Question

# Do students reduce study time in classes where they achieve a higher midterm score? In a...

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.

Table 11.5

 Weekly Study Time Data for Students Who Perform Well on the MidTerm Students 1 2 3 4 5 6 7 8 Before 12 16 16 18 18 12 11 16 After 10 10 6 11 8 6 9 9

Paired T-Test and CI: StudyBefore, StudyAfter

 Paired T for StudyBefore - StudyAfter N Mean StDev SE Mean StudyBefore 8 14.8750 2.7999 .9899 StudyAfter 8 8.6250 1.8468 .6529 Difference 8 6.25000 3.05894 1.08150

95% CI for mean difference: (3.69266, 8.80734)

T-Test of mean difference = 0 (vs not = 0): T-Value = 5.78, 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: 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=

(A) Hypotheses are already completed in part A

(B) Usig the given data

we get

t statistic = 5.78

t critical = T.INV.2T(alpha,df)

where df = n-1 = 8-1 = 7

and alpha= 0.10,0.05 and 0.01

t critical(0.10,7) = T.INV.2T(0.10,7) = 1.89

t critical(0.05,7) = T.INV.2T(0.05,7) = 2.36

t critical(0.01,7) = T.INV.2T(0.01,7) = 3.50

We can see that the t statistic is greater than all the t critical values

So, we can reject the null hypothesis at 0.10,0.05 and 0.01 significance levels

Therefore, we can conclude that true mean study time changed

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