Question

Do students reduce study time in classes where they achieve a higher midterm score? In a Journal of Economic Education article (Winter 2005), Gregory Krohn and Catherine O’Connor studied student effort and performance in a class over a semester. In an intermediate macroeconomics course, they found that “students respond to higher midterm scores by reducing the number of hours they subsequently allocate to studying for the course.” Suppose that a random sample of n = 8 students who performed well on the midterm exam was taken and weekly study times before and after the exam were compared. The resulting data are given in Table 10.6. Assume that the population of all possible paired differences is normally distributed.

Table 10.6

Weekly Study Time Data for Students Who Perform Well on the MidTerm
Students	1	2	3	4	5	6	7	8
Before	17	11	16	18	15	18	17	13
After	9	9	8	11	10	7	10	11

Paired T-Test and CI: Study Before, Study After

   

Paired T for Study Before - Study After
	N	Mean	StDev	SE Mean
StudyBefore	8	15.6250	2.5036	.8851
StudyAfter	8	9.3750	1.4079	.4978
Difference	8	6.25000	3.10530	1.09789

95% CI for mean difference: (3.65391, 8.84609)

T-Test of mean difference = 0 (vs not = 0): T-Value = 5.69, P-Value = .0007

(a) Set up the null and alternative hypotheses to test whether there is a difference in the true mean study time before and after the midterm exam.

H₀: µ_d =  versus H_a: µ_d ≠

(b) Above we present the MINITAB output for the paired differences test. Use the output and critical values to test the hypotheses at the .10, .05, and .01 level of significance. Has the true mean study time changed? (Round your answer to 2 decimal places.)

t =   We have (Click to select)noextremely strongvery strongstrong evidence.

(c) Use the p-value to test the hypotheses at the .10, .05, and .01 level of significance. How much evidence is there against the null hypothesis?

There is (Click to select)no evidencevery strong evidenceextermly strong evidencestrong evidence against the null hypothesis.

Answer 1

(a) Set up the null and alternative hypotheses to test whether there is a difference in the true mean study time before and after the midterm exam.

H₀: µ_d = versus H_a: µ_d ≠

where = 0

(b) critical values to test the hypotheses at the .10, .05, and .01 level of significance. Has the true mean study time changed? (Round your answer to 2 decimal places.)

We need to calculate the test statistic and then compare with the critical values

Test Stat = where D is the difference between the results

=

= 5.69

Decision Criteria: Reject null hypo if the T.S. > C.V

Critical values () at different levels of significance are

Critical value	Decision
at = 0.10	T.S. > C.V Reject
at = 0.05	T.S. > C.V Reject
at = 0.01	T.S. > C.V Reject

There is very strong evidence against the null hypothesis.

We can conclude that at there is a difference between the true mean study time before and after the midterm at 0.10, 0.05 and 0.01 levels of significance.

(c) Use the p-value to test the hypotheses at the .10, .05, and .01 level of significance. How much evidence is there against the null hypothesis?

For two tailed

p - value =

=

= 0.000741

We reject null hypo if p-value <

As we can see that p-value is less than all levels of significance so

There is very strong evidence against the null hypothesis.

I am choosing very strong because at given levels, null hypothesis is rejected but if we decrease the we might not have to reject null hypothesis.