HW7Sol - Homework 8 solution 1 a X = 1 1 1 1 b X X = c We...

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Homework 8 – solution 1. a) X = 1 1 1 2 1 3 1 4 , y = 2 . 0 4 . 2 5 . 9 8 . 3 b) X 0 X = ˆ 4 10 10 30 ! . Therefore C = 1 20 ˆ 30 - 10 - 10 4 ! c) We have ˆ β = CX 0 y = 1 20 ˆ 30 - 10 - 10 4 ! ˆ 1 1 1 1 1 2 3 4 ! 2 . 0 4 . 2 5 . 9 8 . 3 = ˆ - 0 . 5 2 . 06 ! , i.e. ˆ β 0 = - 0 . 05, ˆ β 1 = 2 . 06. d) For the residuals, we obtain e 1 = ˆ y 1 - y 1 = - 0 . 05+2 . 06 · 1 - 2 . 00 = 0 . 01, e 2 = ˆ y 2 - y 2 = - 0 . 05 + 2 . 06 · 2 - 4 . 20 = - 0 . 13, e 3 = ˆ y 3 - y 3 = - 0 . 05 + 2 . 06 · 3 - 5 . 9 = 0 . 23, e 4 = ˆ y 4 - y 4 = - 0 . 05 + 2 . 06 · 4 - 8 . 3 = - 0 . 11. Therefore SS E = 0 . 0868, and ˆ σ 2 = SS E 4 - 2 = 0 . 0434. e) For the confidence interval, we need the quantity x 0 0 Cx 0 = 1 2 . 5 · 1 20 ˆ 30 - 10 - 10 4 ! ˆ 1 2 . 5 ! = 0 . 25 Then the confidence interval can be obtained as x 0 ˆ β ± t 0 . 025 , 2 q ˆ σ 2 x 0 0 Cx 0 = - 0 . 05 + 2 . 06 · 2 . 5 ± 4 . 30 0 . 0434 · 0 . 25 = [4 . 65 , 5 . 55]. 2. a) The fitted line is y = - 105 + 0 . 605 x 1 + 8 . 92 x 2 + 1 . 44 x 3 + 0 . 014 x 4 . b) Substituting the values in the above equation we obtain y = 288. c) We have f 0 = 5 . 11 and the P-value is 0.03. So, the regression is not significant at α = 0 . 01. d) ˆ σ 2 = SS E 7 = 243. e) For x 1 , we have t 0 = 1 . 64 < t 0 . 005 , 7 , so we can’t reject the hypothesis β 1 = 0. Same conclusion is reached for all other regressors. f) The 95% CI on β j has the form CI j = ˆ β j ± t 0 . 025 , 7 se ( ˆ β j ). Substituting the numbers from the Minitab output, we obtain
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