We have that μ e x t 2 1 6 from equation 1 6 b 2 1

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We have that μ = E [ X t ] = 2 / (1 - 0 . 6). From equation (1 - 0 . 6 B 2 )(1 + ψ 1 B + ψ 2 B 2 + ψ 3 B 3 + . . . ) = 1 + 0 . 4 B, we obtain ψ 1 = 0 . 4, ψ 2 - 0 . 6 = 0 and hence ψ 2 = 0 . 6, ψ 3 - 0 . 6 ψ 1 = 0 and hence ψ 3 = 0 . 24. (iii) [15] Find Var( X t ). We have Var( X t ) = Var(0 . 6 X t - 2 + e t + 0 . 4 e t - 1 ) , and hence γ (0) = 0 . 36 γ (0) + 1 + 0 . 4 2 , and hence γ (0) = 1 . 81.
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(iv) [15] Find ρ X ( k ) for k = 1 , 2 , 3. Now from the equation (after removing the mean) E [ X t X t - h ] = 0 . 6 E [ X t - 2 X t - h ] + E [ e t X t - h ] + 0 . 4 E [ e t - 1 X t - h ] , we obtain for h = 1 that γ (1) = 0 . 6 γ (1) + 0 . 4 , and hence γ (1) = 1 and ρ (1) = γ (1) (0) = 0 . 55. For h = 2 we have γ (2) = 0 . 6 γ (0) , and hence ρ (2) = 0 . 6. For h = 3 we have that γ (3) = 0 . 6 γ (1) and hence ρ (3) = 0 . 6 ρ (1) = 0 . 33. 3. Below is a (partial) outprint of a regression analysis (MINITAB). (a) [10] Complete the ANOVA table. Degrees of freedom are 1,6 and 7. SS E = 0 . 221 - 0 . 146 = 0 . 075, MS R = SS R = 0 . 146, MS E = SS E / 6 = 0 . 012, F = MS R /MS E = 11 . 7 (b) [10] Construct the t -test for testing significance of the predictor x . What is the p -value of that test? For simple linear regression F = T 2 . Therefore T = - 11 . 7 = - 3 . 4 with df=6. The p -value is the same as in the F -test, i.e., p -value is 0.014. (c) [10] Construct 95% prediction interval for y at x = 0 . 65. The fitted value is ˆ y = 1 . 04 - 0 . 59 × 0 . 65 = 0 . 657, and 95% prediction interval for y is (note that 0 . 65 = ¯ x ): ˆ y ± t 0 . 25 , 6 p MS E (1 + 1 / 8) = 0 . 657 ± 2 . 45 0 . 012 × 1 . 125 = (0 . 37 , 0 . 94) .
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