Question

Model 1 (MI): Call: 1m(formula = Y - X1 + X2 + factor (X3) Coefficients: Estimate Std. Error t value (Intercept) 7.1745 4.8418 1.482 0.8049 0.2522 3.192 X2 0.6281 0.2460 2.553 factor (X3)B -3.3649 3.8941 -0.864 factor (X3)C -5.9566 4.0353 -1.476 factor (X3)D -6.2820 5.3471 -1.175 Residual standard error: 6.157 on 14 degrees of freedom Multiple R-squared: 0.5566, Adjusted R-squared: 0.3982 F-statistic: 3.514 on (a) and (b) DF, p-value: 0.0288 Model 2 (M2) Call: 1m (formula - Y - X1 + X2) Coefficients: Estimate Std. Error t value (Intercept) 3.9757 4.1709 0.953 0.7991 0.2483 3.218 0.5792 0.2354 2.461

(a) fill the blank (a),(b),(c),(d), and (e)

(b) which model has small test error? justify your answer

(c) compute AIC values for two models. Which model has smaller test error ?

(d) to use F-test, find a F statistic value and degree of freedom for the test

Answer 1

(a)

For Model 1,

Numerator df = Number of predictors in regression model (k) = 5

Denominator df = degree of freedom for residual eror = 14

df total = 5 + 14 = 19

For Model 2,

Numerator df = Number of predictors in regression model (k) = 2

Denominator df = df total - Numerator df = 19 - 2 = 17

Thus,

(a) = 5 , (b) = 14 , (c) = 17 , (d) = 2 (e) = 17

(b)

Standard error for model 1 = 6.157

Standard error for model 2 = 6.081

Model 2 has small test error.

(c)

n = df total + 1 = 19 + 1 = 20

AIC = n log($s^2$) + 2(k + 1)

where s is the standard error.

For Model 1.

AIC = 20 log($6.157^2$) + 2(5 + 1) = 84.70359

For Model 2.

AIC = 20 log($6.081^2$) + 2(2 + 1) = 78.20677

Since AIC for model 2 is less than that of model 1, model 2 has small test error.

(d)

For Model 1,

F statistic value = 3.154 on 5 and 14 degree of freedom

For Model 2,

F statistic value = 7.685 on 2 and 17 degree of freedom