept the null hypothesis when it is false 1m 7 02. In the tradition of hypothesis testing, these two types of errors have acquired the names type I and type II errors. In general, you commit a type I error when you incorrectly reject a null hypothesis that is true. You commit a type II error when you incorrectly accept a null hypothesis that is false. These ideas appear graphically in Figure 9.1.

It is important to realize that the analyst, not the data, determines the type of alternative hypothesis. The hypothesis depends on what the analyst wants to prove, and it should be formed before the data are collected.

Type I error: Switching to new style when it is no better than old style

Type II error: Staying with old style when new style is better

The pizza manager commits a type I error if he concludes, based on sample evidence, that the new-style pizza is better (and switches to it) when in fact the entire customer population would, on average, favor the old-style pizza. In contrast, he commits a type II error if he concludes, again based on sample evidence, that the new style is no better (and discontinues it) when in fact the entire customer population would, on average, favor the new style.

Figure 9.1 Types of Errors in Hypothesis Testing

The traditional hypothesis-testing procedure favors caution in terms of rejecting the null hypothesis. The thinking is that if you reject the null hypothesis and it is really true, then you commit a type I error—which is bad. Given this rather conservative way of think- ing, you are inclined to accept the null hypothesis unless the sample evidence provides strong support for the alternative hypothesis. Unfortunately, you can’t have it both ways. When you accept the null hypothesis, you risk committing a type II error.

This is exactly the dilemma the pizza manager faces. If he wants to avoid a type I error (where he switches to the new style but really shouldn’t), then he will require fairly convincing evidence from the survey that he should switch. If he observes some evidence supporting the new style, such as a sample mean rating of 11.5, this evidence might not be strong enough to make him switch. However, if he decides not to switch, he risks committing a type II error.

9-2d Significance Level and Rejection Region The question, then, is how strong the evidence in favor of the alternative hypothesis must be to reject the null hypothesis. Two approaches to this problem are commonly used. In the first, you prescribe the probability of a type I error that you are willing to tolerate. This type I error probability is usually denoted by a and is most commonly set equal to 0.05, although a 5 0.01 and a 5 0.10 are also frequently used. The value of a, expressed as a percentage like 5%, is called the significance level of the test. Then, given the value of a, you use statistical theory to determine a rejection region. If the sample evidence falls in

Type I errors are usually considered more costly, although this can lead to conservative decision making.

The analyst gets to choose the significance level a. It is traditionally chosen to be 0.05, but it is occasionally chosen to be 0.01 or 0.10.

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the rejection region, you reject the null hypothesis; otherwise, you accept it. The rejection region is chosen precisely so that the probability of a type I error is at most a. Sample evidence that falls into the rejection region is called statistically significant at the a level. For example, if a 5 0.05, the evidence is statistically significant at the 5% level.

The rejection region is the set of sample data that leads to the rejection of the null hypothesis.

The significance level, a, determines the size of the rejection region. Sample results in the rejection region are called statistically significant at the a level.

It is important to understand the effect of varying a. If a is small, such as 0.01, the probability of a type I error is small. Therefore, a lot of sample evidence in favor of the alternative hypothesis is required before the null hypothesis can be rejected. Equivalently, the rejection region in this case is small. In contrast, when a is larger, such as 0.10, the rejection region is larger, and it is easier to reject the null hypothesis.

9-2e Significance from p-values A second more popular approach is to avoid the use of a significance level a and instead simply report how significant the sample evidence is. This is done by means of a p-value. The idea is quite simple—and very important. Suppose in the pizza example that the true mean rating (if it could be observed) is m 5 0. In other words, the customer population, on average, judges the two styles of pizza to be equal. Now suppose the sample mean rating is 12.5. The manager has two options at this point. (Remember that he doesn’t know that m equals 0; he observes only the sample.) He can conclude that (1) the null hypothesis is true—the new-style pizza is not preferred over the old style—and he just observed an unlucky sample, or (2) the null hypothesis is not true—customers do prefer the new-style pizza—and the sample he observed is a typical one for such customers.

The p-value of the sample quantifies this. The p-value is the probability of seeing a random sample at least as extreme as the observed sample, given that the null hypothesis is true. Here, “extreme” is relative to the null hypothesis. For example, a sample mean rat- ing of 13.5 from the pizza customers is more extreme evidence than a sample mean rating of 12.5. Each provides some evidence against the null hypothesis, but the former provides stronger, more extreme evidence.

The p-value of a sample is the probability of seeing a sample with at least as much evidence in favor of the alternative hypothesis as the sample actually observed. The smaller the p-value, the more evidence there is in favor of the alternative hypothesis.

Let’s suppose that the pizza manager collects data from the 100 sampled customers and finds that the p-value for the sample is 0.03. This means that if the entire customer population, on average, judges the two types of pizza to be approximately equal, then only three random samples out of 100 would provide as much evidence in support of the new style as the observed sample. So should he conclude that the null hypothesis is true and he just happened to observe an unlucky sample, or should he conclude that the null hypoth- esis is not true? There is no clear statistical answer to this question; it depends on how convinced the manager must be before switching. But we can say in general that smaller p-values indicate more evidence in support of the alternative hypothesis. If a p-value is sufficiently small, then almost any analyst will conclude that rejecting the null hypothesis (and accepting the alternative) is the more reasonable decision.

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How small is a “small” p-value? This is largely a matter of semantics, but Figure 9.2 indicates the attitude of many analysts. A p-value less than 0.01 is regarded as convincing evidence that the alternative hypothesis is true. After all, fewer than one sample out of 100 would provide this much support for the alternative hypothesis if it weren’t true. If the p-value is between 0.01 and 0.05, there is strong evidence in favor of the alternative hypothesis. Unless the consequences of making a type I error are really serious, this is typically enough evidence to reject the null hypothesis.

Figure 9.2 Evidence in Favor of the Alternative Hypothesis

The interval between 0.05 and 0.10 is a grayer area. If a researcher is trying to prove a research hypothesis and observes a p-value between 0.05 and 0.10, she will probably be reluctant to publish her results as “proof” of the alternative hypothesis, but she will prob- ably be encouraged to continue her research and collect more sample evidence. Finally, p-values larger than 0.10 are generally interpreted as weak evidence (or no evidence) in support of the alternative.

There is a strong connection between the a-level approach and the p-value approach. Specifically, the null hypothesis can be rejected at a specified significance level a only if the p-value from the sample is less than or equal to a. Equivalently, the sample evidence is statistically significant at a given a level only if its p-value is less than or equal to a. For example, if the p-value from a sample is 0.03, the null hypothesis can be rejected at the 10% and the 5% significance levels but not at the 1% level. The p-value essentially states how significant a given sample is.

A p-value measures how unlikely the observed sample results are, given that the null hypothesis is true. Therefore, a low p-value provides evidence for rejecting the null hypothesis and accepting the alternative.

Sample evidence is statistically significant at the a level only if the p-value is less than a.

The advantage of the p-value approach is that you don’t have to choose a significance level a ahead of time. Because it is far from obvious what value of a to choose in any particular situation, this is certainly an advantage. Another compelling advantage is that p-values for standard hypothesis tests are included in virtually all statistical software out- put. In addition, all p-values can be interpreted in basically the same way: A small p-value provides support for the alternative hypothesis.

Key role of p-values

The single most important thing to remember from this chapter is the role of p-values. This is especially important because p-values are listed in virtually all outputs from statistical software. If a p-value is small, the result is statisti- cally significant, meaning that the null hypothesis can be rejected in favor of the alternative.

Analysts don’t always agree on how “small” a p-value needs to be—some say less than 0.01, some say less than 0.05, and some say less than 0.10. But nearly all analysts agree that if a p-value is greater than 0.10, the result is not statistically significant, which means that there is not enough evidence to reject the null hypothesis.

Fundamental Insight

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9-2 concepts in hypothesis testing 3 7 5

9-2f Type II Errors and Power A type II error occurs when the alternative hypothesis is true but there isn’t enough evidence in the sample to reject the null hypothesis. This type of error is traditionally considered less important than a type I error, but it can lead to serious consequences in real situations. For example, in medical trials on a proposed new cancer drug, a type II error occurs if the new drug is really superior to existing drugs but experimental evidence is not sufficiently conclusive to warrant marketing the new drug. For patients suffering from cancer, this is a serious error.

As we stated previously, the alternative hypothesis is typically the hypothesis a researcher wants to prove. If it is in fact true, the researcher wants to be able to reject the null hypothesis and hence avoid a type II error. The probability that she is able to do so is called the power of the test—that is, the power is one minus the probability of a type II error. There are several ways to achieve high power, the most obvious of which is to increase sample size. By sampling more members of the population, you are better able to see whether the alternative is really true and hence avoid a type II error if the alternative is indeed true. As in the previous chapter, there are formulas that specify the sample size required to achieve a certain power for a given set of hypotheses. We will not pursue these in this book, but you should be aware that they exist.

The power of a test is one minus the probability of a type II error. It is the prob- ability of correctly rejecting the null hypothesis when the alternative hypothesis is true.

When using a confidence interval to perform a two-tailed hypothesis test, reject the null hypothesis if and only if the hypothesized value does not lie inside a con- fidence interval for the parameter.

9-2g Hypothesis Tests and Confidence Intervals The results of hypothesis tests are often accompanied by confidence intervals. This pro- vides two complementary ways to interpret the data. However, there is a more formal connection between the two, at least for two-tailed tests. Let a be the stated signifi- cance level of the test. We will state the connection for the most commonly used level, a 5 0.05, although it extends to any a value. The connection is that the null hypothesis can be rejected at the 5% significance level if and only if a 95% confidence interval does not include the hypothesized value of the parameter.

As an example, consider the test of H0:m 5 0 versus Ha:m ? 0. Suppose a 95% con- fidence interval for m extends from 1.35 to 3.42, so that it does not include the hypothe- sized value 0. Then H0 can be rejected at the 5% significance level, and the p-value from the sample must be less than 0.05. On the other hand, if a 95% confidence interval for m extends from 21.25 to 2.31 (negative to positive), the null hypothesis cannot be rejected at the 5% significance level, and the p-value must be greater than 0.05.

There is also a correspondence between one-tailed hypothesis tests and one-sided confidence intervals, but we will not pursue it here.

9-2h Practical Versus Statistical Significance We have stated that statistically significant results are those that produce sufficiently small p-values. In other words, statistically significant results are those that provide strong

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evidence in support of the alternative hypothesis. You frequently hear about studies, par- ticularly in the medical sciences, that produce statistically significant results. For example, you might hear that mice injected with one kind of drug develop significantly more cancer cells than mice injected with a second kind of drug.

The point of this section is that such results are not necessarily significant in terms of importance. They might be significant only in the statistical sense. An example of what could happen is the following. An education researcher wants to see whether quantita- tive SAT scores differ, on average, across gender. He sets up the hypotheses H0:mM 5 mF versus Ha:mM ? mF, where mM and mF are the mean quantitative SAT scores for males and females, respectively. He then randomly samples scores from 4000 males and 4000 females and finds the male and female sample averages to be 521 and 524. He also finds that the sample standard deviation for each group is about 50. Based on these numbers, the p-value for the sample data turns out to be approximately 0.007. (You will learn how to make this calculation later in the chapter.) Therefore, he claims that the results are signifi- cant proof that males do score differently (lower) than females.

If you read these results in a newspaper, your immediate reaction might be, “Who cares?” After all, the difference between 521 and 524 is not very large from a practical point of view. So why does the education researcher get to make his claim? The reasoning is as follows. In all likelihood, the means mM and mF are not exactly equal. There is bound to be some difference between genders over the entire population. If the researcher takes large enough samples—and 4000 is plenty large—he is almost certain to obtain enough evidence to “prove” that the means are not equal. That is, he will almost surely obtain statistically significant results. However, the difference he finds, as in the numbers we quoted, might be of little practical significance. No one really cares whether females score three points higher or lower than males. If the difference were on the order of 30 to 40 points, then the result would be more interesting.

As this example illustrates, there is always a possibility of statistical significance but not practical significance, especially with large sample sizes. To be fair, we should also mention the opposite case, which typically occurs with small sample sizes. Here the results are sometimes not statistically significant even though the truth about the population(s), if it were known, would be of practical significance. Let’s assume that a medical researcher wants to test whether a new form of treatment produces a higher cure rate for a deadly disease than the best treatment currently on the market. Due to expenses, the researcher is able to run a controlled experiment on only a relatively small number of patients with the disease. Unfortunately, the results of the experiment are inconclusive. They show some evidence that the new treatment works better, but the p-value for the test is only 0.25.

In the scientific community these results would not be enough to warrant a switch to the new treatment. However, it is certainly possible that the new treatment, if it were used on a large number of patients, would provide a “significant” improvement in the cure rate—where “significant” now means practical significance. In this type of situation, the researcher could easily fail to discover practical significance because the sample sizes are not large enough to detect it statistically.

From here on, when we use the term “significant,” we mean statistically significant. However, you should always keep the ideas in this section in mind. A statistically signif- icant result is not necessarily of practical importance. Conversely, a result that fails to be statistically significant is not necessarily one that should be ignored.

9-3 Hypothesis Tests for a Population Mean Now that we have covered the general concepts behind hypothesis testing and the prin- cipal sampling distributions (in the previous two chapters), the mechanics of hypothesis testing are fairly straightforward. We discuss in some detail how the procedure works for a population mean. Then in later sections we illustrate similar hypothesis tests for other parameters.

Extremely large samples can easily lead to statistically significant results that are not practically significant. In contrast, small samples can fail to produce statisti- cally significant results that might indeed be practically significant.

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As with confidence intervals, the key to the analysis is the sampling distribution of the sample mean. Recall that if you subtract the true mean m from the sample mean and divide the difference by the standard error s>!n, the result has a t distribution with n 2 1 degrees of freedom. In a hypothesis-testing context, the true mean to use is the null hypothesis value, that is, the borderline value between the null and alternative hypotheses. This value is usually labeled m0, where the subscript indicates that it is based on the null hypothesis.

To run the test, referred to as the t test for a population mean, you calculate the test statistic in Equation (9.1). This t-value indicates how many standard errors the sam- ple mean is from the null value, m0. If the null hypothesis is true, or more specifically, if m 5 m0, this test statistic has a t distribution with n 2 1 degrees of freedom. The p-value for the test is the probability beyond the test statistic in both tails (for a two-tailed alterna- tive) or in a single tail (for a one-tailed alternative) of the t distribution.

Test Statistic for Test of Mean

t-value 5 X 2 m0

s>!n (9.1)

We illustrate the procedure by continuing the pizza manager’s problem in Example 9.1.

EXAMPLE

9.1 A NEW PIZZA STYLE AT PEPPERONI PIZZA RESTAURANT (CONTINUED)

Recall that the manager of Pepperoni Pizza Restaurant is running an experiment to test the hypotheses H0:m # 0 versus Ha:m 7 0, where m is the mean rating in the entire customer population. Here, each customer rates the difference between an old-style pizza and a new-style pizza on a scale from 210 to 110, where negative ratings favor the old style and positive ratings favor the new style. The ratings for 40 randomly selected customers are listed in the file Pizza Ratings.xlsx. Is there sufficient evidence from these sample data for the manager to reject H0?

Objective To use a one-sample t test to see whether consumers prefer the new-style pizza to the old style.

Solution As in the previous chapter, we include a template file, Hypothesis Test Template.xlsx, to calculate the details of most of the tests in this chapter. It is very similar to the template file for confidence intervals. The results for this example are based on the HT Mean template sheet and appear in Figure 9.3. (See the file Pizza Ratings Finished.xlsx.) Again, orange shading is used for the cells you need to fill. The template automatically calculates the formulas in the other cells. In this case, you need to specify the type of test from a dropdown list in cell B2, the hypothesized mean in cell B3, and the sample size, sample mean, and sample standard deviation, based on the data.

