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We have seen that the sample mean is the point estimate of the population mean and the sample proportion is the point estimate of the population proportion. In much of this book we use point estimates to make statistical inferences about populations and processes. As mentioned in Chapter 4, these inferences are based on calculating probabilities. To calculate these probabilities, we use a certain type of probability distribution called a sampling distribution.

In this chapter we discuss the properties of two important sampling distributions—the sampling distribution of the sample mean and the sampling distribution of the sample proportion. In order to help explain these sampling distributions, we consider three previously introduced cases:

The Car Mileage Case: The automaker uses the properties of the sampling distribution of the sample mean and its sample of 50 mileages to provide convincing evidence that the new midsize model’s mean EPA combined city and highway mileage exceeds the tax credit standard of 31 mpg.

The Payment Time Case: The management consulting firm uses the properties of the sampling distribution of the sample mean and its sample of 65 payment times to provide strong evidence that the new electronic billing system has reduced the mean bill payment time by more than 50 percent.

The Cheese Spread Case: The food processing company uses the properties of the sampling distribution of the sample proportion and its survey results to provide extremely strong evidence that fewer than 10 percent of all current purchasers would stop buying the cheese spread if the new spout were used.

We also study a new case, The Risk Analysis Case, which introduces how to deal with risky situations in the contexts of game shows and investment strategy.

7.1: The Sampling Distribution of the Sample Mean

Introductory ideas and basic properties

Suppose that we are about to randomly select a sample of n measurements from a population of measurements having mean μ and standard deviation σ. Before we actually select the sample, there are many different samples of n measurements that we might potentially obtain. Because different samples generally have different sample means, there are many different sample means that we might potentially obtain. It follows that, before we draw the sample, the sample mean is a random variable.

The sampling distribution of the sample mean is the probability distribution of the population of all possible sample means obtained from all possible samples of the same size.

In order to illustrate the sampling distribution of the sample mean, we begin with an intuitive example.

EXAMPLE 7.1: The Risk Analysis Case: Game Shows and Stock Returns

Congratulations! You have just won the question and answer portion of a popular game show and will now be given an opportunity to select a grand prize. The game show host shows you a large revolving drum containing six identical white envelopes that have been thoroughly mixed in the drum. The six identical white envelopes contain six checks made out for grand prizes of 10, 20, 30, 40, 50, and 60 thousand dollars. Usually, a contestant reaches into the drum, selects an envelope, and receives the grand prize in the envelope. Tonight, however, is a special night. You will be given the choice of either selecting one envelope or selecting two envelopes and receiving the average of the grand prizes in the two envelopes. What should you do?

If you select one envelope, you are randomly selecting one grand prize from the population of six grand prizes. The probability that you will receive any particular grand prize is 1/6, and therefore the probability distribution of the population of grand prizes is as shown in Table 7.1 on the next page. Furthermore, a graph of this probability distribution is as shown is Figure 7.1(a). If we calculate the mean μ of the population of grand prizes, we find that

Figure 7.1: A Comparison of Individual Grand Prizes and Sample Mean Grand Prizes

Table 7.1: A Probability Distribution Describing the Population of Six Individual Grand Prizes

If, on the other hand, you select two envelopes, you are randomly selecting a sample of two grand prizes from the population of six grand prizes. The fifteen samples of two grand prizes that you might obtain, as well as the means of these samples, are summarized in Table 7.2(a). For example, if you select an envelope containing 20 thousand dollars and an envelope containing 40 thousand dollars, you will receive a sample mean grand prize of

Table 7.2: The Population of Sample Means

In order to find the probability distribution of the population of sample mean grand prizes, note that different sample mean grand prizes correspond to different numbers of samples. For example, since the sample mean grand prize of 30 thousand dollars corresponds to two out of fifteen samples—the sample (10, 50) and the sample (20, 40)—the probability of obtaining the sample mean grand prize of 30 thousand dollars is 2/15. If we analyze all of the sample mean grand prizes in a similar fashion, we find that the probability distribution of the population of sample mean grand prizes is as shown in Table 7.2(b). This distribution is the sampling distribution of the sample mean. A graph of this distribution is as shown in Figure 7.1(b).

We are now prepared to compare your two game show options. If we examine Figures 7.1(a) and (b), we see that, although the distribution of six individual grand prizes and the distribution of fifteen sample mean grand prizes seem to be centered over the same mean of 35 thousand dollars, the distribution of sample mean grand prizes looks more bell-shaped and less spread out than the distribution of individual grand prizes. In particular, whereas the potential individual grand prizes range from 10 thousand dollars to 60 thousand dollars, the potential sample mean grand prizes range from 15 thousand dollars to 55 thousand dollars. It follows that if you wish to reduce the variability of your potential grand prize winnings, you should select two envelopes rather than one envelope. Doing this will, for example, guarantee a sample mean grand prize of at least 15 thousand dollars and give a 13/15 probability of a sample mean grand prize of at least 25 thousand dollars. On the other hand, whereas the probability of a sample mean grand prize of at least 50 thousand dollars is only 2/15, the probability of an individual grand prize of at least 50 thousand dollars is 2/6.

Therefore,

if you are willing to take on more risk and go for the larger grand prize winnings, you should select one envelope.

Although most of us will never participate in a game show, the game show example illustrates a useful principle. This principle, which is very important in financial investing, is that if you carefully diversify your financial opportunities, you can reduce the variability of your potential financial returns and thus reduce your risk. For example, the year 1987 featured extreme volatility in the stock market, including a loss of over 20 percent of the market’s value on a single day. Figure 7.2(a) shows the percent frequency histogram of the percentage returns for the entire year 1987 for the population of all 1,815 stocks listed on the New York Stock Exchange. The mean of this population of percentage returns is −3.5 percent. Consider drawing a random sample of n = 5 stocks from the population of 1,815 stocks and calculating the mean return, of the sampled stocks. If we use a computer, we can generate all the different samples of five stocks that can be obtained (there are trillions of such samples) and calculate the corresponding sample mean returns. For example, it can be verified that the five stocks in one sample gave 1987 percentage returns of −50 percent, −12 percent, −7 percent, 14 percent, and 30 percent. This implies that the sample mean return was

Figure 7.2: The New York Stock Exchange in 1987: A Comparison of Individual Stock Returns and Sample Mean Returns

Figure 6.2 is adapted with permission from The American Association of Individual Investors Journal, by John K. Ford, “A Method for Grading 1987 Stock Recommendations,” March 1988, pp. 16–17.

A percent frequency histogram describing the population of all possible sample mean returns is given in Figure 7.2(b). Comparing Figures 7.2(a) and (b), we see that, although the histogram of individual stock returns and the histogram of sample mean returns are both bell-shaped and seem to be centered over the same mean of −3.5 percent, the histogram of sample mean returns looks considerably less spread out than the histogram of individual returns. In terms of investing in the stock market, a sample of 5 stocks is a portfolio of 5 stocks, and the sample mean return is the percentage return that an investor would realize if he or she invested equal amounts of money in the stocks in the portfolio. Therefore, Figure 7.2 illustrates that the variation among portfolio returns is considerably less than the variation among individual stock returns. Of course, one would probably not invest in the stock market by randomly selecting stocks. However, we have nevertheless illustrated an important conclusion reached by actual financial research: careful investment diversification can reduce investment return variability and thus reduce the investor’s exposure to risk.

Thus far we have considered a game show example and a stock return example. Together, these examples illustrate several important facts about randomly selecting a sample of n individual measurements from a population of individual measurements having mean μ and standard deviation σ. Specifically, it can be shown that

1 If the population of individual measurements is normally distributed, then the population of all possible sample means is also normally distributed. This is illustrated in Figures 7.2(a) and (b): Because the population of individual stock returns is (approximately) normally distributed, the population of all possible sample mean returns is also (approximately) normally distributed.

2 Even if the population of individual measurements is not normally distributed, there are circumstances when the population of all possible sample means is approximately normally distributed. This result is based on a theorem called the Central Limit Theorem and is discussed more fully in the next subsection. For now, note that the result is intuitively illustrated in Figures 7.1(a) and (b): Although the population of six individual grand prizes does not have a normal distribution (it has a uniform distribution), the population of 15 sample mean grand prizes has a distribution that looks somewhat like a normal distribution.

3 The mean, of the population of all possible sample means equals μ, the mean ofthe population of individual measurements. This is illustrated in both Figures 7.1 and Figure 7.2. That is, in each figure the distribution of individual population measurements and the distribution of all possible sample means are centered over the same mean μ. Note that μ equals 35 thousand dollars in Figure 7.1. in Figure 7.2 μ equals −3.5 percent.

4 The standard deviation, , of the population of all possible sample means is less than σ, the standard deviation of the population of individual measurements. This is also illustrated in both Figures 7.1 and 7.2. That is, in each figure the distribution of all possible sample means is less spread out than the distribution of individual population measurements. Intuitively, is smaller than σ because each possible sample mean is an average of n measurements. Thus, each sample mean averages out high and low sample measurements and can be expected to be closer to the population mean μ than many of the individual population measurements would be. It follows that the different possible sample means are more closely clustered around μ than are the individual population measurements. (Note that we will see that is smaller than σ only if the sample size n is greater than 1.)

In the game show and stock return examples, the populations under consideration are small enough that we can calculate the population mean μ. Usually, however, the population under consideration is so large that it would be either impossible or impractical to calculate the population mean. That is, in most situations the population mean is unknown and our objective is to estimate this mean. To do this, we randomly select a sample of n observations from the population and use the sample mean as the point estimate of the population mean. Furthermore, we can use theoretical properties about the probability distribution of the population of all possible sample means to make statistical inferences about the population mean. We have discussed some of these properties in the stock return case, and we will now explain how these properties relate to estimating the population mean.

To begin, unless we are extremely lucky, the sample mean that we obtain will not equal the population mean μ. However, the mean of the population of all possible sample means, is equal to μ. Therefore, we call the sample mean an unbiased point estimate of the population mean. This unbiasedness property says that, although most of the possible sample means that we might obtain are either above or below the population mean, there is no systematic tendency for the sample mean to overestimate or underestimate the population mean. That is, although we will randomly select only one sample, the unbiased sample mean is “correct on the average” in all possible samples.

In order to assess how close the different possible sample means are to the population mean, we consider the standard deviation of the population of all possible sample means. The following summary box gives a formula for and also summarizes other previously discussed facts about the probability distribution of the population of all possible sample means.

The Sampling Distribution of

Assume that the population from which we will randomly select a sample of n measurements has mean μ and standard deviation σ. Then, the population of all possible sample means

1 Has a normal distribution, if the sampled population has a normal distribution.

2 Has mean

3 Has standard deviation

The formula for in (3) holds exactly if the sampled population is infinite. If the sampled population is finite, this formula holds approximately under conditions to be discussed later in this section.

Stated equivalently, the sampling distribution of has mean has standard deviation (if the sampled population is infinite), and is a normal distribution (if the sampled population has a normal distribution).1

The third result in the summary box says that, if the sampled population is infinite, then

In words, , the standard deviation of the population of all possible sample means, equals σ, the standard deviation of the sampled population, divided by the square root of the sample size n. Furthermore, in addition to showing that is smaller than σ (assuming that the sample size n is larger than one), this formula for also says that decreases as n increases. That is, intuitively, as each possible sample averages more observations, the resulting different possible sample means will differ from each other by less and thus will become more closely clustered around the population mean. It follows that, if we take a larger sample, we are more likely to obtain a sample mean that is near the population mean.

We next use the car mileage case to illustrate the formula for . In this and several other examples we will assume that, although we do not know the true value of the population mean μ, we do know the true value of the population standard deviation σ. Here, knowledge of σ might be based on theory or history related to the population under consideration. For example, because the automaker has been working to improve gas mileages, we cannot assume that we know the true value of the population mean mileage μ for the new midsize model. However, engineering data might indicate that the spread of individual car mileages for the automaker’s midsize cars is the same from model to model and year to year. Therefore, if the mileages for previous models had a standard deviation equal to .8 mpg., it might be reasonable to assume that the standard deviation of the mileages for the new model will also equal .8 mpg. Such an assumption would, of course, be questionable, and in most real-world situations there would probably not be an actual basis for knowing σ. However, assuming that σ is known will help us to illustrate sampling distributions, and in later chapters we will see what to do when σ is unknown.

EXAMPLE 7.2: The Car Mileage Case

Part 1: Basic concepts

Consider the population of the mileages of all cars of the new midsize model type. If we define this population to be the population of the mileages of all cars that could potentially be produced, then, since the automaker could always make “one more car,” the population should be considered to be infinite. If we further assume that this infinite population is normally distributed with mean μ and standard deviation σ = .8 (see Figure 7.3(a)), and if the automaker will randomly select a sample of n cars and test them as prescribed by the EPA, then the population of all possible sample means is normally distributed with mean and standard deviation . In order to show that a larger sample is more likely to give a more accurate point estimate of μ, compare taking a sample of size with n = 5 with taking a sample of size n = 50 If n = 5, then

Figure 7.3: A Comparison of (1) the Population of All Individual Car Mileages, (2) the Sampling Distribution of the Sample Mean When n = 5, and (3) the Sampling Distribution of the Sample Mean When n = 50

and it follows (by the empirical rule) that 95.44 of all possible sample means are within plus or minus mpg of the population mean μ. If n = 50, then

and it follows that 95.44 of all possible sample means are within plus or minus mpg of the population mean μ. Therefore, if n = 50, the different possible sample means that the automaker might obtain will be more closely clustered around μ than they will be if n = 5 (see Figures 7.3(b) and (c)). This implies that the larger sample of size n = 50 is more likely to give a sample mean that is near μ.

Part 2: Statistical inference

Recall from Chapter 3 that the automaker has randomly selected a sample of n = 50 mileages, which has mean = 31.56. We now ask the following question: If the population mean mileage μ exactly equals 31 mpg (the minimum standard for the tax credit), what is the probability of observing a sample mean mileage that is greater than or equal to 31.56 mpg? To find this probability, recall from Chapter 2 that a histogram of the 50 mileages indicates that the population of all individual mileages is normally distributed. Assuming that the population standard deviation σ is known to equal .8 mpg, it follows that the sampling distribution of the sample mean is a normal distribution, with mean and standard deviation Therefore,

To find P(z ≥ 4.96), notice that the largest z value given in Table A.3 (page 863) is 3.99, which gives a right-hand tail area of .00003. Therefore, since P(z ≥ 3.99) = .00003, it follows that P(z ≥ 4.96) is less than .00003 (see Figure 7.4). The fact that this probability is less than .00003 says that, if μ equals 31, then fewer than 3 in 100,000 of all possible sample means are at least as large as the sample mean = 31.56 that we have actually observed. Therefore, if we are to believe that μ equals 31, then we must believe that we have observed a sample mean that can be described as a smaller than 3 in 100,000 chance. Since it is extremely difficult to believe that such a small chance would occur, we have extremely strong evidence that μ does not equal 31 and that μ is, in fact, larger than 31. This evidence would probably convince the federal government that the midsize model’s mean mileage μ exceeds 31 mpg and thus that the midsize model deserves the tax credit.

Figure 7.4: The Probability That ≥ 31.56 When μ = 31 in the Car Mileage Case

To conclude this subsection, it is important to make two comments. First, the formula follows, in theory, from the formula for the variance of the population of all possible sample means. The formula for Second, in addition to holding exactly if the sampled population is infinite, the formula holds approximately if the sampled population is finite and much larger than (say, at least 20 times) the size of the sample. For example, if we define the population of mileages of all new midsize cars to be the population of mileages of all cars that will actually be produced this year, then the popu lation is finite. However, the population would be very large—certainly at least as large as 20 times any reasonable sample size. For example, if the automaker produces 100,000 new midsize cars this year, and if we randomly select a sample of n = 50 of these cars, then the population size of 100,000 is larger than 20 times the sample size of 50 (which is 1,000). It follows that, even though the population is finite and thus the formula would not hold exactly, this formula would hold approximately. The exact formula for when the sampled population is finite is given in a technical note at the end of this section. It is important to use this exact formula if the sampled population is finite and less than 20 times the size of the sample. However, with the exception of the populations considered in the technical note and in Section 8.5, all of the remaining populations to be considered in this book will be either infinite or finite and at least 20 times the size of the sample. Therefore, it will be appropriate to use the formula .

Sampling a nonnormally distributed population: the Central Limit Theorem

We now consider what can be said about the sampling distribution of when the sampled population is not normally distributed. First, as previously stated, the fact that is still true. Second, as also previously stated, the formula is exactly correct if the sampled population is infinite and is approximately correct if the sampled population is finite and much larger than (say, at least 20 times as large as) the sample size. Third, an extremely important result called the Central Limit Theorem tells us that, if the sample size n is large, then the sampling distribution of is approximately normal, even if the sampled population is not normally distributed.

The Central Limit Theorem

If the sample size n is sufficiently large, then the population of all possible sample means is approximately normally distributed (with mean and standard deviation ), no matter what probability distribution describes the sampled population. Furthermore, the larger the sample size n is, the more nearly normally distributed is the population of all possible sample means.

Chapters 7 and 8

The Central Limit Theorem is illustrated in Figure 7.5 for several population shapes. Notice that as the sample size increases (from 2 to 6 to 30), the populations of all possible sample means become more nearly normally distributed. This figure also illustrates that, as the sample size increases, the spread of the distribution of all possible sample means decreases (remember that this spread is measured by which decreases as the sample size increases).

Figure 7.5: The Central Limit Theorem Says that the Larger the Sample Size Is, the More Nearly Normally Distributed Is the Population of All Possible Sample Means

How large must the sample size be for the sampling distribution of to be approximately normal? In general, the more skewed the probability distribution of the sampled population, the larger the sample size must be for the population of all possible sample means to be approximately normally distributed. For some sampled populations, particularly those described by symmetric distributions, the population of all possible sample means is approximately normally distributed for a fairly small sample size. In addition, studies indicate that, if the sample size is at least 30, then for most sampled populations the population of all possible sample means is approxi mately normally distributed. In this book, whenever the sample size n is at least 30, we will assume that the sampling distribution of is approximately a normal distribution. Of course, if the sampled population is exactly normally distributed, the sampling distribution of is exactly normal for any sample size.

