MS Module 21 Multiple regression s2 adjusted R2 practice exam questions

(The attached PDF file has better formatting.)

[The practice problems in the 24 modules explain the statistical procedures; the practice exam questions in this thread shows what you will be asked on the final exam.]

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    0.8    6.1    10.9    9.8
Fitted Value    2    4.0    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) ) ] )