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 11.5. Assume that the population of all possible paired differences is normally distributed.

Weekly Study Time Data for Students Who Perform Well on the MidTerm
Students	1	2	3	4	5	6	7	8
Before	13	14	12	17	19	13	15	18
After	14	5	5	8	9	5	9	7

Paired T for StudyBefore - StudyAfter
	N	Mean	StDev	SE Mean
StudyBefore	8	15.1250	2.5877	.9149
StudyAfter	8	7.7500	3.0589	1.0815
Difference	8	7.37500	3.73927	1.32203

95% CI for mean difference: (4.24889, 10.50111)

T-Test of mean difference = 0 (vs not = 0): T-Value = 5.58, P-Value = .0008

(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.

(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.)

(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?

Answer 1

(a)

H0: Null Hypothesis: = 0 (There is no difference in the true mean study time before and after the mid term exam)

HA: Alternative Hypothesis: 0 (There is difference in the true mean study time before and after the mid term exam)

(b)

(i)

For = 0.10

ndf=n - 1 = 8 - 1 =7

From Table, critical values of t = 1.89

Since the calculated value of t = 5.58 is greater than critical value of t = 1.89, the difference is significant. Reject null hypothesis.

Conclusion:

The data support the claim that the true mean study time has changed.

(ii)

For = 0.05

ndf=n - 1 = 8 - 1 =7

From Table, critical values of t = 2.36

Since the calculated value of t = 5.58 is greater than critical value of t = 2.36, the difference is significant. Reject null hypothesis.

Conclusion:

The data support the claim that the true mean study time has changed.

(iii)

For = 0.01

ndf=n - 1 = 8 - 1 =7

From Table, critical values of t = 3.50

Since the calculated value of t = 5.58 is greater than critical value of t = 3.50, the difference is significant. Reject null hypothesis.

Conclusion:

The data support the claim that the true mean study time has changed.

(c)

(i)

For = 0.10

ndf=n - 1 = 8 - 1 =7

Since the p -value = 0.0008 is less than = 0.10, the difference is significant. Reject null hypothesis.

Conclusion:

The data support the claim that the true mean study time has changed.

(ii)

For = 0.05

ndf=n - 1 = 8 - 1 =7

Since the p -value = 0.0008 is less than = 0.10, the difference is significant. Reject null hypothesis.

Conclusion:

The data support the claim that the true mean study time has changed.

(iii)

For = 0.01

ndf=n - 1 = 8 - 1 =7

Since the p -value = 0.0008 is less than = 0.10, the difference is significant. Reject null hypothesis.

Conclusion:

The data support the claim that the true mean study time has changed.