We can see the shapes of sampling distributions such as those illustrated in Figure 7.5 by using computer simulation. Specifically, for a population having a particular probability distribution, we can have the computer draw a given number of samples of n observations, compute the mean of each sample, and arrange the sample means into a histogram. To illustrate this, consider Figure 7.6(a), which shows the exponential distribution describing the hospital emergency room interarrival times discussed in Example 6.11 (page 267). Figure 7.6(b) gives the results of a simulation in which MINITAB randomly selected 1,000 samples of five interarrival times from this exponential distribution, calculated the mean of each sample, and arranged the 1,000 sample means into a histogram. Figure 7.6(c) gives the results of a simulation in which MINITAB randomly selected 1,000 samples of 30 interarrival times from the exponential distribution, calculated the mean of each sample, and arranged the 1,000 sample means into a histogram. Note that, whereas the histogram in Figure 7.6(b) is somewhat skewed to the right, the histogram in Figure 7.6(c) appears approximately bell-shaped. Therefore, we might conclude that when we randomly select a sample of n observations from an exponential distribution, the sampling distribution of the sample mean is somewhat skewed to the right when n = 5 and is approximately normal when n = 30.

Figure 7.6: Simulating the Sampling Distribution of the Sample Mean When Sampling from an Exponential Distribution

EXAMPLE 7.3: The Payment Time Case

Recall that a management consulting firm has installed a new computer-based billing system in a Hamilton, Ohio, trucking company. Because of the previously discussed advantages of the new billing system, and because the trucking company’s clients are receptive to using this system, the management consulting firm believes that the new system will reduce the mean bill payment time by more than 50 percent. The mean payment time using the old billing system was approximately equal to, but no less than, 39 days. Therefore, if μ denotes the new mean payment time, the consulting firm believes that μ will be less than 19.5days. To assess whether μ is less than 19.5 days, the consulting firm has randomly selected a sample of n = 65 invoices processed using the new billing system and has determined the payment times for these invoices. The mean of the 65 payment times is = 18.1077 days, which is less than 19.5 days. Therefore, we ask the following question: If the population mean payment time is 19.5 days, what is the probability of observing a sample mean payment time that is less than or equal to 18.1077 days? To find this probability, recall from Chapter 2 that a histogram of the 65 payment times indicates that the population of all payment times is skewed with a tail to the right. However, the Central Limit Theorem tells us that, because the sample size n = 65 is large, the sampling distribution of is approximately a normal distribution with mean and standard deviation Assuming that the population standard deviation σ is known to be 4.2 days, equals It follows that

which is the area under the standard normal curve to the left of −2.67. The normal table tells us that this area equals .0038. This probability says that, if μ equals 19.5, then only .0038 of all possible sample means are at least as small as the sample mean = 18.1077 that we have actually observed. Therefore, if we are to believe that μ equals 19.5, we must believe that we have observed a sample mean that can be described as a 38 in 10,000 chance. It is very difficult to believe that such a small chance would occur, so we have very strong evidence that μ does not equal 19.5 and is, in fact, less than 19.5. We conclude that the new billing system has reduced the mean bill payment time by more than 50 percent.

Chapters 7 and 8

Unbiasedness and minimum-variance estimates

Recall that a sample statistic is any descriptive measure of the sample measurements. For instance, the sample mean is a statistic, and so are the sample median, the sample variance s2, and the sample standard deviation s. Not only do different samples give different values of , different samples also give different values of the median, s2, s, or any other statistic. It follows that, before we draw the sample, any sample statistic is a random variable, and

The sampling distribution of a sample statistic is the probability distribution of the population of all possible values of the sample statistic.

In general, we wish to estimate a population parameter by using a sample statistic that is what we call an unbiased point estimate of the parameter.

A sample statistic is an unbiased point estimate of a population parameter if the mean of the population of all possible values of the sample statistic equals the population parameter.

For example, we use the sample mean as the point estimate of the population mean μ because is an unbiased point estimate of μ. That is, . In words, the average of all the different possible sample means (that we could obtain from all the different possible samples) equals μ.

Although we want a sample statistic to be an unbiased point estimate of the population parameter of interest, we also want the statistic to have a small standard deviation (and variance). That is, we wish the different possible values of the sample statistic to be closely clustered around the population parameter. If this is the case, when we actually randomly select one sample and compute the sample statistic, its value is likely to be close to the value of the population parameter. Furthermore, some general results apply to estimating the mean μ of a normally distributed population. In this situation, it can be shown that both the sample mean and the sample median are unbiased point estimates of μ. In fact, there are many unbiased point estimates of μ. However, it can be shown that the variance of the population of all possible sample means is smaller than the variance of the population of all possible values of any other unbiased point estimate of μ. For this reason, we call the sample mean a minimum-variance unbiased point estimate of μ. When we use the sample mean as the point estimate of μ, we are more likely to obtain a point estimate close to μ than if we used any other unbiased sample statistic as the point estimate of μ. This is one reason why we use the sample mean as the point estimate of the population mean.

We next consider estimating the population variance σ2. It can be shown that if the sampled population is infinite, then
s2 is an unbiased point estimate of σ2. That is, the average of all the different possible sample variances that we could obtain (from all the different possible samples) is equal to σ2. This is why we use a divisor equal to n − 1 rather than n when we estimate σ2. It can be shown that, if we used n as the divisor when estimating σ2, we would not obtain an unbiased point estimate of σ2. When the population is finite, s2 may be regarded as an approximately unbiased estimate of σ2 as long as the population is fairly large (which is usually the case).

It would seem logical to think that, because s2 is an unbiased point estimate of σ2, s should be an unbiased point estimate of σ. This seems plausible, but it is not the case. There is no easy way to calculate an unbiased point estimate of σ. Because of this, the usual practice is to use s as the point estimate of σ (even though it is not an unbiased estimate).

This ends our discussion of the theory of point estimation. It suffices to say that in this book we estimate population parameters by using sample statistics that statisticians generally agree are best. Whenever possible, these sample statistics are unbiased point estimates and have small variances.

Technical Note: If we randomly select a sample of size n without replacement from a finite population of size N, then it can be shown that where the quantity is called the finite population multiplier. If the size of the sampled population is at least 20 times the size of the sample (that is, if N ≥ 20n), then the finite population multiplier is approximately equal to one, and approximately equals . However, if the population size N is smaller than 20 times the size of the sample, then the finite population multiplier is substantially less than one, and we must include this multiplier in the calculation of . For instance, in our initial game show example, where the standard deviation σ of the population of N = 6 grand prizes can be calculated to be 17.078, and where N = 6 is only three times the sample size n = 2, it follows that

We will see how this formula can be used to make statistical inferences in Section 8.5.

Exercises for Section 7.1

CONCEPTS

7.1 Suppose that we will randomly select a sample of four measurements from a larger population of measurements. The sampling distribution of the sample mean is the probability distribution of a population. In your own words, describe the units in this population.

7.2 Suppose that we will randomly select a sample of n measurements from a normally distributed population of measurements having mean μ and standard deviation σ. If we consider the sampling distribution of (that is, if we consider the population of all possible sample means):

a Describe the shape of the population of all possible sample means.

b Write formulas that express the central tendency and the variability of the population of all possible sample means. Explain what these formulas say in your own words.

7.3 Explain how the central tendency of the population of all possible sample means compares to the central tendency of the individual measurements in the population from which the sample will be taken.

7.4 Explain how the variability of the population of all possible sample means compares to the variability of the individual measurements in the population from which the sample will be taken. Assume here that the sample size is greater than 1. Intuitively explain why this is true.

7.5 What does the Central Limit Theorem tell us about the sampling distribution of the sample mean?

7.6 In your own words, explain each of the following terms:

a Unbiased point estimate.

b Minimum-variance unbiased point estimate.

METHODS AND APPLICATIONS

7.7 Suppose that we will take a random sample of size n from a population having mean μ and standard deviation σ. For each of the following situations, find the mean, variance, and standard deviation of the sampling distribution of the sample mean :

a μ = 10, σ = 2, n = 25

b μ = 500, σ = .5, n = 100

c μ = 3, σ = .1, n = 4

d μ = 100, σ = 1, n = 1,600

7.8 For each situation in Exercise 7.7, find an interval that contains (approximately or exactly) 99.73 percent of all the possible sample means. In which cases must we assume that the population is normally distributed? Why?

7.9 Suppose that we will randomly select a sample of 64 measurements from a population having a mean equal to 20 and a standard deviation equal to 4.

a Describe the shape of the sampling distribution of the sample mean . Do we need to make any assumptions about the shape of the population? Why or why not?

b Find the mean and the standard deviation of the sampling distribution of the sample mean .

c Calculate the probability that we will obtain a sample mean greater than 21; that is, calculate P( > 21). Hint: Find the z value corresponding to 21 by using and because we wish to calculate a probability about . Then sketch the sampling distribution and the probability.

d Calculate the probability that we will obtain a sample mean less than 19.385; that is, calculate P( < 19.385).

7.10 Suppose that the percentage returns for a given year for all stocks listed on the New York Stock Exchange are approximately normally distributed with a mean of 12.4 percent and a standard deviation of 20.6 percent. Consider drawing a random sample of n = 5 stocks from the population of all stocks and calculating the mean return, , of the sampled stocks. Find the mean and the standard deviation of the sampling distribution of , and find an interval containing 95.44 percent of all possible sample mean returns.

7.11 THE BANK CUSTOMER WAITING TIME CASE WaitTime

Recall that the bank manager wants to show that the new system reduces typical customer waiting times to less than six minutes. One way to do this is to demonstrate that the mean of the population of all customer waiting times is less than 6. Letting this mean be μ, in this exercise we wish to investigate whether the sample of 100 waiting times provides evidence to support the claim that μ is less than 6.

For the sake of argument, we will begin by assuming that μ equals 6, and we will then attempt to use the sample to contradict this assumption in favor of the conclusion that μ is less than 6. Recall that the mean of the sample of 100 waiting times is = 5.46 and assume that σ, the standard deviation of the population of all customer waiting times, is known to be 2.47.

a Consider the population of all possible sample means obtained from random samples of 100 waiting times. What is the shape of this population of sample means? That is, what is the shape of the sampling distribution of ? Why is this true?

b Find the mean and standard deviation of the population of all possible sample means when we assume that μ equals 6.

c The sample mean that we have actually observed is = 5.46. Assuming that μ equals 6, find the probability of observing a sample mean that is less than or equal to = 5.46

d If μ equals 6, what percentage of all possible sample means are less than or equal to 5.46? Since we have actually observed a sample mean of = 5.46, is it more reasonable to believe that (1) μ equals 6 and we have observed one of the sample means that is less than or equal to 5.46 when μ equals 6, or (2) that we have observed a sample mean less than or equal to 5.46 because μ is less than 6? Explain. What do you conclude about whether the new system has reduced the typical customer waiting time to less than six minutes?

7.12 THE VIDEO GAME SATISFACTION RATING CASE VideoGame

Recall that a customer is considered to be very satisfied with his or her XYZ Box video game system if the customer’s composite score on the survey instrument is at least 42. One way to show that customers are typically very satisfied is to show that the mean of the population of all satisfaction ratings is at least 42. Letting this mean be μ, in this exercise we wish to investigate whether the sample of 65 satisfaction ratings provides evidence to support the claim that μ exceeds 42 (and, therefore, is at least 42).

For the sake of argument, we begin by assuming that μ equals 42, and we then attempt to use the sample to contradict this assumption in favor of the conclusion that μ exceeds 42. Recall that the mean of the sample of 65 satisfaction ratings is = 42.95, and assume that σ, the standard deviation of the population of all satisfaction ratings, is known to be 2.64.

a Consider the sampling distribution of for random samples of 65 customer satisfaction ratings. Use the properties of this sampling distribution to find the probability of observing a sample mean greater than or equal to 42.95 when we assume that μ equals 42.

b If μ equals 42, what percentage of all possible sample means are greater than or equal to 42.95? Since we have actually observed a sample mean of = 42.95, is it more reasonable to believe that (1) μ equals 42 and we have observed a sample mean that is greater than or equal to 42.95 when μ equals 42, or (2) that we have observed a sample mean that is greater than or equal to 42.95 because μ is greater than 42? Explain. What do you conclude about whether customers are typically very satisfied with the XYZ Box video game system?

7.13 In an article in the Journal of Management, Joseph Martocchio studied and estimated the costs of employee absences. Based on a sample of 176 blue-collar workers, Martocchio estimated that the mean amount of paid time lost during a three-month period was 1.4 days per employee with a standard deviation of 1.3 days. Martocchio also estimated that the mean amount of unpaid time lost during a three-month period was 1.0 day per employee with a standard deviation of 1.8 days.

Suppose we randomly select a sample of 100 blue-collar workers. Based on Martocchio’s estimates:

a What is the probability that the average amount of paid time lost during a three-month period for the 100 blue-collar workers will exceed 1.5 days?

b What is the probability that the average amount of unpaid time lost during a three-month period for the 100 blue-collar workers will exceed 1.5 days?

c Suppose we randomly select a sample of 100 blue-collar workers, and suppose the sample mean amount of unpaid time lost during a three-month period actually exceeds 1.5 days. Would it be reasonable to conclude that the mean amount of unpaid time lost has increased above the previously estimated 1.0 days? Explain.

7.14 When a pizza restaurant’s delivery process is operating effectively, pizzas are delivered in an average of 45 minutes with a standard deviation of 6 minutes. To monitor its delivery process, the restaurant randomly selects five pizzas each night and records their delivery times.

a For the sake of argument, assume that the population of all delivery times on a given evening is normally distributed with a mean of μ = 45 minutes and a standard deviation of σ = 6 minutes. (That is, we assume that the delivery process is operating effectively.)

(1) Describe the shape of the population of all possible sample means. How do you know what the shape is?

(2) Find the mean of the population of all possible sample means.

(3) Find the standard deviation of the population of all possible sample means.

(4) Calculate an interval containing 99.73 percent of all possible sample means.

b Suppose that the mean of the five sampled delivery times on a particular evening is = 55 minutes. Using the interval that you calculated in a(4), what would you conclude about whether the restaurant’s delivery process is operating effectively? Why?

7.2: The Sampling Distribution of the Sample Proportion

A food processing company markets a soft cheese spread that is sold in a plastic container with an “easy pour” spout. Although this spout works extremely well and is popular with consumers, it is expensive to produce. Because of the spout’s high cost, the company has developed a new, less expensive spout. While the new, cheaper spout may alienate some purchasers, a company study shows that its introduction will increase profits if fewer than 10 percent of the cheese spread’s current purchasers are lost. That is, if we let p be the true proportion of all current purchasers who would stop buying the cheese spread if the new spout were used, profits will increase as long as p is less than .10.

Suppose that (after trying the new spout) 63 of 1,000 randomly selected purchasers say that they would stop buying the cheese spread if the new spout were used. The point estimate of the population proportion p is the sample proportion This sample proportion says that we estimate that 6.3 percent of all current purchasers would stop buying the cheese spread if the new spout were used. Since equals .063, we have some evidence that the population proportion p is less than .10. In order to determine the strength of this evidence, we need to consider the sampling distribution of . In general, assume that we will randomly select a sample of n units from a population, and assume that a proportion p of all the units in the population fall into a particular category (for instance, the category of consumers who would stop buying the cheese spread). Before we actually select the sample, there are many different samples of n units that we might potentially obtain. The number of units that fall into the category in question will vary from sample to sample, so the sample proportion of units falling into the category will also vary from sample to sample. Therefore, we might potentially obtain many different sample proportions. It follows that, before we draw the sample, the sample proportion is a random variable. In the following box we give the properties of the probability distribution of this random variable, which is called the sampling distribution of the sample proportion .

The Sampling Distribution of the Sample Proportion

The population of all possible sample proportions

1 Approximately has a normal distribution, if the sample size n is large.

2 Has mean .

3 Has standard deviation

Stated equivalently, the sampling distribution of has mean has standard deviation and is approximately a normal distribution (if the sample size n is large).2

Property 1 in the box says that, if n is large, then the population of all possible sample proportions approximately has a normal distribution. Here, it can be shown that
n should be considered large if both np and n(1 −
p) are at least 5.3 Property 2, which says that , is valid for any sample size and tells us that , is an unbiased estimate of p. That is, although the sample proportion that we calculate probably does not equal p, the average of all the different sample proportions that we could have calculated (from all the different possible samples) is equal to p. Property 3, which says that

is exactly correct if the sampled population is infinite and is approximately correct if the sampled population is finite and much larger than (say, at least 20 times as large as) the sample size. Property 3 tells us that the standard deviation of the population of all possible sample proportions decrease as the sample size increases. That is, the larger n is, the more closely clustered are all the different sample proportions around the true population proportion. Finally, note that the formula for follows, in theory, from the formula for the variance of the population of all possible sample proportions. The formula for

EXAMPLE 7.4: The Cheese Spread Case

In the cheese spread situation, the food processing company must decide whether p, the proportion of all current purchasers who would stop buying the cheese spread if the new spout were used, is less than .10. In order to do this, remember that when 1,000 purchasers of the cheese spread are randomly selected, 63 of these purchasers say they would stop buying the cheese spread if the new spout were used. Noting that the sample proportion = .063 is less than .10, we ask the following question. If the true population proportion is .10, what is the probability of observing a sample proportion that is less than or equal to .063?

If p equals .10, we can assume that the sampling distribution of is approximately a normal distribution, because both np = 1,000(.10) = 100 and n(1 − p) = 1,000(1 − .10) = 900 are at least 5. Furthermore, the mean and standard deviation of the sampling distribution of are = p = .10 and

Therefore,

which is the area under the standard normal curve to the left of −3.90. The normal table tells us that this area equals .00005. This probability says that, if p equals .10, then only 5 in 100,000 of all possible sample proportions are at least as small as the sample proportion = .063 that we have actually observed. That is, if we are to believe that p equals .10, we must believe that we have observed a sample proportion that can be described as a 5 in 100,000 chance. It follows that we have extremely strong evidence that p does not equal .10 and is, in fact, less than .10. In other words, we have extremely strong evidence that fewer than 10 percent of current purchasers would stop buying the cheese spread if the new spout were used. It seems that introducing the new spout will be profitable.

Exercises for Section 7.2

CONCEPTS

7.15 What population is described by the sampling distribution of ?

7.16 Suppose that we will randomly select a sample of n units from a population and that we will compute the sample proportion of these units that fall into a category of interest. If we consider the sampling distribution of :

a If the sample size n is large, the sampling distribution of is approximately a normal distribution. What condition must be satisfied to guarantee that n is large enough to say that is normally distributed?

b Write formulas that express the central tendency and variability of the population of all possible sample proportions. Explain what each of these formulas means in your own words.

7.17 Describe the effect of increasing the sample size on the population of all possible sample proportions.

METHODS AND APPLICATIONS

7.18 In each of the following cases, determine whether the sample size n is large enough to say that the sampling distribution of is a normal distribution.

a p = .4, n = 100

b p = .1, n = 10

c p = .1, n = 50

d p = .8, n = 400

e p = .98, n = 1,000

f p = .99, n = 400

7.19 In each of the following cases, find the mean, variance, and standard deviation of the sampling distribution of the sample proportion .

a p = .5, n = 250

b p = .1, n = 100

c p = .8, n = 400

d p = .98, n = 1,000

7.20 For each situation in Exercise 7.19, find an interval that contains approximately 95.44 percent of all the possible sample proportions.

