A research team sought to estimate the model E(T)- Ax+ w. The variable Yis a mea

Question

A research team sought to estimate the model E(T)- Ax+ w. The variable Yis a measure of psychological adjustment of a participant observed at age 25 (where a larger number indicates better adjustment); the variable x is the measure of the of the participant's intolerance of deviance at age 15 (where a larger number indicates less tolerance of deviance); and the variable w is the measure of the participant's depression at age 20 (where a larger number indicates greater depression). They observed values of y, x, and w on 550 participants. They found that the variance of Ywas 384.2 The correlation between Y and w was -0.49; the correlation between Yand x was 0.19 and the correlation betweenx and w was -0.41. 9. Compute the partial correlation coefficients r.x and x.v (20 points)- The partial correlation coefficients are rhw-_00137 rtw,--0.4602- 10. Compute the analysis of variance table for the multiple regression analysis of Y. Include the sum of squares due to the regression on w and the sum of squares due to the regression on x after including w. Test the null hypothesis that both A 0 and = 0 at the 0.10, 0.05, and 0.01 levels of significance. +' Analysis of variance table Source- Regression on w Regression on xW Error? Total MS? DF? IH IH 50643.28v 30.08 160252.52v 210925.8+ 292.97 Then SS(w,x)=50673.36,MS(w,x)= 253336.68. F(w,x)= 86.48.+' Reject at the 0.10, 0.05, and 0.01 levels of significance. 11. Is a mediation model or an explanation model a better explanation of the observed results? You must support your choice with results from your analyses to receive credit for this question. (20 points). X(15) Y(25) Mediation Modelw

Explanation / Answer

Solution 1:

rYW.X = [rYW – (rYX*rWX)]/[(1 – rYX^2) (1 – rWX^2)]

rYW.X =[(-0.49) – (0.19)*(-0.41)]/(1 – (0.19)^2) (1 – (-0.41)^2)

rYW.X = -0.4601

rYX.W = [rYX – (rYW*rXW)]/[(1 – rYW^2)(1 – rXW^2)]

rYX.W = [(0.19) – (-0.49)*(-0.41)]/[(1 – (-0.49)^2)(1 – (-0.41)^2)]

rYX.W = -0.0137

Solution 2:

MS (Regression on w) = SS/df = 50643.28/1 = 50643.28

MS (Regression on x|w) = SS/df = 30.08/1 = 30.08

F = MS (Regression on w)/MS (error)

F = 50643.28/292.97 = 172.86

Using F-tables, the p-value is

F (172.86, 1, 547) = 0.000

Since p-value is less than 0.10, 0.05 and 0.01 level of significance, we reject Ho.

F = MS (regression on x|w)/MS (error)

F = 30.08/292.97 = 0.1103

Using F-tables, the p-value is

F (0.1103, 1, 547) = 0.7488

Since p-value is greater than 0.10, 0.05 and 0.01 level of significance, we fail to reject Ho.

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