From the summary statistics, you can see that the sample mean is X 5 2.10 and the sample standard deviation is s 5 4.717. This positive sample mean provides some evidence in favor of the alternative hypothesis, but given the rather large value of s, does it provide enough evidence to reject H0?

To run the test, you calculate the test statistic in cell B9, using the borderline null hypothesis value m0 5 0, and report how much probability is beyond it in the right tail of the appropriate t distribution. The right tail is appropriate because the alterna- tive is one-tailed of the “greater than” variety.

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The t-value indicates that the sample mean is slightly more than 2.8 standard errors to the right of the null value, 0. Intuitively, this provides a lot of evidence in favor of the alternative—it is quite unlikely to see a sample mean 2.8 standard errors to the right of a “true” mean. The probability beyond this value in the right tail of a t distribution with n 2 1 5 39 degrees of free- dom is approximately 0.004, as shown in cell B11. Note how the template uses a nested IF formula for the p-value. This covers all three possible types of alternative hypotheses: greater than, less than, or not equal to.

The p-value for the test indicates that these sample results would be very unlikely if the null hypothesis were true. The manager has two choices at this point. He can conclude that the null hypothesis is true and he obtained a very unlikely sample, or he can conclude that the alternative hypothesis is true—and presumably switch to the new-style pizza. This second conclu- sion certainly appears to be the more reasonable choice.

Another way of interpreting the results of the test is in terms of traditional significance levels. The null hypothesis can be rejected at the 1% significance level because the p-value is less than 0.01. It can also be rejected at the 5% level or the 10% level because the p-value is also less than 0.05 and 0.10. Just remember that the p-value is the preferred way to report the results because it indicates exactly how significant these sample results are.

Figure 9.3 Analysis of Pizza Data 1 2 3 4 5 6 7 8 9

10 11 12 13

A B C D E F H IG Hypothesis test for population mean Type of alternative hypothesis Hypothesized mean

Sample size Sample mean Sample standard deviation Standard error of mean Test statistic (t distribution) Degrees of freedom p-value

=COUNT(Data!B2:B41) =AVERAGE(Data!B2:B41) =STDEV.S(Data!B2:B41) =B7/SQRT(B5) =(B6–B3)/B8 =B5–1 =IF(B2=“One-tailed – greater than”, T.DIST.RT(B9,B10), IF(B2=“One-tailed – less than”, T.DIST(B9,B10,TRUE), T.DIST.2T(ABS(B9),B10)))

One-tailed – greater than 0

40 2.100 4.717 0.746 2.816

39 0.004

test Statistics and r-values

All hypothesis tests are implemented by calculating a test statistic from the data and seeing how far out this test statistic is in one or both tails of some well- known distribution. The details of this procedure might or might not be included in the output from statistical software, but the p-value is always included. The p-value specifies the probability in the tail (or tails) beyond the test statistic. In words, it measures how unlikely such an extreme value of the test statistic is if the null hypothesis is true.

Fundamental Insight

Before leaving this example, we ask one last question. Should the manager switch to the new-style pizza on the basis of these sample results? We would probably recommend “yes.” There is no indication that the new-style pizza costs any more to make than the old-style pizza, and the sample evidence is fairly convincing that customers, on average, prefer the new-style pizza. Therefore, unless there are reasons for not switching that we haven’t mentioned here, we recommend the switch. How- ever, if it costs more to make the new-style pizza (and its price is no higher), hypothesis testing is not the best way to perform the decision analysis.

To complete this example, we mention how the result would change if, for whatever reason, the manager wanted to run a two-tailed test. First, the “Two-tailed - not equal to” option would be selected in cell B2. Then the formula for the p-value would use the T.DIST.2T function, which would double the p-value. This is true in general. The p-value for a two-tailed test is always double the p-value for a corresponding one-tailed test.

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9-3 hypothesis tests for a population Mean 3 7 9

Problems Solutions for problems whose numbers appear within a colored box can be found in the Student Solution Files.

Level A 1. The file P09_01.xlsx contains a random sample of 100

lightbulb lifetimes. The company that produces these light- bulbs wants to know whether it can claim that its lightbulbs typically last more than 1000 burning hours. a. Identify the null and alternative hypotheses. b. Can this lightbulb manufacturer claim that its light-

bulbs typically last more than 1000 hours at the 5% significance level? What about at the 1% significance level? Explain your answers.

2. A manufacturer is interested in determining whether it can claim that the boxes of detergent it sells contain, on average, more than 500 grams of detergent. The firm selects a random sample of 100 boxes and records the amount of detergent (in grams) in each box. The data are provided in the file P09_02.xlsx. a. Identify the null and alternative hypotheses. b. Is there statistical support for the manufacturer’s

claim? Explain. 3. A producer of steel cables wants to know whether

the steel cables it produces have an average breaking strength of 5000 pounds. An average breaking strength of less than 5000 pounds would not be adequate, and to produce steel cables with an average breaking strength in excess of 5000 pounds would unnecessarily increase production costs. The producer collects a random sample of 64 steel cable pieces. The breaking strength for each of these cable pieces is recorded in the file P09_03.xlsx. a. Identify the null and alternative hypotheses. b. Using a 5% significance level, what statistical con-

clusion can the producer reach regarding the average breaking strength of its steel cables? Would the con- clusion be any different at the 1% level? Explain your answers.

4. A U.S. Navy recruiting center knows from past expe- rience that the heights of its recruits have traditionally been normally distributed with mean 69 inches. The recruiting center wants to test the claim that the average height of this year’s recruits is greater than 69 inches. To do this, recruiting personnel take a random sample of 64 recruits from this year and record their heights. The data are provided in the file P09_04.xlsx. a. Identify the null and alternative hypotheses. b. On the basis of the available sample information, do

the recruiters find support for the given claim at the 5% significance level? Explain.

c. Use the sample data to calculate a 95% confidence interval for the average height of this year’s recruits.

Based on this confidence interval, what conclusion should recruiting personnel reach regarding the given claim?

5. Suppose you wish to test H0:m # 10 versus Ha:m 7 10 at the a 5 0.05 significance level. Furthermore, suppose that you observe values of the sample mean and sample standard deviation when n 5 40 that do not lead to the rejection of H0. Is it true that you might reject H0 if you observed the same values of the sample mean and sam- ple standard deviation from a sample with n 7 40? Why or why not?

Level B 6. A study is performed in a large southern town to deter-

mine whether the average amount spent on food per four-person family in the town is significantly differ- ent from the national average. A random sample of the weekly grocery bills of two-person families in this town is given in the file P09_06.xlsx. Assume the national average amount spent on food for a four-person family is $150. a. Identify the null and alternative hypotheses. b. Is the sample evidence statistically significant? If

so, at what significance levels can you reject the null hypothesis?

c. For which values of the sample mean (i.e., average weekly grocery bill) would you reject the null hypoth- esis at the 1% significance level? For which values of the sample mean would you reject the null hypothesis at the 10% level?

7. An aircraft manufacturer needs to buy aluminum sheets with an average thickness of 0.05 inch. The manufac- turer knows that significantly thinner sheets would be unsafe and considerably thicker sheets would be too heavy. A random sample of 100 sheets from a potential supplier is collected. The thickness of each sheet in this sample is measured (in inches) and recorded in the file P09_07.xlsx. a. Identify the null and alternative hypotheses. b. Based on the results of an appropriate hypothesis test,

should the aircraft manufacturer buy aluminum sheets from this supplier? Explain why or why not.

c. For which values of the sample mean (i.e., average thickness) would the aircraft manufacturer decide to buy sheets from this supplier? Assume a significance level of 5% in answering this question.

8. Suppose you observe a random sample of size n from a normally distributed population. If you are able to reject H0:m 5 m0 in favor of a two-tailed alternative hypothesis at the 10% significance level, is it true that you can definitely reject H0 in favor of the appropriate one-tailed alternative at the 5% significance level? Why or why not?

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9-4 Hypothesis Tests for Other Parameters Just as we developed confidence intervals for a variety of parameters, we can develop hypothesis tests for other parameters. They are based on the same sampling distributions discussed in the previous chapter, and they are run and interpreted exactly as the tests for the mean in the previous section. In each case the sample data are used to calculate a test statistic that has a well-known sampling distribution. Then a corresponding p-value mea- sures the support for the alternative hypothesis. Beyond this, only the details change, as we illustrate in this section. Fortunately, the relevant sheets from the template file take care of most of these details.

9-4a Hypothesis Test for a Population Proportion To test a population proportion p, recall that the sample proportion p̂ has a sampling dis- tribution that is approximately normal when the sample size is reasonably large. Specifi- cally, the distribution of the standardized value

p̂ 2 p

!p11 2 p2 >n

is approximately normal with mean 0 and standard deviation 1. This leads to the following z test for a population proportion.

Let p0 be the borderline value of p between the null and alternative hypotheses. Then p0 is substituted for p to obtain the test statistic in Equation (9.2). The p-value of the test is found by seeing how much probability is beyond this test statistic in the tail (or tails) of the standard normal distribution.1 A rule of thumb for checking the large-sample assump- tion of this test is to check whether np0 7 5 and n11 2 p02 7 5.

Test Statistic for Test of Proportion

z-value 5 p̂ 2 p0

!p011 2 p02/n (9.2)

The following example illustrates this test of proportion.

EXAMPLE

9.2 CUSTOMER COMPLAINTS AT WALPOLE APPLIANCE COMPANY

Walpole Appliance Company has a customer service department that handles customer questions and complaints. This department’s processes are set up to respond quickly and accurately to customers who phone in or email their concerns. How- ever, there is a sizable minority of customers who prefer to write letters. Traditionally, the customer service department has not been very efficient in responding to these customers.

Letter writers first receive an email asking them to call customer service (which is exactly what letter writers wanted to avoid in the first place), and when they do call, the customer service representative who answers the phone typically has no knowledge of the customer’s problem. As a result, the department manager estimates that 15% of letter writers have not obtained a satisfactory response within 30 days of the time their letters were first received. The manager’s goal is to reduce this value by at least half, that is, to 7.5% or less.

1 Do not confuse the unknown proportion p with the p-value of the test. They are logically different concepts and just happen, by tradition, to share the same letter p.

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9-4 hypothesis tests for Other parameters 3 8 1

To do so, she changes the process for responding to letter writers. Under the new process, these customers now receive a prompt and courteous form letter that responds to their problem. (This is possible because the vast majority of concerns can be addressed by one of several form letters.) Each form letter states that if the customer still has problems, he or she can call the department. The manager also files the original letters so that if customers do call back, the representative who answers will be able to find their letters quickly and respond intelligently. With this new process in place, the manager has tracked 400 letter writers and has found that only 23 of them are classified as “unsatisfied” after a 30-day period. Does it appear that the manager has achieved her goal?

Objective To use a test for a proportion to see whether the new process of responding to complaint letters results in an acceptably low proportion of unsatisfied customers.

Solution The manager’s goal is to reduce the proportion of unsatisfied customers after 30 days from 0.15 to 0.075 or less. Because the burden of proof is on her to “prove” that she has accomplished this goal, we set up the hypotheses as H0:p $ 0.075 versus Ha:p 6 0.075, where p is the proportion of all letter writers who are still unsatisfied after 30 days. The sample proportion she has observed is p̂ 5 23>400 5 0.0575. This is obviously less than 0.075, but is it enough less to reject the null hypothesis?

The test can be performed with the HT Proportion sheet from the template file, as shown in Figure 9.4. See the file Customer Complaints Finished.xlsx. Note that the standard error and z-value test statistic are calculated according to Equation (9.2), and the p-value is the probability to the left of the test statistic, calculated with the NORM.S.DIST function.

Figure 9.4 Analysis of Customer Complaints

Hypothesis test for proportion Type of alternative hypothesis

Hypothesized proportion

400

IHGFEDCBA

–1.329

0.092

2*(1–NORM.S.DIST(ABS(B9),TRUE))))

IF(B2=“One-tailed – less than”,NORM.S.DIST(B9,TRUE),

=IF(B2=“One-tailed – greater than”,1–NORM.S.DIST(B9,TRUE),

=(B7–B3)/B8

=SQRT(B3*(1–B3)/B5)

=B6/B5

=Data!B4

=Data!B6

0.013

0.058

0.075

One-tailed – less than

Sample size

Number with property of interest

Sample proportion

Standard error of proportion

Test statistic (z value)

p-value

The p-value might not be as low as you expected—or as low as the manager would like. In spite of the fact that the sample proportion appears to be well below the target proportion of 0.075, the evidence in support of the alternative hypothesis is not overwhelming. In statistical terminology, the results are significant at the 10% level, but not at the 5% or 1% levels.

The last sheet of the finished file (not shown here) calculates a 95% confidence interval for the unknown proportion p. This confidence interval extends from 0.035 to 0.080. It includes the target value, 0.075, but just barely. In this sense it also provides some support for the argument that the manager has indeed achieved her goal.2

Analysts might disagree on whether a hypothesis test or a confidence interval is the more appropriate way to present these results. However, we see them as complementary and do not necessarily favor one over the other. The bottom line is that they both provide some, but not totally conclusive, evidence that the manager has achieved her goal.

2 If you are very observant, you will note that the standard error in cell E7 for the hypothesis test uses the target proportion 0.075. In contrast, the standard error for the confidence interval uses the sample proportion 0.0575. The sampling distribution for a hypothesis test always uses the borderline value between H0 and Ha. But because confidence intervals are not connected to any hypotheses, their standard errors must rely on sample data. In most cases, however, the two standard errors are practically the same.

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9-4b Hypothesis Tests for Difference Between Population Means

We now discuss the comparison problem, where the difference between two population means is tested. As in the previous chapter, the form of the analysis depends on whether the two samples are independent or paired. For variety, we begin with the paired case.

If the samples are paired, then the test, referred to as the t test for difference between means from paired samples, proceeds exactly as in Section 9-3, using the differences as the single variable of analysis. That is, if D is the sample mean difference between n pairs, D0 is the hypothesized difference (the borderline value between H0 and Ha), and sD is the sample standard deviation of the differences, then the test is based on the test statistic in Equation (9.3). If D0 is the true mean difference, this test statistic has a t distribution with n 2 1 degrees of freedom. The validity of the test also requires that n be reasonably large and/or the population of differences be approximately normally distributed.

This comparison problem— comparing two population means—is one of the most important problems analyzed with statistical methods. It can be analyzed with confidence intervals, hypothesis tests, or both.

Test Statistic for Paired Samples Test of Difference Between Means

t-value 5 D 2 D0

sD>!n (9.3)

In contrast, if the samples are independent, the test is referred to as the t test for difference between means from independent samples. If the population standard deviations are equal, the two-sample theory discussed in Section 8-7 is relevant. It leads to the test statistic in Equation (9.5). Here, X1 and X2 are the two sample means, D0 is the hypothe- sized difference, n1 and n2 are the sample sizes, and sp is the same pooled estimate of the common population standard deviation as in the previous chapter:

Estimate of Common Standard Deviation

sp 5 Å 1n1 2 12s2

1 1 1n2 2 12s2 2

n1 1 n2 2 2 (9.4)

Test Statistic for Independent Samples Test of Difference Between Means

t-value 5 1X1 2 X22 2 D0

sp!1>n1 1 1>n2

(9.5)

If D0 is the true mean difference, this test statistic has a t distribution with n1 1 n2 2 2 degrees of freedom. The validity of this test again requires that the sample sizes be reason- ably large and/or the populations be approximately normally distributed.

The following example illustrates the paired-sample t test.

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9-4 hypothesis tests for Other parameters 3 8 3

EXAMPLE

9.3 MEASURING THE EFFECTS OF TRADITIONAL AND NEW STYLES OF SOFT-DRINK CANS

Beer and soft-drink companies have become very concerned about the style of their cans. There are cans with fluted and embossed sides and cans with six-color graphics and holograms. Coca-Cola even introduced a contoured can, shaped like the old-fashioned Coke bottle minus the neck. Evidently, these companies believe the style of the can makes a difference to con- sumers, which presumably translates into higher sales.