7.21 Suppose that we will randomly select a sample of n = 100 units from a population and that we will compute the sample proportion of these units that fall into a category of interest. If the true population proportion p equals .9:

a Describe the shape of the sampling distribution of . Why can we validly describe the shape?

b Find the mean and the standard deviation of the sampling distribution of .

c Calculate the following probabilities about the sample proportion . In each case sketch the sampling distribution and the probability.

(1)

(2)

(3)

7.22 In the July 29, 2001, issue of The Journal News (Hamilton, Ohio) Lynn Elber of the Associated Press reported on a study conducted by the Kaiser Family Foundation regarding parents’ use of television set V-chips for controlling their childrens’ TV viewing. The study asked parents who own TVs equipped with V-chips whether they use the devices to block programs with objectionable content.

a Suppose that we wish to use the study results to justify the claim that fewer than 20 percent of parents who own TV sets with V-chips use the devices. The study actually found that 17 percent of the parents polled used their V-chips.4 If the poll surveyed 1,000 parents, and if for the sake of argument we assume that 20 percent of parents who own V-chips actually use the devices (that is, p = .2), calculate the probability of observing a sample proportion of .17 or less. That is, calculate .

b Based on the probability you computed in part a, would you conclude that fewer than 20 percent of parents who own TV sets equipped with V-chips actually use the devices? Explain.

7.23 On February 8, 2002, the Gallup Organization released the results of a poll concerning American attitudes toward the 19th Winter Olympic Games in Salt Lake City, Utah. The poll results were based on telephone interviews with a randomly selected national sample of 1,011 adults, 18 years and older, conducted February 4–6, 2002.

a Suppose we wish to use the poll’s results to justify the claim that more than 30 percent of Americans (18 years or older) say that figure skating is their favorite Winter Olympic event. The poll actually found that 32 percent of respondents reported that figure skating was their favorite event.5 If, for the sake of argument, we assume that 30 percent of Americans (18 years or older) say figure skating is their favorite event (that is, p = .3), calculate the probability of observing a sample proportion of .32 or more; that is, calculate .

b Based on the probability you computed in a, would you conclude that more than 30 percent of Americans (18 years or older) say that figure skating is their favorite Winter Olympic event?

7.24 Quality Progress, February 2005, reports on improvements in customer satisfaction and loyalty made by Bank of America. A key measure of customer satisfaction is the response (on a scale from 1 to 10) to the question: “Considering all the business you do with Bank of America, what is your overall satisfaction with Bank of America?” Here, a response of 9 or 10 represents “customer delight.”

a Historically, the percentage of Bank of America customers expressing customer delight has been 48%. Suppose that we wish to use the results of a survey of 350 Bank of America customers to justify the claim that more than 48% of all current Bank of America customers would express customer delight. The survey finds that 189 of 350 randomly selected Bank of America customers express customer delight. If, for the sake of argument, we assume that the proportion of customer delight is p = .48, calculate the probability of observing a sample proportion greater than or equal to 189/350 = .54. That is, calculate .

b Based on the probability you computed in part (a), would you conclude that more than 48 percent of current Bank of America customers express customer delight? Explain.

7.25 Again consider the survey of 350 Bank of America customers discussed in Exercise 7.24, and assume that 48% of Bank of America customers would currently express customer delight. That is, assume p = .48. Find:

a The probability that the sample proportion obtained from the sample of 350 Bank of America customers would be within three percentage points of the population proportion. That is, find .

b The probability that the sample proportion obtained from the sample of 350 Bank of America customers would be within six percentage points of the population proportion. That is, find

7.26 Based on your results in Exercise 7.25, would it be reasonable to state that the survey’s “margin of error” is ±3 percentage points? ±6 percentage points?Explain.

7.27 A special advertising section in the July 20, 1998, issue of Fortune magazine discusses “outsourcing.” According to the article, outsourcing is “the assignment of critical, but noncore, business functions to outside specialists.” This allows a company to immediately bring operations up to best-in-world standards while avoiding huge capital investments. The article includes the results of a poll of business executives addressing the benefits of outsourcing.

a Suppose we wish to use the poll’s results to justify the claim that fewer than 26 percent of business executives feel that the benefits of outsourcing are either “less or much less than expected.” The poll actually found that 15 percent of the respondents felt that the benefits of outsourcing were either “less or much less than expected.”6 If 1,000 randomly selected business executives were polled, and if for the sake of argument, we assume that 20 percent of all business executives feel that the benefits of outsourcing are either less or much less than expected (that is, p = .20), calculate the probability of observing a sample proportion of .15 or less. That is, calculate .

b Based on the probability you computed in part a, would you conclude that fewer than 20 percent of business executives feel that the benefits of outsourcing are either “less or much less than expected”? Explain.

7.28 The July 20, 1998, issue of Fortune magazine reported the results of a survey on executive training that was conducted by the Association of Executive Search Consultants. The survey showed that 75 percent of 300 polled CEOs believe that companies should have “fast-track training programs” for developing managerial talent.7

a Suppose we wish to use the results of this survey to justify the claim that more than 70 percent of CEOs believe that companies should have fast-track training programs. Assuming that the 300 surveyed CEOs were randomly selected, and assuming, for the sake of argument, that 70 percent of CEOs believe that companies should have fast-track training programs (that is, p = .70), calculate the probability of observing a sample proportion of .75 or more. That is, calculate .

b Based on the probability you computed in part a, would you conclude that more than 70 percent of CEOs believe that companies should have fast-track training programs?Explain.

Chapter Summary

A sampling distribution is the probability distribution that describes the population of all possible values of a sample statistic. In this chapter we studied the properties of two important sampling distributions—the sampling distribution of the sample mean, and the sampling distribution of the sample proportion, .

Because different samples that can be randomly selected from a population give different sample means, there is a population of sample means corresponding to a particular sample size. The probability distribution describing the population of all possible sample means is called the sampling distribution of the sample mean, . We studied the properties of this sampling distribution when the sampled population is and is not normally distributed. We found that, when the sampled population has a normal distribution, then the sampling distribution of the sample mean is a normal distribution. Furthermore, the Central Limit Theorem tells us that, if the sampled population is not normally distributed, then the sampling distribution of the sample mean is approximately a normal distribution when the sample size is large (at least 30). We also saw that the mean of the sampling distribution of always equals the mean of the sampled population, and we presented formulas for the variance and the standard deviation of this sampling distribution. Finally, we explained that the sample mean is a minimum-variance unbiased point estimate of the mean of a normally distributed population.

We also studied the properties of the sampling distribution of the sample proportion . We found that, if the sample size is large, then this sampling distribution is approximately a normal distribution, and we gave a rule for determining whether the sample size is large. We found that the mean of the sampling distri bution of is the population proportion p, and we gave formulas for the variance and the standard deviation of this sampling distribution.

Finally, we demonstrated that knowing the properties of sampling distributions can help us make statistical inferences about population parameters. In fact, we will see that the properties of various sampling distributions provide the foundation for most of the techniques to be discussed in future chapters.

CHAPTER 8: Confidence Intervals

Chapter Outline

8.1 z-Based Confidence Intervals for a Population Mean: σ Known

8.2 t-Based Confidence Intervals for a Population Mean: σ Unknown

8.3 Sample Size Determination

8.4 Confidence Intervals for a Population Proportion

8.5 Confidence Intervals for Parameters of Finite Populations (Optional)

8.6 A Comparison of Confidence Intervals and Tolerance Intervals (Optional)

We have seen that the sample mean is the point estimate of the population mean and the sample proportion is the point estimate of the population proportion. In general, although a point estimate is a reasonable one-number estimate of a population parameter (mean, proportion, or the like), the point estimate will not—unless we are extremely lucky—equal the true value of the population parameter.

In this chapter we study how to use a confidence interval to estimate a population parameter. A confidence interval for a population parameter is an interval, or range of numbers, constructed around the point estimate so that we are very sure, or confident, that the true value of the population parameter is inside the interval.

By computing such an interval, we estimate—with confidence—the possible values that a population parameter might equal. This, in turn, can help us to assess—with confidence—whether a particular business improvement has been made or is needed.

In order to illustrate confidence intervals, we revisit several cases introduced in earlier chapters and also introduce some new cases. Specifically:

In the Car Mileage Case, we use a confidence interval to provide strong evidence that the mean EPA combined city and highway mileage for the automaker’s new midsize model meets the tax credit standard of 31 mpg.

In the Payment Time Case, we use a confidence interval to more completely assess the reduction in mean payment time that was achieved by the new billing system.

In the Marketing Research Case, we use a confidence interval to provide strong evidence that the mean rating of the new bottle design exceeds the minimum standard for a successful design.

In the Cheese Spread Case, we use a confidence interval to provide strong evidence that fewer than 10 percent of all current purchasers will stop buying the cheese spread if the new spout is used, and, therefore, that it is reasonable to use the new spout.

In the Marketing Ethics Case, we use a confidence interval to provide strong evidence that more than half of all marketing researchers disapprove of the actions taken in an “ultraviolet ink scenario.”

Sections 8.1 through 8.4 present confidence intervals for population means and proportions. These intervals are appropriate when the sampled population is either infinite or finite and much larger than (say, at least 20 times as large as) the size of the sample. The appropriate procedures to use when the population is not large compared to the sample are explained in optional Section 8.5. Finally, optional Section 8.6 compares confidence intervals and tolerance intervals.

8.1: z-Based Confidence Intervals for a Population Mean: σ Known

An introduction to confidence intervals for a population mean

We have seen that we use the sample mean as the point estimate of the population mean. A confidence interval for the population mean is an interval constructed around the sample mean so that we are very sure, or confident, that this interval contains the population mean. In order to illustrate a confidence interval, we consider the following intuitive example.

Chapter 8

EXAMPLE 8.1: The Car Mileage Case

Recall that an automaker has introduced a new midsize model and wishes to estimate the mean EPA combined city and highway mileage, μ, that would be obtained by all cars of this type. If the automaker can show that the population mean μ is at least 31 mpg, the government will give the automaker a tax credit. In order to estimate μ, the automaker has conducted EPA mileage tests on a random sample of 50 of its new midsize cars and has obtained the sample of mileages in Table 1.4 (page 10). The mean of this sample of mileages, which is , is the point estimate of μ. However, a sample mean will not—unless we are extremely lucky—equal the true value of a population mean. Therefore, although mpg suggests that μ is at least 31 mpg, it does not make us confident that μ is at least 31 mpg. Later in this section we will find that a confidence interval for μ is [31.34 mpg, 31.78 mpg]. This interval says that we are confident that the true mean mileage μ for the new midsize model is between 31.34 mpg and 31.78 mpg. Furthermore, since this interval estimates that the smallest μ might be is 31.34 mpg, we can be confident that μ exceeds 31 mpg, the minimum standard for the tax credit. As we will see later, the federal government might regard this confidence interval as convincing evidence that the midsize model deserves the tax credit.

When we find a confidence interval for a population mean, we base this interval on what is called a confidence level. This confidence level is a percentage (for example, 95 percent or 99 percent) that, intuitively, expresses how confident we are that the confidence interval contains the population mean. The exact meaning of a confidence level—as well as the confidence level for the interval of the previous example—will be discussed later in this section. First, however, we need to explain how a confidence level is used to find a confidence interval. To do this, we will begin in the car mileage example by showing how a confidence level based on a familiar percentage—95.44 percent—can be used to find a confidence interval. Then, we will generalize our discussions and show how any particular confidence level can be used to find a confidence interval. In addition, since it is simpler to present our initial discussions in terms of a smaller sample size, we will begin by considering a random sample of n = 5 car mileages, and we will then consider using any sample size.

Assume, therefore, that we will randomly select a sample of n = 5 new midsize cars and calculate the mean of the mileages that the cars obtain when tested as prescribed by the EPA. Also, assume that the population of all individual car mileages is normally distributed and that, although we do not know the true value of the population mean mileage μ, we do know that the true value of the population standard deviation σ is .8 mpg (as discussed on pages 285 and 286 of Chapter 7). A confidence interval for the population mean μ is based on the sampling distribution of the sample mean . We have seen in Chapter 7 that the sampling distribution of is the probability distribution of the population of all possible sample means that would be obtained from all possible samples of n = 5 car mileages. We have also seen that, if the assumptions discussed above hold, then the sampling distribution of is a normal distribution, centered at the unknown population mean μ (because ) and having standard deviation

We now reason as follows:

1 The Empirical Rule for a normally distributed population implies that the probability is .9544 that the sample mean will be within plus or minus of the population mean μ. (Note that, although it is generally best to carry more decimal places in intermediate calculations, we have rounded to one decimal place to make it easier to understand the logic to follow.)

2 Saying

is the same as saying

To see this, consider Table 8.1, which gives three samples and the means of these samples. Also, assume that (unknown to any human being) the true value of the population mean μ is 31.5. Then, as illustrated in Figure 8.1, because the sample mean is within .7 of μ = 31.5, the interval [31.3 ± .7] = [30.6, 32.0] contains μ. Similarly, since the sample mean is within .7 of μ = 31.5, the interval [31.7 ± .7] = [31.0, 32.4] contains μ. However, because the sample mean is not within .7 of μ = 31.5, the interval [32.5 ± .7] = [31.8, 33.2] does not contain μ.

Table 8.1: Three Samples of Five Mileages

Figure 8.1: Three 95.44 Percent Confidence Intervals for μ

3 Combining 1 and 2, we have the probability is .9544 that the sample mean will be such that the interval contains the population mean μ.

Statement 3 says that, before we randomly select the sample, there is a .9544 probability that we will obtain an interval that contains μ. In other words, 95.44 percent of all intervals that we might obtain contain μ, and 4.56 percent of these intervals do not contain μ. For this reason, we call the interval a 95.44 percent confidence interval for μ
. To better understand this interval, we must realize that, when we actually select the sample, we will observe one particular sample from the extremely large number of possible samples. Therefore, we will obtain one particular confidence interval from the extremely large number of possible confidence intervals. For example, suppose that when we actually select the sample of five cars and record their mileages, we obtain the leftmost sample of mileages in Table 8.1. Since the mean of our sample is , the 95.44 percent confidence interval for μ that it gives is

Because we do not know the true value of μ, we do not know for sure whether μ is contained in our interval. However, we are 95.44 percent confident that μ is contained in this interval. What we mean by this is that we hope that the interval [30.6, 32.0] is one of the 95.44 percent of all intervals that contain μ and not one of the 4.56 percent of all intervals that do not contain μ. Here we say that 95.44 percent is the confidence level associated with the confidence interval.

To practically interpret the confidence interval [30.6, 32.0], this interval says that we are 95.44 percent confident that μ is between 30.6 mpg and 32.0 mpg. Furthermore, since this interval estimates that μ could be as small as 30.6 mpg, the interval (based on a small sample of 5 mileages) does not make us 95.44 percent confident that μ is at least 31 mpg, the minimum standard for the tax credit. As described in Example 8.1, we will soon see that a confidence interval based on the sample of 50 mileages in Table 1.4 (page 10) makes us confident that μ is at least 31 mpg. In addition, we will see that we need not base a confidence interval on a 95.44 percent confidence level. Rather, we can base a confidence interval on any confidence level less than 100 percent.

A general confidence interval formula

We will now present a general formula for finding a confidence interval for a population mean. To do this, recall from the previous example that, before we randomly select the sample, the probability that the confidence interval

will contain the population mean is .9544. It follows that the probability that this confidence interval will not contain the population mean is .0456. In general, we denote the probability that a confidence interval for a population mean will not contain the population mean by the symbol
α (pronounced alpha). This implies that (1 − α), which we call the confidence coefficient, is the probability that the confidence interval will contain the population mean. We can base a confidence interval for a population mean on any confidence coefficient (1 − α) less than 1. However, in practice, we usually use two decimal place confidence coefficients, such as .95 or .99. To find a general formula for a confidence interval for a population mean μ, we assume that the sampled population is normally distributed, or the sample size n is large. Under these conditions, the sampling distribution of the sample mean is exactly (or approximately, by the Central Limit Theorem) a normal distribution with mean and standard deviation Then, in order to obtain a confidence interval that has a (1 − α) probability of containing μ, we find the normal point zα/2 that gives a right-hand tail area under the standard normal curve equal to α/2, and we find the normal point −zα/2 that gives a left-hand tail area under this curve equal to α/2 (see Figure 8.2). Noting from Figure 8.2 that the area under the standard normal curve between −zα/2 and zα/2 is (1 − α), it can be shown that the probability is (1 − α) that the sample mean will be within plus or minus units of the population mean μ. The quantity is called the margin of error when estimating μ by . If this margin of error is added to and subtracted from to form the interval

then this interval will contain the population mean with probability (1 − α). In other words, this interval is a confidence interval for μ based on a confidence coefficient of (1 − α), and hence we call this interval a 100(1 − α) percent confidence interval for the population mean. Here, 100(1 − α) percent is called the confidence level associated with the confidence interval. This confidence level is the percentage of the time that the confidence interval would contain the population mean if all possible samples were used to calculate this interval. (Note that we will formally justify the confidence interval formula at the end of this section.)

Figure 8.2: The Point zα/2

For example, suppose we wish to find a 95 percent confidence interval for the population mean. Since the confidence level is 95 percent, we have 100(1 − α) = 95. This implies that the confidence coefficient is (1 − α) = .95, which implies that α = .05 and α/2 = .025. Therefore, we need to find the normal point z.025. As shown in Figure 8.3, the area under the standard normal curve between −z.025 and z.025 is .95, and the area under this curve to the left of z.025 is .975. Looking up the area .975 in Table A.3 (page 863), we find that z.025 = 1.96. It follows that the interval

is a 95 percent confidence interval for the population mean μ. This means that if all possible samples were used to calculate this interval, 95 percent of the resulting intervals would contain μ.

Figure 8.3: The Point z.025

As another example, consider a 99 percent confidence interval for the population mean. Because the confidence level is 99 percent, we have 100(1 − α) = 99, and the confidence coefficient is (1 − α) = .99. This implies that α = .01 and α/2 = .005. Therefore, we need to find the normal point z.005. As shown in Figure 8.4, the area under the standard normal curve between −z.005 and z.005 is .99, and the area under this curve to the left of z.005 is .995. Looking up the area .995 in Table A.3, we find that z.005 = 2.575. It follows that the interval

is a 99 percent confidence interval for the population mean μ. This means that if all possible samples were used to calculate this interval, 99 percent of the resulting intervals would contain μ.

