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Chapter 9 Estimation from Sample Data Work Sampling © Dana White/PhotoEdit In the workplace, production experts sometimes conduct studies to find out how much time workers spend doing various job activities. This information can be used in establishing or updating standards for performance, as well as for comparing and evaluating worker performance. Among the approaches for determining how much time a worker spends at various activities is a technique known as work sampling. Compared to alternative methods (such as following the person and timing him or her with a stopwatch), work sampling is unobtrusive in that the behavior observed is not influenced by the observation process itself. In work sampling, a worker is observed at randomly selected points along an interval of time; then the proportion of these observations that involve each selected activity is determined. For example, if we want to determine how much of the time a secretary spends keyboarding, we can observe the person at random times during a typical day or week, then calculate the proportion of time that he or she happens to be keyboarding. If the person were observed to be keyboarding in 100 of 280 random observations, the sample proportion ata mple d would be 100兾280, or 0.357. d on sa e s a b e renc Using this information, along an infe Making with estimation techniques described later in this chapter, we could then arrive at an interval estimate reflecting the likely range of values within which the true population proportion lies. When you finish Section 9.6, you might like to pay a short visit back here and verify that the 90% confidence interval for the population proportion is from 0.310 to 0.404, and that we are 90% confident that the person spends somewhere between 31.0% and 40.4% of his or her time clicking away at the keyboard. 270 Part 3: Sampling Distributions and Estimation learning objectives After reading this chapter, you should be able to: 9.1 • Explain the difference between a point estimate and an interval estimate for a population parameter. • Use the standard normal distribution in constructing a confidence interval for a population mean or proportion. • Use the t distribution in constructing a confidence interval for a population mean. • Decide whether the standard normal distribution or the t distribution should be used in constructing a given confidence interval. • Determine how large a simple random sample must be in order to estimate a population mean or proportion at specified levels of accuracy and confidence. • Use Excel and Minitab to construct confidence intervals. INTRODUCTION In Chapter 8, we began with a population having a known mean () or proportion (); then we examined the sampling distribution of the corresponding sample statistic (x or p) for samples of a given size. In this chapter, we’ll be going in the opposite direction—based on sample data, we will be making estimates involving the (unknown) value of the population mean or proportion. As mentioned previously, the use of sample information to draw conclusions about the population is known as inferential statistics. To repeat a very important point, this chapter completes a key transition discussed at the beginning of Chapter 7: • In Chapter 7, we had access to the population mean and we made probability statements about individual x values taken from the population. • In Chapter 8, we again had access to the population mean, but we invoked the central limit theorem and began making probability statements about the means of samples taken from the population. (Beginning with Chapter 8, the sample mean itself is considered as a random variable.) • In this chapter, we again lack access to the population mean, but we will begin using sample data as the basis from which to make probability statements about the true (but unknown) value of the population mean. As in Chapter 8, we will be relying heavily on the central limit theorem. In the following sections, we will use sample data to make both point and interval estimates regarding the population mean or proportion. While the point estimate is a single number that estimates the exact value of the population parameter of interest (e.g., or ), an interval estimate includes a range of possible values that are likely to include the actual population parameter. When the interval estimate is associated with a degree of confidence that it actually includes the population parameter, it is referred to as a confidence interval. Point and interval estimates can also be made regarding the difference between two population means (1 2) or proportions (1 2). These involve data from two samples, and they will be discussed in the context of the hypothesis-testing procedures of Chapter 11. Chapter 9: Estimation from Sample Data 271 Whenever sample data are used for estimating a population mean or proportion, sampling error will tend to be present because a sample has been taken instead of a census. As a result, the observed sample statistic (x or p) will differ from the actual value of the population parameter ( or ). Assuming a simple random sampling of elements from the population, formulas will be presented for determining how large a sample size is necessary to ensure that such sampling error is not likely to exceed a given amount. exercises 9.1 Differentiate between a point estimate and an interval 9.3 What is necessary for an interval estimate to be a estimate for a population parameter. confidence interval? 9.2 What is meant by inferential statistics, and what role does it play in estimation? 9.2 POINT ESTIMATES An important consideration in choosing a sample statistic as a point estimate of the value of a population parameter is that the sample statistic be an unbiased estimator. An estimator is unbiased if the expected value of the sample statistic is the same as the actual value of the population parameter it is intended to estimate. Three important point estimators introduced in the chapter are those for a population mean (), a population variance (2), and a population proportion (). As Chapter 8 showed, the expected value of the sample mean is the population mean, and the expected value of the sample proportion is the population proportion. As a result, x and p are unbiased estimators of and , respectively. Table 9.1 presents a review of the applicable formulas. Note that the divisor in the formula for the sample variance (s2) is (n 1). Using (n 1) as the divisor in calculating the variance of the sample results in s2 being an unbiased estimate of the (unknown) population variance, 2. The positive square root of s2, the sample standard deviation (s), will not be an unbiased estimate of the population standard deviation (). In practice, however, s is the most frequently used estimator of its population counterpart, . Population Parameter Unbiased Estimator Mean, x Variance, 2 s2 Proportion, p TABLE 9.1 Formula x s2 兺xi n 兺(xi x)2 n1 x successes p n trials An estimator is unbiased if its expected value is the same as the actual value of the corresponding population parameter. Listed here are unbiased point estimators for a population mean, a population variance, and a population proportion. 272 Part 3: Sampling Distributions and Estimation exercises 9.4 What is meant when a sample statistic is said to be an unbiased estimator? 9.5 When calculating the sample variance, what procedure is necessary to ensure that s2 will be an unbiased estimator of 2? Will s be an unbiased estimator of ? 9.6 During the month of July, an auto manufacturer gives its production employees a vacation period so it can tool up for the new model run. In surveying a simple random sample of 200 production workers, the personnel director finds that 38% of them plan to vacation out of state for at least one week during this period. Is this a point estimate or an interval estimate? Explain. days each was absent during the past month was found to be 0, 2, 4, 2, 1, 7, 3, and 2, respectively. a. What is the point estimate for , the mean number of days absent for the firm’s employees? b. What is the point estimate for 2, the variance of the number of days absent? 9.8 The average annual U.S. per capita consumption of 9.7 A simple random sample of 8 employees is selected iceberg lettuce has been estimated as 24.3 pounds. The annual per capita consumption 2 years earlier had been estimated as 21.6 pounds. Could either or both of these consumption figures be considered a point estimate? Could the difference between the two consumption figures be considered an interval estimate? Explain your reasoning in both cases. SOURCE: Bureau of the Census, Statistical from a large firm. For the 8 employees, the number of Abstract of the United States 2002, p. 130. 9.3 A PREVIEW OF INTERVAL ESTIMATES When we know the values of the population mean and standard deviation, we can (if either the population is normally distributed or n is large) use the standard normal distribution in determining the proportion of sample means that will fall within a given number of standard error (x) units of the known population mean. This is exactly what we did in Chapter 8. It is typical of inferential statistics that we must use the mean (x) and standard deviation (s) of a single sample as our best estimates of the (unknown) values of and . However, this does not prevent us from employing x and s in constructing an estimated sampling distribution for all means having this sample size. This is the basis for the construction of an interval estimate for the population mean. When we apply the techniques of this chapter and establish the sample mean as the midpoint of an interval estimate for the population mean, the resulting interval may or may not include the actual value of . For example, in Figure 9.1, six of the seven simple random samples from the same population led to an interval estimate that included the true value of the population mean. In Figure 9.1, the mean of sample number 1 ( x 1) is slightly greater than the population mean (), and the interval estimate based on this sample actually includes . For sample 3, taken from the same population, the estimation interval does not include . In Figure 9.1, we can make these observations because the value of is known. In practice, however, we will not have the benefit of knowing the actual value of the population mean. Therefore, we will not be able to say with complete certainty that an interval based on our sample result will actually include the (unknown) value of . Chapter 9: Estimation from Sample Data 273 FIGURE 9.1 Sampling distribution of the mean for all possible simple random samples with a given sample size x mx = m = 20.0 Does this interval include the actual population mean, m? 15.0 Yes 22.0 29.0 x1 11.0 18.0 25.0 Yes x2 23.5 30.5 37.5 No x3 10.0 17.0 24.0 Yes x4 8.0 15.0 22.0 Yes x5 19.0 26.0 33.0 Yes x6 14.5 21.5 28.5 Yes x7 mx = m = 20.0 The interval estimate for the mean simply describes a range of values that is likely to include the actual population mean. This is also the case for our use of the sample proportion (p) to estimate the population proportion (), as well as for our construction of an interval estimate within which the actual value of is likely to fall. The following terms are of great importance in interval estimation: INTERVAL ESTIMATE A range of values within which the actual value of the population parameter may fall. INTERVAL LIMITS The lower and upper values of the interval estimate. (continued) Examples of seven different interval estimates for a population mean, with each interval based on a separate simple random sample from the population. Six of the seven interval estimates include the actual value of . 274 Part 3: Sampling Distributions and Estimation CONFIDENCE INTERVAL An interval estimate for which there is a specified degree of certainty that the actual value of the population parameter will fall within the interval. CONFIDENCE COEFFICIENT For a confidence interval, the proportion of such intervals that would include the population parameter if the process leading to the interval were repeated a great many times. CONFIDENCE LEVEL Like the confidence coefficient, this expresses the degree of certainty that an interval will include the actual value of the population parameter, but it is stated as a percentage. For example, a 0.95 confidence coefficient is equivalent to a 95% confidence level. ACCURACY The difference between the observed sample statistic and the actual value of the population parameter being estimated. This may also be referred to as estimation error or sampling error. To illustrate these and several other terms discussed so far, we have provided their values in the following example, which is typical of published statistical findings. The methods by which the values were determined will become apparent in the sections to follow. example Interval Estimates “In our simple random sample of 2000 households, we found the average income to be x $65,000, with a standard deviation, s $12,000. Based on these data, we have 95% confidence that the population mean is somewhere between $64,474 and $65,526.” SOLUTION • • • • • • • Point estimate of $65,000 Point estimate of $12,000 Interval estimate of $64,474 to $65,526 Lower and upper interval limits for $64,474 and $65,526 Confidence coefficient 0.95 Confidence level 95% Accuracy For 95% of such intervals, the sample mean would not differ from the actual population mean by more than $526. When constructing a confidence interval for the mean, a key consideration is whether we know the actual value of the population standard deviation (). As Figure 9.2 shows, this will determine whether the normal distribution or the t distribution (see Section 9.5) will be used in determining the appropriate interval. Figure 9.2 also summarizes the procedure for constructing the confidence interval for the population proportion, a technique that will be discussed in Section 9.6. Chapter 9: Estimation from Sample Data 275 FIGURE 9.2 Confidence interval estimate for a Population mean Population proportion s known s unknown z-interval, with limits: s x ± z ––– √n t-interval, with limits: s x ± t ––– √n and df = n – 1 z-interval, with limits: p(1 – p) p ± z ––––––– n Section 9.4 Note 1 Section 9.5 Note 2 Section 9.6 Note 3 This figure provides an overview of the methods for determining confidence interval estimates for a population mean or a population proportion and indicates the chapter section in which each is discussed. Key assumptions are reviewed in the figure notes. √ 1If the population is not normally distributed, n should be at least 30 for the central limit theorem to apply. 