12.68 An operations manager uses five independent variables to predict the produ

Question

12.68 An operations manager uses five independent variables to predict the productivity of a plant. Suppose that for 26 days the manager collects data and uses these 26 observa- tions to fit a regression equation. Assume that the regression sum of squares is 95.6 and that the error sum of squares is 159.0. a. Construct an ANOVA table. Does the overall model contribute to predicting the productivity of the plant? Use a 5% significance level. b. Compute the R2 and interpret its value. What is the value of the adjusted R?

Explanation / Answer

a) here from above information below is ANOVA table:

for 1,24 degree of freedom and 0.05 level ; critical value of F =4.2597

as test statistic is greater then critical value we reject null hypothesis.

we have sufficient evidence to conclude that overall model contribute predicts the productivity of the plant.

b)R2 =SSR/SST   =95.6/254.6 =0.3755 ~ 37.55%

it represent variability explained in dependent variable due to given overall model.

R2adj =1-(SSE/(n-2))/(SST/(n-1)) =1-(159/(26-2))/(254.6/(26-1)) =0.3495

Source DF SS MS F regression 1 95.6000 95.60 14.43 Residual error 24 159.0000 6.63 Total 25 254.6000

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