Figure 8.4: The Point z.005

To compare the 95 percent and 99 percent confidence intervals, notice that the margin of error used to compute the 99 percent interval is larger than the margin of error used to compute the 95 percent interval. Therefore, the 99 percent interval is the longer of these intervals. In general, increasing the confidence level (1) has the advantage of making us more confident that μ is contained in the confidence interval, but (2) has the disadvantage of increasing the margin of error and thus providing a less precise estimate of the true value of μ. Frequently, 95 percent confidence intervals are used to make conclusions. If conclusions based on stronger evidence are desired, 99 percent intervals are sometimes used.

Table 8.2 shows the confidence levels 95 percent and 99 percent, as well as two other confidence levels—90 percent and 98 percent—that are sometimes used to calculate confidence intervals. In addition, this table gives the values of α, α/2, and zα/2 that correspond to these confidence levels. The following box summarizes the formula used in calculating a 100(1 − α) percent confidence interval for a population mean μ.

Table 8.2: The Normal Point zα/2 for Various Levels of Confidence

A Confidence Interval for a Population Mean μ: σ Known

Suppose that the sampled population is normally distributed. Then a 100(1 − α) percent confidence interval for μ
is

This interval is also approximately correct for non-normal populations if the sample size is large (at least 30).

The confidence interval in the summary box is based on the normal distribution and assumes that the true value of the population standard deviation σ is known. Therefore, in the examples to follow we will assume that we know—through theory or history related to the population under consideration—the true value of σ. Of course, in most real-world situations, there would not be a basis for knowing σ. In the next section we will discuss a confidence interval based on the t distribution that does not assume that σ is known. Furthermore, we will revisit the examples of this section assuming that σ is unknown.

EXAMPLE 8.2: The Car Mileage Case

Recall that the federal government will give a tax credit to any automaker selling a midsize model equipped with an automatic transmission that has an EPA combined city and highway mileage estimate of at least 31 mpg. Furthermore, to ensure that it does not overestimate a car model’s mileage, the EPA will obtain the model’s mileage estimate by rounding down—to the nearest mile per gallon—the lower limit of a 95 percent confidence interval for the model’s mean mileage μ. That is, the model’s mileage estimate is an estimate of the smallest that μ might reasonably be. Suppose an automaker conducts mileage tests on a sample of 50 of its new midsize cars and obtains the sample of 50 mileages in Table 1.4, which has mean As illustrated in Figure 8.3, in order to compute a 95 percent confidence interval, we use the normal point zα/2 = z.05/2 = z.025 = 1.96. Assuming that σ is known to equal .8, it follows that the 95 percent confidence interval for μ is

This interval says we are 95 percent confident that the model’s mean mileage μ is between 31.34 mpg and 31.78 mpg. Based on this interval, the model’s EPA mileage estimate is 31 mpg, and the auto maker will receive the tax credit.

If we wish to compute a 99 percent confidence interval for μ, then, as illustrated in Figure 8.4, we use the normal point zα/2 = z.01/2 = z.005 = 2.575. We therefore obtain the interval

This interval says we are 99 percent confident that the model’s mean mileage μ is between 31.27 mpg and 31.85 mpg. Note that increasing the level of confidence to 99 percent has increased the margin of error from .222 to .291, which makes the 99 percent interval longer than the 95 percent interval.

EXAMPLE 8.3: The Payment Time Case

Recall that a management consulting firm has installed a new computer-based, electronic billing system in a Hamilton, Ohio, trucking company. The mean payment time using the trucking company’s old billing system was approximately equal to, but no less than, 39 days. In order to assess whether the mean payment time, μ, using the new billing system is substantially less than 39 days, the consulting firm will use the sample of n = 65 payment times in Table 2.4 to find a 95 percent confidence interval for μ. The mean of the 65 payment times is Using the normal point zα/2 = z.025 = 1.96, and assuming that σ is known to equal 4.2, it follows that the 95 percent confidence interval for μ is

Recalling that the mean payment time using the old billing system is 39 days, this interval says that we are 95 percent confident that the mean payment time using the new billing system is between 17.1 days and 19.1 days. Therefore, we are 95 percent confident that the new billing system reduces the mean payment time by at most 21.9 days and by at least 19.9 days.

Justifying the confidence interval formula

To show why the interval

is a 100(1 − α) percent confidence interval for μ, recall that if the sampled population is normally distributed or the sample size n is large, then the sampling distribution of is (exactly or approximately) a normal distribution with mean and standard deviation It follows that the sampling distribution of

is (exactly or approximately) a standard normal distribution. Therefore, the probability that we will obtain a sample mean such that z is between −zα/2 and zα/2 is 1 − α (see Figure 8.5). That is, we can say that the probability that

equals 1 − α. Using some algebraic manipulations, we can show that this is equivalent to saying that the probability that

equals 1 − α. This probability statement says that the probability is 1 − α (for example, .95) that we will obtain a sample mean such that the interval

contains μ. In other words, this interval is a 100(1 − α) percent confidence interval for μ.

Figure 8.5: A Probability for Deriving a Confidence Interval for the Population Mean

Exercises for Section 8.1

CONCEPTS

8.1 Explain why it is important to calculate a confidence interval in addition to calculating a point estimate of a population parameter.

8.2 Write a paragraph explaining exactly what the term “95 percent confidence” means in the context of calculating a 95 percent confidence interval for a population mean.

8.3 For each of the following changes, indicate whether a confidence interval for μ will have a larger or smaller margin of error:

a An increase in the level of confidence.

b An increase in the sample size.

c A decrease in the level of confidence.

d A decrease in the sample size.

METHODS AND APPLICATIONS

8.4 For each of the following confidence levels, 100(1 − α) percent, find the zα/2 point needed to compute a confidence interval for μ:

a 95%

b 99%

c 99.73%

d 80%

e 97%

f 92%

8.5 Suppose that, for a sample of size n = 100 measurements, we find that Assuming that σ equals 2, calculate confidence intervals for the population mean μ with the following confidence levels:

a 95%
b 99%

c 97%

d 80%

e 99.73%

8.6 THE TRASH BAG CASE TrashBag

Consider the trash bag problem. Suppose that an independent laboratory has tested trash bags and has found that no 30-gallon bags that are currently on the market have a mean breaking strength of 50 pounds or more. On the basis of these results, the producer of the new, improved trash bag feels sure that its 30-gallon bag will be the strongest such bag on the market if the new trash bag’s mean breaking strength can be shown to be at least 50 pounds. The mean of the sample of 40 trash bag breaking strengths in Table 1.9 is If we let μ denote the mean of the breaking strengths of all possible trash bags of the new type and assume that σ equals 1.65:

a Calculate 95 percent and 99 percent confidence intervals for μ.

b Using the 95 percent confidence interval, can we be 95 percent confident that μ is at least 50 pounds? Explain.

c Using the 99 percent confidence interval, can we be 99 percent confident that μ is at least 50 pounds? Explain.

d Based on your answers to parts b and c, how convinced are you that the new 30-gallon trash bag is the strongest such bag on the market?

8.7 THE BANK CUSTOMER WAITING TIME CASE WaitTime

Recall that a bank manager has developed a new system to reduce the time customers spend waiting to be served by tellers during peak business hours. The mean waiting time during peak business hours under the current system is roughly 9 to 10 minutes. The bank manager hopes that the newsystemwill have a mean waiting time that is less than six minutes. The mean of the sample of 100 bank customer waiting times in Table 1.8 is If we let μ denote the mean of all possible bank customer waiting times using the new system and assume that σ equals 2.47:

a Calculate 95 percent and 99 percent confidence intervals for μ.

b Using the 95 percent confidence interval, can the bank manager be 95 percent confident that μ is less than six minutes? Explain.

c Using the 99 percent confidence interval, can the bank manager be 99 percent confident that μ is less than six minutes? Explain.

d Based on your answers to parts b and c, how convinced are you that the new mean waiting time is less than six minutes?

8.8 THE VIDEO GAME SATISFACTION RATING CASE VideoGame

The mean of the sample of 65 customer satisfaction ratings in Table 1.7 is If we let μ denote the mean of all possible customer satisfaction ratings for the XYZ Box video game system, and assume that σ equals 2.64:

a Calculate 95 percent and 99 percent confidence intervals for μ.

b Using the 95 percent confidence interval, can we be 95 percent confident that μ is at least 42 (recall that a very satisfied customer gives a rating of at least 42)? Explain.

c Using the 99 percent confidence interval, can we be 99 percent confident that μ is at least 42? Explain.

d Based on your answers to parts b and c, how convinced are you that the mean satisfaction rating is at least 42?

8.9 In an article in the Journal of Management, Morris, Avila, and Allen studied innovation by surveying firms to find (among other things) the number of new products introduced by the firms. Suppose a random sample of 100 California-based firms is selected and each firm is asked to report the number of new products it has introduced during the last year. The sample mean is found to be Assuming σ equals 8.70:

a Calculate a 98 percent confidence interval for the population mean number of new products introduced in the last year.

b Based on your confidence interval, find a reasonable estimate for the smallest value that the mean number of new products might be. Explain.

8.10 In an article in Marketing Science, Silk and Berndt investigate the output of advertising agencies. They describe ad agency output by finding the shares of dollar billing volume coming from various media categories such as network television, spot television, newspapers, radio, and so forth.

a Suppose that a random sample of 400 U.S. advertising agencies gives an average percentage share of billing volume from network television equal to 7.46 percent, and assume that σ equals 1.42 percent. Calculate a 95 percent confidence interval for the mean percentage share of billing volume from network television for the population of all U.S. advertising agencies.

b Suppose that a random sample of 400 U.S. advertising agencies gives an average percentage share of billing volume from spot television commercials equal to 12.44 percent, and assume that σ equals 1.55 percent. Calculate a 95 percent confidence interval for the mean percentage share of billing volume from spot television commercials for the population of all U.S. advertising agencies.

c Compare the confidence intervals in parts a and b. Does it appear that the mean percentage share of billing volume from spot television commercials for U.S. advertising agencies is greater than the mean percentage share of billing volume from network television? Explain.

8.11 In an article in Accounting and Business Research, Carslaw and Kaplan investigate factors that influence “audit delay” for firms in New Zealand. Audit delay, which is defined to be the length of time (in days) from a company’s financial year-end to the date of the auditor’s report, has been found to affect the market reaction to the report. This is because late reports seem to often be associated with lower returns and early reports seem to often be associated with higher returns.

Carslaw and Kaplan investigated audit delay for two kinds of public companies—owner-controlled and manager-controlled companies. Here a company is considered to be owner controlled if 30 percent or more of the common stock is controlled by a single outside investor (an investor not part of the management group or board of directors). Otherwise, a company is considered manager controlled. It was felt that the type of control influences audit delay. To quote Carslaw and Kaplan:

Large external investors, having an acute need for timely information, may be expected to pressure the company and auditor to start and to complete the audit as rapidly as practicable.

a Suppose that a random sample of 100 public owner-controlled companies in New Zealand is found to give a mean audit delay of days, and assume that σ equals 33 days. Calculate a 95 percent confidence interval for the population mean audit delay for all public owner-controlled companies in New Zealand.

b Suppose that a random sample of 100 public manager-controlled companies in New Zealand is found to give a mean audit delay of days, and assume that σ equals 37 days. Calculate a 95 percent confidence interval for the population mean audit delay for all public manager-controlled companies in New Zealand.

c Use the confidence intervals you computed in parts a and b to compare the mean audit delay for all public owner-controlled companies versus that of all public manager-controlled companies. How do the means compare? Explain.

8.12 In an article in the Journal of Marketing, Bayus studied the differences between “early replacement buyers” and “late replacement buyers” in making consumer durable good replacement purchases. Early replacement buyers are consumers who replace a product during the early part of its lifetime, while late replacement buyers make replacement purchases late in the product’s lifetime. In particular, Bayus studied automobile replacement purchases. Consumers who traded in cars with ages of zero to three years and mileages of no more than 35,000 miles were classified as early replacement buyers. Consumers who traded in cars with ages of seven or more years and mileages of more than 73,000 miles were classified as late replacement buyers. Bayus compared the two groups of buyers with respect to demographic variables such as income, education, age, and so forth. He also compared the two groups with respect to the amount of search activity in the replacement purchase process. Variables compared included the number of dealers visited, the time spent gathering information, and the time spent visiting dealers.

a Suppose that a random sample of 800 early replacement buyers yields a mean number of dealers visited of and assume that σ equals .71. Calculate a 99 percent confidence interval for the population mean number of dealers visited by early replacement buyers.

b Suppose that a random sample of 500 late replacement buyers yields a mean number of dealers visited of and assume that σ equals .66. Calculate a 99 percent confidence interval for the population mean number of dealers visited by late replacement buyers.

c Use the confidence intervals you computed in parts a and b to compare the mean number of dealers visited by early replacement buyers with the mean number of dealers visited by late replacement buyers. How do the means compare? Explain.

8.2: t-Based Confidence Intervals for a Population Mean: σ Unknown

If we do not know σ (which is usually the case), we can use the sample standard deviation s to help construct a confidence interval for μ. The interval is based on the sampling distribution of

If the sampled population is normally distributed, then for any sample size n this sampling distribution is what is called a
t distribution.

Chapter 5

The curve of the t distribution has a shape similar to that of the standard normal curve. Two t curves and a standard normal curve are illustrated in Figure 8.6. A t curve is symmetrical about zero, which is the mean of any t distribution. However, the t distribution is more spread out, or variable, than the standard normal distribution. Since the above t statistic is a function of two random variables, and s, it is logical that the sampling distribution of this statistic is more variable than the sampling distribution of the z statistic, which is a function of only one random variable, . The exact spread, or standard deviation, of the t distribution depends on a parameter that is called the number of degrees of freedom (denoted df). The degrees of freedom df varies depending on the problem. In the present situation the sampling distribution of t has a number of degrees of freedom that equals the sample size minus 1. We say that this sampling distribution is a
t distribution with n
− 1 degrees of freedom. As the sample size n (and thus the number of degrees of freedom) increases, the spread of the t distribution decreases (see Figure 8.6). Furthermore, as the number of degrees of freedom approaches infinity, the curve of the t distribution approaches (that is, becomes shaped more and more like) the curve of the standard normal distribution.

Figure 8.6: As the Number of Degrees of Freedom Increases, the Spread of the t Distribution Decreases and the t Curve Approaches the Standard Normal Curve

In order to use the t distribution, we employ a
t point that is denoted tα. As illustrated in Figure 8.7,
t

α is the point on the horizontal axis under the curve of the t distribution that gives a right-hand tail area equal to α. The value of tα in a particular situation depends upon the right-hand tail area α and the number of degrees of freedom of the t distribution. Values of tα are tabulated in a
t table. Such a table is given in Table A.4 of Appendix A (pages 864 and 865) and a portion of Table A.4 is reproduced in this chapter as Table 8.3. In this t table, the rows correspond to the different numbers of degrees of freedom (which are denoted as df). The values of df are listed down the left side of the table, while the columns designate the right-hand tail area α. For example, suppose we wish to find the t point that gives a right-hand tail area of .025 under a t curve having df = 14 degrees of freedom. To do this, we look in Table 8.3 at the row labeled 14 and the column labeled t.025. We find that this t.025 point is 2.145 (also see Figure 8.8). Similarly, when there are df = 14 degrees of freedom, we find that t.005 = 2.977 (see Table 8.3 and Figure 8.9).

Figure 8.7: An Example of a t Point Giving a Specified Right-Hand Tail Area (This t Point Gives a Right-Hand Tail Area Equal to α).

Figure 8.8: The t Point Giving a Right-Hand Tail Area of .025 under the t Curve Having 14 Degrees of Freedom: t.025 = 2.145

Table 8.3: A t Table

Figure 8.9: The t Point Giving a Right-Hand Tail Area of .005 under the t Curve Having 14 Degrees of Freedom: t.005 = 2.977

Table 8.3 gives t points for degrees of freedom df from 1 to 30. The table also gives t points for 40, 60, 120, and an infinite number of degrees of freedom. Looking at this table, it is useful to realize that the normal points giving the various right-hand tail areas are listed in the row of the t table corresponding to an infinite (∞) number of degrees of freedom. Looking at the row corresponding to ∞, we see that, for example, z.025 = 1.96. Therefore, we can use this row in the t table as an alternative to using the normal table when we need to find normal points (such as zα/2 in Section 8.1).

Table A.4 of Appendix A (pages 864 and 865) gives t points for values of df from 1 to 100. We can use a computer to find t points based on values of df greater than 100. Alternatively, because a t curve based on more than 100 degrees of freedom is approximately the shape of the standard normal curve, t points based on values of df greater than 100 can be approximated by their corresponding z points. That is, when performing hand calculations, it is reasonable to approximate values of tα by zα when df is greater than 100.

We now present the formula for a 100(1 − α) percent confidence interval for a population mean μ based on the t distribution.

A t-Based Confidence Interval for a Population Mean μ: σ Unknown

If the sampled population is normally distributed with mean μ, then a 100(1 − α) percent confidence interval for μ
is

Here s is the sample standard deviation, tα/2 is the t point giving a right-hand tail area of α/2 under the t curve having n − 1 degrees of freedom, and n is the sample size.

Chapter 9

Before presenting an example, we need to make a few comments. First, it has been shown that this confidence interval is approximately valid for many populations that are not exactly normally distributed. In particular, this interval is approximately valid for a mound-shaped, or single-peaked, population, even if the population is somewhat skewed to the right or left. Second, this interval employs the point tα/2, which as shown in Figure 8.10, gives a right-hand tail area equal to α/2 under the t curve having n − 1 degrees of freedom. Here α/2 is determined from the desired confidence level 100(1 − α) percent.

Figure 8.10: The Point tα/2 with n − 1 Degrees of Freedom

EXAMPLE 8.4

One measure of a company’s financial health is its debt-to-equity ratio. This quantity is defined to be the ratio of the company’s corporate debt to the company’s equity. If this ratio is too high, it is one indication of financial instability. For obvious reasons, banks often monitor the financial health of companies to which they have extended commercial loans. Suppose that, in order to reduce risk, a large bank has decided to initiate a policy limiting the mean debt-to-equity ratio for its portfolio of commercial loans to 1.5. In order to estimate the mean debt-to-equity ratio of its loan portfolio, the bank randomly selects a sample of 15 of its commercial loan accounts. Audits of these companies result in the following debt-to-equity ratios:

DebtEq

A stem-and-leaf display of these ratios is given on the page margin, and a box plot of the ratios is given below. The stem-and-leaf display looks reasonably mound-shaped, and both the stem-and-leaf display and the box plot look reasonably symmetrical. Furthermore, the sample mean and standard deviation of the ratios can be calculated to be

Suppose the bank wishes to calculate a 95 percent confidence interval for the loan portfolio’s mean debt-to-equity ratio, μ. Because the bank has taken a sample of size n = 15, we have n − 1 = 15 − 1 = 14 degrees of freedom, and the level of confidence 100(1 − α)% = 95% implies that α = .05. Therefore, we use the t point tα/2 = t.05/2 = t.025 = 2.145 (see Table 8.3). It follows that the 95 percent confidence interval for μ is

This interval says the bank is 95 percent confident that the mean debt-to-equity ratio for its portfolio of commercial loan accounts is between 1.2369 and 1.4497. Based on this interval, the bank has strong evidence that the portfolio’s mean ratio is less than 1.5 (or that the bank is in compliance with its new policy).

