Question

3. Many studies have found a negative relationship between exercise and depression. Suppose we wish to evaluate this claim experimentally. We randomly assign a group of participants suffering from depression into two groups: those who participate in an exercise program 4 days a week for 2 months, and a control group not assigned to an exercise regimen. Depression is measured at the end of two months. The data are below: Exercise No Exercise 19 25 25 17 27 35 15 27 21 (a) Complete the hypothesis test for a non-directional with a = .05 and state your conclusion in APA style (b) Compute the effect size Hedge's g (i.e. estimated Cohen's d) 4. A psychologist wonders whether midsemester warning notices affect performance. She decides to select a sample of delinquent students and, at random times, to send such notices to half of them and no notices to the other half. Experience suggests that among such delinquent students, OGPA = .30. She decides to adopt an a= .05, and power = .95. If the difference between the two groups is equal to or greater that .075 grade points, she would want to know it. (a) State the null and alternative hypothesis for a two-tailed test. (b) Using Cohen's power curves, what should the size of the sample be to test this null hypothesis? All questions taken from King, Rosopa and Minum (2011)

Answer 1

(3)(a) Null hypothesis: H:14 = 42 There is no mean difference between exercise and no exercise. Alternative hypothesis: 1:47 42 There is a mean difference between exercise and no exercise. Rejection rule: If p-value <0.05, then reject the null hypothesis H.

EXCEL (Data Analysis) output: t-Test: Two-Sample Assuming Equal Variances Exercise No Exercise 32.5 34.5 37.42857143 18 Mean Variance Observations Pooled Variance Hypothesized Mean Difference 35.86667 t Stat P(T<=t) one-tail t Critical one-tail PIT<=t) two-tail t Critical two-tail 15 -4.98269 8.19E-05 1.75305 0.000164 2.13145 The value of the test statistic is t= -4.98 p-value = 0.0002. Conclusion:

Use a significance level, a = 0.05. Here, p-value is 0.0002, which is less than the value of a = 0.05. That is, p-value (=0.0002) sal=0.05). Therefore, by the rejection rule, reject the null hypothesis Ho- Interpretation: There is sufficient evidence at level of significance a = 0.05. There is a mean difference between exercise and no exercise. Yes, there is a negative relationship between exercise and depression. (6)

VS pocied 32.5-18 V35.8667 14.5 5.99 = 2.421 The value of the effect size Hedge's is d = 2.421.