2When is unknown, but the underlying population can be assumed to be approximately normally distributed, use of the t distribution is a necessity when n 30. Use of the t distribution is also appropriate, however, when is unknown and the sample sizes are larger. Most computer statistical packages routinely use the t-interval for all sample sizes when s is used to estimate . 3Assumes that both np and n(1 p) are 5. The normal distribution as an approximation to the binomial improves as n becomes larger and for values of p that are closer to 0.5. exercises 9.9 Exactly what is meant by the accuracy of a point estimate? 9.10 A population is approximately normally distributed and the sample size is to be n 40. What additional factor must be considered in determining whether to use the standard normal distribution in constructing the confidence interval for the population mean? 9.11 “In surveying a simple random sample of 1000 em- ployed adults, we found that 450 individuals felt they were underpaid by at least $3000. Based on these results, we have 95% confidence that the proportion of the population of employed adults who share this sentiment is between 0.419 and 0.481.” For this summary statement, identify the a. point estimate of the population proportion. b. confidence interval estimate for the population proportion. c. confidence level and the confidence coefficient. d. accuracy of the sample result. CONFIDENCE INTERVAL ESTIMATES FOR THE MEAN: KNOWN If we don’t know the value of the population mean, chances are that we also do not know the value of the population standard deviation. However, in some cases, usually industrial processes, may be known while is not. If the population 9.4 276 Part 3: Sampling Distributions and Estimation cannot be assumed to be normally distributed, the sample size must be at least 30 for the central limit theorem to apply. When these conditions are met, the confidence interval for the population mean will be as follows: Confidence interval limits for the population mean, known: xz N O T E S 兹n where x sample mean population standard deviation n sample size z z value corresponding to the level of confidence desired (e.g., z 1.96 for the 95% confidence level) /兹n standard error of the sampling distribution of the mean This application assumes that either (1) the underlying population is normally distributed or (2) the sample size is n 30. Also, an alternative way of describing the z value is to refer to it as z␣/2, with ␣兾2 being the area to the right. For example, z0.025 would be 1.96. example z-Interval, Mean From past experience, the population standard deviation of rod diameters produced by a machine has been found to be 0.053 inches. For a simple random sample of n 30 rods, the average diameter is found to be x 1.400 inches. The underlying data are in file CX09RODS. SOLUTION What Is the 95% Confidence Interval for the Population Mean, ? Although we don’t know the value of , x 1.400 inches is our best estimate of the population mean diameter. As Figure 9.3 shows, the sampling distribution of the mean will have a standard error of /兹n, or 0.053/兹30. For the standard normal distribution, 95% of the area will fall between z 1.96 and z 1.96. We are able to use the standard normal distribution table because n 30 and the central limit theorem can be invoked. As a result, the 95% confidence interval for the (unknown) population mean can be calculated as xz 0.053 1.400 1.96 兹n 兹30 or between 1.381 and 1.419 inches Figure 9.3 shows the midpoint (x 1.400 inches) for the 95% confidence interval for the mean, along with the lower and upper limits for the confidence interval. Based on our calculations, the 95% confidence interval for the population mean is from 1.381 to 1.419 inches. Chapter 9: Estimation from Sample Data 277 FIGURE 9.3 Normal distribution: for 95% confidence, z will be ±1.96. Area = 0.95 z = –1.96 0 z = +1.96 The 95% confidence interval for m is s x – z ––– √n 0.053 1.400 – 1.96 ––––– √30 or 1.381 x s x + z ––– √n 1.400 0.053 1.400 + 1.96 ––––– √30 1.400 1.419 More precisely, 95% of such intervals constructed in this way would include the population mean. Since we have taken only one sample and constructed just one interval, it is technically correct to say we have 95% confidence that this particular interval contains the population mean. Although the logic may be tempting, this is not the same as saying the probability is 0.95 that this particular interval will include the population mean. With other factors unchanged, a higher confidence level will require a wider confidence interval. Likewise, a lower confidence level will lead to a narrower confidence interval. In other words, the more certain we wish to be that the interval estimate contains the population parameter, the wider the interval will have to be. Refer to Seeing Statistics Applet 9, at the end of the chapter, to visually demonstrate how the width of the confidence interval changes when higher and lower levels of confidence are specified. Computer Solutions 9.1 shows how we can use Excel or Minitab to generate a confidence interval for the mean when the population standard deviation is known or assumed. In this case, we are replicating the 95% confidence interval shown in Figure 9.3, and the 30 data values are in file CX09RODS. If we use an Excel procedure based on summary statistics, it can be interesting to examine “what-if” scenarios to instantly see how changes in the specified confidence level would change the width of the confidence interval. Construction of the 95% confidence interval for the population mean, based on a sample of 30 rods for which the average diameter is 1.400 inches. From past experience, the population standard deviation is known to be 0.053 inches. Because is known, the normal distribution can be used in determining the interval limits. We have 95% confidence that is between 1.381 and 1.419 inches. 278 Part 3: Sampling Distributions and Estimation computer solutions 9.1 Confidence Interval for Population Mean, Known These procedures show how to construct a confidence interval for the population mean when the population standard deviation is known. EXCEL 1 2 3 4 5 6 7 8 9 A B z-Estimate: Mean Mean Standard Deviation Observations SIGMA LCL UCL C diameter 1.400 0.052 30 0.053 1.381 1.419 Excel confidence interval for based on raw data and known 1. For example, using the 30 rod diameters (file CX09RODS.XLS) on which Figure 9.3 is based: The label and 30 data values are in A1:A31. Click Tools. Click Data Analysis Plus. Click Z-Estimate: Mean. Click OK. 2. Enter A1:A31 into the Input Range box. Enter the known population standard deviation (0.053) into the Standard Deviation (SIGMA) box. Click Labels, since the variable name is in the first cell within the field. The desired confidence level as a decimal fraction is 0.95, so the corresponding alpha value is 1 0.95 0.05. Enter 0.05 into the Alpha box. Click OK. The confidence interval results will be as shown above. Excel confidence interval for based on summary statistics and known 1. For example, with x 1.400, 0.053, and n 30, as in Figure 9.3: Open the ESTIMATORS.XLS workbook, supplied with the text. 2. Using the arrows at the bottom left, select the z-Estimate_Mean worksheet. Enter the sample mean (1.4), the known sigma (0.053), the sample size (30), and the desired confidence level as a decimal fraction (0.95). (Note: As an alternative, you can use Excel worksheet template TMZINT.XLS, supplied with the text. The steps are described within the template.) MINITAB Confidence interval for based on raw data and known One-Sample Z: diameter The assumed standard deviation = 0.053 Variable diameter N 30 Mean 1.40000 StDev 0.05196 SE Mean 0.00968 95% CI (1.38103, 1.41897) 1. For example, using the data (file CX09RODS.MTW) on which Figure 9.3 is based, with the 30 data values in column C1: Click Stat. Select Basic Statistics. Click 1-Sample Z. 2. Select Samples in columns and enter C1 into the box. Enter the known population standard deviation (0.053) into the Standard deviation box. The Perform hypothesis test box should be left blank—we will not be doing hypothesis testing until the next chapter. Chapter 9: Estimation from Sample Data 279 3. Click Options. Enter the desired confidence level as a percentage (95.0) into the Confidence Level box. Within the Alternative box, select not equal. Click OK. Click OK. The printout also includes the sample mean (1.400), the known sigma (0.053), the sample size (30), and the standard error of the mean (calculated as 0.053 ⁄兹30 0.00968). Although the sample standard deviation is shown, it is not used in the construction of this confidence interval. Confidence interval for based on summary statistics and known Follow the procedure in the previous steps for raw data, but in step 2 select Summarized data and enter 30 into the Sample size box and 1.4 into the Mean box. exercises 9.12 What role does the central limit theorem play in the construction of a confidence interval for the population mean? 9.13 In using the standard normal distribution to construct a confidence interval for the population mean, what two assumptions are necessary if the sample size is less than 30? 9.14 The following data values are a simple random sample from a population that is normally distributed, with 2 25.0: 47, 43, 33, 42, 34, and 41. Construct and interpret the 95% and 99% confidence intervals for the population mean. 9.15 A simple random sample of 30 has been collected from a population for which it is known that 10.0. The sample mean has been calculated as 240.0. Construct and interpret the 90% and 95% confidence intervals for the population mean. 9.16 A simple random sample of 25 has been collected from a normally distributed population for which it is known that 17.0. The sample mean has been calculated as 342.0, and the sample standard deviation is s 14.9. Construct and interpret the 95% and 99% confidence intervals for the population mean. 9.17 The administrator of a physical therapy facility has found that postoperative performance scores on a knee flexibility test have tended to follow a normal distribution with a standard deviation of 4. For a simple random sample of ten patients who have recently had knee surgery, the scores are as follows: 101, 92, 94, 88, 52, 93, 76, 84, 72, and 98. Construct and interpret the 90% and 95% confidence intervals for the population mean. 9.18 In testing the heat resistance of electrical components, safety engineers for an appliance manufacturer routinely subject wiring connectors to a temperature of 450 degrees Fahrenheit, then record the amount of time it takes for the connector to melt and cause a short circuit. Past experience has shown the standard deviation of failure times to be 6.4 seconds. In a simple random sample of 40 connectors from a very large production run, the mean time until failure was found to be 35.5 seconds. Construct and interpret the 99% confidence interval for the mean time until failure for all of the connectors from the production run. 9.19 An assembly process includes a torque wrench device that automatically tightens compressor housing bolts; the device has a known process standard deviation of 3 lb-ft in the torque applied. A simple random sample of 35 nuts is selected, and the average torque to which they have been tightened is 150 lb-ft. What is the 95% confidence interval for the average torque being applied during the assembly process? 9.20 A machine that stuffs a cheese-filled snack product can be adjusted for the amount of cheese injected into each unit. A simple random sample of 30 units is selected, and the average amount of cheese injected is found to be x 3.5 grams. If the process standard deviation is known to be 0.25 grams, construct the 95% confidence interval for the average amount of cheese being injected by the machine. 9.21 In Exercise 9.20, if the sample size had been n 5 instead of n 30, what assumption would have to be made about the population distribution of filling weights in order to use z values in constructing the confidence interval? / data set / Note: Exercises 9.22 and 9.23 require a computer and statistical software. 9.22 For one of the tasks in a manufacturing process, the mean time for task completion has historically been 35.0 minutes, with a standard deviation of 2.5 minutes. Workers have recently complained that the machinery used in the task is wearing out and slowing down. In response to the complaints, plant engineers have measured the time required for a sample consisting of 100 task operations. The 100 sample times, in minutes, are in data file XR09022. 280 Part 3: Sampling Distributions and Estimation Using the mean for this sample, and assuming that the population standard deviation has remained unchanged at 2.5 minutes, construct the 95% confidence interval for the population mean. Is 35.0 minutes within the confidence interval? Interpret your “yes” or “no” answer in terms of whether the mean time for the task may have changed. 