Recall that in the two cases discussed in Section 8.1 we calculated z-based confidence intervals for μ by assuming that the population standard deviation σ is known. If σ is actually not known (which would probably be true), we should compute t-based confidence intervals. Furthermore, recall that in each of these cases the sample size is large (at least 30). In general, it can be shown that if the sample size is large, the t-based confidence interval for μ is approximately valid even if the sampled population is not normally distributed (or mound shaped). Therefore, consider the car mileage case and the sample of 50 mileages in Table 1.4, which has mean and standard deviation s = .798. The 95 percent t-based confidence interval for the population mean mileage μ of the new midsize model is

where t.025 = 2.010 is based on n − 1 = 50 − 1 = 49 degrees of freedom—see Table A.4 (page 864). This interval says we are 95 percent confident that the model’s mean mileage μ is between 31.33 mpg and 31.78 mpg. Based on this interval, the model’s EPA mileage estimate is 31 mpg, and the automaker will receive the tax credit.

As another example, the sample of 65 payment times in Table 2.4 has mean and standard deviation s = 3.9612. The 95 percent t-based confidence interval for the population mean payment time using the new electronic billing system is

where t.025 = 1.998 is based on n − 1 = 65 − 1 = 64 degrees of freedom—see Table A.4 (page 865). Recalling that the mean payment time using the old billing system is 39 days, the interval says that we are 95 percent confident that the mean payment time using the new billing system is between 17.1 days and 19.1 days. Therefore, we are 95 percent confident that the new billing system reduces the mean payment time by at most 21.9 days and by at least 19.9 days.

EXAMPLE 8.5: The Marketing Research Case

Recall that a brand group is considering a new bottle design for a popular soft drink and that Table 1.3 (page 8) gives a random sample of n = 60 consumer ratings of this new bottle design. Let μ denote the mean rating of the new bottle design that would be given by all consumers. In order to assess whether μ exceeds the minimum standard composite score of 25 for a successful bottle design, the brand group will calculate a 95 percent confidence interval for μ. The mean and the standard deviation of the 60 bottle design ratings are It follows that a 95 percent confidence interval for μ is

where t.025 = 2.001 is based on n − 1 = 60 − 1 = 59 degrees of freedom—see Table A.4. Since the interval says we are 95 percent confident that the mean rating of the new bottle design is between 29.5 and 31.2, we are 95 percent confident that this mean rating exceeds the minimum standard of 25 by at least 4.5 points and by at most 6.2 points.

Confidence intervals for μ can be computed using MINITAB, Excel, and MegaStat. For example, the MINITAB output in Figure 8.11 tells us that the t-based 95 percent confidence interval for the mean debt-to-equity ratio is [1.2370, 1.4497]. This result is, within rounding, the same interval calculated in Example 8.4. The MINITAB output also gives the sample mean as well as the sample standard deviation s = .1921 and the quantity which is called the standard error of the estimate and denoted “SE Mean” on the MINITAB output. Finally, the MINITAB output gives a box plot of the sample of 15 debt-to-equity ratios and graphically illustrates under the box plot the 95 percent confidence interval for the mean debt-to-equity ratio. Figure 8.12(a) gives the MegaStat output of the 95 percent confidence interval for the mean debt-to-equity ratio, and Figure 8.12(b) gives the Excel output of the information needed to calculate this interval. If we consider the Excel output, we note that (see “Mean”), s = .1921 (see “Standard Deviation”), (see “Standard Error”), and [see “Confidence Level (95.0%)”]. The interval, which must be hand calculated, is [1.3433 ± .1064] = [1.2369, 1.4497].

Figure 8.11: MINITAB Output of a t-Based 95 Percent Confidence Interval for the Mean Debt-to-Equity Ratio

Figure 8.12: MegaStat and Excel Outputs for the Debt-to-Equity Ratio Example

To conclude this section, we note that if the sample size n is small and the sampled population is not mound-shaped or is highly skewed, then the t-based confidence interval for the population mean might not be valid. In this case we can use a nonparametric method—a method that makes no assumption about the shape of the sampled population and is valid for any sample size. In Chapter 18 we further discuss nonparametric methods.

Exercises for Section 8.2

CONCEPTS

8.13 Explain how each of the following changes as the number of degrees of freedom describing a t curve increases:

a The standard deviation of the t curve.

b The points tα and tα/2.

8.14 Discuss when it is appropriate to use the t-based confidence interval for μ.

METHODS AND APPLICATIONS

8.15 Using Table 8.3, find t.10, t.025, and t.001 based on 11 degrees of freedom. Also, find these t points based on 6 degrees of freedom.

8.16 Suppose that for a sample of n = 11 measurements, we find that Assuming normality, compute confidence intervals for the population mean μ with the following levels of confidence:

a 95%
b 99%

c 80%

d 90%

e 98%

f 99.8%

8.17 The bad debt ratio for a financial institution is defined to be the dollar value of loans defaulted divided by the total dollar value of all loans made. Suppose a random sample of seven Ohio banks is selected and that the bad debt ratios (written as percentages) for these banks are 7 percent, 4 percent, 6 percent, 7 percent, 5 percent, 4 percent, and 9 percent. Assuming the bad debt ratios are approximately normally distributed, the MINITAB output of a 95 percent confidence interval for the mean bad debt ratio of all Ohio banks is as follows: BadDebt

a Using the and s on the MINITAB output, verify the calculation of the 95 percent confidence interval, and calculate a 99 percent confidence interval for the mean debt-to-equity ratio.

b Banking officials claim the mean bad debt ratio for all banks in the Midwest region is 3.5 percent and that the mean bad debt ratio for Ohio banks is higher. Using the 95 percent confidence interval, can we be 95 percent confident that this claim is true? Using the 99 percent confidence interval, can we be 99 percent confident that this claim is true?

8.18 In an article in Quality Progress, Blauw and During study how long it takes Dutch companies to complete five stages in the adoption of total quality control (TQC). According to Blauw and During, the adoption of TQC can be divided into five stages as follows: TQC

1 Knowledge: the organization has heard of TQC.

2 Attitude formation: the organization seeks information and compares advantages and disadvantages.

3 Decision making: the organization decides to implement TQC.

4 Implementation: the organization implements TQC.

5 Confirmation: the organization decides to apply TQC as a normal business activity.

Suppose a random sample of five Dutch firms that have adopted TQC is selected. Each firm is asked to report how long it took to complete the implementation stage. The firms report the following durations (in years) for this stage: 2.5, 1.5, 1.25, 3.5, and 1.25. Assuming that the durations are approximately normally distributed, the MegaStat output of a 95 percent confidence interval for the mean duration of the implementation stage for Dutch firms is as follows:

Based on the 95 percent confidence interval, is there conclusive evidence that the mean duration of the implementation stage exceeds one year? Explain. What is one possible reason for the lack of conclusive evidence?

8.19 THE AIR TRAFFIC CONTROL CASE AlertTime

Air traffic controllers have the crucial task of ensuring that aircraft don’t collide. To do this, they must quickly discern when two planes are about to enter the same air space at the same time. They are aided by video display panels that track the aircraft in their sector and alert the controller when two flight paths are about to converge. The display panel currently in use has a mean “alert time” of 15 seconds. (The alert time is the time elapsing between the instant when two aircraft enter into a collision course and when a controller initiates a call to reroute the planes.) According to Ralph Rudd, a supervisor of air traffic controllers at the Greater Cincinnati International Airport, a new display panel has been developed that uses artificial intelligence to project a plane’s current flight path into the future. This new panel provides air traffic controllers with an earlier warning that a collision is likely. It is hoped that the mean “alert time,” μ, for the new panel is less than 8 seconds. In order to test the new panel, 15 randomly selected air traffic controllers are trained to use the panel and their alert times for a simulated collision course are recorded. The sample alert times (in seconds) are: 7.2, 7.5, 8.0, 6.8, 7.2, 8.4, 5.3, 7.3, 7.6, 7.1, 9.4, 6.4, 7.9, 6.2, 8.7.

a Using the fact that and find a 95 percent confidence interval for the mean alert time, μ, for the new panel.

b Can we be 95 percent confident that μ is less than 8 seconds?

8.20 Whole Foods is an all-natural grocery chain that has 50,000 square foot stores, up from the industry average of 34,000 square feet. Sales per square foot of supermarkets average just under $400 per square foot, as reported by USA Today in an article on “A whole new ballgame in grocery shopping.” Suppose that sales per square foot in the most recent fiscal year are recorded for a random sample of 10 Whole Foods supermarkets. The data (sales dollars per square foot) are as follows: 854, 858, 801, 892, 849, 807, 894, 863, 829, 815. Using the fact that find a 95 percent confidence interval for the true mean sales dollars per square foot for all Whole Foods supermarkets during the most recent fiscal year. Are we 95 percent confident that this mean is greater than $800, the historical average for Whole Foods? WholeFoods

8.21 A production supervisor at a major chemical company wishes to determine whether a new catalyst, catalyst XA-100, increases the mean hourly yield of a chemical process beyond the current mean hourly yield, which is known to be roughly equal to, but no more than, 750 pounds per hour. To test the new catalyst, five trial runs using catalyst XA-100 are made. The resulting yields for the trial runs (in pounds per hour) are 801, 814, 784, 836, and 820. Assuming that all factors affecting yields of the process have been held as constant as possible during the test runs, it is reasonable to regard the five yields obtained using the new catalyst as a random sample from the population of all possible yields that would be obtained by using the new catalyst. Furthermore, we will assume that this population is approximately normally distributed. ChemYield

a Using the Excel output in Figure 8.13, find a 95 percent confidence interval for the mean of all possible yields obtained using catalyst XA-100.

b Based on the confidence interval, can we be 95 percent confident that the mean yield using catalyst XA-100 exceeds 750 pounds per hour? Explain.

Figure 8.13: Excel Output for Exercise 8.21

8.22 THE TRASH BAG CASE TrashBag

The mean and the standard deviation of the sample of 40 trash bag breaking strengths in Table 1.9 are Calculate a t-based 95 percent confidence interval for μ, the mean of the breaking strengths of all possible trash bags of the new type. Also, find this interval using the Excel output in Figure 8.14. Are we 95 percent confident that μ is at least 50 pounds?

Figure 8.14: Excel Output for Exercise 8.22

8.23 THE BANK CUSTOMER WAITING TIME CASE WaitTime

The mean and the standard deviation of the sample of 100 bank customer waiting times in Table 1.8 are Calculate a t-based 95 percent confidence interval for μ, the mean of all possible bank customer waiting times using the new system. Are we 95 percent confident that μ is less than six minutes?

8.24 THE VIDEO GAME SATISFACTION RATING CASE VideoGame

The mean and the standard deviation of the sample of n = 65 customer satisfaction ratings in Table 1.7 are Calculate a t-based 95 percent confidence interval for μ, the mean of all possible customer satisfaction ratings for the XYZ Box video game system. Are we 95 percent confident that μ is at least 42, the minimal rating given by a very satisfied customer?

8.3: Sample Size Determination

In Example 8.2 we used a sample of 50 mileages to construct a 95 percent confidence interval for the midsize model’s mean mileage μ. The size of this sample was not arbitrary—it was planned. To understand this, suppose that before the automaker selected the random sample of 50 mileages, it randomly selected the small sample of five mileages that is shown as the leftmost sample in Table 8.1. This sample consists of the mileages

and has mean Assuming that the population of all mileages is normally distributed and that the population standard deviation σ is known to equal .8, it follows that a 95 percent confidence interval for μ is

Although the sample mean is at least 31, the lower limit of the 95 percent confidence interval for μ is less than 31. Therefore, the midsize model’s EPA mileage estimate would be 30 mpg, and the automaker would not receive its tax credit. One reason that the lower limit of this 95 percent interval is less than 31 is that the sample size of 5 is not large enough to make the interval’s margin of error

small enough. We can attempt to make the margin of error in the interval smaller by increasing the sample size. If we feel that the mean of the larger sample will be at least 31.3 mpg (the mean of the small sample we have already taken), then the lower limit of a 100(1 − α) percent confidence interval for μ will be at least 31 if the margin of error is .3 or less.

We will now explain how to find the size of the sample that will be needed to make the margin of error in a confidence interval for μ as small as we wish. In order to develop a formula for the needed sample size, we will initially assume that we know σ. Then, if the population is normally distributed or the sample size is large, the z-based 100(1 − α) percent confidence interval for μ is

To find the needed sample size, we set equal to the desired margin of error and solve for n. Letting E denote the desired margin of error, we obtain

Multiplying both sides of this equation by and dividing both sides by E, we obtain

Squaring both sides of this result gives us the formula for n.

Determining the Sample Size for a Confidence Interval for μ: σ Known

A sample of size

makes the margin of error in a 100(1 − α) percent confidence interval for μ equal to E. That is, this sample size makes us 100(1 − α) percent confident that is within E units of μ. If the calculated value of n is not a whole number, round this value up to the next whole number. (so that the margin of error is at least as small as desired).

If we consider the formula for the sample size n, it intuitively follows that the value E is the farthest that the user is willing to allow to be from μ at a given level of confidence, and the normal point zα/2 follows directly from the given level of confidence. Furthermore, because the population standard deviation σ is in the numerator of the formula for n, it follows that the more variable that the individual population measurements are, the larger is the sample size needed to estimate μ with a specified accuracy.

In order to use this formula for n, we must either know σ (which is unlikely) or we must compute an estimate of σ. We first consider the case where we know σ. For example, suppose in the car mileage situation we wish to find the sample size that is needed to make the margin of error in a 95 percent confidence interval for μ equal to .3. Assuming that σ is known to equal .8, and using z.025 = 1.96, the appropriate sample size is

Rounding up, we would employ a sample of size 28.

In most real situations, of course, we do not know the true value of σ. If σ is not known, we often estimate σ by using a preliminary sample. In this case we modify the above formula for n by replacing σ by the standard deviation s of the preliminary sample and by replacing zα/2 by tα/2.

Thus we obtain

where the number of degrees of freedom for the tα/2 point is the size of the preliminary sample minus 1. Intuitively, using tα/2 compensates for the fact that the preliminary sample’s value of s might underestimate σ.

EXAMPLE 8.6: The Car Mileage Case

Suppose that in the car mileage situation we wish to find the sample size that is needed to make the margin of error in a 95 percent confidence interval for μ equal to .3. Assuming we do not know σ, we regard the previously discussed sample of five mileages (see page 321) as a preliminary sample. Therefore, we replace σ by the standard deviation of the preliminary sample, which can be calculated to be s = .7583, and we replace zα/2 = z.025 = 1.96 by t.025 = 2.776, which is based on n − 1 = 4 degrees of freedom. We find that the appropriate sample size is

Rounding up, we employ a sample of size 50.

When we make the margin of error in our 95 percent confidence interval for μ equal to .3, we can say we are 95 percent confident that the sample mean is within .3 of μ. To understand this, suppose the true value of μ is 31.5. Recalling that the mean of the sample of 50 mileages is we see that this sample mean is within .3 of μ (in fact, it is 31.56 − 31.5 = .06 mpg from u = 31.5). Other samples of 50 mileages would give different sample means that would be different distances from μ. When we say that our sample of 50 mileages makes us 95 percent confident that is within .3 of μ, we mean that 95 percent of all possible sample means based on 50 mileages are within .3 of μ
and 5 percent of such sample means are not. Therefore, when we randomly select one sample of size 50 and compute its sample mean we can be 95 percent confident that this sample mean is within .3 of μ.

In general, the purpose behind replacing zα/2 by tα/2 (when we are using a preliminary sample to obtain an estimate of σ) is to be conservative, so that we compute a sample size that is at least as large as needed. Because of this, as we illustrate in the next example, we often obtain a margin of error that is even smaller than we have requested.

EXAMPLE 8.7: The Car Mileage Case

To see that the sample of 50 mileages has actually produced a 95 percent confidence interval with a margin of error that is as small as we requested, recall that the 50 mileages have mean and standard deviation s = .798. Therefore, the t-based 95 percent confidence interval for μ is

where t.025 = 2.010 is based on n − 1 = 50 − 1 = 49 degrees of freedom—see Table A.4 (page 865). We see that the margin of error in this interval is .227, which is smaller than the .3 we asked for. Furthermore, as the automaker had hoped, the sample mean of the sample of 50 mileages turned out to be at least 31.3. Therefore, since the margin of error is less than .3, the lower limit of the 95 percent confidence interval is higher than 31 mpg, and the midsize model’s EPA mileage estimate is 31 mpg. Because of this, the automaker will receive its tax credit.

Finally, sometimes we do not know σ and we do not have a preliminary sample that can be used to estimate σ. In this case it can be shown that, if we can make a reasonable guess of the range of the population being studied, then a conservatively large estimate of σ is this estimated range divided by 4. For example, if the automaker’s design engineers feel that almost all of its midsize cars should get mileages within a range of 5 mpg, then a conservatively large estimate of σ is 5/4 = 1.25 mpg. When employing such an estimate of σ, it is sufficient to use the z-based sample size formula n = (zα/2σ/E)2, because a conservatively large estimate of σ will give us a conservatively large sample size.

Exercises for Section 8.3

CONCEPTS

8.25 Explain what is meant by the margin of error for a confidence interval. What error are we talking about in the context of an interval for μ?

8.26 Explain exactly what we mean when we say that a sample of size n makes us 99 percent confident that is within E units of μ.

8.27 Why do we often need to take a preliminary sample when determining the size of the sample needed to make the margin of error of a confidence interval equal to E?

METHODS AND APPLICATIONS

8.28 Consider a population having a standard deviation equal to 10. We wish to estimate the mean of this population.

a How large a random sample is needed to construct a 95 percent confidence interval for the mean of this population with a margin of error equal to 1?

b Suppose that we now take a random sample of the size we have determined in part a. If we obtain a sample mean equal to 295, calculate the 95 percent confidence interval for the population mean. What is the interval’s margin of error?

8.29 Referring to Exercise 8.11a, assume that σ equals 33. How large a random sample of public owner-controlled companies is needed to make us

a 95 percent confident that , the sample mean audit delay, is within a margin of error of four days of μ, the true mean audit delay?

b 99 percent confident that is within a margin of error of four days of μ?

