Question

An educational psychologist wants to check the claim that regular physical exercise improves academic achievement. To control for academic aptitude, pairs of college students with similar GPAs are randomly assigned to either a treatment group that attends daily exercise classes or a control group. At the end of the experiment, the following GPAs are reported for the seven pairs of participants:

Pair Number	Physical Exercise	No Physical Exercise
1	4.00	3.75
2	2.67	2.74
3	3.65	3.42
4	2.11	1.67
5	3.21	3.00
6	3.60	3.25
7	2.80	2.65

1) Using t , test the null hypothesis at the .01 level of significance.

a) What is the critical t?

b) What is the value of t?

2) specify the p - value for this test test

3) If appropriate (because the test result is statistically significant), use Cohen’s d to estimate the effect size

4) How might this test result be reported in the literature?

Answer 1

Sr	Sample 2	Sample 2	Difference
1	4	3.75	0.25
-0.07
2	2.67	2.74
3	3.65	3.42	0.23
4	2.11	1.67	0.44
5	3.21	3	0.21
6	3.60	3.25	0.35
7	2.80	2.65	0.15
Average	3.149	2.926	0.223
St. Dev.	0.66	0.674	0.161
n	7	7	7

From the sample data, it is found that

the corresponding sample means are:

Xˉ1=3.149 & Xˉ2=2.926

Also, the provided sample standard deviations are:

s1=0.66 s2 =0.674

and the sample size is n = 7. For the score differences we have

Mean of Dˉ=0.223 standard deviation( D )=0.161

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μD = 0

Ha: Dμ > 0

This corresponds to a right-tailed test, for which a t-test for two paired samples be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.01, and the degrees of freedom are df=6.

Hence, it is found that the critical value for this right-tailed test is tc=3.143, for α=0.01 and df=6.

The rejection region for this right-tailed test is R=t:t>3.143.

(3) Test Statistics

The t-statistic is computed as shown in the following formula:

(4) Decision about the null hypothesis

Since it is observed that t=3.664>tc​=3.143, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0053, and since p=0.0053<0.01, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean μ1 is greater than μ2, at the 0.01 significance level.

Cohen's d is determined by calculating the mean difference between your two groups, and then dividing the result by the pooled standard deviation.

Cohen's d = (M₂ - M₁) ⁄ SD_pooled

where:

SD_pooled = √((SD₁² + SD₂²) ⁄ 2)= 0.6670

Cohen's d = (2.926 - 3.149) ⁄ 0.667037 = 0.334314.