MS Module 21 Multiple regression s-sqrd adjusted R-sqrd practice exam questions
(The attached PDF file has better formatting.)
A multiple regression analysis Y = β0 + β1 X1 + β2 X2 + ε, with 5 data points and independent variables X1 and X2 has the following actual values (yi) and fitted values (ŷi):
Actual Value 2.4 1 6.1 10.9 9.8
Fitted Value 2 4 6 8 10
● The null hypothesis is H0: β1 = β2 = 0
● The alternative hypothesis is Ha: β1 ≠ 0 or β2 ≠ 0
Question 21.1: Residuals
What are the residuals for the five data points?
Answer 21.1: residual = actual value – fitted value:
obs fitted actual residual SST SSE
#1 2 2.4 0.4 12.96 0.16
#2 4 0.8 -3.2 27.04 10.24
#3 6 6.1 0.1 0.01 0.01
#4 8 10.9 2.9 24.01 8.41
#5 10 9.8 -0.2 14.44 0.04
avg 6 6 0
4.44e-16 78.46 18.86
Question 21.2: Total sum of squares
What is the total sum of squares (SST)?
Answer 21.2: average y-value = (2.4 + 0.8 + 6.1 + 10.9 + 9.8) / 5 = 6
SST = (2.4 – 6)2 + (0.8 – 6)2 + (6.1 – 6)2 + (10.9 – 6)2 + (9.8 – 6)2 = 78.46
Question 21.3: Error sum of squares
What is the error sum of squares (SSE)?
Answer 21.3: SSE = (2.4 – 2)2 + (0.8 – 4)2 + (6.1 – 6)2 + (10.9 – 8)2 + (9.8 – 10)2 = 18.86
Question 21.4: Least squares estimate for σ2
What is s2, the least squares estimate for σ2?
Answer 21.4: 18.86 / (5 – 2 – 1) = 9.43
(least squares estimate for σ2 = error sum of squares / degrees of freedom, which are N-k-1)
Question 21.5: R2
What is R2?
Answer 21.5: 1 – 18.86 / 78.46 = 75.96%
(R2 = 1 – error sum of squares / total sum of squares)
Question 21.6: Adjusted R2
What is the adjusted R2?
Answer 21.6: 1 – 18.86 / (5 – 2 – 1) / (78.46 / (5 – 1) ) = 51.92%
(adjust SSE and SST by their degrees of freedom: adjusted R2 = 1 – MSE / MST
= 1 – [ SSE / (n - (k + 1) ] / [ SST / (n - 1) ] )
Question 21.7: F value
What is the test statistic value f to test the null hypothesis?
Answer 21.7: ( (78.46 – 18.86) / 2 ) / (18.86 / (5 – 2 – 1) ) = 3.160
(test statistic f = [ R2 / k ] / [ (1 – R2) / (n - (k + 1) ) ] )
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