# 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

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

**Sample text**

67779. If conclude error variance constant, otherwise error variance not constant. Conclude error variance not constant. 11. a. b. H0 : β1 = β2 = β3 = 0, Ha : not all βk = 0 (k = 1, 2,3). 79806. 79806 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+. 4988 6-2 c. 12. a. 13. 14. a. b. 15. b. c. 4702X3 d&e. i: 1 2 ... 0797 . . 1772 . . 1772 f. No g. 3449. 3449 conclude error variance constant, otherwise error variance not constant. Conclude error variance constant. 16. a. H0 : β1 = β2 = β3 = 0, Ha : not all βk = 0 (k = 1, 2, 3).

B= c. 36. a. Y¯ = XY = 1 n1 − n11 Y11 . .. 29518 × 10−11 x2 b. 6711 for first-order model. c. H0 : β11 = 0, Ha : β11 = 0. 9654. 9654 conclude H0 , otherwise Ha . Conclude Ha . 8628. 8628 conclude H0 , otherwise Ha . Conclude Ha . 37. a. 2485 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 .

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.