Question

6. A researcher wishes to determine whether exercising at least three times a week is more effective than pharmaceuticals (antidepressants) or psychotherapy in addressing mental health issues. He randomly assigns 20 psychiatric patients to one of four treatment conditions and asks their self-reported depression at the mo- ment on a scale of 1 (very de- pressed) to 15 (very happy). Test the null hypothesis that there is no differ- ence between exercise, antidepres- sants, therapy, and the control group using the following data Exer- Antide- cise pressants ap Ther- Con 9 10 8 6 13 10 6 8 6 7 8

Answer 1

The following table is obtained Group 1 Group 2 Group 3 Group 4 10 10 10 14 12 13 46 9.2 450 2.588 26.8 53 10.6 579 2.074 17.2 48 9.6 478 2.074 17.2 38 7.6 306 2.074 17.2 Sum Average- St. Dev. The total sample size is N 20. Therefore, the total degrees of freedom are

d ftotal 20-119 Also, the between-groups degrees of freedom are dfbetwen degrees of freedom are 4-1 3, and the within-groups dfvithin - dftotal - dfbetween -19-3-16 First, we need to compute the total sum of values and the grand mean. The following is obtained ΣΧ,,-46 + 53 + 48 + 38 = 185 i.j Also, the sum of squared values is ΣΧ-450 + 579 + 478 + 306 = 1813 i.j Based on the above calculations, the total sum of squares is computed as follows HEX).-1813 1852 Xij = i.j 101 i.j The within sum of squares is computed as shown in the calculation below: S Swithin withingroups 26.8 +17.217.2+17.2 78.4

The between sum of squares is computed directly as shown in the calculation below Now that sum of squares are computed, we can proceed with computing the mean sum of squares MS -Sbetueen _ 23.35 between between MnSthin78.4 4.9 within 16 Now that sum of squares are computed, we can proceed with computing the mean sum of squares SSbetween23.35 M Sbetween SSwithin78.4 MSuihin dfwithin 4.9 16 Finally, with having already calculated the mean sum of squares, the F-statistic is computed as follows MSbetueen 1,588 4.9 within

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ha: Not all means are equal The above hypotheses will be tested using an F-ratio for a One-Way ANOVA (2) Rejection Region Based on the information provided, the significance level is a 0.05, and the degrees of freedom are dfi 3 and dfe 3, therefore, the rejection region for this F-test is R F:F> Fe 3.239 M Sietween7.783 MSuithin tA.V

(4) Decision about the null hypothesis Since it is observed that F-1.588くF, 3.239, it is then concluded that the null hypothesis is not rejected Using the P-value approach: The p-value is p 0.2313, and since p 0.231320.05, it is concluded that the null hypothesis is not rejected (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that not all 4 population means are equal, at the α-0.05 significance level.