9.23 Sheila Johnson, a state procurement manager, is responsible for monitoring the integrity of a wide range of products purchased by state agencies. She is currently examining a sample of paint containers recently received from a long-time supplier. According to the supplier, the process by which the cans are filled involves a small 9.5 amount of variation from one can to the next, and the standard deviation is 0.25 fluid ounces. The 40 cans in Sheila’s sample were examined to determine how much paint they contained, and the results (in fluid ounces) are listed in data file XR09023. Using the mean for this sample, and assuming that the population standard deviation is 0.25 fluid ounces, construct the 90% confidence interval for the population mean volume for the cans of paint provided by the supplier. If the labels on the paint cans say the mean content for such containers is 100.0 fluid ounces, would your confidence interval tend to support this possibility? CONFIDENCE INTERVAL ESTIMATES FOR THE MEAN: UNKNOWN It is rare that we know the standard deviation of a population but have no knowledge about its mean. For this reason, the techniques of the previous section are much less likely to be used in practice than those discussed here. Whenever the population standard deviation is unknown, it must be estimated by the sample standard deviation, s. For such applications, there is a continuous distribution called the Student’s t distribution. The Student’s t Distribution Description Also referred to as simply the t distribution, this distribution is really a family of continuous, unimodal, bell-shaped distributions. It was developed in the early 1900s by W. S. Gossett, who used the pen name “Student” because his company did not permit employees to publish their research results. The t distribution is the probability distribution for the random variable t (x )兾(s兾兹n). It has a mean of zero, but its shape is determined by what is called the number of degrees of freedom (df ). For confidence interval applications, the specific member of the family is determined by df n 1. The term degrees of freedom refers to the number of values that remain free to vary once some information about them is already known. For example, if four items have a mean of 10.0, and three of these items are known to have values of 8, 12, and 7, there is no choice but for the fourth item to have a value of 13. In effect, one degree of freedom has been lost. The t distribution tends to be flatter and more spread out than the normal distribution, especially for very small sample sizes. Figure 9.4 compares the approximate shape of a standard normal distribution with that of a t distribution for which df 6. The t distribution converges to the normal distribution as the sample size (and df ) increases, and as the number of degrees of freedom approaches infinity, the two distributions are actually identical. As with our use of z previously in this chapter, t represents distance in terms of standard error units. Seeing Statistics Applet 10, at the end of the chapter, can be used in seeing how the shape of the t distribution responds to different df values. Chapter 9: Estimation from Sample Data 281 FIGURE 9.4 Standard normal distribution t distribution with df = 6 example t Distribution Table Using the t Distribution Table. A table for t values that correspond to selected areas beneath the t distribution appears on the pages immediately preceding the back cover. A portion of the table is reproduced as Table 9.2 on page 282. In general, it is used in the same way as the standard normal table, but there are two exceptions: (1) the areas provided are for the right tail only, and (2) it is necessary to refer to the appropriate degrees of freedom (df ) row in finding the appropriate t value. SOLUTION For a Sample Size of n ⴝ 15, What t Values Would Correspond to an Area Centered at t ⴝ 0 and Having an Area beneath the Curve of 95%? The area of interest beneath the curve can be expressed as 0.95, so the total area in both tails combined will be (1.00 0.95), or 0.05. Since the curve is symmetrical, the area in just one tail will be 0.05兾2, or 0.025. The number of degrees of freedom will be the sample size minus 1, or df n 1, or 15 1 14. Referring to the 0.025 column and the df 14 row of the table, we find that the value of t corresponding to a right-tail area of 0.025 is t 2.145. Because the curve is symmetrical, the value of t for a left-tail area of 0.025 will be t 2.145. Note that these values of t (t 2.145) are farther apart than the z values (z 1.96) that would have led to a 95% area beneath the standard normal curve. Remember that the shape of the t distribution tends to be flatter and more spread out than that of the normal distribution, especially for small samples. For a Sample Size of n ⴝ 99, What t Values Would Correspond to an Area Centered at t ⴝ 0 and Having an Area beneath the Curve of 90%? In this case, the proportion of the area beneath the curve is 0.90, so each tail will have an area of (1.00 0.90)兾2, or 0.05. Therefore, we will refer to the 0.05 A comparison of the approximate shape of the standard normal distribution with that of a t distribution having 6 degrees of freedom. The shape of the t distribution is flatter and more spread out, but approaches that of the standard normal distribution as the number of degrees of freedom increases. 282 Part 3: Sampling Distributions and Estimation TABLE 9.2 A portion of the Student’s t distribution table. The t distribution is really a family of symmetric, continuous distributions with a mean of t 0. The specific member of the distribution depends on the number of degrees of freedom, or df. As df increases, the t distribution approaches the normal distribution, and the t values in the infinity row are identical to the z values for the standard normal distribution. a = right-tail area (For a right-tail area of 0.025 and df = 15, the t value is 2.131.) 0 ␣ t 0.10 0.05 0.025 df ⴝ 1 2 3 4 5 3.078 1.886 1.638 1.533 1.476 6.314 2.920 2.353 2.132 2.015 12.706 4.303 3.182 2.776 2.571 31.821 6.965 4.541 3.747 3.365 63.657 9.925 5.841 4.604 4.032 6 7 8 9 10 1.440 1.415 1.397 1.383 1.372 1.943 1.895 1.860 1.833 1.812 2.447 2.365 2.306 2.262 2.228 3.143 2.998 2.896 2.821 2.764 3.707 3.499 3.355 3.250 3.169 11 12 13 14 15 1.363 1.356 1.350 1.345 1.341 1.796 1.782 1.771 1.761 1.753 2.201 2.179 2.160 2.145 2.131 2.718 2.681 2.650 2.624 2.602 3.106 3.055 3.012 2.977 2.947 ⯗ ⯗ ⯗ ⯗ ⯗ ⯗ 1.290 1.290 1.290 1.282 1.661 1.660 1.660 1.645 1.984 1.984 1.984 1.960 2.365 2.365 2.364 2.326 2.627 2.626 2.626 2.576 98 99 100 “Infinity” 0.01 0.005 column in the t table. Subtracting 1 from the sample size of 99, df 99 1, or 98. Using the 0.05 column and the df 98 row, the corresponding t value is t 1.661. Since t 1.661 corresponds to a right-tail area of 0.05, t 1.661 will correspond to a left-tail area of 0.05. This is due to the symmetry of the distribution, and the distance from t 1.661 to t 1.661 will include 90% of the area beneath the curve. You can use Seeing Statistics Applet 11, at the end of the chapter, to further examine areas beneath the t distribution curve. Should you encounter a situation in which the number of degrees of freedom exceeds the df 100 limit of the t distribution table, just use the corresponding z value for the desired level of confidence. These z values are listed in the df infinity row in the t distribution table. Chapter 9: Estimation from Sample Data Using the t table instead of the standard normal table to which you’ve become accustomed may seem cumbersome at first. As we mentioned previously, however, the t-interval is the technically appropriate procedure whenever s has been used to estimate . This is also the method you will either use or come into contact with when dealing with computer statistical packages and their construction of confidence intervals. When using computer statistical packages, it’s easy to routinely (and correctly) use the t distribution for constructing confidence intervals whenever is unknown and being estimated by s. Confidence Intervals Using the t Distribution Aside from the use of the t distribution, the basic procedure for estimating confidence intervals is similar to that of the previous section. The appropriate t value is used instead of z, and s replaces . The t distribution assumes the underlying population is approximately normally distributed, but this assumption is important only when the sample is small — that is, n 30. The interval estimate is summarized as follows: Confidence interval limits for the population mean, unknown: xt s 兹n where x sample mean s sample standard deviation n sample size t t value corresponding to the level of confidence desired, with df n 1 (e.g., t 2.201 for 95% confidence, n 12, and df 12 1 11) s/兹n estimated standard error of the sampling distribution of the mean (If n 30, it must be assumed that the underlying population is approximately normally distributed.) example t-Interval, Mean A simple random sample of n 90 manufacturing employees has been selected from those working throughout a state. The average number of overtime hours worked last week was x 8.46 hours, with a sample standard deviation of s 3.61 hours. The underlying data are in file CX09OVER. SOLUTION What Is the 98% Confidence Interval for the Population Mean, ? The first step in determining the appropriate value of t is to identify the column of the t distribution table to which we must refer. Since the confidence level is 98%, the right-tail area of interest is (1.00 0.98)兾2, or 0.01. For this sample size, the number of degrees of freedom will be 90 1, or df 89. Referring to 283 AN IMPORTANT NOTE 284 Part 3: Sampling Distributions and Estimation FIGURE 9.5 Although the sample size in this example is relatively large (n 90), the t distribution was used in constructing the 98% confidence interval for the population mean. This is because the population standard deviation is unknown and is being estimated by the sample standard deviation (s 3.61 hours). The t distribution: with df = n – 1 = 89. For 98% confidence, t will be ±2.369. Area = 0.98 t = –2.369 0 t = +2.369 The 98% confidence interval for m is s x – t ––– √n 3.61 8.46 – 2.369 ––––– √90 or s x + t ––– √n x 7.56 8.46 3.61 8.46 + 2.369 ––––– √90 8.46 9.36 the 0.01 column and the df 89 row, we determine t 2.369. Due to the symmetry of the t distribution, 98% of the area beneath the curve will be between t 2.369 and t 2.369. For the results summarized in Figure 9.5, the underlying calculations for the 98% confidence interval are xt A REMINDER s 兹n 8.46 2.369 3.61 兹90 or between 7.56 and 9.36 hours If n is so large (e.g., n 101) that df exceeds the finite limits of the t table, just use the infinity row of the table. In the preceding example, if n were 101, we would refer to the 0.05 column and the df infinity row and obtain a t value of 1.645. This is the same as using z instead of t. For such large samples, the z and t distributions are similar enough that the z will be a very close approximation. Computer Solutions 9.2 shows how we can use Excel or Minitab to generate a confidence interval for the mean when the population standard deviation is unknown. In this case, we are replicating the 98% confidence interval shown in Figure 9.5, and the 90 data values are in file CX09OVER. Once again, if we use an Excel procedure based on summary statistics, it can be interesting to examine “what-if” scenarios to instantly see how changes in the specified confidence level would change the width of the confidence interval. Chapter 9: Estimation from Sample Data 285 computer solutions 9.2 Confidence Interval for Population Mean, Unknown These procedures show how to construct a confidence interval for the population mean when the population standard deviation is unknown. EXCEL 1 2 3 4 5 6 7 A B t-Estimate: Mean C D hours 8.4603 3.61 7.559 9.362 Mean Standard Deviation LCL UCL Excel confidence interval for based on raw data and unknown 1. For example, using the 90 overtime data values (file CX09OVER.XLS) on which Figure 9.5 is based: The label and 90 data values are in A1:A91. Click Tools. Click Data Analysis Plus. Click t-Estimate: Mean. Click OK. 2. Enter A1:A91 into the Input Range box. Click Labels. The desired confidence level as a decimal fraction is 0.98, so enter the corresponding alpha value (0.02) into the Alpha box. Click OK. The lower portion of the printout lists the lower and upper limits for the 98% confidence interval. Excel confidence interval for based on summary statistics and unknown 1. Open the ESTIMATORS.XLS workbook, supplied with the text. 2. Using the arrows at the bottom left, select the t-Estimate_Mean worksheet. Enter the sample mean (8.46), the sample standard deviation (3.61), the sample size (90), and the desired confidence level as a decimal fraction (0.98). (Note: As an alternative, you can use the Excel worksheet template TMTINT.XLS, supplied with the text. The steps are described within the template.) MINITAB Confidence interval for based on raw data and unknown One-Sample T: hours Variable hours N 90 Mean 8.460 StDev 3.610 SE Mean 0.381 98% CI (7.559, 9.362) 1. For example, using the data (file CX09OVER.MTW) on which Figure 9.5 is based, with the 90 data values in column C1: Click Stat. Select Basic Statistics. Click 1-Sample t. 2. Select Samples in columns and enter C1 into the box. The Perform hypothesis test box should be left blank—we will not be doing hypothesis testing until the next chapter. 3. Click Options. Enter the desired confidence level as a percentage (98.0) into the Confidence Level box. Within the Alternative box, select not equal. Click OK. Click OK. Confidence interval for based on summary statistics and unknown Follow the procedure in the previous steps for raw data, but in step 2 select Summarized data, enter 90 into the Sample size box, 8.46 into the Mean box, and 3.61 into the Standard deviation box. 286 Part 3: Sampling Distributions and Estimation exercises 9.24 When the t distribution is used in constructing a confidence interval based on a sample size of less than 30, what assumption must be made about the shape of the underlying population? 9.25 Why are the t values listed in the df infinity row of the t distribution table identical to the z values that correspond to the same right-tail areas of the standard normal distribution? What does this indicate about the relationship between the t and standard normal distributions? 9.26 In using the t distribution table, what value of t would correspond to an upper-tail area of 0.025 for 19 degrees of freedom? 9.27 In using the t distribution table, what value of t would correspond to an upper-tail area of 0.10 for 28 degrees of freedom? 9.29 For df 85, determine the value of A that corresponds to each of the following probabilities: a. P(t A) 0.10 b. P(t A) 0.025 c. P(A t A) 0.98 55 54 9.30 Given the following observations in a simple random sample from a population that is approximately normally distributed, construct and interpret the 90% and 95% confidence intervals for the mean: 79 71 98 74 70 59 102 92 96 9.31 Given the following observations in a simple random sample from a population that is approximately normally distributed, construct and interpret the 95% and 99% confidence intervals for the mean: 66 50 34 34 59 42 56 61 51 53 45 48 38 57 58 47 52 50 9.33 The service manager of Appliance Universe has recorded the times for a simple random sample of 50 refrigerator service calls taken from last year’s service records. The sample mean and standard deviation were 25 minutes and 10 minutes, respectively. a. Construct and interpret the 95% confidence interval for the mean. b. It’s quite possible that the population of such times is strongly skewed in the positive direction — that is, some jobs, such as compressor replacement, might take 3 or 4 hours. If this were true, would the interval constructed in part (a) still be appropriate? Explain your answer. 