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 16 students in the first section, he computes a mean and a standard deviation of 75.9 and 21.5, respectively. In the second section, a random sample of 20 students results in a mean of 74.2 and a standard deviation of 1.03.

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 assumption regarding the population variances is used to conduct the test?
	Unknown population standard deviations that are not equal Unknown population standard deviations that are equal Known population standard deviations

c.	Implement the test at ? = 0.01 using the critical value approach.
	Reject H0; scores are higher in the first section. Do not reject H0; scores are higher in the first section. Reject H0; scores are not higher in the first section. Do not reject H0; scores are not higher in the first section.

Answer 1

a) OPtion - B) H₀: < 0

                       H_A: > 0

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

                                     = (75.9 - 74.2)/sqrt((21.5)^2/16 + (1.03)^2/20)

                                     = 0.32

b-2) Option - A) 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))

         = ((21.5)^2/16 + (1.03)^2/20)^2/(((21.5)^2/16)^2/15 + ((1.03)^2/20)^2/19)

         = 15

At alpha = 0.01, the critical value is t_{0.99, 15} = 2.602

As the test statistic value is not greater than the critical value (0.32 < 2.602), so the null hypothesis is not rejected.

Option - D) Do not reject H₀; scores are not higher in the first section.