Question

Professor Francis has implemented a new book for his Fall 2013 statistics course. He wants to see if the new book has an effect on the exam average for Exam 3. He knows that the Exam 3 mean for all students who have taken his statistic course over the past 10 years is 77.2. He collects some data from a sample of 9 students and finds that the Exam 3 mean for his Fall 2013 statistics students is 81.7 with a variance of 32.49.

a)Conduct a t-test to see is his theory is correct.
(Be sure to show all 4 steps of a hypothesis test. Also be sure to consider if this is a one-tailed or two-tailed test.)

b)Measure effect size using Cohen's D.

c)Measure effect size using percentage of variance explained (r)

d)Estimate effect size by constructing 90% an confidence interval.

Answer 1

3a]

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μ = 77.2

Ha: μ ≠ 77.2

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.1, and the critical value for a two-tailed test is tc​=1.86.

The rejection region for this two-tailed test is R={t:∣t∣>1.86}

(3) Test Statistics

The t-statistic is computed as follows:

(4) Decision about the null hypothesis: Since it is observed that ∣t∣=2.368>tc​=1.86, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0454, and since p=0.0454<0.1, it is concluded that the null hypothesis is rejected.

(5) Conclusion: It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean μ is different than 77.2, at the 0.1 significance level.

b] Effect size Cohens D = 81.7-77.2/5.7= 0.7895

c] r^2= t^2/t^2+df= (2.368)^2/(2.368)^2+8= 5.61/5.61+8= 0.4122=

41.22% of variance explained.

d] Confidence Interval: The 90% confidence interval is 78.167<μ<85.233.