9.34 An automobile rental agency has the following mileages for a simple random sample of 20 cars that were rented last year. Given this information, and assuming the data are from a population that is approximately normally distributed, construct and interpret the 90% confidence interval for the population mean: 9.28 For df 25, determine the value of A that corresponds to each of the following probabilities: a. P(t A) 0.025 b. P(t A) 0.10 c. P(A t A) 0.99 67 c. Given that the population standard deviation is not known, which of these two confidence intervals should be used as the interval estimate for ? 52 54 9.32 A consumer magazine has contacted a simple random sample of 33 owners of a certain model of automobile and asked each owner how many defects had to be corrected within the first 2 months of ownership. The average number of defects was x 3.7, with a standard deviation of 1.8 defects. a. Use the t distribution to construct a 95% confidence interval for the average number of defects for this model. b. Use the z distribution to construct a 95% confidence interval for the average number of defects for this model. 35 50 65 74 64 92 69 59 37 50 88 38 80 59 39 29 61 60 miles 9.35 One of the most popular products sold by a manufacturer of electrical fuses is the standard 30-ampere fuse used by electrical contractors in the construction industry. The company has tested a simple random sample of 16 fuses from its production output and found the amperages at which they “blew” to be as shown here. Given this information, and assuming the data are from a normally distributed population, construct and interpret the 95% confidence interval for the population mean amperage these fuses will withstand. 30.6 30.2 27.7 28.5 29.0 27.5 28.9 28.1 30.3 30.8 28.5 30.3 30.0 28.5 29.0 28.2 amperes 9.36 The author of an entry-level book on using Microsoft Word has carried out a test in which 35 novices were provided with the book, a computer, and the assignment of converting a complex handwritten document with text and tables into a Microsoft Word file. The novices required an average time of 105 minutes, with a standard deviation of 20 minutes. Construct and interpret the 90% confidence interval for the population mean time it would take all such novices to complete this task. 9.37 An office equipment manufacturer has developed a new photocopy machine and would like to estimate the average number of 81/2-by-11 copies that can be made using a single bottle of toner. For a simple random sample of 20 bottles of toner, the average was 1535 pages, with a Chapter 9: Estimation from Sample Data standard deviation of 30 pages. Making and stating whatever assumptions you believe are necessary, construct and interpret the 95% confidence interval for the population mean. 9.38 Researchers have estimated that office workers in Germany receive an average of 15.0 fax messages per day. Assuming this finding to be based on a simple random sample of 80 German office workers, with a sample standard deviation of s 3.5 messages, construct and interpret the 95% confidence interval for the population mean. Given this confidence interval, would it seem very unusual if another sample of this size were to have a mean of 16.5 faxes? SOURCE: Anne R. Carey and Genevieve Lynn, “Message Overload?” USA Today, September 13, 1999, p. 1B. 9.39 According to Nielsen//NetRatings, the average visitor to the American Greetings website spends 11.85 minutes at the site. Assuming this finding to be based on a simple random sample of 20 visitors to the site, with a sample standard deviation of s 3.0 minutes, and from a population of visiting times that is approximately normally distributed, construct and interpret the 98% confidence interval for the population mean. Given this confidence interval, would it seem very unusual if another sample of this size were to have a mean visiting time of 13.0 minutes? SOURCE: 287 / data set / Note: Exercises 9.40 and 9.41 require a computer and statistical software. 9.40 In an article published in a British medical journal, Japanese researchers found that adults who were videotaped in a relaxed setting blinked an average of 15.0 times per minute. Under the assumption that this finding was based on the sample data in file XR09040, construct and interpret the 99% confidence interval for the population mean. Based on this confidence interval, would it seem very unusual if another sample of this size were to exhibit a mean blink rate of 16.0 times per minute? SOURCE: “Blink Factor,” USA Today, August 9, 1999, p. 3D. 9.41 Automotive researchers have reported that building the Ford Mustang required an average of 22.3 labor hours. Under the assumption that this finding was based on the sample data in file XR09041, construct and interpret the 95% confidence interval for the population mean number of labor hours required for this model. Based on this confidence interval, would it seem very unusual if another sample of this size were to require an average of 22.9 labor hours for production of a Ford Mustang? SOURCE: Michael Woodyard, “U.S. Makers Narrow Efficiency Gap,” Automotive News, June 21, 1999, p. 8. “Top Web Properties,” USA Today, April 27, 2000, p. 3D. CONFIDENCE INTERVAL ESTIMATES FOR THE POPULATION PROPORTION Determining a confidence interval estimate for the population proportion requires that we use the sample proportion (p) for two purposes: (1) as a point estimate of the (unknown) population proportion, , and (2) in combination with the sample size (n) in estimating the standard error of the sampling distribution of the sample proportion for samples of this size. The technique of this section uses the normal distribution as an approximation to the binomial distribution. This approximation is considered satisfactory whenever np and n(1 p) are both 5, and becomes better for large values of n and whenever p is closer to 0.5. The midpoint of the confidence interval is the sample proportion, and the lower and upper confidence limits are determined as follows: Confidence interval limits for the population proportion: 兹 pz p(1 p) n where p sample proportion 兹 number of successes number of trials n sample size z z value corresponding to desired level of confidence (e.g., z 1.96 for 95% confidence) p(1 p) estimated standard error of the n sampling distribution of the proportion 9.6 288 Part 3: Sampling Distributions and Estimation example z-Interval, Proportion In a USA Today/CNN poll, 1406 adults were randomly selected from across the United States. In response to the question, “Do you agree that the current system discourages the best candidates from running for president?” 22% responded “strongly agree.”1 SOLUTION What Is the 95% Confidence Interval for the Population Proportion Who Would Have Answered “Strongly Agree” to the Question Posed? The sample proportion p 0.22 is our point estimate of and the midpoint of the interval. Since the confidence level is to be 95%, z will be 1.96. The resulting confidence interval, shown in Figure 9.6, is 兹 pz 兹 p(1 p) 0.22 1.96 n 0.22(1 0.22) 0.198 to 0.242 1406 From these results, we have 95% confidence that the population proportion is somewhere between 0.198 and 0.242. Expressed in terms of percentage points, the interval would be from 19.8% to 24.2% for the percentage of the population who would have “strongly agreed,” and the interval width would be (24.2 19.8), or 4.4 percentage points. 1Source: Jean Becker, “Voters Favor a National Primary,” USA Today, February 5, 1988, p. 8A. FIGURE 9.6 The 95% confidence interval for a population proportion, based on a political poll having a sample proportion of p 0.22 and a sample size of n 1406. We have 95% confidence that is between 0.198 and 0.242. Normal distribution: for 95% confidence, z will be ±1.96. Area = 0.95 z = –1.96 0 z = +1.96 The 95% confidence interval for p is p–z (1 – p) √ p–––––– n 0.22 – 1.96 or – 0.22) ––––––––––– √ 0.22(1 1406 0.198 p 0.22 0.220 p+z – p) –––––– √ p(1 n √ – 0.22) 0.22 + 1.96 0.22(1 ––––––––––– 1406 0.242 Chapter 9: Estimation from Sample Data 289 Computer Solutions 9.3 shows how we can use Excel or Minitab to generate a confidence interval for a population proportion. In this case, we are replicating the 95% confidence interval shown in Figure 9.6. As always, if we use an Excel procedure based on summary statistics, it can be interesting to examine “what-if” scenarios to instantly see how changes in the specified confidence level would change the width of the confidence interval. computer solutions 9.3 Confidence Interval for Population Proportion These procedures show how to construct a confidence interval for the population proportion. EXCEL 1 2 3 4 5 6 A B z-Estimate of a Proportion Sample proportion Sample size Confidence level 0.22 1406 0.95 C Confidence Interval Estimate 0.22 Lower confidence limit Upper confidence limit D E plus/minus 0.022 0.198 0.242 Excel confidence interval for based on summary statistics 1. For example, with n ⫽ 1406 and p ⫽ 0.22, as in Figure 9.6: Open the ESTIMATORS.XLS workbook, supplied with the text. 2. Using the arrows at the bottom left, select the z-Estimate_Proportion worksheet. Enter the sample proportion (0.22), the sample size (1406), and the desired confidence level as a decimal fraction (0.95). The confidence interval appears as shown here. (Note: As an alternative, you can use Excel worksheet template TMPINT.XLS, supplied with the text. The steps are described within the template.) Excel confidence interval for based on raw data 1. For example, if we had 20 data values that were coded (1 ⫽ female, 2 ⫽ male), with the label and data values in A1:A21: Click Tools. Click Data Analysis Plus. Click Z-Estimate: Proportion. Click OK. 2. Enter A1:A21 into the Input Range box. Enter 1 into the Code for Success box. Click Labels. The desired confidence level as a decimal fraction is 0.95, so enter the corresponding alpha value (0.05) into the Alpha box. Click OK. MINITAB Minitab confidence interval for based on summary statistics Test and CI for One Proportion Sample 1 X 309 N 1406 Sample p 0.219772 95% CI (0.198128, 0.241417) Using the normal approximation. 1. This interval is based on the summary statistics for Figure 9.6: Click Stat. Select Basic Statistics. Click 1 Proportion. Select Summarized Data. Enter the sample size (in this case, 1406) into the Number of Trials box. Multiply the sample proportion (0.22) times the sample size (1406) to get the number of “successes” or “events” as (0.22)(1406) ⫽ 309.32. Round to the nearest integer and enter the result (309) into the Number of Events box. The Perform hypothesis test box should be left blank. (continued) 290 Part 3: Sampling Distributions and Estimation 2. Click Options. Enter the desired confidence level as a percentage (95.0) into the Confidence Level box. Within the Alternative box, select not equal. Click to select Use test and interval based on normal distribution. Click OK. Click OK. Minitab confidence interval for based on raw data 1. For example, if column C1 contains 20 data values that are coded (1 female, 2 male): Click Stat. Select Basic Statistics. Click 1 Proportion. Select Samples in columns and enter C1 into the dialog box. 2. Follow step 2 in the summary-information procedure. Note: Minitab will select the larger of the two codes (i.e., 2 male) as the “success” or “event” and provide the sample proportion and the confidence interval for the population proportion of males. To obtain the results for females, just recode the data so females will have the higher code number: Click Data. Select Code. Click Numeric to Numeric. Enter C1 into both the Code data from columns box and the Into columns box. Enter 1 into the Original values box. Enter 3 into the New box. Click OK. The new codes will be (3 female, 2 male). exercises 9.42 Under what conditions is it appropriate to use the normal approximation to the binomial distribution in constructing the confidence interval for the population proportion? 9.43 A pharmaceutical company found that 46% of 1000 U.S. adults surveyed knew neither their blood pressure nor their cholesterol level. Assuming the persons surveyed to be a simple random sample of U.S. adults, construct a 95% confidence interval for the population proportion of U.S. adults who would have given the same answer if a census had been taken instead of a survey. 9.44 An airline has surveyed a simple random sample of air travelers to find out whether they would be interested in paying a higher fare in order to have access to e-mail during their flight. Of the 400 travelers surveyed, 80 said e-mail access would be worth a slight extra cost. Construct a 95% confidence interval for the population proportion of air travelers who are in favor of the airline’s e-mail idea. 9.45 In response to media inquiries and concerns expressed by groups opposed to violence, the president of a university with over 25,000 students has agreed to survey a simple random sample of her students to find out whether the student body thinks the school’s “Plundering Pirate” mascot should be changed to one that is less aggressive in name and appearance. Of the 200 students selected for participation in the survey, only 20% believe the school should select a new and more kindly mascot. Construct a 90% confidence interval for the population proportion of students who believe the mascot should be changed. Based on the sample findings and associated confidence interval, comment on the credibility of a local journalist’s comment that “over 50%” of the students would like a new mascot. 9.46 In examining a simple random sample of 100 sales invoices from several thousand such invoices for the previous year, a researcher finds that 65 of the invoices involved customers who bought less than $2000 worth of merchandise from the company during that year. Construct a 90% confidence interval for the proportion of all sales invoices that were for customers buying less than $2000 worth of merchandise during the year. 9.47 Survey researchers estimate that 40% of U.S. women age 18–29 save in a 401(k) or individual retirement account. Assuming the persons surveyed to be a simple random sample of 1000 U.S. women in this age group, construct a 95% confidence interval for the population proportion of U.S. women in this age group who would have given the same answer if a census had been taken instead of a survey. SOURCE: Anne R. Carey and Sam Ward, “Women Saving for Retirement,” USA Today, September 5, 1996, p. 1B. 9.48 A study by the Society of Human Resource Manage- ment found 23% of U.S. business executives surveyed believe that an employer has no right to read employees’ e-mail. Assuming that the survey included a simple random sample of 1200 executives, construct a 90% confidence interval for the population proportion of U.S. business executives who believe that employers have no right to read employees’ e-mail. SOURCE: Anne R. Carey and Marcy E. Mullins, “Bosses OK E-Mail Oversight,” USA Today, April 22, 1996, p. 1A. Chapter 9: Estimation from Sample Data 9.49 According to Nielsen Media Research viewership data, the top television broadcast of all time was the last episode of M*A*S*H, which aired on February 28, 1983, and was viewed by an estimated 60.2% of all TV households. Assuming this estimate was based on a simple random sample of 1800 TV households, what is the 95% confidence interval for the proportion of all TV households who viewed the last episode of M*A*S*H? SOURCE: The World Almanac and Book of Facts 2003, p. 283. 