8.30 Referring to Exercise 8.12b, assume that α equals .66. How large a sample of late replacement buyers is needed to make us

a 99 percent confident that the sample mean number of dealers visited, is within a margin of error of .04 of μ, the true mean number of dealers visited?

b 99.73 percent confident that is within a margin of error of .05 of μ?

8.31 Referring to Exercise 8.21, regard the sample of five trial runs for which s = 19.65 as a preliminary sample. Determine the number of trial runs of the chemical process needed to make us

a 95 percent confident that the sample mean hourly yield, is within a margin of error of eight pounds of the true mean hourly yield μ when catalyst XA-100 is used.

b 99 percent confident that is within a margin of error of five pounds of μ. ChemYield

8.32 Referring to Exercise 8.20, regard the sample of 10 sales figures for which s = 32.866 as a preliminary sample. How large a sample of sales figures is needed to make us 95 percent confident that , the sample mean sales dollars per square foot, is within a margin of error of $10 of μ, the true mean sales dollars per square foot for all Whole Foods supermarkets. WholeFoods

8.33 THE AIR TRAFFIC CONTROL CASE AlertTime

Referring to Exercise 8.19, regard the sample of 15 alert times for which s = 1.026 as a preliminary sample. Determine the sample size needed to make us 95 percent confident that , the sample mean alert time, is within a margin of error of .3 seconds of μ, the true mean alert time using the new display panel.

8.4: Confidence Intervals for a Population Proportion

In Chapter 7, the soft cheese spread producer decided to replace its current spout with the new spout if p, the true proportion of all current purchasers who would stop buying the cheese spread if the new spout were used, is less than .10. Suppose that when 1,000 current purchasers are randomly selected and are asked to try the new spout, 63 say they would stop buying the spread if the new spout were used. The point estimate of the population proportion p is the sample proportion . This sample proportion says we estimate that 6.3 percent of all current purchasers would stop buying the cheese spread if the new spout were used. Since equals .063, we have some evidence that p is less than .10.

In order to see if there is strong evidence that p is less than .10, we can calculate a confidence interval for p. As explained in Chapter 7, if the sample size n is large, then the sampling distribution of the sample proportion is approximately a normal distribution with mean and standard deviation By using the same logic we used in developing confidence intervals for μ, it follows that a 100(1 − α) percent confidence interval for p is

Estimating p(1 − p) by it follows that a 100(1 − α) percent confidence interval for p can be calculated as summarized below.

A Large Sample Confidence Interval for a Population Proportion p

If the sample size n is large, a 100(1 − α) percent confidence interval for the population proportion p
is

Here n should be considered large if both are at least 5.1

EXAMPLE 8.8: The Cheese Spread Case

In the cheese spread situation, consider calculating a confidence interval for p, the population proportion of purchasers who would stop buying the cheese spread if the new spout were used. In order to see whether the sample size n = 1,000 is large enough to enable us to use the confidence interval formula just given, recall that the point estimate of p is Therefore, because are both greater than 5, we can use the confidence interval formula. For example, a 95 percent confidence interval for p is

This interval says that we are 95 percent confident that between 4.79 percent and 7.81 percent of all current purchasers would stop buying the cheese spread if the new spout were used. Below we give the MegaStat output of this interval.

A 99 percent confidence interval for p is

The upper limits of both the 95 percent and 99 percent intervals are less than .10. Therefore, we have very strong evidence that the true proportion p of all current purchasers who would stop buying the cheese spread is less than .10. Based on this result, it seems reasonable to use the new spout.

In the cheese spread example, a sample of 1,000 purchasers gives us a 95 percent confidence interval for p—[.063 ± .0151]—with a reasonably small margin of error of .0151. Generally speaking, quite a large sample is needed in order to make the margin of error in a confidence interval for p reasonably small. The next two examples demonstrate that a sample size of 200, which most people would consider quite large, does not necessarily give a 95 percent confidence interval for p with a small margin of error.

EXAMPLE 8.9

Antibiotics occasionally cause nausea as a side effect. Scientists working for a major drug company have developed a new antibiotic called Phe-Mycin. The company wishes to estimate p, the proportion of all patients who would experience nausea as a side effect when being treated with Phe-Mycin. Suppose that a sample of 200 patients is randomly selected. When these patients are treated with Phe-Mycin, 35 patients experience nausea. The point estimate of the population proportion p is the sample proportion . This sample proportion says that we estimate that 17.5 percent of all patients would experience nausea as a side effect of taking Phe-Mycin. Furthermore, because are both at least 5, we can use the previously given formula to calculate a confidence interval for p. Doing this, we find that a 95 percent confidence interval for p is

This interval says we are 95 percent confident that between 12.2 percent and 22.8 percent of all patients would experience nausea as a side effect of taking Phe-Mycin. Notice that the margin of error (.053) in this interval is rather large. Therefore, this interval is fairly long, and it does not provide a very precise estimate of p.

EXAMPLE 8.10: The Marketing Ethics Case: Estimating Marketing Researchers’ Disapproval Rates

In the book Essentials of Marketing Research, William R. Dillon, Thomas J. Madden, and Neil H. Firtle discuss a survey of marketing professionals, the results of which were originally published by Ishmael P. Akoah and Edward A. Riordan in the Journal of Marketing Research. In the study, randomly selected marketing researchers were presented with various scenarios involving ethical issues such as confidentiality, conflict of interest, and social acceptability. The marketing researchers were asked to indicate whether they approved or disapproved of the actions described in each scenario. For instance, one scenario that involved the issue of confidentiality was described as follows:

Use of ultraviolet ink A project director went to the marketing research director’s office and requested permission to use an ultraviolet ink to precode a questionnaire for a mail survey. The project director pointed out that although the cover letter promised confidentiality, respondent identification was needed to permit adequate cross-tabulations of the data. The marketing research director gave approval.

Of the 205 marketing researchers who participated in the survey, 117 said they disapproved of the actions taken in the scenario. It follows that a point estimate of p, the proportion of all marketing researchers who disapprove of the actions taken in the scenario, is Furthermore, because are both at least 5, a 95 percent confidence interval for p is

This interval says we are 95 percent confident that between 50.29 percent and 63.85 percent of all marketing researchers disapprove of the actions taken in the ultraviolet ink scenario. Notice that since the margin of error (.0678) in this interval is rather large, this interval does not provide a very precise estimate of p. Below we show the MINITAB output of this interval.

In order to find the size of the sample needed to estimate a population proportion, we consider the theoretically correct interval

To obtain the sample size needed to make the margin of error in this interval equal to E, we set

and solve for n. When we do this, we get the following result:

Determining the Sample Size for a Confidence Interval for p

A sample of size

makes the margin of error in a 100(1 − α) percent confidence interval for p equal to E. That is, this sample size makes us 100(1 − α) percent confident that is within E units of p. If the calculated value of n is not a whole number, round this value up to the next whole number.

Looking at this formula, we see that, the larger p(1 − p) is, the larger n will be. To make sure n is large enough, consider Figure 8.15, which is a graph of p(1 − p) versus p. This figure shows that p(1 − p) equals .25 when p equals .5. Furthermore, p(1 − p) is never larger than .25. Therefore, if the true value of p could be near .5, we should set p(1 − p) equal to .25. This will ensure that n is as large as needed to make the margin of error as small as desired. For example, suppose we wish to estimate the proportion p of all registered voters who currently favor a particular candidate for President of the United States. If this candidate is the nominee of a major political party, or if the candidate enjoys broad popularity for some other reason, then p could be near .5. Furthermore, suppose we wish to make the margin of error in a 95 percent confidence interval for p equal to .02. If the sample to be taken is random, it should consist of

registered voters. The MegaStat output of the results of this calculation is shown in Figure 8.16. In reality, a list of all registered voters in the United States is not available to polling organizations. Therefore, it is not feasible to take a (technically correct) random sample of registered voters. For this reason, polling organizations actually employ other (more complicated) kinds of samples. We have explained some of the basic ideas behind these more complex samples in optional Section 1.4. For now, we consider the samples taken by polling organizations to be approximately random. Suppose, then, that when the sample of voters is actually taken, the proportion of sampled voters who favor the candidate turns out to be greater than .52. It follows, because the sample is large enough to make the margin of error in a 95 percent confidence interval for p equal to .02, that the lower limit of such an interval is greater than .50. This says we have strong evidence that a majority of all registered voters favor the candidate. For instance, if the sample proportion equals .53, we are 95 percent confident that the proportion of all registered voters who favor the candidate is between .51 and .55.

Figure 8.15: The Graph of p(1 − p) versus p

Figure 8.16: MegaStat Output of a Sample Size Calculation

Major polling organizations conduct public opinion polls concerning many kinds of issues. Whereas making the margin of error in a 95 percent confidence interval for p equal to .02 requires a sample size of 2,401, making the margin of error in such an interval equal to .03 requires a sample size of only

or 1,068 (rounding up). Of course, these calculations assume that the proportion p being estimated could be near .5. However, for any value of p, increasing the margin of error from .02 to .03 substantially decreases the needed sample size and thus saves considerable time and money. For this reason, although the most accurate public opinion polls use a margin of error of .02, the vast majority of public opinion polls use a margin of error of .03 or larger.

When the news media report the results of a public opinion poll, they express the margin of error in a 95 percent confidence interval for p in percentage points. For instance, if the margin of error is .03, the media would say the poll’s margin of error is 3 percentage points. The media seldom report the level of confidence, but almost all polling results are based on 95 percent confidence. Sometimes the media make a vague reference to the level of confidence. For instance, if the margin of error is 3 percentage points, the media might say that “the sample result will be within 3 percentage points of the population value in 19 out of 20 samples.” Here the “19 out of 20 samples” is a reference to the level of confidence, which is 100(19/20) = 100(.95) = 95 percent.

As an example, suppose a news report says a recent poll finds that 34 percent of the public favors military intervention in an international crisis, and suppose the poll’s margin of error is reported to be 3 percentage points. This means the sample taken is large enough to make us 95 percent confident that the sample proportion is within .03 (that is, 3 percentage points) of the true proportion p of the entire public that favors military intervention. That is, we are 95 percent confident that p is between .31 and .37.

If the population proportion we are estimating is substantially different from .5, setting p equal to .5 will give a sample size that is much larger than is needed. In this case, we should use our intuition or previous sample information—along with Figure 8.17—to determine the largest reasonable value for p(1 − p). Figure 8.17 implies that as p gets closer to .5, p(1 − p) increases. It follows that p(1 − p) is maximized by the reasonable value of p that is closest to .5. Therefore, when we are estimating a proportion that is substantially different from .5, we use the reasonable value of p that is closest to .5 to calculate the sample size needed to obtain a specified margin of error.

Figure 8.17: As p Gets Closer to .5, p(1 − p) Increases

EXAMPLE 8.11

Again consider estimating the proportion of all patients who would experience nausea as a side effect of taking the new antibiotic Phe-Mycin. Suppose the drug company wishes to find the size of the random sample that is needed in order to obtain a 2 percent margin of error with 95 percent confidence. In Example 8.9 we employed a sample of 200 patients to compute a 95 percent confidence interval for p. This interval, which is [.122, .228], makes us very confident that p is between .122 and .228. Because .228 is the reasonable value of p that is closest to .5, the largest reasonable value of p(1 − p) is .228(1 − .228) = .1760, and thus the drug company should take a sample of

patients.

Finally, as a last example of choosing p for sample size calculations, suppose that experience indicates that a population proportion p is at least .75. Then, .75 is the reasonable value of p that is closest to .5, and we would use the largest reasonable value of p(1 − p), which is .75(1 − .75) = .1875.

Exercises for Section 8.4

CONCEPTS

8.34 a What does a population proportion tell us about the population?

b Explain the difference between p and .

c What is meant when a public opinion poll’s margin of error is 3 percent?

8.35 Suppose we are using the sample size formula in the box on page 327 to find the sample size needed to make the margin of error in a confidence interval for p equal to E. In each of the following situations, explain what value of p would be used in the formula for finding n:

a We have no idea what value p is—it could be any value between 0 and 1.

b Past experience tells us that p is no more than .3.

c Past experience tells us that p is at least .8.

METHODS AND APPLICATIONS

8.36 In each of the following cases, determine whether the sample size n is large enough to use the large sample formula presented in the box on page 325 to compute a confidence interval for p.

a

b

c

d

e

f

8.37 In each of the following cases, compute 95 percent, 98 percent, and 99 percent confidence intervals for the population proportion p.

a

b

c

d

8.38 Quality Progress, February 2005, reports on the results achieved by Bank of America in improving customer satisfaction and customer loyalty by listening to the ‘voice of the customer.’ A key measure of customer satisfaction is the response on a scale from 1 to 10 to the question: “Considering all the business you do with Bank of America, what is your overall satisfaction with Bank of America?”2 Suppose that a random sample of 350 current customers results in 195 customers with a response of 9 or 10 representing ‘customer delight.’ Find a 95 percent confidence interval for the true proportion of all current Bank of America customers who would respond with a 9 or 10. Are we 95 percent confident that this proportion exceeds .48, the historical proportion of customer delight for Bank of America?

8.39 THE MARKETING ETHICS CASE: CONFLICT OF INTEREST

Consider the marketing ethics case described in Example 8.10. One of the scenarios presented to the 205 marketing researchers is as follows:

A marketing testing firm to which X company gives most of its business recently went public. The marketing research director of X company had been looking for a good investment and proceeded to buy some $20,000 of their stock. The firm continues as X company’s leading supplier for testing.

Of the 205 marketing researchers who participated in the ethics survey, 111 said that they disapproved of the actions taken in the scenario. Use this sample result to show that the 95 percent confidence interval for the proportion of all marketing researchers who disapprove of the actions taken in the conflict of interest scenario is as given in the MINITAB output below. Interpret this interval.

b On the basis of this interval, is there convincing evidence that a majority of all marketing researchers disapprove of the actions taken in the conflict of interest scenario? Explain.

8.40 In a news story distributed by the Washington Post, Lew Sichelman reports that a substantial fraction of mortgage loans that go into default within the first year of the mortgage were approved on the basis of falsified applications. For instance, loan applicants often exaggerate their income or fail to declare debts. Suppose that a random sample of 1,000 mortgage loans that were defaulted within the first year reveals that 410 of these loans were approved on the basis of falsified applications.

a Find a point estimate of and a 95 percent confidence interval for p, the proportion of all first-year defaults that are approved on the basis of falsified applications.

b Based on your interval, what is a reasonable estimate of the minimum percentage of first-year defaults that are approved on the basis of falsified applications?

8.41 On January 7, 2000, the Gallup Organization released the results of a poll comparing the lifestyles of today with yesteryear. The survey results were based on telephone interviews with a randomly selected national sample of 1,031 adults,18 years and older, conducted December 20–21, 1999.3

a The Gallup poll found that 42 percent of the respondents said that they spend less than three hours watching TV on an average weekday. Based on this finding, calculate a 99 percent confidence interval for the proportion of U.S. adults who say that they spend less than three hours watching TV on an average weekday. Based on this interval, is it reasonable to conclude that more than 40 percent of U.S. adults say they spend less than three hours watching TV on an average weekday?

b The Gallup poll found that 60 percent of the respondents said they took part in some form of daily activity (outside of work, including housework) to keep physically fit. Based on this finding, find a 95 percent confidence interval for the proportion of U.S. adults who say they take part in some form of daily activity to keep physically fit. Based on this interval, is it reasonable to conclude that more than 50 percent of U.S. adults say they take part in some form of daily activity to keep physically fit?

c In explaining its survey methods, Gallup states the following: “For results based on this sample, one can say with 95 percent confidence that the maximum error attributable to sampling and other random effects is plus or minus 3 percentage points.” Explain how your calculations for part b verify that this statement is true.

8.42 In an article in the Journal of Advertising, Weinberger and Spotts compare the use of humor in television ads in the United States and the United Kingdom. They found that a substantially greater percentage of U.K. ads use humor.

a Suppose that a random sample of 400 television ads in the United Kingdom reveals that 142 of these ads use humor. Show that the point estimate and 95 percent confidence interval for the proportion of all U.K. television ads that use humor are as given in the MegaStat output below.

b Suppose a random sample of 500 television ads in the United States reveals that 122 of these ads use humor. Find a point estimate of and a 95 percent confidence interval for the proportion of all U.S. television ads that use humor.

c Do the confidence intervals you computed in parts a and b suggest that a greater percentage of U.K. ads use humor? Explain. How might an ad agency use this information?

8.43 In an article in CA Magazine, Neil Fitzgerald surveyed Scottish business customers concerning their satisfaction with aspects of their banking relationships. Fitzgerald reports that, in 418 telephone interviews conducted by George Street Research, 67 percent of the respondents gave their banks a high rating for overall satisfaction.

a Assuming that the sample is randomly selected, calculate a 99 percent confidence interval for the proportion of Scottish business customers who give their banks a high rating for overall satisfaction.

b Based on this interval, can we be 99 percent confident that more than 60 percent of Scottish business customers give their banks a high rating for overall satisfaction?

8.44 In the March 16, 1998, issue of Fortune magazine, the results of a survey of 2,221 MBA students from across the United States conducted by the Stockholm-based academic consulting firm Universum showed that only 20 percent of MBA students expect to stay at their first job five years or more.4 Assuming that a random sample was employed, find a 95 percent confidence interval for the proportion of all U.S. MBA students who expect to stay at their first job five years or more. Based on this interval, is there strong evidence that fewer than one-fourth of all U.S. MBA students expect to stay?

8.45 Consumer Reports (January 2005) indicates that profit margins on extended warranties are much greater than on the purchase of most products.5 In this exercise we consider a major electronics retailer that wishes to increase the proportion of customers who buy extended warranties on digital cameras. Historically, 20 percent of digital camera customers have purchased the retailer’s extended warranty. To increase this percentage, the retailer has decided to offer a new warranty that is less expensive and more comprehensive. Suppose that three months after starting to offer the new warranty, a random sample of 500 customer sales invoices shows that 152 out of 500 digital camera customers purchased the new warranty. Find a 95 percent confidence interval for the proportion of all digital camera customers who have purchased the new warranty. Are we 95 percent confident that this proportion exceeds .20?

8.46 The manufacturer of the ColorSmart-5000 television set claims 95 percent of its sets last at least five years without needing a single repair. In order to test this claim, a consumer group randomly selects 400 consumers who have owned a ColorSmart-5000 television set for five years. Of these 400 consumers, 316 say their ColorSmart-5000 television sets did not need a repair, whereas 84 say their ColorSmart-5000 television sets did need at least one repair.

a Find a 99 percent confidence interval for the proportion of all ColorSmart-5000 television sets that have lasted at least five years without needing a single repair.

b Does this confidence interval provide strong evidence that the percentage of ColorSmart-5000 television sets that last at least five years without a single repair is less than the 95 percent claimed by the manufacturer? Explain.

