SECTION 10.3 Comparing Two Variances 557 Extending Concepts Finding Left-Tailed Critical F-Values In this section, you only needed to calculate the right-tailed critical F-value for a two-tailed test. For other applications of the F-distribution, you will need to calculate the left-tailed critical F-value. To calculate the left-tailed critical F-value, perform the steps below. (1) Interchange the values for d.f.N and d.f.D. (2) Find the corresponding F-value in Table 7. (3) Calculate the reciprocal of the F-value to obtain the left-tailed critical F-value. In Exercises 27 and 28, find the right- and left-tailed critical F-values for a two-tailed test using the level of significance a and degrees of freedom d.f.N and d.f.D. 27. a = 0.05, d.f.N = 6, d.f.D = 3 28. a = 0.10, d.f.N = 20, d.f.D = 15 Confidence Interval for S 2 1 S 2 2 When s 2 1 and s 2 2 are the variances of randomly selected, independent samples from normally distributed populations, then a confidence interval for s 2 1 s 2 2 is s2 1 s2 2 # 1 FR 6 s 2 1 s 2 2 6 s2 1 s2 2 # 1 FL where FR is the right-tailed critical F-value and FL is the left-tailed critical F-value. In Exercises 29 and 30, construct the confidence interval for s 2 1 s 2 2. Assume the samples are random and independent, and the populations are normally distributed. 29. Cholesterol Contents In a recent study of the cholesterol contents of grilled chicken sandwiches served at fast food restaurants, a nutritionist found that random samples of sandwiches from Restaurant A and from Restaurant B had the sample statistics shown in the table. Construct a 95% confidence interval for s 2 1 s 2 2, where s 2 1 and s 2 2 are the variances of the cholesterol contents of grilled chicken sandwiches from Restaurant A and Restaurant B, respectively. Cholesterol contents of grilled chicken sandwiches Restaurant A Restaurant B s2 1 = 10.89 s 2 2 = 9.61 n1 = 16 n2 = 12 30. Carbohydrate Contents In a recent study of the carbohydrate contents of grilled chicken sandwiches served at fast food restaurants, a nutritionist found that random samples of sandwiches from Restaurant A and from Restaurant B had the sample statistics shown in the table. Construct a 95% confidence interval for s 2 1 s 2 2, where s 2 1 and s 2 2 are the variances of the carbohydrate contents of grilled chicken sandwiches from Restaurant A and Restaurant B, respectively. Carbohydrate contents of grilled chicken sandwiches Restaurant A Restaurant B s2 1 = 5.29 s 2 2 = 3.61 n1 = 16 n2 = 12
RkJQdWJsaXNoZXIy NjM5ODQ=