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 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 excel workbook Homework 4 page 1. Assume that the population of all possible paired differences is normally distributed. Can it be confirmed that students reduce their study time after earning a high midterm grade? Use α =0.05. Complete the hypothesis test in excel and write out the hypotheses being tested, the value of the test statistic, critical value, p-value and the complete conclusion including your alpha level, test statistic value and p-value.
Students | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
Before | 15 | 14 | 17 | 17 | 19 | 14 | 13 | 16 |
After | 9 | 9 | 11 | 10 | 19 | 10 | 14 | 10 |
Before | After | Difference |
15 | 9 | 6 |
14 | 9 | 5 |
17 | 11 | 6 |
17 | 10 | 7 |
19 | 19 | 0 |
14 | 10 | 4 |
13 | 14 | -1 |
16 | 10 | 6 |
Sample mean of the difference using excel function AVERAGE(), x̅d = 4.1250
Sample standard deviation of the difference using excel function STDEV.S(), sd = 2.9970
Sample size, n = 8
Null and Alternative hypothesis:
Ho : µd ≤ 0 ; H1 : µd > 0
Test statistic:
t = (x̅d)/(sd/√n) = (4.125)/(2.997/√8) = 3.8930
df = n-1 = 7
Critical value :
Right tailed critical value, t-crit = ABS(T.INV(0.05, 7)) = 1.895
Reject Ho if t > 1.895
p-value :
Right tailed p-value = T.DIST.RT(3.893, 7) = 0.0030
Decision:
p-value < α, Reject the null hypothesis
Conclusion:
There is enough evidence to conclude that the students reduce their study time after earning a high midterm grade at α = 0.05.
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