I need detailed solution about this problem. Here is problem and answer A resear

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I need detailed solution about this problem.

Here is problem and answer

A research team sought to estimate the model E (Y) = beta_0 + beta_1 x + beta_2 w. The variable Y is 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 Y was 384.2. The correlation between Y and w was -0.49; the correlation between Y and x was 0.19; and the correlation between x and w was -0.41. Compute the partial correlation coefficients r_Y w middot x and r_Y x middot w The partial correlation coefficients are r_Y x middot w = -0.0137 r_Y w.middot x = -0.4602 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 beta_1 = 0 and beta_2 = 0 at the 0.10, 0.05, and 0.01 levels of significance. Then SS (w, x) = 50673.36, MS (w, x) = 25336.68, F(w, x) = 86.48. Reject at the 0.010, 0.05 and 0.01 levels of significance.

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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