# Get Applied Linear Statistical Models 5th Edition - Instructor's PDF

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

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Extra resources for Applied Linear Statistical Models 5th Edition - Instructor's Solutions Manual

Example text

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.

08306X5 {2002} b. 42. a. H0 : β3 = β4 = β5 = 0, Ha : not all βk = 0 (k = 3, 4, 5). 9223. 9223 conclude H0 , otherwise Ha . Conclude H0 . 8-7 c. H0 : β2 = β5 = β6 = β7 = 0, Ha : not all βk = 0 (k = 2, 5, 6, 7). 71408. 71408 conclude H0 , otherwise Ha . Conclude H0 . 9. 10. b. X1 X2 X3 X4 c. 11. a. 12. 969 Note: Variable numbers for predictors are those in the appendix. 13. b. 423   X3 1 c. 14. a. 15. b. 068   X3 1 c. 16. a. 17. a. 8638 X1 , X3 b. 10 c. X1 , X3 d. 18. a. 19 a. x1 , x2 , x3 , x1 x2 b.

H0 : β11 = β33 = β13 = 0, Ha : not all βk = 0 (k = 11, 33, 13). 8267. 8267 conclude H0 , otherwise Ha . Conclude H0 . 1444 c. 38. a. b. 6139 for first-order model. 8-6 c. 39. a. H0 : β11 = 0, Ha : β11 = 0. 621. 621 conclude H0 , otherwise Ha . Conclude Ha . 871. 871 conclude H0 , otherwise Ha . Conclude Ha . 2X5 b. 2693 c. H0 : β3 = β4 = β5 = 0, Ha : not all βk = 0 (k = 3, 4, 5). 09645. 09645 conclude H0 , otherwise Ha . Conclude H0 . 121. 40. a. b. c. 02406X3 X4 H0 : β5 = β6 = 0, Ha : not both β5 = 0 and β6 = 0.