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440 CHAPTER 8 CONFIDENCE INTERVALS FOR ONE POPULATION MEAN Actually, like the z-interval procedure, the t-interval procedure works reasonably well even when the variable under consideration is not normally distributed and the sample size is small or moderate, provided the variable is not too far from being normally distributed. In other words, the t-interval procedure is robust to moderate violations of the normality assumption. When considering the t-interval procedure, it is also important to watch for outliers. The presence of outliers calls into question the normality assumption. And even for large samples, outliers can sometimes unduly affect a t-interval because the sample mean and sample standard deviation are not resistant to outliers. Guidelines for when to use the t-interval procedure are the same as those given for the z-interval procedure in Key Fact 8.1 on page 421. And remember, always look at the data before applying the t-interval procedure to ensure that it is reasonable to use it. EXAMPLE 8.9 Illustrates Procedure 8.2 The U.S. Federal Bureau of Investigation (FBI) compiles data on robbery and property crimes and publishes the information in Population-at-Risk Rates and Selected Crime Indicators. A sample of last year’s pick-pocket offenses yields the values lost shown in Table 8.5. Use the data to obtain a 95% confidence interval for the mean value lost, µ, of all last year’s pick-pocket offenses. TABLE 8.5 Value lost ($) for a sample of 25 pick-pocket offenses Solution 447 313 217 833 649 207 844 768 277 554 627 253 1064 805 570 430 397 26 653 223 883 214 587 549 443 Since the sample size, n H 25, is moderate, we first need to consider questions of normality and outliers. (See the second bulleted item in Key Fact 8.1 on page 421.) To do that, we constructed a normal probability plot for the data in Table 8.5, as shown in Fig. 8.8. The normal probability plot in Fig. 8.8 shows no outliers and falls roughly in a straight line. Thus we can apply Procedure 8.2 to obtain the required confidence interval. Step 1 For a confidence level of 1 − α, use Table IV to find tα/2 with df H n − 1, where n is the sample size. We want a 95% confidence interval, so α H 1 − 0.95 H 0.05. Since n H 25, df H 25 − 1 H 24. Consulting Table IV, we find that tα/2 H t0.05/2 H t0.025 H 2.064. 8.4 441 3 2 Normal score FIGURE 8.8 Normal probability plot of the value-lost data in Table 8.5 CONFIDENCE INTERVALS FOR ONE POPULATION MEAN WHEN σ IS UNKNOWN 1 0 21 22 23 0 200 400 600 800 1000 1200 Value lost ($) Step 2 The confidence interval for µ is from s s x − tα/2 · √ to x + tα/2 · √ . n n From Step 1, tα/2 H 2.064. Applying the usual formulas for x and s to the data in Table 8.5, we find that x H 513.32 and s H 262.23. Consequently, a 95% confidence interval for µ is from 262.23 262.23 513.32 − 2.064 · √ to 513.32 + 2.064 · √ , 25 25 or 405.07 to 621.57. We can be 95% confident that the mean value lost, µ, of all last year’s pick-pocket offenses is somewhere between $405.07 and $621.57. EXAMPLE 8.10 Illustrates Procedure 8.2 The U.S. Department of Agriculture publishes data on chicken consumption in Food Consumption, Prices, and Expenditures. Last year’s chicken consumptions, in pounds, for 17 randomly selected people are displayed in Table 8.6. Use the data to obtain a 90% confidence interval for last year’s mean chicken consumption, µ. TABLE 8.6 Sample of last year’s chicken consumptions (lbs) Solution 47 59 53 39 53 51 62 55 50 49 0 72 50 65 45 70 63 A normal probability plot of the data in Table 8.6 is displayed in Fig. 8.9(a). The plot reveals an outlier—the observation of 0 lb. Since the sample size is only moderate, it is inappropriate to apply Procedure 8.2 to the data in Table 8.6.