Question

In order to determine how many hours per week freshmen college students watch television, a random sample of 25 students was selected. It was determined that the students in the sample spent an average of 19.5 hours with a sample standard deviation of 3.9 hours watching TV per week. Please answer the following questions: (a) Provide a 95% confidence interval estimate for the average number of hours that all college freshmen spend watching TV per week. (b) Assume that a sample of 36 students was selected (with the same mean and the sample standard deviation). Provide a 95% confidence interval estimate for the average number of hours that all college freshmen spend watching TV per week.

Answer #1

a)

sample mean, xbar = 19.5

sample standard deviation, s = 3.9

sample size, n = 25

degrees of freedom, df = n - 1 = 24

Given CI level is 95%, hence α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.06

ME = tc * s/sqrt(n)

ME = 2.06 * 3.9/sqrt(25)

ME = 1.607

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))

CI = (19.5 - 2.06 * 3.9/sqrt(25) , 19.5 + 2.06 *
3.9/sqrt(25))

CI = (17.89 , 21.11)

b)

sample mean, xbar = 19.5

sample standard deviation, s = 3.9

sample size, n = 36

degrees of freedom, df = n - 1 = 35

Given CI level is 95%, hence α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.03

ME = tc * s/sqrt(n)

ME = 2.03 * 3.9/sqrt(36)

ME = 1.32

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))

CI = (19.5 - 2.03 * 3.9/sqrt(36) , 19.5 + 2.03 *
3.9/sqrt(36))

CI = (18.18 , 20.82)

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