Question

QUESTION 5 In order to determine how many hours per week freshmen college students watch television, a random sample of 256 students was selected. It was determined that the students in the sample spent an average of 14 hours with a standard deviation of 3.6 hours watching TV per week. a. Provide a 95% confidence interval estimate for the average number of hours that all college freshmen spend watching TV per week. Assume that a sample of 66 students was selected that resulted in the same mean and standard b. deviation. Provide a 95% confidence interval estimate for the average number of hours that all college freshmen spend watching TV week. per

Answer 1

a)

95% Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 256- 1 ) = 1.969
14 ± t(0.05/2, 256 -1) * 3.6/√(256)
Lower Limit = 14 - t(0.05/2, 256 -1) 3.6/√(256)
Lower Limit = 13.557
Upper Limit = 14 + t(0.05/2, 256 -1) 3.6/√(256)
Upper Limit = 14.443
95% Confidence interval is ( 13.557 , 14.443 )

b)

95% Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 66- 1 ) = 1.997
14 ± t(0.05/2, 66 -1) * 3.6/√(66)
Lower Limit = 14 - t(0.05/2, 66 -1) 3.6/√(66)
Lower Limit = 13.115
Upper Limit = 14 + t(0.05/2, 66 -1) 3.6/√(66)
Upper Limit = 14.885
95% Confidence interval is ( 13.115 , 14.885 )