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 21 students in the first section, he computes a mean and a standard deviation of 88.9 and 20.7, respectively. In the second section, a random sample of 19 students results in a mean of 85.2 and a standard deviation of 1.02.

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: ?1 ? ?2 ? 0; HA: ?1 ? ?2 > 0 H0: ?1 ? ?2 ? 0; HA: ?1 ? ?2 < 0 H0: ?1 ? ?2 = 0; HA: ?1 ? ?2 ? 0

b-1.	Calculate the value of the test statistic. (Round intermediate calculations to 4 decimal places and final answer to 2 decimal places.)

Test statistic

b-2.	What is assumption regarding the population variances used to conduct the test?
	Known population standard deviations Unknown population standard deviations that are not equal Unknown population standard deviations that are equal

c.	Implement the test at ? = 0.10 using the critical value approach.
	Do not reject H0; scores are higher in the first section. Do not reject H0; scores are not higher in the first section. Reject H0; scores are not higher in the first section. Reject H0; scores are higher in the first section.

Answer 1

a) OPtion - A) H₀: < 0

                       H_A: > 0

b-1) The test statistic t = ()/sqrt(s1^2/n1 + s2^2/n2)

                                     = (88.9 - 85.2)/sqrt((20.7)^2/21 + (1.02)^2/19)

                                     = 0.82

b-2) Option - B) Unknown population standard deviations that are not equal.

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

         = ((20.7)^2/21 + (1.02)^2/19)^2/(((20.7)^2/21)^2/20 + ((1.02)^2/19)^2/18)

         = 20

At alpha = 0.10, the critical value is t_{0.9, 20} = 1.325

As the test statistic value is not greater than the critical value (0.82 < 1.325), so the null hypothesis is not rejected.

Option - B) Do not reject H₀, scores are not higher in the first section.