Assume that a soft-drink company is considering a style change to its current can, which has been the company’s trade- mark for many years. To determine whether this new style is popular with consumers, the company runs a number of focus group sessions around the country. At each of these sessions, randomly selected consumers are allowed to examine the new and traditional styles, exchange ideas, and offer their opinions. Eventually, they fill out a form where, among other questions, they are asked to respond to the following items, each on a scale of 1 to 7, 7 being the best:

• Rate the attractiveness of the traditional-style can (AO). • Rate the attractiveness of the new-style can (AN). • Rate the likelihood that you would buy the product with the traditional-style can (WBO). • Rate the likelihood that you would buy the product with the new-style can (WBN).

(A and WB stand for “attractiveness” and “would buy,” and O and N stand for “old” and “new.”) The data are listed in the file Soft-Drink Cans.xlsx. What can the company conclude from these data? Are hypothesis tests appropriate?

Objective To use paired-sample t tests for differences between means to see whether consumers rate the attractiveness, and their likeli- hood to purchase, higher for a new-style can than for the traditional-style can.

Solution First, it is a good idea to examine summary statistics for the data. The averages from each survey item are shown at the bottom of Figure 9.5. They indicate some support for the new-style can. Also, you might expect the ratings for a given consumer to be correlated. This turns out to be the case, as shown by the relatively large positive correlations in Figure 9.5. These large posi- tive correlations indicate that if you want to examine differences between survey items, a paired-sample procedure will make the most efficient use of the data. Of course, a paired-sample procedure also makes sense because each consumer answers each item on the form. If this is confusing, think about the following alternative setup, where there are four separate groups of con- sumers. The first group responds to item 1 only, the second group responds to item 2 only, and so on. Then the responses to the various items are in no way paired, and an independent-sample procedure would be used instead. However, this experimental design is not as efficient as the paired design in terms of making the best use of a given amount of data.

Figure 9.5 Data and Summary Measures for Soft-Drink Cans 1

2 3 4 5 6

179 180 181 182 183 184 185 186 187 188 189

A B C D E Consumer AO AN WBO WBN

1 2 3 4 5 178 179 180

Averages

Correlations AO AN WBO WBN

5 7 6 1 3 5 3 3

4.41

AO 1.000 0.740 0.746 0.594

7 7 7 3 4 4 4 5

4.95

AN 0.740 1.000 0.595 0.401

4 6 7 1 1 4 1 6

3.86

WBO 0.746 0.595 1.000 0.774

1 6 6 1 1 3 3 7

4.34

WBN 0.594 0.401 0.774 1.000

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There are several differences of interest. The two most obvious are the difference between the attractiveness ratings of the two styles and the difference between the likelihoods of buying the two styles—that is, column B minus column C and col- umn D minus column E. A third difference of interest is the difference between the attractiveness ratings of the new style and the likelihoods of buying the new can—that is, column C minus column E. This difference indicates whether perceptions of the new-style can are likely to translate into actual sales. Finally, a fourth difference that might be of interest is the difference between the third difference (column C minus column E) and the similar difference for the old style (column B minus column D). This checks whether the translation of perceptions into sales is any different for the two styles of cans.

All of these differences appear next to the original data in Figure 9.6.

Figure 9.6 Original and Difference Variables for Soft-Drink Can Data

1 2 3 4 5 6

179 180 181

A B C D E F G H I J Consumer AO AN WBO WBN AO−AN WBO−WBN AN−WBN AO−WBO (AN−WBN)−(AO−WBO)

1 4 1 –2 3 6 1 5 2 6 6 0 0 1 1 0 3 7 6 –1 1 1 2 4 1 1 –2 0 2 0 2 5 1 1 –1

–1

0 3 2 1 178 4 3 1 1 1 1 0 179 1 3 –1

–1 1 2

180 6 7 –2 –2

–2 –3 1

5 7 7 7 6 7 1 3 3 4 5 4 3 4 3 5

Figure 9.7 Analysis of Differences

A B C D E F

AO–AN Two-tailed - not equal to

180 –0.539

1.351 0.101

–5.351 179

0.000

WBO–WBN Two-tailed – not equal to

180 –0.478

1.347 0.100

–4.758 179

0.000

AN–WBN Two-tailed – not equal to

180 0.611 2.213 0.165 3.705

179 0.000

AO–WBO Two-tailed – not equal to

180 0.550 1.309 0.098 5.639

179 0.000

(AN–WBN)–( AO–WBO) Two-tailed – not equal to

180 0.061 2.045 0.152 0.401

179 0.689

Hypothesis test for difference Type of alternative hypothesis Hypothesized difference

Sample size Sample mean diff Sample std dev of diff Standard error Test statistic (t-value) Degrees of freedom p-value

For each of the differences, you can test the mean difference over all potential consumers with a paired-sample analysis. (You actually run the one-sample procedure on the difference variables.) Exactly as in the previous chapter, each difference variable is treated as a single sample and the same t test as in Section 9-3 is run on this sample. In each case the hypothesized difference, D0, is 0. The only question is whether to run one-tailed or two-tailed tests. This depends on the prior beliefs of the company. We have run them all as two-tailed tests. In any case, you can change two-tailed p-values to one-tailed p-values by dividing by 2.

The results for all five differences appear in Figure 9.7. The calculations were made with the HT Paired Mean Diff sheet from the template file and were then rearranged to all be on the same sheet. Here are some comments on the results.

• From the output for AO 2 AN, there is overwhelming evidence that consumers, on average, rate the attractiveness of the new design higher than the attractiveness of the current design. The t-distributed test statistic is 25.351, calculated as

20.539 2 0

0.101 5 25.351

The corresponding p-value for a two-tailed test of the mean difference is (to four decimal places) 0.0000. A 99% confidence interval (not shown) for the mean difference extends from 20.801 to 20.277. Note that this 99% confidence interval does not include the hypothesized value 0. This is consistent with the fact that the two-tailed p-value is less than 0.01. (Recall the relationship between confidence intervals and two-tailed hypothesis tests from Section 9-2g.)

• The results are basically the same for the difference between consumers’ likelihoods of buying the product with the two styles, WBO 2 WBN. Again, consumers are definitely more likely, on average, to buy the product with the new-style can. A 99% confidence interval for the mean difference (not shown) extends from 20.739 to 20.216, which is again all negative.

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9-4 hypothesis tests for Other parameters 3 8 5

• The company’s hypothesis that consumers’ ratings of attractiveness of the new-style can are greater, on average, than their likelihoods of buying the product with this style can is confirmed. In this case, the relevant difference is AN 2 WBN. Although the results are shown for a two-tailed test, the results for a one-tailed “greater than” test are easy to infer. You sim- ply divide the p-value in cell D11 by 2.

• There is no evidence that the difference between attractiveness ratings and the likelihood of buying is any different for the new-style can than for the current-style can. Here the relevant difference is 1AN 2 WBN2 2 1AO 2 WBO2. The test sta- tistic for a two-tailed test of this difference is 0.401 and the corresponding p-value, 0.689, isn’t even close to any of this traditional significance levels. Furthermore, a 99% confidence interval for this mean difference (not shown) extends from a negative value, 20.336, to a positive value, 0.458.

These results are further supported by histograms of the difference variables, such as those shown in Figures 9.8 and 9.9. (Box plots could be used instead.) The histogram in Figure 9.8 shows many more negative differences than positive differ- ences. This leads to the large negative test statistic and the all-negative confidence interval. In contrast, the histogram in Figure 9.9 is almost perfectly symmetric around 0 and hence provides no evidence that this mean difference is not zero.

Figure 9.8 Histogram of the AO-AN Variable

60 Histogram of AO−AN

≤–3 (–2, –1)(–3, –2) (–1, 0) (0, 1) (1, 2) >2

Figure 9.9 Histogram of the (AN-WBN)- (AO-WBO) Variable

40 45 50

Histogram of (AN−WBN)−(AO−WBO)

5 10

20 25

30 35

≤ –5 (–5, –4) (–4, –3) (–3, –2) (–2, –1) (–1, 0) (0, 1) (1, 2) (2, 3) (3, 4) > 4

This example illustrates once again how hypothesis tests and confidence intervals provide complementary information, although the confidence intervals are arguably more useful here. The hypothesis test for the first difference, for example, shows that the average rating for the new style is undoubtedly larger than for the current style. This is useful information, but it is even more useful to know how much larger the average for the new style is. A confidence interval provides this information.

We conclude this example by recalling the distinction between practical significance and statistical significance. Due to the extremely low p-values, the results in columns B to E of Figure 9.7, leave no doubt as to statistical significance. But this could be due to the large sample size. That is, if the true mean differences are even slightly different from 0, large samples will almost surely discover this and report small p-values. The soft-drink company, on the other hand, is more interested in know- ing whether the observed differences are of any practical importance. This is not a statistical question. It is a question of what differences are important for the business. We suspect that the company would indeed be quite impressed with the observed differences in the sample—and might very well switch to the new-style can.

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The following example illustrates the independent two-sample t test. You can tell that a paired-sample procedure is not appropriate because there is no attempt to match the observations in the two samples in any way. Indeed, this is obvious because the sample sizes are not equal.

EXAMPLE

9.4 PRODUCTIVITY DUE TO EXERCISE AT INFORMATRIX SOFTWARE COMPANY

Many companies have installed exercise facilities at their plants. The goal is not only to provide a bonus (free use of exercise equipment) for their employees, but to make the employees more productive by getting them in better shape. One such (fictional) company, Informatrix Software Company, installed exercise equipment on site a year ago. To check whether it has had a beneficial effect on employee productivity, the company gathered data on a sample of 80 randomly chosen employees, all between the ages of 30 and 40 and all with similar job titles and duties. The company observed which of these employees use the exercise facility regularly (at least three times per week on average). This group included 23 of the 80 employees in the sample. The other 57 employees were asked whether they exercise regularly elsewhere, and 6 of them replied that they do. The remaining 51, who admitted to being nonexercisers, were then compared to the combined group of 29 exercisers.

The comparison was based on the employees’ productivity over the year, as rated by their supervisors. Each rating was on a scale of 1 to 25, 25 being the best. To increase the validity of the study, neither the employees nor the supervisors were told that a study was in progress. In particular, the supervisors did not know which employees were involved in the study or which were exercisers. The data from the study are listed in the file Exercise & Productivity.xlsx. It includes a Yes/No column on whether the person is an exerciser and a column of ratings. Do these data support the company’s (alternative) hypothesis that exercisers outperform nonexercisers on average? Can the company infer that any difference between the two groups is due to exercise?

Objective To use a two-sample t test for the difference between means to see whether regular exercise increases worker productivity.

Solution To see whether there is any indication of a difference between the two groups, a good first step is to create side-by-side box plots of the Rating variable. These appear in Figure 9.10.

Side-by-side box plots are typically a good way to begin the analysis when comparing two populations.

Significance and Sample Size in tests of Differences

The contrast between statistical and practical significance is especially evident in tests of differences between means. If the sample sizes are relatively small, it is likely that no statistical significance will be found, even though the real differ- ence between means, if they could be estimated more accurately with more data, might be practically significant. On the other hand, if the sample sizes are very large, then just about any difference between sample means is likely to be statis- tically significant, even though the real difference between means might be of no practical importance.

Fundamental Insight

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Although there is a great deal of overlap between the two distributions, the distribution for the exercisers is somewhat higher than that for the nonexercisers. Also, the variances of the two distributions appear to be roughly the same, although there is slightly more variation in the nonexerciser distribution.

A formal test of the mean difference uses the hypotheses H0:m1 2 m2 # 0 versus Ha:m1 2 m2 7 0, where m1 and m2 are the mean ratings for the exerciser and nonexerciser populations. It makes sense to use a one-tailed test, with the alterna- tive of the “greater than” variety, because the company expects higher ratings, on average, for the exercisers. The output for this test appears in Figure 9.10. The calculations were performed by the HT Mean Diff sheet from the template file. If you compare these formulas to the two-sample confidence interval formulas in the previous chapter, you will see that they are very similar. Again, array formulas are used for the sample standard deviations (because Excel has no STDEVIF function), a test for equality of the two variances is shown in the bottom section, and the standard error of the difference and the degrees of freedom depend on the p-value of this test for equal variances.

The output shows that the observed sample mean difference, 2.725, is indeed positive. That is, the exercisers in the sample outperformed the nonexercisers by 2.725 rating points on average. The output also shows that (1) the standard error of the sample mean difference is 1.142, (2) the test statistic is 2.387, and (3) the p-value for a one-tailed test is slightly less than 0.01. In words, the data provide enough evidence to reject the null hypothesis at the 1% significance level, as well as at the 5% and 10% levels. It is clear that exercisers perform better, in terms of mean ratings, than nonexercisers. A 95% confidence interval for this mean difference (not shown) is all negative; it extends from 24.988 to 20.452.

This answers the first question we posed, but it doesn’t answer the second. There is no way to be sure that the higher ratings for the exercisers are a direct result of exercise. It is possible that employees who exercise are naturally more ambi- tious and hard-working people, and that this extra drive is responsible for both their exercising and their higher ratings. This study is an observational study. The company observes two randomly selected groups of employees and analyzes the results. It does not explicitly control for other factors, such as personality, that might be responsible for differences in ratings. Therefore, the company can never be sure that there is a causal relationship between exercise and performance ratings. All the company can state is that exercisers appear, on average, to be more productive than nonexercisers—for whatever reason.

The test reported at the bottom of Figure 9.11 is a formal test of the hypothesis H0:s 2 1>s2

2 5 1 versus Ha:s 2 1>s2

2 ? 1, where the parameter being tested is the ratio of the two population variances. (The details behind this test are explained in the following subsection.) If this null hypothesis can be rejected on the basis of a low p-value in cell B22, then the equal- variance assumption is almost certainly not valid, and the procedure that does not assume equal variances is used. The p-value in cell B11, 0.145, suggests that the evidence against equal population variances is far from overwhelming, so it is safe to use equal-variance procedure for this example.

Figure 9.10 Box Plots for Exercise Data

Box Plot of Ratings by Exerciser 25

0 Yes No

9-4 hypothesis tests for Other parameters 3 8 7

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9-4c Hypothesis Test for Equal Population Variances As we just explained, the two-sample procedure for a difference between population means depends on whether population variances are equal. Therefore, it is natural to test first for equal variances. This test, referred to as the F test for equality of two variances, is phrased in terms of the ratio of population variances, s2

1>s2 2. The

null hypothesis is that this ratio is 1 (equal variances), whereas the alternative is that it is not 1 (unequal variances). The test statistic for this test is the ratio of sample variances:

F-value 5 s2 1>s2

Assuming that the population variances are equal, this test statistic has an F distribution with n1 2 1 and n2 2 1 degrees of freedom.

The F distribution, named after the famous statistician R. A. Fisher, is another sampling distribution that arises frequently in statistical studies. (It will appear again in the next two chapters on regression.) Because it always describes a ratio, there are two degrees of freedom parameters, one for the numerator and one for the denominator, and the numer- ator degrees of freedom is always quoted first.

To run the test, you first calculate the ratio of variances. (See cell B21 in Figure 9.11.) It then implements the F test to calculate the corresponding p-value (in cell B22). For our purposes, the most important thing is the p-value from the test. A small p-value provides strong evidence that the population variances are not equal. Other- wise, an equal-variance assumption is reasonable. The p-value for the exercise data, 0.145, provides some evidence of unequal variances, but the evidence is certainly not overwhelming.