8.47 In the book Cases in Finance, Nunnally and Plath present a case in which the estimated percentage of uncollectible accounts varies with the age of the account. Here the age of an unpaid account is the number of days elapsed since the invoice date.

Suppose an accountant believes the percentage of accounts that will be uncollectible increases as the ages of the accounts increase. To test this theory, the accountant randomly selects 500 accounts with ages between 31 and 60 days from the accounts receivable ledger dated one year ago. The accountant also randomly selects 500 accounts with ages between 61 and 90 days from the accounts receivable ledger dated one year ago.

a If 10 of the 500 accounts with ages between 31 and 60 days were eventually classified as uncollectible, find a point estimate of and a 95 percent confidence interval for the proportion of all accounts with ages between 31 and 60 days that will be uncollectible.

b If 27 of the 500 accounts with ages between 61 and 90 days were eventually classified as uncollectible, find a point estimate of and a 95 percent confidence interval for the proportion of all accounts with ages between 61 and 90 days that will be uncollectible.

c Based on these intervals, is there strong evidence that the percentage of accounts aged between 61 and 90 days that will be uncollectible is higher than the percentage of accounts aged between 31 and 60 days that will be uncollectible? Explain.

8.48 Consider Exercise 8.41b and suppose we wish to find the sample size n needed in order to be 95 percent confident that , the sample proportion of respondents who said they took part in some sort of daily activity to keep physically fit, is within a margin of error of .02 of p, the true proportion of all U.S. adults who say that they take part in such activity. In order to find an appropriate value for p(1 − p), note that the 95 percent confidence interval for p that you calculated in Exercise 8.41b was [.57, .63]. This indicates that the reasonable value for p that is closest to .5 is .57, and thus the largest reasonable value for p(1 − p) is .57(1 − .57) = .2451. Calculate the required sample size n.

8.49 Referring to Exercise 8.46, determine the sample size needed in order to be 99 percent confident that , the sample proportion of ColorSmart-5000 television sets that last at least five years without a single repair, is within a margin of error of .03 of p, the true proportion of sets that last at least five years without a single repair.

8.50 Suppose we conduct a poll to estimate the proportion of voters who favor a major presidential candidate. Assuming that 50 percent of the electorate could be in favor of the candidate, determine the sample size needed so that we are 95 percent confident that , the sample proportion of voters who favor the candidate, is within a margin of error of .01 of p, the true proportion of all voters who are in favor of the candidate.

8.5: Confidence Intervals for Parameters of Finite Populations (Optional)

It is best to use the confidence intervals presented in Sections 8.1 through 8.4 when the sampled population is either infinite or finite and much larger than (say, at least 20 times as large as) the sample. Although these previously discussed intervals are sometimes used when a finite population is not much larger than the sample, better methods exist for handling such situations. We present these methods in this section.

As we have explained, we often wish to estimate a population mean. Sometimes we also wish to estimate a population total.

A population total is the sum of the values of all the population measurements.

For example, companies in financial trouble have sometimes falsified their accounts receivable invoices in order to mislead stockholders. For this reason, independent auditors are often asked to estimate a company’s true total sales for a given period. The auditor randomly selects a sample of invoices from the population of all invoices, and then independently determines the actual amount of each sale by contacting the purchasers. The sample results are used to estimate the company’s total sales, and this estimate can then be compared with the total sales reported by the company.

In order to estimate a population total, which we denote as τ (pronounced “tau”), we note that the population mean μ is the population total divided by the number, N, of population measurements. That is, we have μ = τ/N, which implies that τ = Nμ. It follows, because a point estimate of the population mean μ is the sample mean , that

A point estimate of a population total τ is N
, where N is the size of the population.

EXAMPLE 8.12

A company sells and installs satellite dishes and receivers for both private individuals and commercial establishments (bars, restaurants, and so forth). The company accumulated 2,418 sales invoices during last year. The total of the sales amounts listed on these invoices (that is, the total sales claimed by the company) is $5,127,492.17. In order to estimate the true total sales, τ, for last year, an independent auditor randomly selects 242 of the invoices and determines the actual sales amounts by contacting the purchasers. When the sales amounts are averaged, the mean of the actual sales amounts for the 242 sampled invoices is This says that a point estimate of the true total sales τ is

This point estimate is considerably lower than the claimed total sales of $5,127,492.17. However, we cannot expect the point estimate of τ to exactly equal the true total sales, so we need to calculate a confidence interval for τ before drawing any unwarranted conclusions.

In order to find a confidence interval for the mean and total of a finite population, we consider the sampling distribution of the sample mean . It can be shown that, if we randomly select a large sample of n measurements without replacement from a finite population of N measurements, then the sampling distribution of is approximately normal with mean and standard deviation

It can also be shown that the appropriate point estimate of is , where s is the sample standard deviation. This point estimate of is used in the confidence intervals for μ and τ, which we summarize as follows:

Confidence Intervals for the Population Mean and Population Total for a Finite Population

Suppose we randomly select a sample of n measurements without replacement from a finite population of N measurements. Then, if n is large (say, at least 30)

1 A 100(1 − α) percent confidence interval for the population mean μ
is

2 A 100(1 − α) percent confidence interval for the population total τ
is found by multiplying the lower and upper limits of the 100(1 − α) percent confidence interval for μ by N.

The quantity in the confidence intervals for μ and τ is called the finite population correction. If the population size N is much larger than (say, at least 20 times as large as) the sample size n, then the finite population correction is approximately equal to 1. For example, if we randomly select (without replacement) a sample of 1,000 from a population of 1 million, then the finite population correction is In such a case, many people believe it is not necessary to include the finite population correction in the confidence interval calculations. This is because the correction is not far enough below 1 to meaningfully shorten the confidence intervals for μ and τ. However, if the population size N is not much larger than the sample size n (say, if n is more than 5 percent of N), then the finite population correction is substantially less than 1 and should be included in the confidence interval calculations.

EXAMPLE 8.13

Recall that the satellite dish dealer claims that its total sales τ for last year were $5,127,492.17. Since the company accumulated 2,418 invoices during last year, the company is claiming that μ, the mean sales amount per invoice, is$5,127,492.17/2,418 = $2,120.55. Suppose when the independent auditor randomly selects a sample of n = 242 invoices, the mean and standard deviation of the actual sales amounts for these invoices are Here the sample size n = 242 is (242/2,418)100 = 10.008 percent of the population size N = 2,418. Because n is more than 5 percent of N, we should include the finite population correction in our confidence interval calculations. It follows that a 95 percent confidence interval for the mean sales amount μ per invoice is

The upper limit of this interval is less than the mean amount of $2,120.55 claimed by the company, so we have strong evidence that the company is overstating its mean sales per invoice for last year. A 95 percent confidence interval for the total sales τ last year is found by multiplying the lower and upper limits of the 95 percent confidence interval for μ by N = 2,418. Therefore, this interval is [1,782.21(2,418), 1,905.65(2,418)], or [4,309,383.8, 4,607,861.7]. Because the upper limit of this interval is more than $500,000 below the total sales amount of $5,127,492.17 claimed by the company, we have strong evidence that the satellite dealer is substantially overstating its total sales for last year.

We sometimes estimate the total number, τ, of population units that fall into a particular category. For instance, the auditor of Examples 8.12 and 8.13 might wish to estimate the total number of the 2,418 invoices having incorrect sales amounts. Here the proportion, p, of the population units that fall into a particular category is the total number, τ, of population units that fall into the category divided by the number, N, of population units. That is, p = τ/N, which implies that τ = Np. Therefore, since a point estimate of the population proportion p is the sample proportion , a point estimate of the population total τ is . For example, suppose that 34 of the 242 sampled invoices have incorrect sales amounts. Because the sample proportion is , a point estimate of the total number of the 2,418 invoices that have incorrect sales amounts is

We now summarize how to find confidence intervals for p and τ.

Confidence Intervals for the Proportion of and Total Number of Units in a Category on When Sampling a Finite Population

Suppose that we randomly select a sample of n units without replacement from a finite population of N units. Then, if n is large

1 A 100(1 − α) percent confidence interval for the population proportion p
is

2 A 100(1 − α) percent confidence interval for the population total τ
is found by multiplying the lower and upper limits of the 100(1 − α) percent confidence interval for p by N.

EXAMPLE 8.14

Recall that in Examples 8.12 and 8.13 we found that 34 of the 242 sampled invoices have incorrect sales amounts. Since , a 95 percent confidence interval for the proportion of the 2,418 invoices that have incorrect sales amounts is

This interval says we are 95 percent confident that between 9.89 percent and 18.21 percent of the invoices have incorrect sales amounts. A 95 percent confidence interval for the total number of the 2,418 invoices that have incorrect sales amounts is found by multiplying the lower and upper limits of the 95 percent confidence interval for p by N = 2,418. Therefore, this interval is [.0989(2,418), .1821(2,418)], or [239.14, 440.32], and we are 95 percent confident that between (roughly) 239 and 440 of the 2,418 invoices have incorrect sales amounts.

Finally, we can determine the sample size that is needed to make the margin of error in a confidence interval for μ, p, or τ equal to a desired size E by setting the appropriate margin of error formula equal to E and by solving the resulting equation for the sample size n. We will not carry out the details in this book, but the procedure is the same as illustrated in Sections 8.3 and 8.4. Exercise 8.57 gives the reader an opportunity to use the sample size formulas that are obtained.

Exercises for Section 8.5

CONCEPTS

8.51 Define a population total. Give an example of a population total that will interest you in your career when you graduate from college.

8.52 Explain why the finite population correction is unnecessary when the population is at least 20 times as large as the sample. Give an example using numbers.

METHODS AND APPLICATIONS

8.53 A retailer that sells home entertainment systems accumulated 10,451 sales invoices during the previous year. The total of the sales amounts listed on these invoices (that is, the total sales claimed by the company) is $6,384,675. In order to estimate the true total sales for last year, an independent auditor randomly selects 350 of the invoices and determines the actual sales amounts by contacting the purchasers. The mean and the standard deviation of the 350 sampled sales amounts are

a Find a 95 percent confidence interval for μ, the true mean sales amount per invoice on the 10,451 invoices.

b Find a point estimate of and a 95 percent confidence interval for τ, the true total sales for the previous year.

c What does this interval say about the company’s claim that the true total sales were $6,384,675? Explain.

8.54 A company’s manager is considering simplification of a travel voucher form. In order to assess the costs associated with erroneous travel vouchers, the manager must estimate the total number of such vouchers that were filled out incorrectly in the last month. In a random sample of 100 vouchers drawn without replacement from the 1,323 travel vouchers submitted in the last month, 31 vouchers were filled out incorrectly.

a Find a point estimate of and a 95 percent confidence interval for the true proportion of travel vouchers that were filled out incorrectly in the last month.

b Find a point estimate of and a 95 percent confidence interval for the total number of travel vouchers that were filled out incorrectly in the last month.

c If it costs the company $10 to correct an erroneous travel voucher, find a reasonable estimate of the minimum cost of correcting all of last month’s erroneous travel vouchers. Would it be worthwhile to spend $5,000 to design a simplified travel voucher that could be used for at least a year?

8.55 A personnel manager is estimating the total number of person-days lost to unexcused absences by hourly workers in the last year. In a random sample of 50 employees drawn without replacement from the 687 hourly workers at the company, records show that the 50 sampled workers had an average of days of unexcused absences over the past year with a standard deviation of s = 1.26.

a Find a point estimate of and a 95 percent confidence interval for the total number of unexcused absences by hourly workers in the last year.

b Can the personnel manager be 95 percent confident that more than 2,500 person-days were lost to unexcused absences last year? Can the manager be 95 percent confident that more than 3,000 person-days were lost to unexcused absences last year? Explain.

8.56 An auditor randomly samples 32 accounts receivable without replacement from a firm’s 600 accounts and checks to verify that all documents for the accounts comply with company procedures. Ten of the 32 accounts are found to have documents not in compliance. Find a point estimate of and a 95 percent confidence interval for the total number of accounts having documents that do not comply with company procedures.

8.57 SAMPLE SIZES WHEN SAMPLING FINITE POPULATIONS

a Estimating μ and τ

Consider randomly selecting a sample of n measurements without replacement from a finite population consisting of N measurements and having variance σ2. Also consider the sample size given by the formula

Then, it can be shown that this sample size makes the margin of error in a 100(1 − α) percent confidence interval for μ equal to E if we set D equal to (E/zα/2)2. It can also be shown that this sample size makes the margin of error in a 100(1 − α) percent confidence interval for τ equal to E if we set D equal to [E/(zα/2N)]2. Now consider Exercise 8.55. Using s2 = (1.26)2, or 1.5876, as an estimate of σ2, determine the sample size that makes the margin of error in a 95 percent confidence interval for the total number of person-days lost to unexcused absences last year equal to 100 days.

b Estimating p and τ

Consider randomly selecting a sample of n units without replacement from a finite population consisting of N units and having a proportion p of these units fall into a particular category. Also, consider the sample size given by the formula

It can be shown that this sample size makes the margin of error in a 100(1 − α) percent confidence interval for p equal to E if we set D equal to (E/zα/2)2. It can also be shown that this sample size makes the margin of error in a 100(1 − α) percent confidence interval for tequal to E if we set D equal to [E/(zα/2N)]2. Now consider Exercise 8.54. Using as an estimate of p, determine the sample size that makes the margin of error in a 95 percent confidence interval for the proportion of the 1,323 vouchers that were filled out incorrectly equal to .04.

8.6: A Comparison of Confidence Intervals and Tolerance Intervals (Optional)

In this section we compare confidence intervals with tolerance intervals. We saw in Chapter 3 that a tolerance interval is an interval that is meant to contain a specified percentage (often 68.26 percent, 95.44 percent, or 99.73 percent) of the individual population measurements. By contrast, a confidence interval for the population mean μ is an interval that is meant to contain one thing—the population mean μ—and the confidence level associated with the confidence interval expresses how sure we are that this interval contains μ. Often we choose the confidence level to be 95 percent or 99 percent because such a confidence level is usually considered high enough to provide convincing evidence about the true value of μ.

EXAMPLE 8.15: The Car Mileage Case

Recall in the car mileage case that the mean and the standard deviation of the sample of 50 mileages are Also, recall that we have concluded in Example 3.9 (page 129) that the estimated tolerance intervals imply that approximately (1) 68.26 percent of all individual cars will obtain mile ages between 30.8 mpg and 32.4 mpg; (2) 95.44 percent of all individual cars will obtain mileages between 30.0 mpg and 33.2 mpg; and (3) 99.73 percent of all individual cars will obtain mileages between 29.2 mpg and 34.0 mpg. By contrast, we have seen in Section 8.2 (page 317) that a 95 percent t-based confidence interval for the mean, μ, of the mileages of all individual cars is This interval says that we are 95 percent confident that μ is between 31.33 mpg and 31.79 mpg. Figure 8.18 graphically depicts the three estimated tolerance intervals and the 95 percent confidence interval, which are shown below a MegaStat histogram of the 50 mileages. Note that the estimated tolerance intervals, which are meant to contain the many mileages that comprise specified percentages of all individual cars, are longer than the 95 percent confidence interval, which is meant to contain the single population mean μ.

Figure 8.18: A Comparison of Confidence Intervals and Tolerance Intervals

Exercises for Section 8.6

CONCEPTS

8.58 What is a tolerance interval meant to contain?

8.59 What is a confidence interval for the population mean meant to contain?

8.60 Intuitively, why is a tolerance interval longer than a confidence interval?

METHODS AND APPLICATIONS

In Exercises 8.61 through 8.63 we give the mean and the standard deviation of a sample that has been randomly selected from a population. For each exercise, find estimated tolerance intervals that contain approximately 68.26 percent, 95.44 percent, and 99.73 percent of the individual population measurements. Also, find a 95 percent confidence interval for the population mean. Interpret the estimated tolerance intervals and the confidence interval in the context of the situation related to the exercise.

8.61 THE TRASH BAG CASE TrashBag

The mean and the standard deviation of the sample of 40 trash bag breaking strengths are

8.62 THE BANK CUSTOMER WAITING TIME CASE WaitTime

The mean and the standard deviation of the sample of 100 bank customer waiting times are

8.63 THE VIDEO GAME SATISFACTION RATING CASE VideoGame

The mean and the standard deviation of the sample of 65 customer satisfaction ratings are

Chapter Summary

In this chapter we discussed confidence intervals for population means, proportions, and totals. We began by assuming that the population is either infinite or much larger than (say, at least 20 times as large as) the sample. First, we studied how to compute a confidence interval for a population mean. We saw that when the population standard deviation σ is known, we can use the normal distribution to compute a confidence interval for a population mean. When σ is not known, if the population is normally distributed (or at least mound-shaped) or if the sample size n is large, we use the
t distribution to compute this interval. We also studied how to find the size of the sample needed if we wish to compute a confidence interval for a mean with a prespecified confidence level and with a prespecified margin of error. Figure 8.19 is a flowchart summarizing our discussions concerning how to compute an appropriate confidence interval for a population mean.

Figure 8.19: Computing an Appropriate Confidence Interval for a Population Mean

Next we saw that we are often interested in estimating the proportion of population units falling into a category of interest. We showed how to compute a large sample confidence interval for a population proportion, and we saw how to find the sample size needed to estimate a population proportion with a prespecified confidence level and with a prespecified margin of error.

In optional Section 8.5 we continued by studying how to compute confidence intervals for parameters of finite populations that are not much larger than the sample. We saw how to compute confidence intervals for a population mean and total when we are sampling without replacement. We also saw how to compute confidence intervals for a population proportion and for the total number of units in a category when sampling a finite population. In optional section 8.6 we concluded this chapter by comparing confidence intervals with tolerance intervals.

