HW Solutions Stat 86

HW Solutions Stat 86 - S SE (F ) = 49.4933 Reduced model:...

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Unformatted text preview: S SE (F ) = 49.4933 Reduced model: ˆ Y = 61.15167 − 9.20167X2 − 8.89000X3 − 1.09333X1 X2 + 1.28500X1 X3 −1.54333X2 X3 − 1.03500X1 X2 X3 , SSE (R) = 667.8413 H0 : α1 = 0, Ha : α1 = 0. F ∗ = (618.348/1) ÷ (49.4933/14) = 174.91, F (.975; 1, 14) = 6.298. d. 24.17. If F ∗ ≤ 6.298 conclude H0 , otherwise Ha . Conclude Ha . P -value = 0+ ˆ ˆ D = µ2.. − µ1.. = α2 − α1 = −2ˆ 1 = 11.35000, s2 {α1 } = .18413, s{D} = .8582, ˆ ˆ ˆ ˆ α ˆ t(.975; 14) = 2.145, 11.35000 ± 2.145(.8582), 9.509 ≤ D ≤ 13.191 √ 2n = 4.1475, n = 14 1.8 24-3 ...
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This note was uploaded on 12/21/2011 for the course STA 2014 taught by Professor Davehatley during the Fall '11 term at UNF.

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