9.50 In a major industry where well over 100,000 manufacturing employees are represented by a single union, a simple random sampling of n 100 union members finds that 57% of those in the sample intend to vote for the new labor contract negotiated by union and management representatives. a. What is the 99% confidence interval for the population proportion of union-represented employees who intend to vote for the labor contract? b. Based on your response to part (a), does contract approval by the union appear to be a “sure thing”? Why or why not? 9.51 Repeat Exercise 9.50, but assume that the sample size was n 900 instead of n 100. 9.52 Based on its 1999 survey, Student Monitor reports that 20% of U.S. college students used the Internet for job hunting during the month preceding the survey. Assuming this finding to be based on a simple random sample of 1600 college students, construct and interpret the 90% confidence interval for the population proportion of college students who used the Internet for job hunting during this period. SOURCE: Julie Stacey, “Online Extracurricular Activities,” USA Today, March 13, 2000, p. 10D. 9.53 According to Keynote Systems, Wal-Mart’s website was available 95% of the time during a holiday shopping season. Assuming this finding to be based on a simple random sample of 200 attempts, construct and interpret the 90% confidence interval for the population proportion of the time the Wal-Mart site was available during this period. SOURCE: “Measuring How Key Web Sites Handle Holiday Shopping Rush,” USA Today, November 17, 1999, p. 3B. 9.54 A Pathfinder Research Group survey estimates that, of U.S. adults who have a favorite among The Three Stooges, Moe is the favorite of 31% of them. Assuming this finding 291 to be based on a simple random sample of 800 Stooge fans who have a favorite Stooge, construct and interpret the 95% confidence interval for the population proportion whose favorite Stooge is Moe. SOURCE: Anne R. Carey and Marcy E. Mullins, “Favorite Stooges,” USA Today, November 22, 1996, p. 1D. 9.55 Estelle McCarthy, a candidate for state office in New Jersey, has been advised that she must get at least 65% of the union vote in her state. A recent political poll of likely voters included 800 union-member respondents, and 60% of them said they intended to vote for Ms. McCarthy. Based on the survey, construct and interpret the 95% confidence interval for the population proportion of likely-voter union members who intend to vote for Ms. McCarthy. Is the 65% level of support within the confidence interval? Given your answer to the preceding question, comment on the possibility that Ms. McCarthy might not succeed in obtaining the level of union support she needs. / data set / Note: Exercises 9.56 and 9.57 require a computer and statistical software. 9.56 In the documentation that accompanies its products that are returned for in-warranty service, a manufacturer of electric can openers asks the customer to indicate the reason for the return. The codes and return problem categories are: (1) “doesn’t work,” (2) “excessive noise,” and (3) “other.” Data file XR09056 contains the problem codes for a simple random sample of 300 product returns. Based on this sample, construct and interpret the 95% confidence interval for the population proportion of returns that were because the product “doesn’t work.” 9.57 An investment counselor has purchased a large mailing list consisting of 50,000 potential investors. Before creating a brochure to send to members of the list, the counselor mails a questionnaire to a small simple random sampling of them. In one of the questions, the respondent is asked, “Do I think of myself as someone who enjoys taking risks?” The response codes are (1) “Yes” and (2) “No.” The results for the 200 investors who answered this question are represented by the codes listed in data file XR09057. Based on this sample, construct and interpret the 99% confidence interval for the population proportion of investors on the counselor’s mailing list who think of themselves as someone who enjoys taking risks. SAMPLE SIZE DETERMINATION In our interval estimates to this point, we have taken our results, including a stated sample size, then constructed a confidence interval. In this section, we’ll proceed in the opposite direction—in other words, we will decide in advance the desired confidence interval width, then work backward to find out how large a 9.7 292 Part 3: Sampling Distributions and Estimation sample size is necessary to achieve this goal. Central to our discussion in this section is the fact that the maximum likely sampling error (accuracy) is one-half the width of the confidence interval. Estimating the Population Mean To show how the necessary sample-size equation is obtained, we will consider a case in which is known. Of especial importance is that the distance from the midpoint to the upper confidence limit can be expressed as either (1) the maximum likely sampling error (e) or (2) z times the standard error of the sampling distribution. Since the two quantities are the same, we can set up the following equation and solve for n: ez z2 ⴢ 2 ⎯→ Solving for n ⎯→ n e2 兹n Required sample size for estimating a population mean: n z2 ⴢ 2 e2 where n required sample size z z value for which z corresponds to the desired level of confidence known (or, if necessary, estimated) value of the population standard deviation e maximum likely error that is acceptable One way of estimating an unknown is to use a relatively small-scale pilot study from which the sample standard deviation is used as a point estimate of the population standard deviation. A second approach is to estimate by using the results of a similar study done at some time in the past. A third method is to estimate as 1兾6 the approximate range of data values. example Sample Size, Estimating a Mean A state politician would like to determine the average amount earned during summer employment by state teenagers during the past summer’s vacation period. She wants to have 95% confidence that the sample mean is within $50 of the actual population mean. Based on past studies, she has estimated the population standard deviation to be $400. SOLUTION What Sample Size Is Necessary to Have 95% Confidence That x– Will Be Within $50 of the Actual Population Mean? For this situation, 95% confidence leads to a z value of 1.96, e the $50 maximum likely error that is acceptable, and the estimated value of is $400. The necessary sample size will be n z2 ⴢ 2 1.962 ⴢ 4002 245.9 persons, rounded up to 246 e2 502 Chapter 9: Estimation from Sample Data Since we can’t include a fraction of a person in the sample, we round up to n 246 to ensure 95% confidence in being within $50 of the population mean. Whenever the calculated value of n is not an integer, it is a standard (though slightly conservative) practice to round up to the next integer value. Note that if we cut the desired maximum likely error in half, the necessary sample size will quadruple. This is because the e term in the denominator is squared. The desire for extremely accurate results can lead to sample sizes that grow very rapidly in size (and expense) as the specified value for e is reduced. Estimating the Population Proportion As when estimating the population mean, the maximum likely error (e) in estimating a population proportion will be one-half of the eventual confidence interval width. Likewise, the distance from the midpoint of the confidence interval to the upper confidence limit can be described in two ways. Setting them equal and solving for n gives the following result: 兹 ez p(1 p) z2p(1 p) ⎯→ Solving for n ⎯→ n n e2 Required sample size for estimating a population proportion: n z2p(1 p) e2 where n required sample size z z value for which z corresponds to the desired level of confidence p the estimated value of the population proportion (As a conservative strategy, use p 0.5 if you have no idea as to the actual value of .) e maximum likely error that is acceptable In applying the preceding formula, we should first consider whether the true population proportion is likely to be either much less or much greater than 0.5. If we have absolutely no idea as to what might be, using p 0.5 is the conservative strategy to follow. This is because the required sample size, n, is proportional to the value of p(1 p), and this value is largest whenever p 0.5. If we are totally uncertain regarding the actual population proportion, we may wish to conduct a pilot study to get a rough idea of its value. If we can estimate the population proportion as being either much less or much more than 0.5, we can obtain the desired accuracy with a smaller sample than would have otherwise been necessary. If the population proportion is felt to be within a range, such as “between 0.20 and 0.40,” we should use as our estimate the value that is closest to 0.5. For example, if we believe the population proportion is somewhere between 0.20 and 0.40, it should be estimated as 0.40 when calculating the required sample size. 293 statistics statistics in in action action 9.1 9.1 Sampling Error in Survey Research When survey results are published, they are sometimes accompanied by an explanation of how the survey was conducted and how much sampling error could have been present. Persons who have had a statistics course need only know the size of the sample and the sample proportion or percentage for a given question. For the general public, however, the explanation of survey methods and sampling error needs a bit of rephrasing. The following statement accompanied the results of a survey commissioned by the Associated Press. this description. For 95% confidence, z 1.96. Since some questions may have a population proportion of 0.5 (the most conservative value to use when determining the required sample size), this is used in the following calculations: n The first sentence in this description briefly describes the size and nature of the sample. The second sentence describes the confidence level as 19兾20, or 95%, and the sampling error as plus or minus 3 percentage points. Using the techniques of this chapter, we can verify the calculations in where p estimated population proportion and e How Poll Was Conducted The Associated Press poll on taxes was taken Feb. 14–20 using a random sample of 1009 adult Americans. No more than one time in 20 should chance variations in the sample cause the results to vary by more than 3 percentage points from the answers that would be obtained if all Americans were polled. z 2p(1 p) e2 兹 兹 z2p(1 p) n (1.96)2(0.5)(1 0.5) 0.0309 1009 As these calculations indicate, the Associated Press rounded down to the nearest full percentage point (from 3.09 to 3.0) in its published explanation of the sampling error. This is not too unusual, since the general public would probably have enough to digest in the description quoted above without having to deal with decimal fractions. Source: “How Poll Was Conducted,’’ in Howard Goldberg, “Most Not Ready to Scrap Tax System,” Indiana Gazette, February 26, 1996, p. 3. example Sample Size, Estimating a Proportion A tourist agency researcher would like to determine the proportion of U.S. adults who have ever vacationed in Mexico and wishes to be 95% confident that the sampling error will be no more than 0.03 (3 percentage points). SOLUTION Assuming the Researcher Has No Idea Regarding the Actual Value of the Population Proportion, What Sample Size Is Necessary to Have 95% Confidence That the Sample Proportion Will Be within 0.03 (3 Percentage Points) of the Actual Population Proportion? For the 95% level of confidence, the z value will be 1.96. The maximum acceptable error is e 0.03. Not wishing to make an estimate, the researcher will use p 0.5 in calculating the necessary sample size: n z2p(1 p) 1.962(0.5)(1 0.5) 1067.1 persons, rounded up to 1068 e2 0.032 Chapter 9: Estimation from Sample Data 295 If the Researcher Believes the Population Proportion Is No More Than 0.3, and Uses p ⴝ 0.3 as the Estimate, What Sample Size Will Be Necessary? Other factors are unchanged, so z remains 1.96 and e is still specified as 0.03. However, the p(1 p) term in the numerator will be reduced due to the assumption that the population proportion is no more than 0.3. The required sample size will now be n z2p(1 p) 1.962(0.3)(1 0.3) 896.4 persons, rounded up to 897 e2 0.032 As in determining the necessary size for estimating a population mean, lower values of e lead to greatly increased sample sizes. For example, if the researcher estimated the population proportion as being no more than 0.3, but specified a maximum likely error of 0.01 instead of 0.03, he would have to include nine times as many people in the sample (8068 instead of 897). Computer Solutions 9.4 (page 296) shows how we can use Excel to determine the necessary sample size for estimating a population mean or proportion. With these procedures, it is very easy to examine “what-if” scenarios and instantly see how changes in confidence level or specified maximum likely error will affect the required sample size. exercises 9.58 “If we want to cut the maximum likely error in half, we’ll have to double the sample size.” Is this statement correct? Why or why not? 9.59 In determining the necessary sample size in making an interval estimate for a population mean, it is necessary to first make an estimate of the population standard deviation. On what bases might such an estimate be made? 9.60 From past experience, a package-filling machine has been found to have a process standard deviation of 0.65 ounces of product weight. A simple random sample is to be selected from the machine’s output for the purpose of determining the average weight of product being packed by the machine. For 95% confidence that the sample mean will not differ from the actual population mean by more than 0.1 ounces, what sample size is required? 