As Figure 9.11 indicates, this F test for equal variances is automatically run in the HT Mean Diff sheet from the template file because it is a natural part of the two-sample test for a mean difference. However, this F test can also be important in its own right, aside from any interest in mean differences. Therefore, it is implemented in a separate HT Ratio of Variances sheet in the template file.

9-4d Hypothesis Test for Difference Between Population Proportions

One of the most common uses of hypothesis testing is to test whether two population proportions are equal. The following z test for difference between proportions can then be used. Let p1 and p2 be the two population proportions, and let p̂1 and p̂2 be

The F distribution is a distri- bution of positive values and is always skewed to the right. It typically appears in tests of equal variances.

3 8 8 c h a p t e r 9 h y p o t h e s i s te s t i n g

Figure 9.11 Analysis of Exercise Data

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19 20 21 22

A B C D E F H I J KG Hypothesis test for difference between means Type of alternative hypothesis Hypothesized difference

Category 1 Category 2 Sample size 1 Sample size 2 Sample mean 1 Sample mean 2 Sample mean difference (1 minus 2) Sample std dev 1 Sample std dev 2 Pooled std dev Standard error of diff Test statistic (t-value) Degrees of freedom p-value

Test for equal variances Ratio of sample variances p-value

One-tailed � greater than 0.00

Yes No 29 51

16.862 14.137

2.725 4.103 5.307 4.909 1.142 2.387

78 0.010

0.598 0.145

Yes No =COUNTIF(Data!$B2:$B$81,B5) =COUNTIF(Data!$B$2:$B$81,B6) =AVERAGEIF(Data!$B$2:$B$81,B5,Data!$C$2:$C$81) =AVERAGEIF(Data!$B$2:$B$81,B6,Data!$C$2:$C$81) =B9 � B10 =STDEV.S(IF(Data!$B$2:$B$81=B5,Data!$C$2:$C$81)) =STDEV.S(IF(Data!$B$2:$B$81=B6,Data!$C$2:$C$81)) =SQRT(((B7 � 1)*B12^2+(B8 � 1)*B13^2)/B17) =IF(B22>0.1,B14*SQRT(1/B7+1/B8),SQRT(B12^2/B7+B13^2/B8)) =(B11 � B3)/B15 =IF(B22>0.1,B7+B8 � 2,ROUND(B15^4/((B12^2/B7)^2/(B7 � 1)+(B13^2/B8)^2/(B8 � 1)),0)) =IF($B$2=“One-tailed � greater than”, T.DIST.RT(B16,B17), IF($B$2=“One-tailed � less than”, T.DIST(B16,B17,TRUE),T.DIST.2T(ABS(B16),B17)))

=(B12/B13)^2 =2*(0.5 � ABS(0.5 � F.DIST(B21,B7 � 1,B8 � 1,TRUE)))

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9-4 hypothesis tests for Other parameters 3 8 9

the corresponding sample proportions, based on sample sizes n1 and n2. The goal is to test whether the sample proportions differ enough to conclude that the population proportions are not equal. As usual, a test on the difference p̂1 2 p̂2, requires a stan- dard error. If the null hypothesis is true, so that p1 5 p2, then it can be shown that the standard error of p̂1 2 p̂2 is given by Equation (9.6), where p̂c is the pooled propor- tion from the two samples combined. For example, if p̂1 5 20>85 and p̂2 5 34>115, then p̂c 5 120 1 342 > 185 1 1152 5 54>200. The reason for using this pooled estimate is that if the null hypothesis is true and the two population proportions are equal, it makes sense to base an estimate of this common proportion on the combined sample of data.

Standard Error for Difference between Sample Proportions

SE1 p̂1 2 p̂22 5 "p̂c11 2 p̂c2 11/n1 1 1/n22 (9.6)

Test Statistic for Difference between Proportions

z-value 5 p̂1 2 p̂2

SE1 p̂1 2 p̂22 (9.7)

Given this standard error, the rest is straightforward. Assuming that the sample sizes are reasonably large, the test statistic in Equation (9.7) has (approximately) a standard nor- mal distribution. The test can be run as illustrated in the following example.

EXAMPLE

9.5 EMPLOYEE EMPOWERMENT AT ARMCO COMPANY ArmCo Company, a large manufacturer of automobile parts, has several plants in the United States. For years, ArmCo employees have complained that their suggestions for improvements in the manufacturing processes have been ignored by upper management. In the spirit of employee empowerment, ArmCo management at the Midwest plant decided to initiate a number of policies to respond to employee suggestions. For example, a mailbox was placed in a central location, and employees were encouraged to drop suggestions into this box. No such initiatives were taken at the other ArmCo plants. As expected, there was a great deal of employee enthusiasm at the Midwest plant shortly after the new policies were implemented, but the question was whether life would revert to normal and the enthusiasm would dampen with time.

To check this, 100 randomly selected employees at the Midwest plant and 300 employees from other plants were asked to fill out a questionnaire six months after the implementation of the new policies at the Midwest plant. Employees were instructed to respond to each item on the questionnaire by checking either a “yes” box or a “no” box. Two specific items on the questionnaire were the following:

• Management at this plant is generally responsive to employee suggestions for improvements in the manufacturing processes. • Management at this plant is more responsive to employee suggestions now than it used to be.

The results of the questionnaire for these two items appear in Figure 9.12. (See the file Empowerment.xlsx.) Note that the given data for this example are counts, not long Yes/No columns of individual observations. Does it appear that the policies at the Midwest plant are appreciated? Should ArmCo implement these policies in its other plants?

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Objective To use a test for the difference between proportions to see whether a program of accepting employee suggestions is appreci- ated by employees.

Solution For either questionnaire item, let p1 be the proportion of “yes” responses that would be obtained at the Midwest plant if the questionnaire were given to all of its employees, and define p2 similarly for the other plants. Management certainly hopes to find a larger proportion of “yes” responses (to either item) at the Midwest plant than at the other plants, so the appropriate test is one-tailed, with the hypotheses set up as H0: p1 2 p2 # 0 versus Ha: p1 2 p2 7 0. (These could also be written as H0: p1 # p2 versus H0: p1 7 p2, but this has no effect on the test.)

Using the counts in Figure 9.12, the test for either difference follows directly from Equations (9.5) and (9.6). The results are shown in Figure 9.13, with calculations made by the HT Proportion Diff sheet from the template file. (See the file Empowerment Finished.xlsx.)

Figure 9.12 Data for Employee Empowerment Example

A B

Category Yes No

Totals

Category Yes No

Totals

Midwest 39 61

100

Other 93

207 300

Midwest 68 32

100

Other 159 141 300

Employee empowerment results

Item 1: Management responds

Item 2: Things have improved

Figure 9.13 Analysis of Empowerment Data

1 2 3 4 5 6 7 8 9

10 11 12 13 14 15 16 17 18 19

A B C D E F H I JG Hypothesis test for difference between proportions Type of alternative hypothesis

Category 1 Category 2 Sample size 1 Sample size 2 Number 1 with property of interest Number 2 with property of interest Sample proportion 1 Sample proportion 2 Sample proportion diff Pooled sample proportion Standard error of difference Test statistic (z distribution) p-value

=Data!B13 =Data!C13 =Data!B11 =Data!C11 =C9/C7 =C10/C8 =C11 – C12 =(C9+C10)/(C7+C8) =SQRT(C14*(1 – C14)*(1/C7+1/C8)) =C13/C15 =IF(C2=“One-tailed – greater than”, 1 – NORM.S.DIST(C16,TRUE), IF(C2=“One-tailed – less than”, NORM.S.DIST(C16,TRUE), 2*(1 – NORM.S.DIST(ABS(C16),TRUE))))

One-tailed – greater than One-tailed – greater than

100 300

39 93

0.390 0.310 0.080 0.330 0.054 1.473 0.070

100 300

68 159

0.680 0.530 0.150 0.568 0.057 2.622 0.004

Management responds Midwest

Other

Things have improved Midwest

Other

The p-values for the two tests (row 17) are 0.070 and 0.004. These results should be fairly good news for management. There is moderate, but not overwhelming, support for the hypothesis that management at the Midwest plant is more responsive than at the other plants, at least as perceived by employees. There is convincing support for the hypothesis that things have improved more at the Midwest plant than at the other plants. Corresponding 95% confidence intervals for the differences between proportions (not shown here) are included in the finished version of the file. Because they are almost completely positive, they support the hypothesis-test findings. Moreover, they provide a range of plausible values for the differences between the population proportions.

The only real downside to these findings, from Midwest management’s point of view, is the sample proportion p̂1 for the first item. Only 39% of the sampled employees at the Midwest plant believe that management generally responds to their suggestions, even though 68% believe things are better than they used to be. A reasonable conclusion by ArmCo management is that they are on the right track at the Midwest plant, and the policies initiated there ought to be initiated at other plants, but more must be done at all plants.

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9-4 hypothesis tests for Other parameters 3 9 1

Problems Solutions for problems whose numbers appear within a colored box can be found in the Student Solution Files.

Level A 9. In the past, 60% of all undergraduate students enrolled at

State University earned their degrees within four years of matriculation. A random sample of 95 students from the class that matriculated in the fall of 2012 was recently selected to test whether there has been a change in the proportion of students who graduate within four years. Administrators found that 40 of these 95 students gradu- ated in the spring of 2016 (i.e., four academic years after matriculation). a. Given the sample outcome, calculate a 95% confi-

dence interval for the relevant population proportion. Does this interval estimate suggest that there has been a change in the proportion of students who graduate within four years? Why or why not?

b. Suppose now that State University administrators want to test the claim made by faculty that the proportion of students who graduate within four years at State University has fallen below the historical value of 60% this year. Use this sample proportion to test their claim. Report a p-value and interpret it.

10. Suppose a well-known baseball player states that, at this stage of his career, he is a “300 hitter” or better. That is, he claims that he gets a hit in at least 30% of his at-bats. Over the next month of the baseball season, this player has 105 at-bats and gets 33 hits. a. Identify the null and alternative hypotheses from the

player’s point of view. b. Is there enough evidence from this month’s data to

reject the null hypothesis at the 5% significance level? c. We might raise two issues with this test. First, does

the data come from a random sample from some population? Second, what is the relevant population? Discuss these issues. Do you think the test in part b is valid? Is it meaningful?

11. The director of admissions of a distinguished (i.e., top-20) MBA program is interested in studying the proportion of entering students in similar graduate business programs who have achieved a composite score on the Graduate Management Admissions Test (GMAT) in excess of 630. In particular, the admissions director believes that the proportion of students entering top-rated programs with such composite GMAT scores is now 50%. To test this hypothesis, he has collected a random sample of MBA candidates entering his program in the fall of 2019. He believes that these students’ GMAT scores are indicative of the scores earned by their peers in his program and in competitors’ programs. The GMAT scores for these 125 individuals are given in the Data 2019 sheet of the file P09_11.xlsx. (You can ignore the data in the Data

2009 sheet for now.) Test the admission director’s claim at the 5% significance level and report your findings. Does your conclusion change when the significance level is increased to 10%?

12. A market research consultant hired by a leading soft-drink company wants to determine the proportion of consumers who favor its low-calorie drink over the leading compet- itor’s low-calorie drink in a particular urban location. A random sample of 250 consumers from the market under investigation is provided in the file P08_17.xlsx. a. Calculate a 95% confidence interval for the propor-

tion of all consumers in this market who prefer this company’s drink over the competitor’s. What does this confidence interval tell us?

b. Does the confidence interval in part a support the claim made by one of the company’s marketing man- agers that more than half of the consumers in this urban location favor its drink over the competitor’s? Explain your answer.

c. Comment on the sample size used in this study. Specifically, is the sample unnecessarily large? Is it too small? Explain your reasoning.

13. The CEO of a medical supply company is committed to expanding the proportion of highly qualified women in the organization’s staff of salespersons. He claims that the proportion of women in similar sales positions across the country in 2019 is less than 50%. Hoping to find support for his claim, he directs his assistant to collect a random sample of salespersons employed by his com- pany, which is thought to be representative of sales staffs of competing organizations in the industry. These data are listed in the Data 2019 sheet of the file P09_13.xlsx. (You can ignore the data in the Data 2014 sheet for now.) Test this manager’s claim using the given sample data and report a p-value. Is there statistical support for his hypothesis that the proportion of women in similar sales positions across the country is less than 50%?

14. Management of a software development firm would like to establish a wellness program during the lunch hour to enhance the physical and mental health of its employees. Before introducing the wellness program, management must first be convinced that a sufficiently large majority of its employees are not already exercising at lunchtime. Spe- cifically, it plans to initiate the program only if less than 40% of its personnel take time to exercise prior to eating lunch. To make this decision, management has surveyed a random sample of 100 employees regarding their midday exercise activities. The results of the survey are given in the Before sheet of the file P09_14.xlsx. Is there sufficient evidence at the 10% significance level for managers of this organization to initiate a corporate wellness program? Why or why not? What about at the 1% significance level?

15. The managing partner of a major consulting firm is trying to assess the effectiveness of expensive computer skills training given to all new entry-level professionals. In an effort to make such an assessment, she administers

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3 9 2 c h a p t e r 9 h y p o t h e s i s te s t i n g

a computer skills test immediately before and after the training program to each of 40 randomly chosen employ- ees. The pretraining and posttraining scores of these 40 individuals are recorded in the file P09_15.xlsx. Do the given sample data support the claim at the 10% signif- icance level that the organization’s training program is increasing the new employee’s working knowledge of computing? What about at the 1% significance level?

16. A large buyer of household batteries wants to decide which of two equally priced brands to purchase. To do this, he takes a random sample of 100 batteries of each brand. The lifetimes, measured in hours, of the randomly chosen batteries are recorded in the file P09_16.xlsx. a. Using the given sample data, calculate a 95%

confidence interval for the difference between the mean lifetimes of brand 1 and brand 2 batteries. Based on this confidence interval, which brand would you advise the buyer to purchase? Would you even need a confidence interval to make this recommendation? Explain.

b. Repeat part a with a 99% confidence interval. c. How are your results in parts a and b related to

hypothesis testing? Be specific. 17. The managers of a chemical manufacturing plant want

to determine whether recent safety training workshops have reduced the weekly number of reported safety violations at the facility. The management team has ran- domly selected weekly safety violation reports for each of 25 weeks prior to the safety training and 25 weeks after the safety workshops. These data are provided in the file P09_17.xlsx. Given this evidence, is it possible to conclude that the safety workshops have been effec- tive in reducing the number of safety violations reported per week? Report a p-value and interpret your findings for the management team.

18. A real estate agent has collected a random sample of 75 houses that were recently sold in a suburban com- munity. She is particularly interested in comparing the appraised value and recent selling price of the houses in this particular market. The values of these two variables for each of the 75 randomly chosen houses are provided in the file P08_21.xlsx. Using these sample data, test whether there is a statistically significant mean differ- ence between the appraised values and selling prices of the houses sold in this suburban community. Report a p-value. For which levels of significance is it appropri- ate to conclude that no difference exists between these two values? Which is more appropriate, a one-tailed test or a two-tailed test? Explain your reasoning.

19. The owner of two submarine sandwich shops located in a particular city would like to know how the mean daily sales of the first shop (located in the downtown area) compares to that of the second shop (located on the southwest side of town). In particular, he would like to determine whether the mean daily sales levels of these two restaurants are essentially equal. He records the sales (in dollars) made at each location for 30 randomly

chosen days. These sales levels are given in the file P09_19.xlsx. Calculate a 95% confidence level for the mean difference between the daily sales of restaurant 1 and restaurant 2. Based on this confidence interval, is it possible to conclude that there is a statistically signif- icant mean difference at the 5% level of significance? Explain why or why not. Can you infer from this confi- dence interval whether there is a statistically significant mean difference at the 10% level? What about at the 1% level? Again, explain why or why not.

