Question

SECTION 7 - 2 1. A simple random sample of 39 salaries of NCAA football coaches has a mean of $416,953 with standard deviation $462,364. Construct a 95% confidence interval estimate of the mean salary of an NCAA football coach. 2. In a test of weight loss programs, 27 adults used the Atkins weight loss program. After 12 months, their mean weight loss was found to be 2.4 lb, with a standard deviation of 4.7 lb. Construct a 90% confidence interval estimate of the mean weight loss for all such subjects.

Answer 1

Solution: 1. Given that sample sizen 39 football coaches has a mean = 416953 SD = 462,364. So 96confidence interval for the mean is. X-tol ( 3 ) <le at taller) c. I = II to 12 (5) (from t-table to/222024) - 416953 – 2.024 (462364) = 416953 – 2.024 6.24499 = 416953 + 14985118872 416953-149851.88 72 242416953+149851-8872 267101.1 <ll< 566804.9

2. Given that Sample size n=27 mean t = 2.4 S.D=4.7 Degree of freedom e df = null = 27-1 - 26. At 90% confidence level the t test is do 1-0.90 = 0.1 212= Ook 0.05 tdizi df = +0.05,26 = 1.706 (from Table] Margin of error E = txp, df Some = 1.706 ( 41 ) = 1.706 (5019615) 3.1543103 confidence interval estimate of the population The 90% mean is ae <ulante 2.4-1.543103 cu gu+1.543103

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