9.61 Based on a pilot study, the population standard deviation of scores for U.S. high school graduates taking a new version of an aptitude test has been estimated as 3.7 points. If a larger study is to be undertaken, how large a simple random sample will be necessary to have 99% confidence that the sample mean will not differ from the actual population mean by more than 1.0 points? 9.62 A consumer agency has retained an independent testing firm to examine a television manufacturer’s claim that its 25-inch console model consumes just 110 watts of electricity. Based on a preliminary study, the population standard deviation has been estimated as 11.2 watts for these sets. In undertaking a larger study, and using a simple random sample, how many sets must be tested for the firm to be 95% confident that its sample mean does not differ from the actual population mean by more than 3.0 watts? 9.63 A national political candidate has commissioned a study to determine the percentage of registered voters who intend to vote for him in the upcoming election. To have 95% confidence that the sample percentage will be within 3 percentage points of the actual population percentage, how large a simple random sample is required? 9.64 Suppose that Nabisco would like to determine, with 95% confidence and a maximum likely error of 0.03, the proportion of first graders in Pennsylvania who had Nabisco’s Spoon-Size Shredded Wheat for breakfast at least once last week. In determining the necessary size of a simple random sample for this purpose: a. Use 0.5 as your estimate of the population proportion. b. Do you think the population proportion could really be as high as 0.5? If not, repeat part (a) using an estimated proportion that you think would be more likely to be true. What effect does your use of this estimate have on the sample size? 296 Part 3: Sampling Distributions and Estimation computer solutions 9.4 Sample Size Determination These procedures determine the necessary sample size for estimating a population mean or a population proportion. EXCEL 1 2 3 4 5 6 7 8 9 10 11 A B C D Sample size required for estimating a population mean: Estimate for sigma: Maximum likely error, e: 400.00 50.00 Confidence level desired: alpha = (1 - conf. level desired): The corresponding z value is: 0.95 0.05 1.960 The required sample size is n = 245.9 Sample size for estimating a population mean, using Excel worksheet template TMNFORMU.XLS that accompanies the text To determine the necessary sample size for estimating a population mean within $50 and with 95% confidence, assuming a population standard deviation of $400: Open Excel worksheet TMNFORMU.XLS. Enter the estimated sigma (400), the maximum likely error (50), and the specified confidence level as a decimal fraction (0.95). The required sample size (in cell D11) should then be rounded up to the nearest integer (246). This procedure is also described within the worksheet template. Caution Do not save any changes when exiting Excel. Sample size for estimating a population proportion, using Excel worksheet template TMNFORPI.XLS that accompanies the text To determine the necessary sample size for estimating a population proportion within 0.03 (3 percentage points) and with 95% confidence: Open Excel worksheet TMNFORPI.XLS. Enter the estimate for pi (0.50), the maximum likely error (0.03), and the specified confidence level as a decimal fraction (0.95). The required sample size (in cell D11) should then be rounded up to the nearest integer (1068). (Note: If you have knowledge about the population and can estimate pi as either less than or greater than 0.50, use your estimate and the necessary sample size will be smaller. Otherwise, be conservative and use 0.50 as your estimate.) This procedure is also described within the worksheet template. Caution Do not save any changes when exiting Excel. 9.65 The Chevrolet dealers of a large county are conduct- ing a study to determine the proportion of car owners in the county who are considering the purchase of a new car within the next year. If the population proportion is believed to be no more than 0.15, how many owners must be included in a simple random sample if the dealers want to be 90% confident that the maximum likely error will be no more than 0.02? 9.66 In Exercise 9.65, suppose that (unknown to the dealers) the actual population proportion is really 0.35. If they use their estimated value ( 0.15) in determining the sample size and then conduct the study, will their maximum likely error be greater than, equal to, or less than 0.02? Why? 9.67 In reporting the results of their survey of a simple random sample of U.S. registered voters, pollsters claim 95% confidence that their sampling error is no more than 4 percentage points. Given this information only, what sample size was used? Chapter 9: Estimation from Sample Data 297 9.8 WHEN THE POPULATION IS FINITE Whenever sampling is without replacement and from a finite population, it may be necessary to modify slightly the techniques for confidence-interval estimation and sample size determination in the preceding sections. As in Chapter 8, the general idea is to reduce the value of the standard error of the estimate for the sampling distribution of the mean or proportion. As a rule of thumb, the methods in this section should be applied whenever the sample size (n) is at least 5% as large as the population. When n 0.05N, there will be very little difference in the results. Confidence-Interval Estimation Whether we are dealing with interval estimation for a population mean () or a population proportion (), the confidence intervals will be similar to those in Figure 9.2. The only difference is that the “” term will be multiplied by the “finite population correction factor” shown in Table 9.3. As in Chapter 8, this correction depends on the sample size (n) and the population size (N). As an example of how this works, we’ll consider a situation such as that in Section 9.5, where a confidence interval is to be constructed for the population mean, and the sample standard deviation (s) is used as an estimate of the population standard deviation (). In this case, however, the sample will be relatively large compared to the size of the population. example Interval Estimates, Finite Population According to the Bureau of the Census, the population of Kent County, Texas, is 812 persons.2 For purposes of our example, assume that a researcher has interviewed a simple random sample of 400 persons and found that their average TABLE 9.3 Confidence Interval Estimate for the Population Mean, Known Infinite Population Finite Population xz xz 兹n ( 兹 ) 兹n Nn N1 ⴢ Confidence Interval Estimate for the Population Mean, Unknown Infinite Population xt Finite Population s xt 兹n ( 兹 ) s 兹n ⴢ Nn N1 Confidence Interval Estimate for the Population Proportion Infinite Population 兹 pz 2Source: p(1 p) n The World Almanac and Book of Facts 2003, p. 459. Finite Population pz (兹 p(1 p) ⴢ n 兹 Nn N1 ) Summary of confidence interval formulas when sampling without replacement from a finite population. As a rule of thumb, they should be applied whenever the sample is at least 5% as large as the population. The formulas and terms are similar to those in Figure 9.2 but include a “finite population correction factor,” the value of which depends on the relative sizes of the sample (n) and population (N). 298 Part 3: Sampling Distributions and Estimation number of years of formal education is x 11.5 years, with a standard deviation of s 4.3 years. SOLUTION Considering That the Population Is Finite and n ⱖ 0.05N, What Is the 95% Confidence Interval for the Population Mean? Since the number of degrees of freedom (df n 1 400 1, or 399) exceeds the limits of our t distribution table, the t distribution and normal distribution can be considered to be practically identical, and we can use the infinity row of the t table. The appropriate column in this table will be 0.025 for a 95% confidence interval, and the entry in the infinity row of this column is a t value of 1.96. Since s is being used to estimate , and the sample is more than 5% as large as the population, we will use the “ unknown” formula of the finite population expressions in Table 9.3. The 95% confidence interval for the population mean can be determined as The finite population correction term: this will be smaller than 1.0 so the standard error will be less than if an infinite population were involved. xt ( ( 兹 s ⴢ 兹n Nn N1 兹 ) ) 4.3 812 400 ⴢ 812 1 兹400 11.5 1.96(0.215 ⴢ 0.713) 11.5 0.300 or from 11.200 to 11.800 11.5 1.96 As expected, the finite correction term (0.713) is less than 1.0 and leads to a 95% confidence interval that is narrower than if an infinite population had been assumed. (Note: If the population had been considered infinite, the resulting interval would have been wider, with lower and upper limits of 11.079 and 11.921 years, respectively.) Sample Size Determination As in confidence-interval estimation, the rule of thumb is to change our sample size determination procedure slightly whenever we are sampling without replacement from a finite population and the sample is likely to be at least 5% as large as the population. Although different in appearance, the following formulas are applied in the same way that we used their counterparts in Section 9.7. N O T E If you were to substitute an N value of infinity into each of the following equations, you would find that the right-hand term in the denominator of each would be eliminated, and the result would be an expression exactly the same as its counterpart in Section 9.7. Chapter 9: Estimation from Sample Data Required sample size for estimating the mean of a finite population: n 2 e2 2 z2 N where n required sample size N population size z z value for which z corresponds to the desired level of confidence known (or, if necessary, estimated) value of the population standard deviation e maximum likely error that is acceptable Required sample size, estimating the proportion for a finite population: n p(1 p) e2 p(1 p) z2 N where n required sample size N population size z z value for which z corresponds to the desired level of confidence p the estimated value of the population proportion (As a conservative strategy, use p 0.5 if you have no idea as to the actual value of .) e maximum likely error that is acceptable example Sample Size, Finite Population The Federal Aviation Administration (FAA) lists 8586 pilots holding commercial helicopter certificates.3 Suppose the FAA wishes to question a simple random sample of these individuals to find out what proportion are interested in switching jobs within the next 3 years. Assume the FAA wishes to have 95% confidence that the sample proportion is no more than 0.04 (i.e., 4 percentage points) away from the true population proportion. SOLUTION Considering That the Population Is Finite, What Sample Size Is Necessary to Have 95% Confidence That the Sample Proportion Will Not Differ from the Population Proportion by More Than 0.04? Since the actual population proportion who are interested in switching jobs has not been estimated, we will be conservative and use p 0.5 in deciding on the necessary sample size. For the 95% confidence level, z will be 1.96. Applying the 3Source: General Aviation Manufacturers Association, General Aviation Statistical Database, 2005 Edition, p. 37. 299 300 Part 3: Sampling Distributions and Estimation finite population formula, with N 8586, the number of pilots who should be included in the sample is n p(1 p) 0.5(1 0.5) 561.0 e2 p(1 p) 0.042 0.5(1 0.5) z2 N 1.962 8586 Had the population been infinite, the required sample size would have been calculated as in Section 9.7. This would have resulted in n 600.25, rounded up to 601. By recognizing that the population is finite, we are able to achieve the desired confidence level and maximum error with a sample size that includes only 561 pilots instead of 601. exercises 9.68 As a rule of thumb, under what conditions should the finite population correction be employed in determining confidence intervals and calculating required sample sizes? homes might exceed the Environmental Protection Agency’s recommended limit of 15 parts per billion of lead. 9.69 Compared to situations where the population is either ulation of 800. In order to have 95% confidence that the sampling error in estimating is no more than 0.03, what sample size will be necessary? infinite or very large compared to the sample size, what effect will the finite population correction tend to have on a. the width of a confidence interval? b. the required size of a sample? 9.70 The personnel manager of a firm with 200 employ- ees has selected a simple random sample of 40 employees and examined their health-benefit claims over the past year. The average amount claimed during the year was $260, with a standard deviation of $80. Construct and interpret the 95% confidence interval for the population mean. Was it necessary to make any assumptions about the shape of the population distribution? Explain. 9.71 Of 1200 undergraduates enrolled at a univer- sity, a simple random sample of 600 have been surveyed to measure student support for a $5 activities fee increase to help fund women’s intercollegiate athletics at the NCAA division 1A level. Of those who were polled, 55% supported the fee increase. Construct and interpret the 95% and 99% confidence intervals for the population proportion. Based on your results, comment on the possibility that the fee increase might lose when it is voted on at next week’s university-wide student referendum. 9.72 A local environmental agency has selected a simple random sample of 16 homes to be tested for tap-water lead. Concentrations of lead were found to have a mean of 12 parts per billion and a standard deviation of 4 parts per billion. Considering that the homes were selected from a community in which there are 100 homes, construct and interpret the 95% confidence interval for the population mean. Based on your results, comment on the possibility that the average lead concentration in this community’s 9.73 A simple random sample is to be drawn from a pop- 9.74 A simple random sample is to be drawn from a pop- ulation of 2000. The population standard deviation has been estimated as being 40 grams. In order to have 99% confidence that the sampling error in estimating is no more than 5 grams, what sample size will be necessary? 