20. Suppose that an investor wants to compare the risks associated with two different stocks. One way to mea- sure the risk of a given stock is to measure the variation in the stock’s daily price changes. The investor obtains a random sample of 25 daily price changes for stock 1 and 25 daily price changes for stock 2. These data are pro- vided in the file P09_20.xlsx. Explain why this inves- tor can compare the risks associated with the two stocks by testing the null hypothesis that the variances of the stocks’ price changes are equal. Perform this test, using a 10% significance level, and interpret the results.

21. A manufacturing company wants to determine whether there is a difference between the variance of the number of units produced per day by one machine operator and the similar variance for another machine operator. The file P09_21.xlsx contain the number of units produced by operator 1 and operator 2, respectively, on each of 25 days. Note that these two sets of days are not necessarily the same, so you can assume that the two samples are independent of one another. a. Identify the null and alternative hypotheses in this

situation. b. Do these sample data indicate a statistically signifi-

cant difference at the 10% level? Explain your answer. With your conclusion, which possible error could you be making, a type I or type II error?

c. At which significance levels could you not reject the null hypothesis?

22. A large buyer of household batteries wants to decide which of two equally priced brands to purchase. To do this, he takes a random sample of 100 batteries of each brand. The lifetimes, measured in hours, of the batteries are recorded in the file P09_16.xlsx. Before testing for the difference between the mean lifetimes of these two batteries, he must first determine whether the underlying population variances are equal. a. Perform a test for equal population variances. Report

a p-value and interpret its meaning. b. Based on your conclusion in part a, which test statistic

should be used in performing a test for the difference between population means? Perform this test and interpret the results.

23. Do undergraduate business students who major in finance earn, on average, higher annual starting sala- ries than their peers who major in marketing? Before addressing this question through a statistical hypothesis

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9-4 hypothesis tests for Other parameters 3 9 3

test, you should determine whether the variances of annual starting salaries of the two types of majors are equal. The file P09_23.xlsx contains (hypothetical) starting salaries of 50 randomly selected finance majors and 50 randomly chosen marketing majors. a. Perform a test for equal population variances. Report

a p-value and interpret its meaning. b. Based on your conclusion in part a, which test statistic

should you use in performing a test for the existence of a difference between population means? Perform this test and interpret the results.

24. The CEO of a medical supply company is committed to expanding the proportion of highly qualified women in the organization’s large staff of salespersons. Given the recent hiring practices of his human resources director, he claims that the company has increased the proportion of women in sales positions throughout the organization between 2014 and 2019. Hoping to find support for his claim, he directs his assistant to collect random samples of the salespersons employed by the company in 2014 and 2019. These data are listed in the file P09_13.xlsx. Test the CEO’s claim using the sample data and report a p-value. Is there statistical support for the claim that his strategy is effective?

25. The director of admissions of a top-20 MBA program is interested in studying the proportion of entering students in similar graduate business programs who have achieved a composite score on the Graduate Management Admissions Test (GMAT) in excess of 630. In particular, the admissions director believes that the proportion of students entering top-rated programs with such composite GMAT scores is higher in 2019 than it was in 2009. To test this hypothesis, he has collected ran- dom samples of MBA candidates entering his program in the fall of 2019 and in the fall of 2009. He believes that these students’ GMAT scores are indicative of the scores earned by their peers in his program and in com- petitors’ programs. The GMAT scores for the randomly selected students entering in each year are listed in the file P09_11.xlsx. Test the admission director’s claim at the 5% significance level and report your findings. Does your conclusion change when the significance level is increased to 10%?

26. Managers of a software development firm have estab- lished a wellness program during the lunch hour to enhance the physical and mental health of their employ- ees. Now, they would like to see whether the wellness program has increased the proportion of employees who exercise regularly during the lunch hour. To make this assessment, the managers surveyed a random sample of 100 employees about their noontime exercise habits before the wellness program was initiated. Later, after the program was initiated, another 100 employees were independently chosen and surveyed about their lunch- time exercise habits. The results of these two surveys are given in the file P09_14.xlsx.

a. Calculate a 95% confidence interval for the differ- ence in the proportions of employees who exer- cise regularly during their lunch hour before and after the implementation of the corporate wellness program.

b. Does the confidence interval found in part a support the claim that the wellness program has increased the proportion of employees who exercise regularly during the lunch hour? If so, at which levels of signif- icance is this claim supported?

c. Would your results in parts a and b differ if the same 100 employees surveyed before the program were also surveyed after the program? Explain.

27. An Environmental Protection Agency official asserts that more than 80% of the plants in the northeast region of the United States meet air pollution standards. An anti- pollution advocate is not convinced by the EPA’s claim. She takes a random sample of 64 plants in the northeast region and finds that 56 meet the federal government’s pollution standards. a. Does the sample information support the EPA’s claim

at the 5% level of significance? b. For which values of the sample proportion (based on a

sample size of 64) would the sample data support the EPA’s claim, using a 5% significance level?

c. Would the conclusion found in part a change if the sample proportion remained constant but the sample size increased to 124? Explain why or why not.

Level B 28. A television network decides to cancel one of its shows

if it is convinced that less than 14% of the viewing pub- lic are watching this show. a. If a random sample of 1500 households with televi-

sions is selected, what sample proportion values will lead to this show’s cancellation, assuming a 5% sig- nificance level?

b. What is the probability that this show will be can- celled if 13.4% of all viewing households are watch- ing it? That is, what is the probability that a sample will lead to rejection of the null hypothesis? You can assume that 13.4% is the population proportion (even though it wouldn’t be known to the network).

29. An economic researcher wants to know whether he can reject the null hypothesis, at the 10% significance level, that no more than 20% of the households in Pennsylvania make more than $70,000 per year. a. If 200 Pennsylvania households are chosen at random,

how many of them would have to be earning more than $70,000 per year to allow the researcher to reject the null hypothesis?

b. Assuming that the true proportion of all Pennsylvania households with annual incomes of at least $70,000 is 0.217, find the probability of not rejecting a false null hypothesis when the sample size is 200.

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3 9 4 c h a p t e r 9 h y p o t h e s i s te s t i n g

30. Senior partners of an accounting firm are concerned about recent complaints by some female managers that they are paid less than their male counterparts. In response to these charges, the partners ask their human resources director to record the salaries of female and male managers with equivalent education, work experi- ence, and job performance. A random sample of these pairs of managers is provided in the file P09_30.xlsx. That is, each male-female pair is matched in terms of education, work experience, and job performance. a. Do these data support the claim made by the female

managers? Report and interpret a p-value. b. Assuming a 5% significance level, which values of

the sample mean difference between the female and male salaries would support the claim of discrimina- tion against female managers?

31. Do undergraduate business students who major in finance earn, on average, higher annual starting sala- ries than their peers who major in marketing? Address this question through a statistical hypothesis test. The file P09_23.xlsx contains the (hypothetical) starting salaries of 50 randomly selected finance majors and 50 randomly selected marketing majors. a. Is it appropriate to perform a paired-comparison t test

with these data? Explain why or why not. b. Perform an appropriate hypothesis test with a 5%

significance level. Summarize your findings. c. How large would the difference between the mean

starting salaries of finance and marketing majors have to be before you could conclude that finance majors earn more on average? Employ a 5% significance level in answering this question.

32. The file P02_35.xlsx contains data from a survey of 500 randomly selected households. Test for the existence of a significant difference between the mean debt levels of the households in the first (i.e., SW) and second (i.e., NW) sectors of this community. Perform similar hypothesis tests for the differences between the mean debt levels of households from all other pairs of locations (i.e., first and third, first and fourth, second and third, second and fourth, and third and fourth). Summarize your findings.

33. Elected officials in a Florida city are preparing the annual budget for their community. They want to deter- mine whether their constituents living across town are typically paying the same amount in real estate taxes each year. Given that there are over 20,000 homeowners in this city, they have decided to sample a representa- tive subset of taxpayers and thoroughly study their tax payments. A randomly selected set of 170 homeowners is given in the file P09_33.xlsx. Specifically, the offi- cials want to test whether there is a difference between the mean real estate tax bill paid by residents of the first neighborhood of this town and each of the remaining five neighborhoods. That is, each pair referenced below is from neighborhood 1 and one of the other neighbor- hoods.

a. Before conducting any hypothesis tests on the difference between various pairs of mean real estate tax payments, perform a test for equal population variances for each pair of neighborhoods. For each pair, report a p-value and interpret its meaning.

b. Based on your conclusions in part a, which test statis- tic should be used in performing a test for a difference between population means in each pair?

c. Given your conclusions in part b, perform an appro- priate test for the difference between mean real estate tax payments in each pair of neighborhoods. For each pair, report a p-value and interpret its meaning.

34. Suppose that you sample two normal populations inde- pendently. The variances of these two populations are s2

1 and s2 2. You take random samples of sizes n1 and n2

and observe sample variances of s2 1 and s2

2. a. If n1 5 n2 5 21, how large must the fraction s1>s2 be

before you can reject the null hypothesis that s2 1 is no

greater than s2 2 at the 5% significance level?

b. Answer part a when n1 5 n2 5 41. c. If s1 is 25% greater than s2, approximately how large

must n1 and n2 be if you are able to reject the null hypothesis in part a at the 5% significance level? Assume that n1 and n2 are equal.

35. Two teams of workers assemble automobile engines at a manufacturing plant in Michigan. Quality control personnel inspect a random sample of the teams’ assemblies and judge each assembly to be acceptable or unacceptable. A random sample of 127 assemblies from team 1 shows 12 unacceptable assemblies. A similar random sample of 98 assemblies from team 2 shows 5 unacceptable assemblies. a. Calculate a 95% confidence interval for the difference

between the proportions of unacceptable assemblies from the two teams.

b. Based on the confidence interval found in part a, is there sufficient evidence to conclude, at the 5% signif- icance level, that the two teams differ with respect to their proportions of unacceptable assemblies?

c. For which values of the difference between these two sample proportions could you conclude that a statisti- cally significant difference exists at the 5% level?

36. A market research consultant hired by a leading soft- drink company is interested in determining whether there is a difference between the proportions of female and male consumers who favor the company’s low- calorie brand over the leading competitor’s low-calorie brand in a particular urban location. A random sample of 250 consumers from the market under investigation is provided in the file P08_17.xlsx. a. After separating the 250 randomly selected consum-

ers by gender, perform the statistical test and report a p-value. At which levels of a will the market research consultant conclude that there is essentially no dif- ference between the proportions of female and male consumers who prefer this company’s brand to the competitor’s brand in this urban area?

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9-5 tests for Normality 3 9 5

b. Marketing managers at this company have asked their market research consultant to explore further the poten- tial differences in the proportions of women and men who prefer drinking the company’s brand to the compet- itor’s brand. Specifically, the company’s managers wants to know whether the potential difference between the proportions of female and male consumers who favor the company’s brand varies by the age of the consum- ers. Using the same random sample of consumers as in part a, assess whether this difference varies across the four given age categories: under 20, between 20 and 40, between 40 and 60, and over 60. Specifically, run the test in part a four times, one for each age group. Are the results the same for each age group?

37. The employee benefits manager of a large public uni- versity wants to determine whether differences exist in the proportions of various groups of full-time employ- ees who prefer adopting the second (i.e., plan B) of three available health care plans in the coming annual enrollment period. A random sample of the university’s employees and their tentative health care preferences is given in the file P08_16.xlsx. a. Perform tests for differences in the proportions of

employees within respective classifications who favor plan B in the coming year. For instance, the first such test should examine the difference between the propor- tion of administrative employees who favor plan B and the proportion of the support staff who prefer plan B.

b. Report a p-value for each of your hypothesis tests and interpret your results. How might the benefits man- ager use the information you have derived from these tests?

38. The file P02_35.xlsx contains data from a survey of 500 randomly selected households. Researchers would like to use the available sample information to test whether home ownership rates vary by household location. For example, is there a nonzero difference between the proportions of individuals who own their homes (as opposed to those who rent their homes) in households located in the first (i.e., SW) and sec- ond (i.e., NW) sectors of this community? Use the sample data to test for a difference in home owner- ship rates in these two sectors as well as for those of other pairs of household locations. In each test, use a 5% significance level. Interpret and summarize your results. (You should perform and interpret a total of six hypothesis tests.)

39. For testing the difference between two proportions, !p̂c11 2 p̂c2 11/n1 1 1/n22 is used as the approximate standard error of p̂1 2 p̂2, where p̂c is the pooled sam- ple proportion. Explain why this is reasonable when the null-hypothesized value of p1 2 p2 is zero. Why would this not be a good approximation when the null-hypoth- esized value of p1 2 p2 is a nonzero number? What would you recommend using for the standard error of p̂1 2 p̂2 in that case?

9-5 Tests for Normality In this section we discuss several tests for normality.3 As you have already seen, many statistical procedures are based on the assumption that population data are normally dis- tributed. The tests in this section allow you to test this assumption. The null hypothesis is that the population is normally distributed, whereas the alternative is that the population distribution is not normal. Therefore, the burden of proof is on showing that the population distribution is not normal. Unless there is sufficient evidence to this effect, the normal assumption will continue to be accepted.

The first test we discuss is called a chi-square goodness-of-fit test. It is quite intu- itive. A histogram of the sample data is compared to the expected bell-shaped histogram that would be observed if the data were normally distributed with the same mean and standard deviation as in the sample. If the two histograms are sufficiently similar, the null hypothesis of normality is accepted. Otherwise, it can be rejected.

The test is based on a numerical measure of the difference between the two histograms. Let C be the number of categories in the histogram, and let Oi be the observed number of observations in category i. Also, let Ei be the expected number of observations in category i if the population were normal with the same mean and standard deviation as in the sam- ple. Then the goodness-of-fit measure in Equation (9.8) is used as a test statistic. If the null hypothesis of normality is true, this test statistic has (approximately) a chi-square distribu- tion with C 2 3 degrees of freedom. Because large values of the test statistic indicate a poor fit—the Oi’s do not match up well with the Ei’s—the p-value for the test is the probability to the right of the test statistic in the chi-square distribution with C 2 3 degrees of freedom.

3 The tests in this section could (with a lot of work) be implemented with Excel-only formulas. However, it is much easier to implement them with StatTools, as we will do here.

The chi-square test for normality makes a compar- ison between the observed histogram and a histogram based on normality.

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(Here, x is the Greek letter chi.) Although it is possible to perform this test with Excel-only formulas, it is certainly

preferable to use StatTools, as illustrated in Example 9.6.

EXAMPLE

9.6 DISTRIBUTION OF METAL STRIP WIDTHS IN MANUFACTURING

A company manufactures strips of metal that are supposed to have width of 10 centimeters. For purposes of quality control, the manager plans to run some statistical tests on these strips. However, realizing that these statistical procedures assume normally distributed widths, he first tests this normality assumption on 90 randomly sampled strips. How should he proceed?

Objective To use the chi-square goodness-of-fit test to see whether a normal distribution of the metal strip widths is reasonable.

Solution The sample data appear in Figure 9.14, where each width is measured to three decimal places. (See the file Testing Normality Finished.xlsx.) A histogram of the widths is also shown.

Test Statistic for Chi-Square Test of Normality

x 2-value 5 aC i5 1

1Oi 2 Ei22/Ei (9.8)

Figure 9.14 Data for Testing Normality

1 2 3 4 5 6 7 8 9

10 11 12 13 14

Part 1 2 3 4 5 6 7 8 9

10 11 12 13

Width 9.990

10.031 9.985 9.983

10.004 10.000

9.992 9.996 9.997 9.993 9.991 9.991

10.006

B C D E F G H I J

35 Histogram of Width

(9.970, 9 .978)

20 15

25 30

10 5

(9.978, 9 .985)

(9.985, 9 .993)

(9.993, … )

(10.000, … )

(10.008, … )

(10.016, … )

(10.023, … )

(10.031, … )

To run the test, select Chi-Square Test from the StatTools Normality Tests dropdown list, which leads to a dialog box where you can either specify the bins for a histogram or you can accept StatTools’s default bins. For now, do the latter.4 The resulting histograms in Figure 9.15 provide visual evidence of the goodness of fit. The left bars represent the observed frequen- cies (the Ois), and the right bars represent the expected frequencies for a normal distribution (the Eis). The normal fit to the data appears to be quite good.

