Online courses: A sample of 266 students who were taking online courses were asked to describe...

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

Solutions

Expert Solution

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

orchestra answered 1 year ago

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