9.75 There are 100 members in the United States Senate. A political scientist wants to estimate, with 95% confidence and within 3 percentage points, the percentage who own stock in foreign companies. How many senators should be interviewed? Explain any assumptions you used in obtaining your recommended sample size. 9.76 A transportation company operates 200 trucks and would like to use a hidden speed monitor device to record the maximum speed at which a truck is operated during the period that the device is installed. The trucks are driven primarily on interstate highways, and the company wants to estimate the average maximum speed for its fleet with 90% confidence and within 2 miles per hour. Using (and explaining) your own estimate for the population standard deviation, determine the number of trucks on which the company should install the hidden speed-recording device. 9.77 A research firm supports a consumer panel of 2000 households that keep written diaries of their weekly grocery expenditures. The firm would like to estimate, with 95% confidence and within 4 percentage points, the percentage of its panel households who would be interested in providing more extensive information in return for an extra $50 per week remuneration. How many of the Chapter 9: Estimation from Sample Data 301 households should be surveyed? Explain any assumptions you used in obtaining your recommended sample size. (and explain) your own estimate for the population standard deviation. 9.78 A university official wants to estimate, with 99% 9.79 A quality-management supervisor believes that no confidence and within $2, the average amount that members of fraternities and sororities spend at local restaurants during the first week of the semester. If the total fraternity/sorority membership is 300 people, how many members should be included in the sample? Use more than 5% of the items in a recent shipment of 2000 are defective. If she wishes to determine, within 1 percentage point and with 99% confidence, the percentage of defective items in the shipment, how large a simple random sample would be necessary? SUMMARY • Inferential statistics: point and interval estimates for a population parameter Chapter 8 examined the sampling distribution of a sample mean or a sample proportion from a known population. In this chapter, the emphasis has been on the estimation of an unknown population mean () or proportion () on the basis of sample statistics. Point estimates involve using the sample mean (x) or proportion (p) as the single best estimate of the value of the population mean or proportion. Interval estimates involve a range of values that may contain the actual value of the population parameter. When interval estimates are associated with a degree of certainty that they really do include the true population parameter, they are referred to as confidence intervals. a confidence interval for a population mean or a population • Constructing proportion The procedure appropriate to constructing an interval estimate for the population mean depends largely on whether the population standard deviation is known. Figure 9.2 summarizes these procedures and their underlying assumptions. Although the t-interval is often associated with interval estimates based on small samples, it is appropriate for larger samples as well. Using computer statistical packages, we can easily and routinely apply the t distribution for interval estimates of the mean whenever is unknown, even for very large sample sizes. A trade-off exists between the degree of confidence that an interval contains the population parameter and the width of the interval itself. The more certain we wish to be that the interval estimate contains the parameter, the wider the interval will have to be. • Sample size determination Accuracy, or sampling error, is equal to one-half of the confidence interval width. The process of sample size determination anticipates the width of the eventual confidence interval, then determines the required sample size that will limit the maximum likely sampling error to an acceptable amount. • When the sample is large compared to the population As in Chapter 8, when sampling is without replacement from a finite population, it is appropriate to use a finite population correction factor whenever the sample is at least 5% of the size of the population. Such corrections are presented for both interval estimation and sample size determination techniques within the chapter. • Computer-generated confidence intervals Most computer statistical packages are able to construct confidence interval estimates of the types discussed in the chapter. Examples of Excel and Minitab outputs are provided for a number of chapter examples in which such confidence intervals were developed. 9.9 302 Part 3: Sampling Distributions and Estimation equations Confidence Interval Limits for the Population Mean, Known xz 兹n where x sample mean population standard deviation n sample size z z value for desired confidence level 兾兹n standard error of the sampling distribution of the mean (Assumes that either (1) the underlying population is normally distributed or (2) the sample size is n 30.) Confidence Interval Limits for the Population Mean, Unknown s where x sample mean xt 兹n s sample standard deviation n sample size t t value corresponding to the level of confidence desired, with df n 1 s兾兹n estimated standard error of the sampling distribution of the mean (If n 30, this requires the assumption that the underlying population is approximately normally distributed.) Confidence Interval Limits for the Population Proportion 兹 pz p(1 p) n where p sample proportion 兹 number of successes number of trials n sample size z z value corresponding to desired level of confidence (e.g., z 1.96 for 95% confidence) p(1 p) estimated standard error of the sampling n distribution of the proportion Required Sample Size for Estimating a Population Mean n z2 ⴢ 2 e2 where n required sample size z z value for which z corresponds to the desired level of confidence known (or, if necessary, estimated) value of the population standard deviation e maximum likely error that is acceptable Required Sample Size for Estimating a Population Proportion n z2p(1 p) e2 where n required sample size z z value for desired level of confidence p estimated value of the population proportion (if not estimated, use p 0.5) e maximum likely error that is acceptable Chapter 9: Estimation from Sample Data 303 Confidence Interval Estimates When the Population Is Finite • For the population mean, known: xz • ( 兹 ⴢ 兹n Nn N1 where n sample size N population size For the population mean, unknown: xt • ) ( 兹 s ⴢ 兹n Nn N1 ) For the population proportion: pz (兹 p(1 p) ⴢ n 兹 Nn N1 ) Required Sample Size for Estimating the Mean of a Finite Population n 2 2 2 z N e2 where n required sample size N population size z z value for desired level of confidence known (or estimated) value of the population standard deviation e maximum likely error that is acceptable Required Sample Size for Estimating the Proportion for a Finite Population n p(1 p) p(1 p) z2 N e2 where n required sample size N population size z z value for desired level of confidence p the estimated population proportion (if not estimated, use p 0.5) e maximum likely error that is acceptable chapter exercises 9.80 In a destructive test of product quality, a briefcase manufacturer places each of a simple random sample of the day’s production in a viselike device and measures how many pounds it takes to crush the case. From past experience, the standard deviation has been found to be 21.5 pounds. For 35 cases randomly selected from today’s production, the average breaking strength was 341.0 pounds. Construct and interpret the 99% confidence interval for the mean breaking strength of the briefcases produced today. 9.81 Working independently, each of two researchers has devised a sampling plan to be carried out for the purpose of constructing a 90% confidence interval for the mean of a certain population. What is the probability that neither of their confidence intervals will include the population mean? 9.82 The accompanying data represent one-way commuting times (minutes) for a simple random sample of 15 persons who work at a large assembly plant. The data are also in file XR09082. Assuming an approximately normal distribution of commuting times for those who work at the plant, construct and interpret the 90% and 95% confidence intervals for the mean. 21.7 39.0 30.0 26.8 28.0 33.6 33.1 24.7 33.3 27.9 28.4 34.1 23.5 28.9 35.1 304 Part 3: Sampling Distributions and Estimation 9.83 A torque wrench used in the final assembly of cylinder heads has a process standard deviation of 5.0 lb-ft. The engineers have specified that a process average of 135 lb-ft is desirable. For a simple random sample of 30 nuts that the machine has recently tightened, the sample mean is 137.0 lb-ft. Construct and interpret the 95% confidence interval for the current process mean. Discuss the possibility that the machine may be in need of adjustment to correct the process mean. 9.84 There are approximately 109 million television households in the United States. A ratings service would like to know, within 5 percentage points and with 95% confidence, the percentage of these households who tune in to the first episode of a network miniseries. How many television households must be included in the sample? SOURCE: The World Almanac and Book of Facts 2006, p. 278. 9.85 In Exercise 9.84, a small-scale preliminary survey has indicated that no more than 20% of the television households will tune in to the first episode of the miniseries. Given this information, how large must the sample be? 9.86 In a survey of 500 U.S. adults, 45% of them said that lounging at the beach was their “dream vacation.” Assuming this to be a simple random sample of U.S. adults, construct and interpret the 95% and 99% confidence intervals for the proportion of U.S. adults who consider lounging at the beach to be their dream vacation. 9.87 For the following simple random sample of household incomes (thousands of dollars) from a large county, construct and interpret the 90% and 95% confidence intervals for the population mean. The data are also in file XR09087. 58.3 50.0 58.1 33.5 51.1 38.1 42.3 60.4 55.8 46.2 40.4 52.5 51.3 47.5 48.5 59.3 40.9 37.1 39.1 43.6 55.3 42.3 48.2 42.8 61.1 34.7 35.5 52.9 44.7 51.5 9.88 For a new process with which the production personnel have little experience, neither the standard deviation nor the mean of the process is known. Twenty different simple random samples, each with n 50, are to be drawn from the process, and a 90% confidence interval for the mean is to be constructed for each sample. What is the probability that at least 2 of the confidence intervals will not contain the population mean? 9.89 There were 904 new Subway Restaurants franchises opened during 2002. Suppose that Subway wished to survey a simple random sample of the new franchisees to find out what percentage of them were totally pleased with their relationship with the company. If Subway wanted to have 90% confidence in being within 3 percentage points of the population percentage who are pleased, how many of the new franchisees would have to be included in the sample? SOURCE: Subway.com. June 13, 2003. 9.90 In Exercise 9.89, suppose Subway has carried out the study, using the sample size determined in that exercise, and 27.5% of the franchisees say they are pleased with their relationship with Subway. Construct and interpret the 95% confidence interval for the population percentage. 9.91 A research firm wants to be 90% confident that a population percentage has been estimated to within 3 percentage points. The research manager calculates the necessary sample size with 0.5 as his estimate of the population proportion. A new business school graduate who has just joined the firm questions the research manager further, and they agree that the population proportion is no more than 0.3. If interviews cost $10 each, how much money has the new graduate just saved the company? 9.92 The activities director of a large university has surveyed a simple random sample of 100 students for the purpose of determining approximately how many students to expect at next month’s awards ceremony to be held in the gymnasium. Forty of the students said they plan to attend. What are the upper and lower 95% confidence limits for the number of the university’s 10,000 students who plan to attend the awards ceremony? 9.93 A research firm has found that 39% of U.S. adults in the over-$75,000 income category work at least 51 hours per week. Assuming this was a simple random sample of 500 adults in this income group, construct and interpret the 95% and 99% confidence intervals for the proportion who work at least 51 hours per week. For each of the confidence intervals, identify and explain the maximum likely error in the study. 9.94 For a process having a known standard deviation, a simple random sample of 35 items is selected. If the width of the 95% confidence interval is identified as y, express the width of the 99% confidence interval as a multiple of y. 9.95 The makers of Count Chocula breakfast cereal would like to determine, within 2 percentage points and with 99% confidence, the percentage of U.S. senior citizens who have Count Chocula for breakfast at least once a week. What sample size would you recommend? 9.96 In a work-sampling study, an industrial engineer has observed the activities of a clerical worker on 121 randomly selected times during a workweek. On 32 of these occasions, the employee was talking on the telephone. For an 8-hour day, what are the upper and lower 95% confidence limits for the number of minutes this employee talks on the phone? 9.97 A researcher would like to determine, within 3 percentage points and with 90% confidence, the percentage of Americans who have a certain characteristic. If she feels certain that the percentage is somewhere between 20% and 40%, how many persons should be included in the sample? Chapter 9: Estimation from Sample Data 9.98 In a survey of 1320 executives who oversee corporate data systems, 24% said they had experienced losses caused by computer viruses during the past year. Assuming the executives were a simple random sample of all such executives, construct and interpret the 90% confidence interval for the population proportion who were monetarily harmed by computer viruses that year. 9.99 An airline would like to determine, within 3 percentage points and with 95% confidence, the percentage of next month’s customers who judge the courtesy of its employees as being “very good to excellent.” What sample size would you recommend? 