4 You might try defining the bins differently and rerunning the test. The category definitions can make a difference in the results. This is a minor disadvan- tage of the chi-square test.

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9-5 tests for Normality 3 9 7

The StatTools output in Figure 9.16 confirms this statistically. Each value in column E is an Ei, calculated as the total number of observations multiplied by the normal probability of being in the corresponding category. Column F contains the individual 1Oi 2 Ei22>Ei terms, and cell B11 contains their sum, the chi-square test statistic. The corresponding p-value in cell B12 is 0.5206.

Figure 9.15 Observed and Normal Histograms` 25

Bi n

O cc

up at

io n

Bi n

# 1

Bi n

# 2

Bi n

# 3

Bi n

# 4

Bi n

# 5

Bi n

# 6

Bi n

# 7

Bi n

# 8

Chi-Square Test for Width/Width Data

Width Normal

Figure 9.16 Chi-Square Test of Normality

7 8

A B C D E F Width

Width DataChi-Square Test 9

10 11 12 13 14 15

Mean 9.999256 Std Dev 0.009728 Chi-Square Stat. 4.2027 p-Value 0.5206

Chi-Squared Bins Bin Min Bin Max Actual Normal Distance

9.983000 0.127416 17 18 19 20 21 22 23

Bin #1 –Inf 5 4.2630 Bin #2 9.983000 9.988167 6 7.1827 0.1948 Bin #3 9.988167 9.993333 14 12.9751 0.0810 Bin #4 9.993333 9.998500 20 17.7934 0.2736 Bin #5 9.998500 10.003667 13 18.5249 1.6477 Bin #6 10.003667 10.008833 19 14.6421 1.2970 Bin #7 10.008833 10.014000 9 8.7859 0.0052 Bin #8 10.014000 +Inf 4 5.8328 0.5759

Remember that the burden of proof is on showing that the distribution is not normal. The high p-value here indicates that there is not enough evidence to reject the null hypothesis of normality, and this finding is confirmed on the next two sheets.

This large p-value provides no evidence whatsoever of non-normality. It implies that if this procedure were repeated on many random samples, each taken from a population known to be normal, a fit at least this poor would occur in over 50% of the samples. Stated differently, fewer than 50% of the fits would be better than the one observed here. Therefore, the manager can feel comfortable in making a normal assumption for this population.

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We make several comments about this chi-square procedure. First, the test does depend on which (and how many) bins you use for the histogram. Reasonable choices are likely to lead to the same conclusion, but this is not guaranteed. Second, the test is not very effective unless the sample size is large: at least 80 or 100, say. Only then can you begin to see the true shape of the histogram and judge accurately whether it is normal. Finally, the test tends to be too sensitive if the sample size is really large. In this case any little “bump” on the observed histogram is likely to lead to a conclusion of non-normality. This is one more example of practical versus statistical significance. With a large sample size you might be able to reject normality with a high degree of certainty, but the practical difference between the observed and normal histograms could very well be unimportant.

The chi-square test of normality is an intuitive one because it is based on histo- grams. However, it suffers from the first two points discussed in the previous para- graph. In particular, it is not as powerful as other available tests. This means that it is often unable to distinguish between normal and non-normal distributions, and hence it often fails to reject the null hypothesis of normality when it should be rejected. A more powerful test is called the Lilliefors test.5 This test is based on the cumulative distri- bution function 1cdf2 , which shows the probability of being less than or equal to any particular value. Specifically, the Lilliefors test compares two cdfs: the cdf from a normal distribution and the cdf corresponding to the given data. This latter cdf, called the empirical cdf , shows the fraction of observations less than or equal to any partic- ular value. If the data come from a normal distribution, the normal and empirical cdfs should be quite close. Therefore, the Lilliefors test compares the maximum vertical distance between the two cdfs and compares it to specially tabulated values. If this maximum vertical distance is sufficiently large, the null hypothesis of normality can be rejected.

To run the Lilliefors test for the Width variable in Example 9.6, select Lilliefors Test from the StatTools Normality Tests dropdown list. StatTools then shows the numerical outputs in Figure 9.17 and the corresponding graph in Figure 9.18 of the normal and empirical cdfs. The numeric output indicates that the maximum vertical distance between the two curves is 0.0513. It also provides a number of “CVal” values for comparison. If the test statistic is larger than any of these, the null hypothesis of normality can be rejected at the corresponding significance level. In this case, however, the test statistic is relatively small—not nearly large enough to reject the normal hypothesis at any of the usual signifi- cance levels. This conclusion agrees with the one based on the chi-square goodness-of-fit test (as well as the closeness of the two curves in Figure 9.18). Nevertheless, you should be aware that the two tests do not agree on all data sets.

5 This is actually a special case of the more general and widely known Kolmogorov-Smirnoff (or K-S) test.

The Lilliefors test is based on a comparison of the cdf from the data and a normal cdf.

Figure 9.17 Lilliefors Test Results 7

A B Width

Lilliefors Test Results Width Data 9

10 11 12 13 14 15

Sample Size 90 Sample Mean 9.999256 Sample Std Dev 0.009728 Test Statistic 0.0513 CVal (15% Sig. Level) 0.0810 CVal (10% Sig. Level) 0.0856 CVal (5% Sig. Level) 0.0936 CVal (2.5% Sig. Level) 0.099816

17 CVal (1% Sig. Level) 0.1367

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9-5 tests for Normality 3 9 9

We conclude this section with a popular, but informal, test of normality. This is based on a plot called a quantile-quantile (or Q-Q) plot. Although the technical details for forming this plot are somewhat complex, it is basically a scatterplot of the standardized values from the data set versus the values that would be expected if the data were perfectly normally distributed (with the same mean and standard deviation as in the data set). If the data are, in fact, normally distributed, the points in this plot tend to cluster around a 45° line. Any large deviation from a 45° line signals some type of non-normality. Again, however, this is not a formal test of normality. A Q-Q plot is usually used only to obtain a general idea of whether the data are normally distributed and, if they are not, what type of non-normality exists. For example, if points on the right of the plot are well above a 45° line, this is an indication that the largest observations in the data set are larger than would be expected from a normal distribution. Therefore, these points might be high-end outliers and/or a signal of positive skewness.

To obtain a Q-Q plot for the Width variable in Example 9.6, select Q-Q Normal Plot from the StatTools Normality Tests dropdown list and check each option at the bottom of the dialog box. The Q-Q plot for the Width data in Example 9.6 appears in Figure 9.19. Although the points in this Q-Q plot do not all lie exactly on a 45° line, they are about as close to doing so as can be expected from real data. Therefore, there is no reason to question the normal hypothesis for these data—the same conclusion as from the chi-square and Lilliefors tests. (Note that in the StatTools Q-Q plot dialog box, you can elect to plot standardized Q-values. This option was used in Figure 9.19. The plot with unstandardized Q-values, not shown here, provides virtually the same information. The only difference is in the scale of the vertical axis.)

If data are normally distrib- uted, the points on the corresponding Q-Q plot should be close to a 45° line.

Figure 9.18 Normal and Empirical Cumulative Distribution Functions

0.7

0.8

0.9

1.0 Normal and Empirical Cumulative Distributions of Width/Width Data

0.0

0.1

0.2

0.3

0.4

0.5

0.6

–3.0 –2.0 –1.0 0.0 1.0 2.0 3.0

Figure 9.19 Q-Q Plot with Standardized Q-Values

1.5

2.5

3.5 Q-Q Normal Plot of Width/Width Data

–3.5

–2.5

–1.5

–0.5

0.5

–3.5 –2.5 –1.5 –0.5 0.5 1.5 2.5 3.5

St an

da rd

ize d

Q -V

al ue

Z-Value

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4 0 0 c h a p t e r 9 h y p o t h e s i s te s t i n g

Problems Solutions for problems whose numbers appear within a colored box can be found in the Student Solution Files.

Level A 40. The file P02_11.xlsx contains data on 148 houses sold

in a certain suburban region. a. Create a histogram of the selling prices. Is there any

visual evidence that the distribution of selling prices is not normal?

b. Test the selling prices for normality using the chi- square test. Is there enough evidence at the 5% sig- nificance level to conclude that selling prices are not normally distributed? If so, what is there about the distribution that is not normal?

c. Use the Lilliefors test and the Q-Q plot to check for normality of selling prices. Do these suggest the same conclusion as in part b? Explain.

41. The file P09_33.xlsx contains real estate taxes paid by a sample of 170 homeowners in a Florida city. a. Create a histogram of the taxes paid. Is there any visual

evidence that the distribution of taxes paid is not normal? b. Test the taxes paid for normality using the chi-square

test. Is there enough evidence at the 5% significance level to conclude that taxes paid are not normally dis- tributed? If so, what is there about the distribution that is not normal?

c. Use the Lilliefors test and the Q-Q plot to check for normality of taxes paid. Do these suggest the same conclusion as in part b? Explain.

42. The file P09_42.xlsx contains many years of monthly percentage changes in the Dow Jones Industrial Aver- age (DJIA). (This is the same data set that was used for Example 2.6 in Chapter 2.) a. Create a histogram of the percentage changes in the

DJIA. Is there any visual evidence that the distribution of the Dow percentage changes is not normal?

b. Test the percentage changes of the DJIA for normal- ity using the chi-square test. Is there enough evidence at the 5% significance level to conclude that the Dow percentage changes are not normally distributed? If so, what is there about the distribution that is not normal?

c. Use the Lilliefors test and the Q-Q plot to check for normality of percentage changes. Do these suggest the same conclusion as in part b? Explain.

d. Repeat parts a–c, but use data only from 2000 on. Do you get the same results as for the full data set?

Level B 43. Will the chi-square test ever conclude, at the 5% signif-

icance level, that data are not normally distributed when you know that they are? Check this with simulation.

Specifically, generate n normally distributed numbers with mean 100 and standard deviation 15. You can do this with the formula 5NORM.INV(RAND(),100,12). Do not freeze them; keep them random. Then run the chi-square normality test on the random numbers. Because the chi-square results are linked to the data, you will get new chi-square results every time you press F9 to recalculate. a. Using n 5 150, do you ever get a p-value less than

0.05? If so, what does such a p-value mean? Would you expect to get a few such p-values? Explain.

b. Repeat part a using n 5 1000. Do the results change in any qualitative way?

c. Repeat parts a and b, but use the Lilliefors test instead of the chi-square test. Do you get the same basic results?

44. Repeat the previous problem but with a non-normal population. Specifically, generate n random num- bers from a fifty-fifty mixture of two normal dis- tributions with respective means 90 and 110 and common standard deviation 10. You can do this wi th the formula 5 IF(RAND()<0.5,NORM. INV(RAND(),90,10),NORM.INV(RAND(),110,10)) This is not a normal distribution because it has two peaks.)

45. The file P09_45.xlsx contains measurements of ounces in randomly selected cans from a soft-drink filling machine. These cans reportedly contain 12 ounces, but because of natural variation, the actual amounts differ slightly from 12 ounces. a. Can the company legitimately state that the amounts

in cans are normally distributed? b. Assuming that the distribution is normal with the mean

and standard deviation found in this sample, calculate the probability that at least half of the next 100 cans filled will contain less than 12 ounces.

c. If the test in part a indicated that the data are not nor- mally distributed, how might you calculate the proba- bility requested in part b?

46. The chi-square test for normality discussed in this sec- tion is far from perfect. If the sample is too small, the test tends to accept the null hypothesis of normality for any population distribution even remotely bell-shaped; that is, it is not powerful in detecting non-normality. On the other hand, if the sample is very large, it will tend to reject the null hypothesis of normality for any data set.6 Check this by using simulation. First, simulate data from a normal distribution using a large sample size. Is there a good chance that the null hypothesis will (wrongly) be rejected? Then simulate data from a non-normal dis- tribution (uniform, say, or the mixture in Problem 44) using a small sample size. Is there a good chance that the null hypothesis will (wrongly) not be rejected? Sum- marize your findings in a short report.

6 Actually, all of the tests for normality suffer from this latter problem.

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9-6 chi-Square test for Independence 4 0 1

9-6 Chi-Square Test for Independence The test we discuss in this section, like one of the tests for normality from the previous section, uses the name “chi-square.” However, this test, called the chi-square test for independence, has an entirely different objective. It is used in situations where a popu- lation is categorized in two different ways. For example, people might be characterized by their smoking habits and their drinking habits. The question then is whether these two attributes are independent in a probabilistic sense. They are independent if information on a person’s drinking habits is of no use in predicting the person’s smoking habits (and vice versa). In this particular example, however, you might suspect that the two attributes are dependent. In particular, you might suspect that heavy drinkers are more likely (than non- heavy drinkers) to be heavy smokers, and you might suspect that nondrinkers are more likely (than drinkers) to be nonsmokers. The chi-square test for independence enables you to test this empirically.

The null hypothesis for this test is that the two attributes are independent. Therefore, statistically significant results are those that indicate some sort of dependence. As always, this puts the burden of proof on the alternative hypothesis. In the smoking–drinking exam- ple, we will continue to believe that smoking and drinking habits are unrelated—that is, independent—unless there is sufficient evidence from the data that they are dependent. Furthermore, even if we are able to conclude that they are dependent, the test itself does not indicate the form of dependence. It could be that heavy drinkers tend to be nonsmok- ers, and nondrinkers tend to be heavy smokers. Although this is unlikely, it is definitely a form of dependence. The only way we can decide which form of dependence exists is to look closely at the data.

The data for this test consist of counts in various combinations of categories. These are usually arranged in a rectangular contingency table, also called a cross-tabs, or, using Excel terminology, a pivot table. For example, if there are three smoking categories and three drinking categories, the table will have three rows and three columns, for a total of nine cells. The count in a cell is the number of observations in that particular combination of categories. The following example illustrates this data setup and the resulting analysis.

Rejecting independence does not indicate the form of dependence. To see this, you must look more closely at the data.

The chi-square test for independence is based on the counts in a contingency (or cross-tabs) table. It tests whether the row variable is probabilistically indepen- dent of the column variable.

EXAMPLE

9.7 RELATIONSHIP BETWEEN DEMANDS FOR DESKTOPS AND LAPTOPS AT BIG OFFICE

Big Office, a chain of large office supply stores, sells a variety of Windows and Mac laptops. Company executives want to know whether the demands for these two types of computers are related in any way. They might act as complementary products, where high demand for Windows laptops accompanies high demand for Mac laptops (computers in general are hot), they might act as substitute products (demand for one takes away demand for the other), or their demands might be unrelated. Because of limitations in its information system, Big Office does not have the exact demands for these products. However, it does have daily information on categories of demand, listed in aggregate (that is, over all stores). These data appear in Figure 9.20. (See the file Laptop Demand.xlsx.) Each day’s demand for each type of computer is categorized as Low, Medium Low, Medium High, or High. The table is based on 250 days, so that the counts add to 250. The individual counts show, for example, that demand was high for both Windows and Mac laptops on 11 of the 250 days. For convenience, the row and column totals are provided in the margins. Based on these data, can Big Office conclude that demands for these two products are independent?

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4 0 2 c h a p t e r 9 h y p o t h e s i s te s t i n g

Figure 9.20 Laptop Demand Data 1

2 3 4 5 6 7 8 9

A B C D E F G Demands at Big Office

Windows laptops Low Medium Low Medium High High

Mac laptops Low 4 17 17 5 43 Medium Low 8 23 22 27 80 Medium High 16 20 14 20 70 High 10 17 19 11 57

38 77 72 63 250

Objective To use the chi-square test of independence to test whether demand for Windows laptops is independent of demand for Mac laptops.

