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	18	15	11	17	16	15	12	19
After	5	8	6	6	5	9	14	5

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

   

Paired T for Study Before - Study After
	N	Mean	StDev	SE Mean
StudyBefore	8	15.3750	2.7742	.9808
StudyAfter	8	7.2500	3.1053	1.0979
Difference	8	8.12500	5.24915	1.85585

95% CI for mean difference: (3.73660, 12.51340)

T-Test of mean difference = 0 (vs not = 0): T-Value = 4.38, P-Value = .0032

(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)strongvery strongextremely strongno 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 evidencestrong evidenceextermly strong evidence against the null hypothesis.

Answer 1

A) paired t test is to be performed in minitab

H0=null hypothesis :there is no significant difference in mean study time before and after midterm i.e miu(d) =0

H1=alternative hypothesis : there is significant difference in the meam study time before and after midterm. i.e miu(d) not equal to 0

B) the given output corresponds to 95% confidence interval. i.e alpha=0.05. The p-value in this case is 0.0032 <0.05 so we reject the null hypothesis and conclude there is significant difference among mean study times before and after midterm.

At 0.1 and 0.01 level of significance, we do a paired t test in minitab using following steps : stat, basic statistics, paired t test.

Output :

conclusion : at alpha=0.01, t value(obs) =4.38, critical value of t for df =8-1=7 is 0.399 as tobs>tcri and it is very large so we reject the null hypothesis. Rejection is extremely strong as the difference of tobs and tcrit os very high.

At alpha=0.10, t value (obs) =4.38 and tcrit =1.895.

As tobs >tcri we reject the null hypothesis. The difference between tobs and tcrit is less strong than the previous value so the rejection is not extremely strong but very strong.

C) at every alpha we reject the null hypothesis and conclude there is significant difference among the mean study times. And as the difference between tobs and tcrit is very high, there is extremely strong evidence against the null hypothesis.