9.100 A consultant conducts a pilot study to estimate a population standard deviation, then determines how large a simple random sample will be necessary to have a given level of confidence that the difference between x and will be within the maximum error specified by her client. The necessary sample size has been calculated as n 100. If the client suddenly decides that the maximum error must be only one-fourth that originally specified, what sample size will now be necessary? 9.101 There are 1733 machinery rebuilding and repairing companies in the United States. A tool manufacturer wishes to survey a simple random sample of these firms to find out what proportion of them are interested in a new tool design. If the tool manufacturer would like to be 95% confident that the sample proportion is within 0.01 of the actual population proportion, how many machinery rebuilding and repairing companies should be included in the sample? SOURCE: American Business Information, Sales Leads & Mailing Lists, August 1999, p. 22. 9.102 In Exercise 9.101, suppose the tool manufacturer has carried out the study, using the sample size determined in that exercise, and 39.0% of the machinery rebuilding and repairing companies are interested in the new tool design. Construct and interpret the 95% confidence interval for the population percentage. 9.103 The Colgate-Palmolive Company has 37,700 employees. If the company wishes to estimate, within 2 percentage points and with 99% confidence, the percentage of employees who are interested in participating in a new stock option benefits program, how large a simple random sample will be necessary? SOURCE: Colgate-Palmolive Company, 2002 Annual Report, p. 8. 9.104 To gain information about competitors’ products, companies sometimes employ “reverse engineering,” which consists of buying the competitor’s product, then taking it apart and examining the parts in great detail. Engaging in this practice, a bicycle manufacturer intends to buy two or more of a leading competitor’s mountain bikes and measure the tensile strength of the crossbar portion of the frame. Past experience has shown these strengths to be approximately normally distributed with a standard deviation of 20 pounds per square inch (psi). 305 If the bike purchaser wants to have 90% confidence that the sampling error will be no more than 5 psi, how many of the competitor’s mountain bikes should be purchased for destructive testing? 9.105 A survey of business travelers found that 40% of those surveyed utilize hotel exercise facilities during their stay. Under the assumption that a simple random sample of 1000 business travelers were surveyed, construct and interpret the 90% and 95% confidence intervals for the proportion of business travelers who use their hotel’s exercise facilities. 9.106 A researcher, believing to be no more than 0.40, calculates the necessary sample size for the confidence level and maximum likely error he has specified. Upon completing the study, he finds the sample proportion to be 0.32. Is the maximum likely error greater than, equal to, or less than that originally specified? Explain. 9.107 A truck loaded with 8000 electronic circuit boards has just pulled into a firm’s receiving dock. The supplier claims that no more than 3% of the boards fall outside the most rigid level of industry performance specifications. In a simple random sample of 300 boards from this shipment, 12 fall outside these specifications. Construct the 95% confidence interval for the percentage of all boards in this shipment that fall outside the specifications, then comment on whether the supplier’s claim would appear to be correct. 9.108 A researcher has estimated that U.S. college students spend an average of 17.2 hours per week on the Internet. Assuming a simple random sample of 500 college students and a sample standard deviation of 1.4 hours per week, construct and interpret the 99% confidence interval for the population mean. / data set / Note: Exercises 9.109–9.111 require a computer and statistical software. 9.109 According to the National Restaurant Association, the average check for a sit-down dinner is $25. Such a finding could have been based on data like the 800 sample checks in file XR09109. Using the data in this file, construct and interpret the 95% confidence interval for the population mean. SOURCE: “U.S. Dining-Out Tab: $1B a Day,” USA Today, May 25, 2000, p. 1D. 9.110 For taxpayers having an adjusted gross income of $1 million or more, the Internal Revenue Service reports that the average deduction for gifts to charity was $144,700. Curious to see how his state compares, a legislator surveys a simple random sample of 200 taxpayers from his state who are in this gross income category, with the data as shown in file XR09110. Using the data in this file, construct and interpret the 90% confidence interval for the mean charitable-gifts deduction for all of the state’s taxpayers who are in the $1 million or more adjusted gross income category. Is $144,700 within the 306 confidence interval? Given the answer to the preceding question, comment on whether the state’s taxpayers who are in this income group might not be typical of those in the nation as a whole in terms of their tax-deductible charitable contributions. SOURCE: “Brilliant Deductions, Taxing Questions,” USA Today, March 3, 2000, p. 3B. 9.111 To avoid losing part of their federal highway fund allocation, state safety administrators must ensure that interstate speed limits are adequately enforced within their state. In an upcoming test, federal researchers will be randomly selecting and clocking a very large sample of vehicles on a given section of the state’s portion of an Part 3: Sampling Distributions and Estimation interstate highway that has historically had a relatively high accident rate. In anticipation of the upcoming study, state administrators randomly select and clock 100 vehicles along this route, obtaining the speeds shown in data file XR09111. Construct and interpret the 95% confidence interval for the population mean vehicle speed along this stretch of highway. Based on this interval, comment on whether the mean speed for the population of vehicles using this part of the highway might be 70 mph, the cutoff above which federal highway funds become endangered. integrated cases Thorndike Sports Equipment (Thorndike Video Unit Four) Seeing the fishing pole in his grandfather’s office, Ted Thorndike’s first thought is that old Luke is going to go fishing again and leave him to manage the store. He is quite surprised to learn the fishing pole is actually an inspiration for a new series of ads that Luke has in mind. The elder Thorndike explains, “Ted, this fishing pole is made of graphite, the same stuff that goes into our GrafPro racquetball racquets. It’s so flexible and strong that it can be bent so the two ends actually touch each other. They even show this in the ads.” Although Luke realizes that you can’t do exactly the same thing with a racquetball racquet, he’d like to put some of his racquets into a horizontal mounting device, then see how much weight they’ll take before they break. If the amount of weight is impressive enough, Luke plans to include this kind of test in the television advertisements he’s planning for the firm’s racquetball racquets. However, he wants to be careful not to brag about the racquet being able to hold too much weight, since the firm could get into trouble with the government and other truth-in-advertising advocates. He asks Ted to set up a test in which racquets are mounted horizontally, then the weight on the end is gradually increased until they break. Based on the test results, a weight value would be selected such that the average racquet would almost certainly be able to withstand this amount. Although accuracy is important, Ted has been instructed not to break more than 15 or 20 racquets in coming up with an average for all the racquets. For 20 racquets subjected to this severe test, the weight (in pounds) at which each one failed was as follows. The data are also in file THORN09. 221 208 224 217 228 220 222 230 223 217 229 236 218 224 215 222 218 225 221 234 Ted believes it’s reasonable to assume the population of breaking strengths is approximately normally distributed. Because of Luke’s concern about being able to support the advertising claim, he wants to be very conservative in estimating the population mean for these breaking strengths. Ted needs some help in deciding how conservative he would like to be, and in coming up with a number that can be promoted in the ads. Springdale Shopping Survey The case in Chapter 2 listed 30 questions asked of 150 respondents in the community of Springdale. The coding key for these responses was also provided in this earlier exercise. The data are in file SHOPPING. In this exercise, some of the estimation techniques presented in the chapter will be applied to the survey results. You may assume that (continued) seeing statistics: applet 9 Chapter 9: Estimation from Sample Data 307 Confidence Interval Size This applet allows us to construct and view z-intervals for the population mean by using the slider to specify the confidence level. As in Figure 9.3, the sample mean is 1.400 inches, the sample size is 30, and the population standard deviation is known to be 0.053 inches. Note that the confidence interval limits shown in the graph may sometimes differ slightly from those we would calculate using the pocket calculator and our standard normal distribution table. This is because the applet is using more exact values for z than we are able to show within printed tables like the one in the text. Applet Exercises 9.1 With the slider positioned so as to specify a 95% confidence interval for , what are the upper and lower confidence limits? 9.2 Move the slider so that the confidence interval is now 99%. Describe how the increase in the confidence level has changed the width of the confidence interval. 9.3 Move the slider so that the confidence interval is now 80%. Describe how the decrease in the confidence level has changed the width of the confidence interval. 9.4 Position the slider at its extreme left position, then gradually move it to the far right. Describe how this movement changes the confidence level and the width of the confidence interval. these respondents represent a simple random sample of all potential respondents within the community and that the population is large enough that application of the finite population correction would not make an appreciable difference in the results. Managers associated with shopping areas like these find it useful to have point estimates regarding variables describing the characteristics and behaviors of their customers. In addition, it is helpful for them to have some idea as to the likely accuracy of these estimates. Therein lies the benefit of the techniques presented in this chapter and applied here. 1. Item C in the description of the data collection instrument lists variables 7, 8, and 9, which represent the respondent’s general attitude toward each of the three shopping areas. Each of these variables has numerically equal distances between the possible responses, and for purposes of analysis they may be considered to be of the interval scale of measurement. a. Determine the point estimate, then construct the 95% confidence interval for 7 the average attitude toward Springdale Mall. What is the maximum likely error in the point estimate of the population mean? b. Repeat part (a) for 8 and 9, the average attitudes toward Downtown and West Mall, respectively. 2. Given the breakdown of responses for variable 26 (sex of respondent), determine the point estimate, then construct the 95% confidence interval for 26 the population proportion of males. What is the maximum likely error in the point estimate of the population proportion? 3. Given the breakdown of responses for variable 28 (marital status of respondent), determine the point estimate, then construct the 95% confidence interval for 28 the population proportion in the “single or other” category. What is the maximum likely error in the point estimate of the population proportion? seeing statistics: applet 10 Comparing the Normal and Student t Distributions In this applet, we use a slider to change the number of degrees of freedom and shape for the Student t distribution and then observe how the resulting shape compares to that of the standard normal distribution. The standard normal distribution is fixed and shown in red, and the Student t distribution is displayed in blue. Applet Exercises 10.1 Move the slider so that df 5. Describe the shape of the t distribution compared to that of the standard normal distribution. 10.2 Move the slider downward so that df 2. How has this decrease changed the shape of the t distribution? 10.3 Gradually move the slider upward so that df increases from 2 to 10. Describe how the shape of the t distribution changes along the way. 10.4 Position the slider so that df 2, then gradually move it upward until df 100. Describe how the shape of the t distribution changes along the way. seeing statistics: applet 11 Student t Distribution Areas In this applet, we use a slider to change the number of degrees of freedom for the t distribution, and text boxes allow us to change the t value or the two-tail probability for a given df. When changing a text-box entry, be sure the cursor is still within the box before pressing the enter or return key. Applet Exercises 11.1 With the slider set so that df 9 and the left text box containing t 3.25, what is the area beneath the curve between t 3.25 and t 3.25? 11.2 Gradually move the slider upward until df 89. What effect does this have on the t value shown in the text box? 11.3 Position the slider so that df 2, then gradually move it upward until df 100. Describe how the value in the t text box and the shape of the t distribution change along the way. 11.4 With the slider set so that df 9, enter 0.10 into the two-tail probability text box at the right. What value of t now appears in the left text box? To what right-tail area does this correspond? Verify the value of t for df 9 and this right-tail area by using the t table immediately preceding the back cover of the book.