Solution The idea of the test is to compare the actual counts in the table with what would be expected under independence. If the actual counts are sufficiently far from the expected counts, the null hypothesis of independence can be rejected. The distance measure used to check how far apart they are, shown in Equation (9.9), is essentially the same chi-square statistic used in the chi-square test for normality. Here, Oij is the actual count in cell i, j (row i, column j), Eij is the expected count for this cell assuming inde- pendence, and the sum is over all cells in the table. If this test statistic is sufficiently large, the independence hypothesis can be rejected. (We provide more details of the test shortly.)

Test Statistic for Chi-Square Test for Independence

chi-square test statistic 5 a ij 1Oij 2 Eij22>Eij (9.9)

Expected Counts Assuming Row and Column Independence

Eij 5 RiCj > N (9.10)

What is expected under independence? The totals in row 9 indicate that demand for Windows laptops was low on 38 of the 250 days. Therefore, if you had to estimate the probability of low demand for Windows laptops, your estimate would be 38>250 5 0.152. Now, if demands for the two products were independent, you should arrive at this same estimate from the data in any of rows 5 through 8. That is, a probability estimate for Windows laptops should be the same regardless of the demand for Mac laptops. The probability estimate of low Windows demand from row 5, for example, is 4>43 5 0.093. Sim- ilarly, for rows 6, 7, and 8 it is 8>80 5 0.100, 16>70 5 0.229, and 10>57 5 0.175, respectively. These calculations provide some evidence that Windows and Mac laptops act as substitute products—the probability of low Windows demand is larger when Mac demand is medium high or high than when it is low or medium low.

This reasoning is the basis for calculating the Eijs. Specifically, it can be shown that the relevant formula for Eij is given by Equation (9.10), where Ri is the row total in row i, Ci is the total in column j, and N is the number of observations. For example, E11 for these data is 431382 >250 5 6.536, which is slightly larger than the corresponding observed count, O11 5 4.

The results appear in Figure 9.21. (Neither a template nor StatTools is required.) The formulas shown in the figure follow directly from Equations (9.9) and (9.10). The p-value of the test, 0.045, can be interpreted in the usual way. Specifically, the null hypothesis of independence can be rejected at the 5% or 10% significance levels, but not at the 1% level. There is a fairly strong evidence that the demands for the two products are not independent.

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9-6 chi-Square test for Independence 4 0 3

If the alternative hypothesis of dependence is accepted, the output in Figure 9.22 can be used to examine its form. These two tables show the counts as percentages of row totals and as percentages of column totals. If the demands were independent, the rows of this first table should be identical, and the columns of the second table should be identical. This is because each row in the first table shows the distribution of Windows demand for a given category of Mac demand, whereas each column in the second table shows the distribution of Mac demand for a given category of Windows demand. A close study of these percentages again provides some evidence that the two products act as substitutes, but the evidence is not overwhelming.

Figure 9.21 Output for Chi-Square Test

Test of independence

Expected counts assuming independence

Low Medium Low Medium High High

Low 6.54

12.16 10.64

8.66

Medium High 12.38 23.04 20.16 16.42

High 10.84 20.16 17.64 14.36

Formula in cell C15 (copied to table) =$G5*C$9/$G$9

Formula in cell C22 (copied to table) =(C15�C5)^2/C15

Medium Low 13.24 24.64 21.56 17.56

Low 0.98 1.42 2.70 0.21

Medium High 1.72 0.05 1.88 0.41

High 3.14 2.32 0.32 0.79

Medium Low 1.07 0.11 0.11 0.02

Low Medium Low Medium High High

17.242 9

0.045

=SUM(C22:F25) =(4�1)*(4�1) =CHISQ.DIST.RT(B27,B28)

Table of (O�E)^2/E values

Test statistic Degrees of freedom p-value

J K L M N O P Q

Tables of counts expressed as percentages of rows or of columns are useful for judging the form (and extent) of any possible dependence.

Figure 9.22 Counts Shown as Percentages 11

Counts as % of rows

Counts as % of columns

Low Medium Low Medium High High

9.3%

10.0% 22.9% 17.5%

39.5% 27.5% 20.0% 33.3%

11.6% 33.8% 28.6% 19.3%

39.5% 28.8% 28.6% 29.8%

B C D E F

Low High

10.5% 26.3%

23.6% 26.4%

7.9%

17.5%

22.1% 22.1%

High

HighMedium High

Medium High

Medium Low

Low

Finally, it is worth noting that the table of counts necessary for the chi-square test of independence can be a pivot table. For example, the pivot table in Figure 9.23 shows counts of the Married and Own Home attributes. (For Married, 1 means married, 0 means unmarried, and for Own Home, 1 means a home owner, 0 means not a home owner. This pivot table is based on the data in the Catalog Marketing.xlsx file from Chapter 2.) To see whether these two attributes are independent, the chi-square test would be performed on the table of counts. You might want to check that the p-value for the test is 0.000 (to three decimals), so that Married and Own Home are definitely not independent.

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4 0 4 c h a p t e r 9 h y p o t h e s i s te s t i n g

Problems Solutions for problems whose numbers appear within a colored box can be found in the Student Solution Files.

Level A 47. The file P08_46.xlsx contains data on 400 orders

placed to ElecMart company over a period of several months. For each order, the file lists the time of day, the type of credit card used, the region of the country where the customer resides, and others. (This is the same data set used in Example 3.4 of Chapter 3.) Use a chi-square test for independence to see whether the following variables are independent. If the variables appear to be related, discuss the form of dependence you see. a. Time and Region b. Region and Buy Category c. Gender and Card Type

48. The file P08_17.xlsx categorizes 250 randomly selected consumers on the basis of their gender, their age, and their preference for our brand or a competitor’s brand of a low-calorie soft drink. Use a chi-square test for inde- pendence to see whether the drink preference is inde- pendent of gender, and then whether it is independent of age. If you find any dependence, discuss its nature.

49. The file P02_11.xlsx contains data on 148 houses that were recently sold. Two variables in this data set are the selling price of the house and the number of bedrooms in the house. We want to use a chi-square test for indepen- dence to see whether these two variables are indepen- dent. However, this test requires categorical variables, and Selling Price is essentially continuous. Therefore, to run the test, first divide the prices into several categories: less than 120, 120 to 130, 130 to 140, and greater than 140. Then run the test and report your results.

Level B 50. The file P03_50.xlsx contains annual salaries for all

NBA basketball players in each of five seasons.

a. Using only the data for the most recent season (2008– 2009), check whether there is independence between position and salary. To do this, first change any hyphenated position such as C-F to the first listed, in this case C. (Presumably, this is the player’s primary position.) Then make Salary categorical with four cat- egories: the first is all salaries below the first quartile, the second is all salaries from the first quartile to the median, and so on. Explain your findings.

b. Repeat part a but with a Yes/No playoff team cate- gorization instead of position. The playoff teams in that season were Atlanta, Boston, Chicago, Cleve- land, Dallas, Denver, Detroit, Houston, Los Angeles Lakers, Miami, New Orleans, Orlando, Philadelphia, Portland, San Antonio, and Utah.

51. The file P09_51.xlsx contains data on 1000 randomly selected Walmart customers. The data set includes demographic variables for each customer as well as their salaries and the amounts they have spent at Walmart during the past year. a. A lookup table in the file suggests a way to categorize

the salaries. Use this categorization and chi-square tests of independence to see whether Salary is inde- pendent of (1) Age, (2) Gender, (3) Home, or (4) Mar- ried. Discuss any types of dependence you find.

b. Repeat part a, replacing Salary with Amount Spent. First you must categorize Amount Spent. Create four catego- ries for Amount Spent based on the four quartiles. The first category is all values of Amount Spent below the first quartile of Amount Spent, the second category is between the first quartile and the median, and so on.

52. The file P09_52.xlsx contains data on close to 10,000 customers from several large cities in the United States. The variables include the customers’ gender and their first choice among several types of movies. Perform chi- square tests of independence to test whether the follow- ing variables are related. If they are, discuss the form of dependence you see. a. State and First Choice b. City and First Choice c. Gender and First Choice

Figure 9.23 Using a Pivot Table for a Chi-Square Test

Count Married 0 1 Grand Total

Grand Total

Own Home

0 1

307 177 484

191 325 516

498 502

1000

9-7 Conclusion The concepts and procedures in this chapter form a cornerstone in both applied and theoretical statistics. Of particular impor- tance is the interpretation of a p-value, especially because p-values are reported in virtually all statistical software packages. A p-value summarizes the evidence in support of an alternative hypothesis, which is usually the hypothesis an analyst is trying to prove. Small p-values provide support for the alternative hypothesis, whereas large p-values provide little or no support for it.

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9-7 conclusion 4 0 5

Although hypothesis testing continues to be an important tool for analysts, it is important to note its limitations, particularly in business applications. First, given a choice between a confidence interval for some population parameter and a test of this parameter, we generally favor the confidence interval. For example, a confidence interval not only indicates whether a mean difference is 0, but it also provides a plausible range for this difference. Second, many business decision problems cannot be han- dled adequately with hypothesis-testing procedures. Either they ignore important cost information or they treat the consequences of incorrect decisions (type I and type II errors) in an inappropriate way. Finally, the statistical significance at the core of hypoth- esis testing is sometimes quite different from the practical significance that is of most interest to business managers.

Summary of Key Terms TERM EXPLANATION EXCEL PAGES EQUATION Null hypothesis Hypothesis that represents the current

thinking or status quo 365

Alternative hypothesis Typically, the hypothesis the analyst is try- ing to prove or research hypothesis

365

One-tailed test Test where values in only one direction will lead to rejection of the null hypothesis

367

Two-tailed test Test where values in both directions will lead to rejection of the null hypothesis

367

Type I error Error committed when null hypothesis is true but is rejected

367

Type II error Error committed when null hypothesis is false but is not rejected

367

Significance level The probability of a type I error an analyst chooses

368

Rejection region Sample results that lead to rejection of null hypothesis

368

Statistically significant results

Sample results that lead to rejection of null hypothesis

368

p-value Probability of observing a sample result at least as extreme as the one actually observed

369

Power Probability of correctly rejecting the null when it is false

371

t test for a population mean

Test for a mean from a single population Excel formulas in hypothe- sis test template.xlsx file

372 9.1

z test for a population proportion

Test for a proportion from a single population

Excel formulas in hypothe- sis test template.xlsx file

377 9.2

t test for difference between means from paired samples

Test for the difference between two pop- ulation means when samples are paired in a natural way

Excel formulas in hypothe- sis test template.xlsx file

379 9.3

t test for difference between means from independent samples

Test for the difference between two population means when samples are independent

Excel formulas in hypothe- sis test template.xlsx file

379 9.4, 9.5

F test for equality of two variances

Test for equality of two population variances, used to check an assumption of two-sample t test for difference between means

Excel formulas in hypothe- sis test template.xlsx file

387

F distribution Skewed distribution useful for testing equality of variances

F.DIST, F.INV, and other F functions

387

z test for difference between proportions

Test for difference between similarly defined proportions from two populations

Excel formulas in hypothe- sis test template.xlsx file

388 9.6, 9.7

Tests for normality Tests to check whether a population is normally distributed; possibilities include chi-square test, Lilliefors test, and Q-Q plot

StatTools/Normality Tests 395 9.8

Chi-square test for independence

Test to check whether two attributes are probabilistically independent

Excel formulas 402 9.9, 9.10

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4 0 6 c h a p t e r 9 h y p o t h e s i s te s t i n g

Problems Solutions for problems whose numbers appear within a colored box can be found in the Student Solution Files.

Conceptual Questions C.1. Suppose you are testing the null hypothesis that a

mean equals 75 versus a two-tailed alternative. If the true (but unknown) mean is 80, what kind of error might you make? When will you not make this error?

C.2. Suppose you hear the claim that a given test, such as the chi-square test for normality, is not very power- ful. What exactly does this mean? If another test, such as the Lilliefors test, is claimed to be more powerful, how is it better than the less powerful test?

C.3. Explain exactly what it means for a test statistic to fall in the rejection region.

C.4. Give an example of when a one-sided test on a population mean would make more sense than a two-tailed test. Give an example of the opposite. In general, why do we say that there is no statistical way to decide whether a test should be run as a one-tailed test or a two-tailed test?

C.5. For any given hypothesis test, that is, for any specifi- cation of the null and alternative hypotheses, explain why you could make only a type I error or a type II error, but not both. When would you make a type I error? When would you make a type II error? Answer as generally as possible.

C.6. What are the null and alternative hypotheses in the chi- square or Lilliefors test for normality? Where is the burden of proof? Might you argue that it should go in the other direction? Explain.

C.7. We didn’t discuss the role of sample size in this chapter as thoroughly as we did for confidence intervals in the previous chapter, but more advanced books do include sample size formulas for hypothesis testing. Consider the situation where you are testing the null hypothesis that a population mean is less than or equal to 100 versus a one- tailed alternative. A sample size formula might indicate the sample size needed to make the power at least 0.90 when the true mean is 103. What are the trade-offs here? Essentially, what is the advantage of a larger sample size?

C.8. Suppose that you wish to test a researcher’s claim that the mean height in meters of a normally distributed population of rosebushes at a nursery has increased from its commonly accepted value of 1.60. To carry out this test, you obtain a random sample of size 150 from this population. This sample yields a mean of 1.80 and a standard deviation of 1.30. What are the appropriate null and alternative hypotheses? Is this a one-tailed or two-tailed test?

C.9. Suppose that you wish to test a manager’s claim that the proportion of defective items generated by a par- ticular production process has decreased from its long- run historical value of 0.30. To carry out this test, you

obtain a random sample of 300 items produced through this process. The test indicates a p-value of 0.01. What exactly is this p-value telling you? At what levels of significance can you reject the null hypothesis?

C.10. Suppose that a 99% confidence interval for the propor- tion p of all Lakeside residents whose annual income exceeds 80,000 extends from 0.10 to 0.18. The con- fidence interval is based on a random sample of 150 Lakeside residents. Using this information and a 1% significance level, you wish to test H0: p 5 0.08 versus Ha: p ? 0.08. Based on the given information, are you able to reject the null hypothesis? Why or why not?

C.11. Suppose that you are performing a one-tailed hypoth- esis test. “Assuming that everything else remains constant, a decrease in the test’s level of significance 1a2 leads to a higher probability of rejecting the null hypothesis.” Is this statement true or false? Explain your reasoning.

C.12. Can pleasant aromas help people work more effi- ciently? Researchers conducted an investigation to answer this question. Fifty students worked a paper- and-pencil maze ten times. On five attempts, the stu- dents wore a mask with floral scents. On the other five attempts, they wore a mask with no scent. The 10 tri- als were performed in random order and each used a different maze. The researchers found that the subjects took less time to complete the maze when wearing the scented mask. Is this an example of an observational study or a controlled experiment? Explain.

Level A 53. The file P09_53.xlsx contains the number of days 44

mothers spent in the hospital after giving birth (in the year 2005). Before health insurance rules were changed (the change was effective January 1, 2005), the average number of days spent in a hospital by a new mother was two days. For a 5% level of significance, do the data in the file indicate (the research hypothesis) that women are now spending less time in the hospital after giving birth than they were prior to 2005? Explain your answer in terms of the p-value for the test.

54. Eighteen readers took a speed-reading course. The file P09_54.xlsx contains the number of words that they could read before and after the course. Test the alterna- tive hypothesis at the 5% significance level that reading speeds have increased, on average, as a result of the course. Explain your answer in terms of the p-value. Do you need to assume that reading speeds (before and after) are normally distributed? Why or why not?

