The following table shows the regression output of a multiple regression model r

Question

The following table shows the regression output of a multiple regression model relating the beginninng salaraies in dollars of 30 employees is a given company to the following predictor variables:

·Gender (a dummy variable (1=male, 0=female) . Education (Years of schooling at time of the hire) ·Experience (Number of months of previous work experienc) . Months (number of months with the company) ANOVA Table Sum of Squares Mean Square Source df Regression 23665352 Error 22657938 Coefficients Table Estimate Std. Error t valus POS ) 0.000 0.000 0.000 0.034 0.000 3526.4 722.5 327.7 Intercept Gender Education 90.02 Experience 1.269 Months 24.69 0.5877 5.201 10.76 6.13 3.65 2.16 4.50 23.406 (a) Construct a 95% confidence interval for 1 and interpret your result (b) Test Ho : 1 = 2 As = 0 Use = 0.05. (c) Compute r2, r2 and s (d) Test H0 : 1-700 against Ha : 1700. Use = 0.05

Explanation / Answer

here, k= 4 (4 variables)

n= 30

df of regression = 4

df of residuals = 30-4-1 = 25

(a) t distributio has df for this question = 25

thus critical t is equal to 2.06

thus confidence interval = 722.5 + 2.06*117.8 , 722.5 + 2.06*117.8

= (965.168, 479.832)

(b) Fstatistic = (23665332/4)/(22657938/25) = 6.52

F as degrees of freedom 4, 25 and critical value is 2.76

Since our F is greater than critical value, therefore null hypothesis is rejected.

(c) r^2 = SSR/SST = SSR / (SSR+SSE) = 23665332/(23665332+22657938) = 0.51

r^a = 1- [(1-r^2)((n-1)/(n-k-1)] = 1- [(1-0.51^2)(29/25)] = 1-0.85 = 0.15

s = square root (SSE/28) = square root (22657938/28) = 899.56

(d) test statistic = (722-700)/117.8 = 0.1867 , critical t = 2.06

Thus, Null hypothesis can't be rejected.

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