Question

Online courses: A sample of 266
students who were taking online courses were asked to describe their overall impression of online learning on a scale of 1–7, with 7 representing the most favorable impression. The average score was 5.65 , and the standard deviation was 0.95.

(a) Construct a 99% confidence interval for the mean score.

(b) Assume that the mean score for students taking traditional courses is 5.58 . A college that offers online courses claims that the mean scores for online courses and traditional courses are the same. Does the confidence interval contradict this claim? Explain

Answer 1

a)
sample mean, xbar = 5.65
sample standard deviation, s = 0.95
sample size, n = 266
degrees of freedom, df = n - 1 = 265

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, tc = t(α/2, df) = 2.595

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (5.65 - 2.595 * 0.95/sqrt(266) , 5.65 + 2.595 * 0.95/sqrt(266))
CI = (5.5 , 5.8)

b)
As the mean score 5.58 is include in the CI, we can not conclude that the average score using online courses is different than the average score using traditional courses