55. Statistics have shown that a child 0 to 4 years of age has a 0.0002 probability of getting cancer in any given year. Assume that during each of the last seven years there have been 100 children ages 0 to 4 years whose parents work in a university’s business school. Four of these children have gotten cancer. Use this evidence to

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9-7 conclusion 4 0 7

test whether the incidence of childhood cancer among children ages 0 to 4 years whose parents work at this business school exceeds the national average. State your hypotheses and determine the appropriate p-value.

56. African Americans in a St. Louis suburb sued the city claiming they were discriminated against in school- teacher hiring. Of the city’s population, 5.7% were Afri- can American; of 405 teachers in the school system, 15 were African American. Set up appropriate hypotheses and determine whether African Americans are underrep- resented. Does your answer depend on whether you use a one-tailed or two-tailed test? In discrimination cases, the Supreme Court always uses a two-tailed test at the 5% significance level. (Source: U.S. Supreme Court Case, Hazlewood v. City of St. Louis)

57. We hear that teenagers in today’s world spend too much time playing video games. Does this have a significant effect on the grades they earn at school? You could test by this dividing students into two groups, those whose current high school grade-point average (GPA) is 3.0 or above and those whose GPA is below 3.0. Then you could sample students from each of these groups, dis- cover the numbers of hours per week spent playing video games, and run a test to see whether high-GPA students average less time playing video games than low-GPA students. Run such a test on the (fictional) data in the file P09_57.xlsx and report the results. If the results are significant, does this prove that too many hours spent playing video games causes lower GPAs? Explain.

58. Sixty people have rated a new beer on a taste scale of 0 to 100. Their ratings are in the file P09_58.xlsx. Mar- keting has determined that the beer will be a success if the average taste rating exceeds 75. Using a 5% signifi- cance level, is there sufficient evidence to conclude that the beer will be a success? Discuss your result in terms of a p-value. Assume ratings are at least approximately normally distributed.

59. Fifty people were asked to rate a competitive beer on a taste scale of 0 to 100. Another 50 people were asked to rate our beer on the same taste scale. The file P09_59.xlsx contains the results. Do these data provide sufficient evidence to conclude, at the 1% significance level, that people believe our beer tastes better than the competi- tor’s? Assume ratings are at least approximately nor- mally distributed. Would you reach the same conclusion if only the data from the first 10 people in each group were used?.

60. Callaway is thinking about entering the golf ball market. The company will make a profit if its market share is more than 20%. A market survey indicates that 140 of 624 golf ball purchasers will buy a Callaway golf ball. a. Is this enough evidence to persuade Callaway to enter

the golf ball market? b. How would you make the decision if you were Cal-

laway management? Would you use hypothesis testing?

61. Sales of a new product will be profitable if the average of sales per store exceeds 100 per week. The product was test-marketed for one week at 10 stores, with the results listed in the file P09_61.xlsx. Assume that sales at each store are at least approximately normally distributed. a. Is this enough evidence to persuade the company to

market the new product? b. How would you make the decision if you were decid-

ing whether to market the new product? Would you use hypothesis testing?

62. A recent study concluded that children born to mothers who take Prozac tend to have more birth defects than children born to mothers who do not take Prozac. a. What do you think the null and alternative hypotheses

were for this study? b. If you were a spokesperson for Eli Lilly (the com-

pany that produces Prozac), how might you rebut the conclusions of this study?

Level B 63. Suppose that you are the state superintendent of a state’s

public schools. You want to know whether decreasing the class size in grades 1 through 3 will improve stu- dent performance. Explain how you would set up a test to determine whether decreased class size improves stu- dent performance. What hypotheses would you use in this experiment? (This was actually done and smaller class size did help, particularly with minority students.)

64. The file P02_35.xlsx contains data from a survey of 500 randomly selected households. Economic researchers would like to test for a significant difference between the mean annual income levels of the first household wage earners in the first (i.e., SW) and second (i.e., NW) sectors of this community. In fact, they intend to per- form similar hypothesis tests for the differences between the mean annual income levels of the first household wage earners from all other pairs of locations (i.e., first and third, first and fourth, second and third, second and fourth, and third and fourth). a. Before conducting any hypothesis tests on the differ-

ence between various pairs of mean income levels, perform a test for equal population variances in each pair of locations. For each pair, report a p-value and interpret its meaning.

b. Based on your conclusions in part a, which test sta- tistic should be used in performing a test for the exis- tence of a difference between population means?

c. Given your conclusions in part b, perform a test for the existence of a difference in mean annual income levels in each pair of locations. For each pair, report a p-value and interpret its meaning.

65. A group of 25 husbands and wives were chosen ran- domly. Each person was asked to indicate the most he or she would be willing to pay for a new car (assum- ing each had decided to buy a new car). The results are

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4 0 8 c h a p t e r 9 h y p o t h e s i s te s t i n g

shown in the file P09_65.xlsx. Can you accept the alter- native hypothesis that the husbands are willing to spend more, on average, than the wives at the 5% significance level? What is the associated p-value? Is it appropri- ate to use a paired-sample or independent-sample test? Does it make a difference? Explain your reasoning.

66. A company is concerned with the high cholesterol levels of many of its employees. To help combat the problem, it opens an exercise facility and encourages its employ- ees to use this facility. After a year, it chooses a random 100 employees who claim they use the facility regularly, and another 200 who claim they don’t use it at all. The cholesterol levels of these 300 employees are checked, with the results shown in the file P09_66.xlsx. a. Is this sample evidence “proof” that the exercise facil-

ity, when used, tends to lower the mean cholesterol level? Phrase this as a hypothesis-testing problem and do the appropriate analysis. Do you feel comfortable that your analysis answers the question definitively (one way or the other)? Why or why not?

b. Repeat part a, but replace “mean cholesterol level” with “percentage with level over 215.” (The company believes that any level over 215 is dangerous.)

c. What can you say about causality? Could you ever conclude from such a study that the exercise causes low cholesterol? Why or why not?

67. Suppose that you are trying to compare two popula- tions on some variable (GMAT scores of men versus women, for example). Specifically, you are testing the null hypothesis that the means of the two populations are equal versus a two-tailed hypothesis. Are the follow- ing statements correct? Why or why not? a. A given difference (such as five points) between sam-

ple means from these populations will probably not be considered statistically significant if the sample sizes are small, but it will probably be considered statisti- cally significant if the sample sizes are large.

b. Virtually any difference between the population means will lead to statistically significant sample results if the sample sizes are sufficiently large.

68. Continuing the previous problem, analyze part b in Excel as follows. Start with hypothetical population mean GMAT scores for men and women, along with population standard deviations. Enter these at the top of a worksheet. You can make the two means as close as you like, but not identical. Simulate a sample of men’s GMAT scores with your mean and standard deviation in column A. Do the same for women in column B. The sample sizes do not have to be the same, but you can make them the same. Then run the test for the difference between two means. (The point of this problem is that if the population means are fairly close and you pick rel- atively small sample sizes, the sample mean differences probably won’t be significant. If you find this, generate new samples of a larger sample size and redo the test. Now they might be significant. If not, try again with a

still larger sample size. Eventually, you should get sta- tistically significant differences.)

69. This problem concerns course scores (on a 02100 scale) for a large undergraduate computer programming course. The class is composed of both underclassmen (freshmen and sophomores) and upperclassmen (juniors and seniors). Also, the students can be categorized according to their previous mathematical background from previous courses as “low” or “high” mathematical background. The data for these students are in the file P09_69.xlsx. The variables are:

• Score: score on a 02100 scale

• Upper Class: 1 for an upperclassman, 0 otherwise

• High Math: 1 for a high mathematical background, 0 otherwise

For the following questions, assume that the students in this course represent a random sample from all college students who might take the course. This latter group is the population. a. Calculate a 90% confidence interval for the population

mean score for the course. Do the same for the mean of all upperclassmen. Do the same for the mean of all upperclassmen with a high mathematical background.

b. The professor believes he has enough evidence to “prove” the research hypothesis that upperclassmen score at least five points better, on average, than lowerclassmen. Do you agree? Answer by running the appropriate test.

c. If a “good” grade is one that is at least 80, is there enough evidence to reject the null hypothesis that the fraction of good grades is the same for students with low math backgrounds as those with high math back- grounds? Which do you think is more appropriate, a one-tailed or two-tailed test? Explain your reasoning.

70. A cereal company wants to see which of two promotional strategies, supplying coupons in a local newspaper or including coupons in the cereal package itself, is more effective. (In the latter case, there is a label on the pack- age indicating the presence of the coupon inside.) The company randomly chooses 80 Kroger’s stores around the country—all of approximately the same size and overall sales volume—and promotes its cereal one way at 40 of these sites, and the other way at the other 40 sites. (All are at different geographical locations, so local news- paper ads for one of the sites should not affect sales at any other site.) Unfortunately, as in many business exper- iments, there is a factor beyond the company’s control, namely, whether its main competitor at any particular site happens to be running a promotion of its own. The file P09_70.xlsx has 80 observations on three variables:

• Sales: number of boxes sold during the first week of the company’s promotion

• Promotion Type:1 if coupons are in local paper, 0 if coupons are inside box

• Competitor Promotion:1 if main competitor is run- ning a promotion, 0 otherwise

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9-7 conclusion 4 0 9

a. Based on all 80 observations, calculate (1) the differ- ence in sample mean sales between stores running the two different promotional types (and indicate which sample mean is larger), (2) the standard error of this difference, and (3) a 90% confidence interval for the population mean difference.

b. Test whether the population mean difference is zero (the null hypothesis) versus a two-tailed alternative. State whether you should accept or reject the null hypothesis, and why.

c. Repeat part b, but now restrict the population to stores where the competitor is not running a promotion of its own.

d. Based on data from all 80 observations, can you accept the (alternative) hypothesis, at the 5% level, that the mean company sales drop by at least 30 boxes when the competitor runs its own promotion (as opposed to not running its own promotion)?

e. We often use the term population without really think- ing what it means. For this problem, explain in words exactly what the population mean refers to. What is the relevant population?

71. There is a lot of concern about “salary compression” in universities. This is the effect of paying huge salaries to attract newly-minted Ph.D. graduates to university tenure-track positions and not having enough left in the budget to compensate tenured faculty as fully as they might deserve. In short, it is very possible for a new hire to make a larger salary than a person with many years of valuable experience. The file P09_71.xlsx contains (fictional but realistic) salaries for a sample of business school professors, some already tenured and some not yet through the tenure process. Formulate reasonable null and alternative hypotheses and then test them with this data set. Write a short report of your findings.

CASE 9.1 Regression Toward the Mean In Chapters 10 and 11, you will study regression, a method for relating one variable to other explanatory variables. However, the term regression has sometimes been used in a slightly different way, meaning “regression toward the mean.” The example often cited is of male heights. If a father is unusually tall, for example, his son will typically be taller than average but not as tall as the father. Similarly, if a father is unusually short, the son will typically be shorter than aver- age but not as short as the father. We say that the son’s height tends to regress toward the mean. This case illustrates how regression toward the mean can occur.

Suppose a company administers an aptitude test to all of its job applicants. If an applicant scores below some value, he or she cannot be hired immediately but is allowed to retake a similar exam at a later time. In the interim the applicant can presumably study to prepare for the second exam. If we focus on the applicants who fail the exam the first time and then take it a second time, we would probably expect them to score bet- ter on the second exam. One plausible reason is that they are more familiar with the exam the second time. However, we will rule this out by assuming that the two exams are suffi- ciently different from one another. A second plausible reason is that the applicants have studied between exams, which has a beneficial effect. However, we will argue that even if studying has no beneficial effect whatsoever, these applicants will tend to do better the second time around. The reason is regression toward the mean. All of these applicants scored unusually low on the first exam, so they will tend to regress toward the mean on the second exam—that is, they will tend to score higher.

You can employ simulation to demonstrate this phenom- enon, using the following model. Assume that the scores of all potential applicants are normally distributed with mean m and standard deviation s. Because we are assuming that any study- ing between exams has no beneficial effect, this distribution of scores is the same on the second exam as on the first. An

applicant fails the first exam if his or her score is below some cutoff value L. Now, we would certainly expect scores on the two exams to be positively correlated, with some correlation r. (This is the Greek letter “rho,” often used for a population cor- relation.) That is, if everyone took both exams, applicants who scored high on the first would tend to score high on the second, and those who scored low on the first would tend to score low on the second. (This isn’t regression to the mean, but simply that some applicants are better than others.)

Given this model, you can proceed by simulating many pairs of scores, one pair for each applicant. The scores for each exam should be normally distributed with parame- ters m and s, but the trick is to make them correlated. You can use our Binormal function to do this. (Binormal is short for bivariate normal.) This function is supplied in the file C09_01 .xlsm. (Binormal is not a built-in Excel function.) It takes a pair of means (both equal to m), a pair of standard deviations (both equal to s), and a correlation r as arguments, with the syntax = BINORMAL (means, stdevs, correlation). To enter the formula, highlight two adjacent cells such as B5 and C5, type the formula, and press Ctrl1Shift1Enter. (This is because Binormal is an array function.) Then copy and paste to generate similar values for other applicants.

Once you have generated pairs of scores for many appli- cants, you should ignore all pairs except for those where the score on the first exam is less than L. (Sorting is suggested here, but freeze the random numbers first.) For these pairs, test whether the mean score on the second exam is higher than on the first, using a paired-samples test. If it is, you have demonstrated regression toward the mean. As you will probably discover, however, the results will depend on the values of the parameters you choose for m, s, r, and L. You should experiment with these. Assuming that you are able to demonstrate regression toward the mean, can you explain intuitively why it occurs?

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4 1 0 c h a p t e r 9 h y p o t h e s i s te s t i n g

CASE 9.2 Friday Effect in the Stock Market Many people believe that there is a “Friday effect” in the stock market. They don’t necessarily spell out exactly what they mean by this, but there is a sense that stock prices tend to be lower on Fridays than on other days. Because stock prices are readily available on the Web, it should be fairly easy to test this (alternative) hypothesis empirically. Before

collecting data and running a test, however, you must decide exactly which hypotheses you want to test because there are several possibilities. Formulate at least two sets of null/alter- native hypotheses. Then gather some stock price data and test your hypotheses. Can you conclude that there is a statis- tically significant Friday effect in the stock market?

CASE 9.3 Removing Vioxx from the Market For years, the drug Vioxx, developed and marketed by Merck, was one of the blockbuster drugs on the market. One of a number of so-called Cox-2 anti-inflammatory drugs, Vioxx was considered by many people a miracle drug for alleviat- ing the pain from arthritis and other painful afflictions. Vioxx was marketed heavily on television, prescribed by most phy- sicians, and used by an estimated two million Americans.

All of that changed in October 2004, when the results of a large study were released. The study, which followed approximately 2600 subjects over a period of about 18 months, concluded that Vioxx use over a long period of time caused a significant increase in the risk of developing seri- ous heart problems. Merck almost immediately pulled Vioxx from the American market and doctors stopped prescribing it. On the basis of the study, Merck faced not only public embarrassment but the prospect of huge financial losses.

More specifically, the study had 1287 patients use Vioxx for an 18-month period, and it had another 1299 patients use a placebo over the same period. After 18 months, 45 of

the Vioxx patients had developed serious heart problems, whereas only 25 patients on the placebo developed such problems.

Given these results, would you agree with the conclu- sion that Vioxx caused a significant increase in the risk of developing serious heart problems? First, answer this from a purely statistical point of view, where significant means sta- tistically significant. What hypothesis should you test, and how should you run the test? When you run the test, what is the corresponding p-value? Next, look at it from the point of view of patients. If you were a Vioxx user, would these results cause you significant worry? After all, some of the subjects who took placebos also developed heart problems, and 45 might not be considered that much larger than 25. Finally, look at it from Merck’s point of view. Are the results practically significant to the company? What does it stand to lose? Develop an estimate, no matter how wild it might be, of the financial losses Merck might incur. Just think of all of those American Vioxx users and what they might do.

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