Question

David Anderson has been working as a lecturer at Michigan State University for the last three years. He teaches two large sections of introductory accounting every semester. While he uses the same lecture notes in both sections, his students in the first section outperform those in the second section. He believes that students in the first section not only tend to get higher scores, they also tend to have lower variability in scores. David decides to carry out a formal test to validate his hunch regarding the difference in average scores. In a random sample of 22 students in the first section, he computes a mean and a standard deviation of 84.4 and 12.8, respectively. In the second section, a random sample of 25 students results in a mean of 81.9 and a standard deviation of 1.22. Use Table 2. Sample 1 consists of students in the first section and Sample 2 represents students in the second section. a. Construct the null and the alternative hypotheses to test David’s hunch. H0: μ1 − μ2 = 0; HA: μ1 − μ2 ≠ 0 H0: μ1 − μ2 ≥ 0; HA: μ1 − μ2 < 0 H0: μ1 − μ2 ≤ 0; HA: μ1 − μ2 > 0 b-1. Calculate the value of the test statistic. (Round all intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) Test statistic b-2. What assumption regarding the population variances is used to conduct the test? Known population standard deviations. Unknown population standard deviations that are equal Unknown population standard deviations that are not equal c. Implement the test at α = 0.05 using the critical value approach. Reject H0; there is evidence that scores are higher in the first section. Reject H0; there is no evidence that scores are higher in the first section. Do not reject H0; there is evidence that scores are higher in the first section. Do not reject H0; there is no evidence that scores are higher in the first section.

Answer 1

From the given data

     

     

a.

Null Hypothesis:

versus

Alternative Hypothesis

b1.

We use the t-test for difference of means and the test statistic is calculated as

Where

Under the Null Hypothesis, the t-test statistic is calculated as

b2. Answer:Unknown population standard deviations that are equal

The assumption regarding the population variances to conduct the test is that the population variances are equal i.e.,

c. Answer: Do not reject H0; there is evidence that scores are higher in the first section

At α = 0.05, t-tabulated value at 45 (= 22+25-2) degrees of freedom for a right tail test is, t-tabulated = 1.69

Since the calculated t (=0.95) < t-tabulated (=1.69), we do not reject the null hypothesis and infer that the first section outperform those in the second section in scores