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

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

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Conclude H0 . 121. 40. a. b. c. 02406X3 X4 H0 : β5 = β6 = 0, Ha : not both β5 = 0 and β6 = 0. 353. 353 conclude H0 , otherwise Ha . Conclude H0 . 41. a. 63348X7 b. H0 : β2 = 0, Ha : β2 = 0. 983. 983 conclude H0 , otherwise Ha . Conclude Ha . 932. 932 conclude H0 , otherwise Ha . Conclude Ha . c. 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).

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 . 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.

Conclude H0 . 5713. c. 1 = 142, 092. 4. Yes. 5. a. 84, df : 1, 1, 1, 42 b. H0 : β3 = 0, Ha : β3 = 0. 4039. 4039 conclude H0 , otherwise Ha . Conclude H0 . 065. 6. H0 : β2 = β3 = 0, Ha : not both β2 and β3 = 0. 0327. 0327 conclude H0 , otherwise Ha . Conclude Ha . 022. 7. a. 2306, df : 1, 1, 1, 1, 76. b. H0 : β3 = 0, Ha : β3 = 0. 9806. 9806 conclude H0 , otherwise Ha . Conclude H0 . 5704. 8. H0 : β2 = β3 = 0, Ha : not both β2 and β3 = 0. 8958. 8958 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+.