A survey about duration of a job task done by a person with or without a course?

No course group


Sample size n1 is 15, sample mean x1 is 5.3 hours, sample standard deviation S1 is 2 hours





With course group


Sample size n2 is 13, mean x2 is 4.9 hours, sample std deviation S2 is 2.8 hours





Based on data, what should be concluded about the average time it takes to complete the job with and without the course offered?


Use alpha of 0.01


(Perform the 6 steps hypothesis test)

A survey about duration of a job task done by a person with or without a course?
If you assume equal variances, the calculation is described at:





http://en.wikipedia.org/wiki/Student%27s...





n = n1 + n2 = 15 + 13 = 28


df = n - 1 = 27 degrees of freedom


s^2 = ((n1 - 1)s1^2 + (n2 - 1)s2^2)(1/n1 + 1/n2) / (n1 + n2 - 2)


= (14 * 2^2 + 12 * 2.8^2) * (1/15+ 1/13) / (15 + 13 - 2)


= 0.8288


s = 0.9104


t = (x1-bar - x2-bar) / s


= (5.3 - 4.9) / 0.9104


= 0.43936





By the table on that page, Prob(T {v=27} %26lt; 2.771) = 0.995 so by the symmetry of the distribution, P(T %26lt; 2.771) = P(T %26gt; 2.771) = 1 - 0.995 = 0.005, meaning P(-2.771 %26lt; T %26lt; 2.771) = 0.99. The measured t = 0.43936 is well within that range. So assuming equal variances, at alpha = 0.01 we do not reject the null hypothesis of equal means.





If you do not assume equal variances, then see Welch's t test





http://en.wikipedia.org/wiki/Welch%27s_t...





s^2 = (s1^2 / n1) + (s2^2 / n2)


t = (x1-bar - x2-bar)/s


df = s^4 / (s1^4 * n1^2 * (n1 - 1) + s2^4 * n2^2 * (n2 - 1))





Use the F-test to test the equal variance assumption.





http://rkb.home.cern.ch/rkb/AN16pp/node8...





Dan