Glossary of Terms

confidence coefficient:

The (before sampling) probability that a confidence interval for a population parameter will contain the population parameter. (page 307)

confidence interval:

An interval of numbers computed so that we can be very confident (say, 95 percent confident) that a population parameter is contained in the interval. (page 305)

confidence level:

The percentage of time that a confidence interval would contain a population parameter if all possible samples were used to calculate the interval. (pages 306 and 308)

degrees of freedom (for a t curve):

A parameter that describes the exact spread of the curve of a t distribution. (page 314)

margin of error:

The quantity that is added to and subtracted from a point estimate of a population parameter to obtain a confidence interval for the parameter. It gives the maximum distance between the population parameter of interest and its point estimate when we assume the parameter is inside the confidence interval. (page 308)

population total:

The sum of the values of all the population measurements. (page 332)

standard error of the estimate :

The point estimate of . (page 318)

t distribution:

A commonly used continuous probability distribution that is described by a distribution curve similar to a normal curve. The t curve is symmetrical about zero and is more spread out than a standard normal curve. (pages 313 and 314)

t point, tα:

The point on the horizontal axis under a t curve that gives a right-hand tail area equal to α. (page 314)

t table:

A table of t point values listed according to the area in the tail of the t curve and according to values of the degrees of freedom. (pages 314–315)

Important Formulas

A z-based confidence interval for a population mean μ with σ known: page 309

A t-based confidence interval for a population mean μ with σ unknown: page 316

Sample size when estimating μ: page 322

A large sample confidence interval for a population proportion p: page 325

Sample size when estimating p: page 327

Estimation of a mean and a total for a finite population: page 333

Estimation of a proportion and a total for a finite population: page 335

Supplementary Exercises

8.64 In an article in the Journal of Accounting Research, Ashton, Willingham, and Elliott studied audit delay (the length of time from a company’s fiscal year-end to the date of the auditor’s report) for industrial and financial companies. In the study, a random sample of 250 industrial companies yielded a mean audit delay of 68.04 days with a standard deviation of 35.72 days, while a random sample of 238 financial companies yielded a mean audit delay of 56.74 days with a standard deviation of 34.87 days. Use these sample results to do the following:

a Calculate a 95 percent confidence interval for the mean audit delay for all industrial companies. Note: t.025 = when df = 249.

b Calculate a 95 percent confidence interval for the mean audit delay for all financial companies. Note: t.025 = 1.97 when df = 237.

c By comparing the 95 percent confidence intervals you calculated in parts a and b, is there strong evidence that the mean audit delay for financial companies is shorter than the mean audit delay for industrial companies? Explain.

8.65 In an article in Accounting and Business Research, Beattie and Jones investigate the use and abuse of graphic presentations in the annual reports of United Kingdom firms. The authors found that 65 percent of the sampled companies graph at least one key financial variable, but that 30 percent of the graphics are materially distorted (nonzero vertical axis, exaggerated trend, or the like). Results for U.S. firms have been found to be similar.

a Suppose that in a random sample of 465 graphics from the annual reports of United Kingdom firms, 142 of the graphics are found to be distorted. Find a point estimate of and a 95 percent confidence interval for the proportion of U.K. annual report graphics that are distorted.

b Based on this interval, can we be 95 percent confident that more than 25 percent of all graphics appearing in the annual reports of U.K. firms are distorted? Explain. Does this suggest that auditors should understand proper graphing methods?

c Determine the sample size needed in order to be 95 percent confident that , the sample proportion of U.K. annual report graphics that are distorted, is within a margin of error of .03 of p, the true proportion of U.K. annual report graphics that are distorted.

8.66 On January 4, 2000, the Gallup Organization released the results of a poll dealing with the likelihood of computer-related Y2K problems and the possibility of terrorist attacks during the New Year’s holiday at the turn of the century.6 The survey results were based on telephone interviews with a randomly selected national sample of 622 adults, 18 years and older, conducted December 28, 1999.

a The Gallup poll found that 61 percent of the respondents believed that one or more terrorist attacks were likely to happen on the New Year’s holiday. Based on this finding, calculate a 95 percent confidence interval for the proportion of all U.S. adults who believed that one or more terrorist attacks were likely to happen on the 2000 New Year’s holiday. Based on this interval, is it reasonable to conclude that fewer than two-thirds of all U.S. adults believed that one or more terrorist attacks were likely?

b In explaining its survey methods, Gallup states the following: “For results based on this sample, one can say with 95 percent confidence that the maximum error attributable to sampling and other random effects is plus or minus 4 percentage points.” Explain how your calculations for part a verify that this statement is true.

8.67 The manager of a chain of discount department stores wishes to estimate the total number of erroneous discounts allowed by sales clerks during the last month. A random sample of 200 of the chain’s 57,532 transactions for the last month reveals that erroneous discounts were allowed on eight of the transactions. Use this sample information to find a point estimate of and a 95 percent confidence interval for the total number of erroneous discounts allowed during the last month.

8.68 THE DISK BRAKE CASE

National Motors has equipped the ZX-900 with a new disk brake system. We define the stopping distance for a ZX-900 to be the distance (in feet) required to bring the automobile to a complete stop from a speed of 35 mph under normal driving conditions using this new brake system. In addition, we define μ to be the mean stopping distance of all ZX-900s. One of the ZX-900’s major competitors is advertised to achieve a mean stopping distance of 60 feet. National Motors would like to claim in a new advertising campaign that the ZX-900 achieves a shorter mean stopping distance.

Suppose that National Motors randomly selects a sample of n = 81 ZX-900s. The company records the stopping distance of each automobile and calculates the mean and standard deviation of the sample of n = 81 stopping distances to be ft and s = 6.02 ft.

a Calculate a 95 percent confidence interval for μ. Can National Motors be 95 percent confident that μ is less than 60 ft? Explain.

b Using the sample of n = 81 stopping distances as a preliminary sample, find the sample size necessary to make National Motors 95 percent confident that is within a margin of error of one foot of μ.

8.69 A large construction contractor is building 257 homes, which are in various stages of completion. For tax purposes, the contractor needs to estimate the total dollar value of its inventory due to construction in progress. The contractor randomly selects (without replacement) a sample of 40 of the 257 houses and determines the accumulated costs (the amount of money tied up in inventory) for each sampled house. The contractor finds that the sample mean accumulated cost is and that the sample standard deviation is s = $28,865.04.

a Find a point estimate of and a 99 percent confidence interval for the total accumulated costs (total amount of money tied up in inventory) for all 257 homes that are under construction.

b Using the confidence interval as the basis for your answer, find a reasonable estimate of the largest possible total dollar value of the contractor’s inventory due to construction in progress.

8.70 In an article in the Journal of Retailing, J. G. Blodgett, D. H. Granbois, and R. G. Walters investigated negative word-of-mouth consumer behavior. In a random sample of 201 consumers, 150 reported that they engaged in negative word-of-mouth behavior (for instance, they vowed never to patronize a retailer again). In addition, the 150 respondents who engaged in such behavior, on average, told 4.88 people about their dissatisfying experience (with a standard deviation equal to 6.11).

a Use these sample results to compute a 95 percent confidence interval for the proportion of all consumers who engage in negative word-of-mouth behavior. On the basis of this interval, would it be reasonable to claim that more than 70 percent of all consumers engage in such behavior? Explain.

b Use the sample results to compute a 95 percent confidence interval for the mean number of people who are told about a dissatisfying experience by consumers who engage in negative word-of-mouth behavior. On the basis of this interval, would it be reasonable to claim that these dissatisfied consumers tell, on average, at least three people about their bad experience? Explain. Note: t.025 = 1.98 when df = 149.

8.71 THE CIGARETTE ADVERTISEMENT CASE ModelAge

A random sample of 50 perceived age estimates for a model in a cigarette advertisement showed that years and that s = 3.7432 years.

a Use this sample to calculate a 95 percent confidence interval for the population mean age estimate for all viewers of the ad.

b Remembering that the cigarette industry requires that models must appear at least 25 years old, does the confidence interval make us 95 percent confident that the mean perceived age estimate is at least 25? Is the mean perceived age estimate much more than 25? Explain.

8.72 In an article in the Journal of Management Information Systems, Mahmood and Mann investigate how information technology (IT) investment relates to company performance. In particular, Mahmood and Mann obtain sample data concerning IT investment for companies that effectively use information systems. Among the variables studied are the company’s IT budget as a percentage of company revenue, percentages of the IT budget spent on staff and training, and number of PCs and terminals as a percentage of total employees.

a Suppose a random sample of 15 companies considered to effectively use information systems yields a sample mean IT budget as a percentage of company revenue of with a standard deviation of s = 1.64. Assuming that IT budget percentages are approximately normally distributed, calculate a 99 percent confidence interval for the mean IT budget as a percentage of company revenue for all firms that effectively use information systems. Does this interval provide evidence that a firm can successfully use information systems with an IT budget that is less than 5 percent of company revenue? Explain.

b Suppose a random sample of 15 companies considered to effectively use information systems yields a sample mean number of PCs and terminals as a percentage of total employees of with a standard deviation of s = 25.37. Assuming approximate normality, calculate a 99 percent confidence interval for the mean number of PCs and terminals as a percentage of total employees for all firms that effectively use information systems. Why is this interval so wide? What can we do to obtain a narrower (more useful) confidence interval?

8.73 THE INVESTMENT CASE InvestRet

Suppose that random samples of 50 returns for each of the following investment classes give the indicated sample mean and sample standard deviation:

a For each investment class, compute a 95 percent confidence interval for the population mean return.

b Do these intervals suggest that the current mean return for each investment class differs from the historical (1970 to 1994) mean return given in Table 3.11 (page 159)? Explain.

8.74 THE INTERNATIONAL BUSINESS TRAVEL EXPENSE CASE

Recall that the mean and the standard deviation of a random sample of 35 one-day travel expenses in Moscow are Find a 95 percent confidence interval for the mean, μ, of all one-day travel expenses in Moscow.

8.75 THE UNITED KINGDOM INSURANCE CASE

Assume that the U.K. insurance survey is based on 1,000 randomly selected U.K. households and that 640 of these households spent money for life insurance in 1993. Find a 95 percent confidence interval for the proportion, p, of all U.K. households that spent money for life insurance in 1993.

8.76 How safe are child car seats? Consumer Reports (May 2005) tested the safety of child car seats in 30 mph crashes. They found “slim safety margins” for some child car seats. Suppose that Consumer Reports simulates the safety of the market-leading child car seat. Their test consists of placing the maximum claimed weight in the car seat and simulating crashes at higher and higher miles per hour until a problem occurs. The following data identifies the speed at which a problem with the car seat first appeared; such as the strap breaking, seat shell cracked, strap adjuster broke, detached from the base, etc.: 31.0, 29.4, 30.4, 28.9, 29.7, 30.1, 32.3, 31.7, 35.4, 29.1, 31.2, 30.2. Using the fact that , find a 95 percent confidence interval for the true mean speed at which a problem with the car seat first appears. Are we 95 percent confident that this mean is at least 30 mph? CarSeat

8.77 In Exercise 2.85 (page 91), we briefly described a series of international quality standards called ISO 9000. In the results of a Quality Systems Update/Deloitte & Touche survey of ISO 9000 registered companies published by CEEM Information Systems, 515 of 620 companies surveyed reported that they are encouraging their suppliers to pursue ISO 9000 registration.7

a Using these survey results, compute a 95.44 percent confidence interval for the proportion of all ISO 9000 registered companies that encourage their suppliers to pursue ISO 9000 registration. Assume here that the survey participants have been randomly selected.

b Based on this interval, is there conclusive evidence that more than 75 percent of all ISO 9000 registered companies encourage their suppliers to pursue ISO 9000 registration?

8.78: Internet Exercise

What is the average selling price of a home? The Data and Story Library (DASL) contains data, including the sale price, for a random sample of 117 homes sold in Albuquerque, New Mexico. Go to the DASL website (http://lib.stat.cmu.edu/DASL/) and retrieve the home price data set (http://lib.stat.cmu.edu/DASL/Datafiles/homedat.html.) Use Minitab, Excel, or MegaStat to produce appropriate graphical (histogram, stem-and-leaf, box plot) and numerical summaries of the price data. Identify, from your numerical summaries, the sample mean and standard deviation. Use these summaries to construct a 99% confidence interval for μ, the mean sale price. Use statistical software (MINITAB, Excel, or MegaStat) to compute a 99% confidence interval for μ. Do the results of your hand calculations agree with those from your statistical software?

Technical note: There are many ways to capture the home price data from the DASL site. One simple way is to select just the rows containing the data values (and not the labels), copy, paste directly into an Excel or MINITAB worksheet, add your own variable labels, and save the resulting worksheet. It is possible to copy the variable labels from DASL as well, but the differences in alignment and the intervening blank line add to the difficulty. AlbHome

Appendix 8.1: Confidence Intervals Using MINITAB

The instruction blocks in this section each begin by describing the entry of data into the MINITAB data window. Alternatively, the data may be loaded directly from the data disk included with the text. The appropriate data file name is given at the top of each instruction block. Please refer to Appendix 1.1 for further information about entering data, saving data, and printing results when using MINITAB.

Confidence interval for a population mean in Figure 8.11 on page 318 (data file: Ratio.MTW):

• In the Data window, enter the debt-to-equity ratio data from Example 8.4 (pages 316–317) into a single column with variable name Ratio.

• Select Stat : Basic Statistics : 1-Sample t.

• In the “1-Sample t (Test and Confidence Interval)” dialog box, select “Samples in columns.”

• Select the variable name Ratio into the “Samples in columns” window.

• Click the Options… button.

• In the “1-Sample t—Options” dialog box, enter the desired level of confidence (here 95.0) into the “Confidence level” window.

• Select “not equal” from the Alternative drop-down menu, and click OK in the “1-Sample t—Options” dialog box.

• To produce a boxplot of the data with a graphical representation of the confidence interval, click the Graphs … button, check the “Boxplot of data” checkbox, and click OK in the “1-Sample t—Graphs” dialog box.

• Click OK in “1-Sample t (Test and Confidence Interval)” dialog box.

• The confidence interval is given in the Session window, and the boxplot appears in a graphics window.

A “1-Sample Z” interval is also available in MINITAB under Basic Statistics. It requires a user-specified value of the population standard deviation, which is rarely known.

Confidence interval for a population proportion in the marketing ethics situation of Example 8.10 on pages 326 and 327:

• Select Stat : Basic Statistics : 1 Proportion

• In the “1 Proportion (Test and Confidence Interval)” dialog box, select “Summarized data.”

• Enter the number of trials (here equal to 205) and the number of successes—or events—(here equal to 117) into the appropriate windows.

• Click on the Options … button.

• In the “1 Proportion—Options” dialog box, enter the desired level of confidence (here 95.0) into the “Confidence level” window.

• Select “not equal” from the Alternative drop-down menu.

• Check the “Use test and interval based on normal distribution” checkbox.

• Click OK in the “1 Proportion—Options” dialog box.

• Click OK in the “1 Proportion (Test and Confidence Interval)” dialog box.

• The confidence interval will be displayed in the Session window.

Appendix 8.2: Confidence Intervals Using Excel

The instruction block in this section begins by describing the entry of data into an Excel spreadsheet. Alternatively, the data may be loaded directly from the data disk included with the text. The appropriate data file name is given at the top of the instruction block. Please refer to Appendix 1.2 for further information about entering data, saving data, and printing results when using Excel.

Confidence interval for a population mean in Figure 8.12(b) on page 319 (data file: DebtEq.xlsx):

• Enter the debt-to-equity ratio data from Example 8.4 (page 317) into cells A2 to A16 with the label Ratio in cell A1.

• Select Data : Data Analysis : Descriptive Statistics.

• Click OK in the Data Analysis dialog box.

• In the Descriptive Statistics dialog box, enter A1.A16 into the Input Range window.

• Place a checkmark in the “Labels in first row” checkbox.

• Under output options, select “New Worksheet Ply” to have the output placed in a new worksheet and enter the name Output for the new worksheet.

• Place checkmarks in the Summary Statistics and “Confidence Level for Mean” checkboxes. This produces a t-based margin of error for a confidence interval.

• Type 95 in the “Confidence Level for Mean” box.

• Click OK in the Descriptive Statistics dialog box.

• A descriptive statistics summary will be displayed in cells A3 through B16 in the Output worksheet. Drag the column borders to reveal complete labels for all of the descriptive statistics.

• Type the heading “95% Confidence Interval” into cells D13 to E13.

• Compute the lower bound of the interval by typing the formula = B3 − B16 into cell D15. This subtracts the margin of error of the interval (labeled “Confidence Level (95%)”) from the sample mean.

• Compute the upper bound of the interval by typing the formula = B3 + B16 into cell E15.

Appendix 8.3: Confidence Intervals Using MegaStat

Confidence interval for the population mean debt-to-equity ratio in Figure 8.12(a) on page 319:

• Select Add-Ins : MegaStat : Confidence Intervals / Sample Size

• In the “Confidence Intervals / Sample Size” dialog box, click on the “Confidence Interval—mean” tab.

• Enter the sample mean (here equal to 1.3433) into the Mean window.

• Enter the sample standard deviation (here equal to .1921) into the “Std Dev” window.

• Enter the sample size (here equal to 15) into the “n” window.

• Select a level of confidence from the pull-down menu or type a desired percentage.

• Select a t-based or z-based interval by clicking on “t” or “z.” Here we request a t-based interval.

• Click OK in the “Confidence Intervals / Sample Size” dialog box.

Confidence interval for a population proportion in the cheese spread situation of Example 8.8 on page 325:

• In the “Confidence Intervals / Sample Size” dialog box, click on the “Confidence interval—p” tab.

• Enter the sample proportion (here equal to .063) into the “p” window.

• Enter the sample size (here equal to 1000) into the “n” window.

• Select a level of confidence from the pull-down menu or type a desired percentage.
• Click OK in the “Confidence Intervals / Sample Size” dialog box.

Sample size determination for a proportion problem in Figure 8.16 on page 328:

• In the “Confidence Intervals / Sample Size” dialog box, click on the “Sample size—p” tab.

• Enter the desired margin of error (here equal to 0.02) into the “E” window and enter an estimate of the population proportion into the “p” window.

• Select a level of confidence from the pull-down menu or type a desired percentage.
• Click OK in the “Confidence Intervals / Sample Size” dialog box.

Sample size determination for a population mean problem is done by clicking on the “Sample Size—mean” tab. Then enter a desired margin of error, an estimate of the population standard deviation, and the desired level of confidence. Click OK.

1 Some statisticians suggest using the more conservative rule that both must be at least 10. Furthermore, because is an unbiased point estimate of p(1 − p) / n, a more correct 100(1 − α) percent confidence interval for p is . However, because n is large, there is little difference between intervals obtained by using this formula and those obtained by using the formula in the above box.

2 Source: “Driving Organic Growth at Bank of America” Quality Progress (February 2005), pp. 23–27.

3 Source: World Wide Web, http://www.gallup.com/poll/releases/, The Gallup Organization, January 7, 2000.

4 Source: Shelly Branch, “MBAs: What Do They Really Want,” Fortune (March 16, 1998), p. 167.

5 Consumer Reports, January 2005, page 51.

6 Source: World Wide Web, http://www.gallup.com/poll/releases/, The Gallup Organization, January 4, 2000.

7 Source: Is ISO 9000 for You? (Fairfax, VA: CEEM Information Services).

(Bowerman 304)

Bowerman, Bruce L. Business Statistics in Practice, 5th Edition. McGraw-Hill Learning Solutions, 022008. .