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 15 19 12 17 16 15 11 16 After 11 18 9 10 8 9 11 10 Paired T-Test and CI: Study Before, Study After Paired T for Study Before - Study After N Mean StDev SE Mean StudyBefore 8 15.1250 2.5877 .9149 StudyAfter 8 10.7500 3.1053 1.0979 Difference 8 4.37500 2.87539 1.01660 95% CI for mean difference: (1.97112, 6.77888) T-Test of mean difference = 0 (vs not = 0): T-Value = 4.30, P-Value = .0036 (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: µd = versus Ha: µd ≠ (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 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 against the null hypothesis. 02_16_2018_QC_CS-118501

Answer 1

A) To analyze the data, we perform a paired t test to test the mean difference. Hence we set the hypothesis as :

H0 : miu(d) =0 vs H1 : not H0

B) at alpha=0.05 we see, critical value =4.30 and observed value for T(df=8-1=7) statistic is 2.365. So observed T>tabulated T hence we reject the null hypothesis at 5% level of significance.

At alpha=0.10 we see, critical value=4.30 and observed value for T(df=7) statistic is 1.895. So observed T>tabulated T, hence we reject the null hypothesis at 10% level of significance.

At alpha=0.01 we see, critical value=4.30 and observed value for T(df=7) statistic is 3.499. So observed T>tabulated T, hence we reject the null hypothesis at 1% level of significance.

C) Also in each case the p-value is 0.004 which is < alpha= 0.01, 0.05, 0.1 hence in each case we reject the null hypothesis and conclude there is significant difference in mean values i.e the students significantly reduce study time if they score good marks in midsem.