# hw8sol - Solutions to Homework 8 Problem 8.3 a The 95...

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Solutions to Homework 8 Problem 8.3 a. The 95% confidence interval for the population mean is ¯ x ± z α/ 2 σ n = 80 ± 1 . 96 15 60 = 80 ± 3 . 80 = (76 . 20 , 83 . 80) b. If there were 120 items in the sample the 95% confidence interval is ¯ x ± z α/ 2 σ n = 80 ± 1 . 96 15 120 = 80 ± 2 . 68 = (77 . 32 , 82 . 68) c. The larger the sample size, the narrower the confidence interval. Problem 8.5 a. The margin of error is z α/ 2 σ n = 1 . 96 5 49 = 1 . 4 . b. The 95% confidence interval is ¯ x ± MOE = 24 . 80 ± 1 . 40 = (23 . 40 , 26 . 20) . Problem 8.12 (a,b,c) a. t = 2 . 179 b. Here t = 1 . 676 for an upper tail area of 0.05 with 50 degrees of freedom. Since the t distribution is symmetric the value of t with a lower tail area of 0.05 with 50 degrees of freedom is t = - 1 . 676. c. t = 2 . 457 Problem 8.13 a. The point estimate of the population mean is ¯ x : ¯ x = 10 + 8 + · · · + 5 8 = 10 . 1

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b. The point estimate of the population standard deviation is s : s = s (10 - 10) 2 + (8 - 10) 2 + · · · + (5 - 10) 2 7 = 3 . 46 . c. First, t α/ 2 = 2 . 365 since α/ 2 = 0 . 025 and the degrees of freedom are n - 1 = 7. So the margin of error is t α/ 2 s n = 2 . 365 3 . 46 8 = 2 . 89 . d. The 95% confidence interval is ¯ x ± MOE = 10 ± 2 . 89 = (7 . 11 , 12 . 89) . Problem 8.15 The formula for a confidence interval is
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