This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 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 ....
View
Full
Document
This note was uploaded on 01/24/2011 for the course STATISTICS 19897 taught by Professor Jager,abigaill during the Spring '10 term at Kansas State University.
 Spring '10
 Jager,AbigailL

Click to edit the document details