# Applied Linear Statistical Models 5th Edition - Instructor's by Michael Kutner, Christopher Nachtsheim, John Neter, William

By Michael Kutner, Christopher Nachtsheim, John Neter, William Li

Read or Download Applied Linear Statistical Models 5th Edition - Instructor's Solutions Manual PDF

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Example text

A. 89 x2  1 −    3 x 1 Yˆ = b0 + b1 x + b11 x2 ¯ + b11 (X − X) ¯ 2 = b0 + b1 (X − X) ¯ 2 − 2b11 X X ¯ ¯ + b11 X 2 + b11 X = b0 + b1 X − b1 X ¯ + b11 X ¯ 2 ) + (b1 − 2b11 X)X ¯ = (b0 − b1 X + b11 X 2 Hence: ¯ + b11 X ¯2 b0 = b0 − b1 X ¯ b = b1 − 2b11 X 1 b11 = b11 8-4  ¯ X ¯2  1 −X ¯  1 −2X b. A =   0  0 0 1   σ02 σ01 σ02  σ 2 {b} =   σ01 σ12 σ12  σ02 σ12 σ22 where σ02 = σ 2 {b0 }, σ01 = σ{b0 , b1 }, etc. for the regression coefficients in the transformed x variables. 32. 33. a. b.

17. a. b. 18. b. c. 1524 d&e. 2 ... 5138 . . 5857 . . 0407 f. No g. 9905. 9905 conclude error variance constant, otherwise error variance not constant. Conclude error variance constant. 19. a. H0 : β1 = β2 = β3 = β4 = 0, Ha : not all βk = 0 (k = 1, 2, 3, 4). 4920. 4920 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+ b. 00000138) c. 20. 21. 22. a. Yes b. 2 No, yes, Yi = loge Yi = β0 + β1 Xi1 + β2 Xi2 + εi , where εi = loge εi c. Yes d. No, no e. 23. a. Q = (Yi − β1 Xi1 − β2 Xi2 )2 ∂Q = −2 (Yi − β1 Xi1 − β2 Xi2 )Xi1 ∂β1 ∂Q = −2 (Yi − β1 Xi1 − β2 Xi2 )Xi2 ∂β2 Setting the derivatives equal to zero, simplifying, and substituting the least squares estimators b1 and b2 yields: Yi Xi1 − b1 2 − b2 Xi1 Yi Xi2 − b1 Xi1 Xi2 − b2 Xi1 Xi2 = 0 2 Xi2 =0 and: b1 = 2 Yi Xi2 Xi1 Xi2 − Yi Xi1 Xi2 2 2 ( Xi1 Xi2 )2 − Xi1 Xi2 2 Yi Xi1 Xi1 Xi2 − Yi Xi2 Xi1 2 2 ( Xi1 Xi2 )2 − Xi1 Xi2 n 1 1 √ L= exp − 2 (Yi − β1 Xi1 − β2 Xi2 )2 2σ i=1 2πσ 2 It is more convenient to work with loge L: n 1 loge L = − loge (2πσ 2 ) − 2 (Yi − β1 Xi1 − β2 Xi2 )2 2 2σ ∂ loge L 1 = 2 (Yi − β1 Xi1 − β2 Xi2 )Xi1 ∂β1 σ ∂ loge L 1 = 2 (Yi − β1 Xi1 − β2 Xi2 )Xi2 ∂β2 σ Setting the derivatives equal to zero, simplifying, and substituting the maximum likelihood estimators b1 and b2 yields the same normal equations as in part (a), and hence the same estimators.

01669; Bonferroni c. 8. a. 181 b. 016, yes c. 9. d. 016, yes a. 168 b. 340, no c. 10. a. 881 b. 46556, no c. 843 d. 12. a. c. d. 50. 718. 718 conclude H0 , otherwise Ha . Conclude Ha . 13. a. 170 No b. 14. a. b. c. H0 : E{Y } = β1 X, Ha : E{Y } = β1 X. 5. 5 conclude H0 , otherwise Ha . Conclude H0 . 15. b. i: 1 2 ... 1820 . . 4580 No c. 16. a. b. c. H0 : E{Y } = β1 X, Ha : E{Y } = β1 X. 22939. 22939 conclude H0 , otherwise Ha . Conclude Ha . 17. b. i: 1 2 ... 21108 . . 2639 c. H0 : E{Y } = β1 X, Ha : E{Y } = β1 X.