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HW Solutions Stat 72

# HW Solutions Stat 72 - 42 If F ∗ ≤ 3 42 conclude H...

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Chapter 22 ANALYSIS OF COVARIANCE 22.7. a. e ij : i j = 1 j = 2 j = 3 j = 4 j = 5 j = 6 1 . 5281 . 4061 . 0089 . 4573 . 1140 . 1911 2 . 2635 . 2005 . 3196 . 2995 . 1662 . 0680 3 . 1615 . 2586 . 0099 . 3044 . 0472 . 1700 i j = 7 j = 8 j = 9 j = 10 j = 11 j = 12 1 . 0660 . 0939 . 0112 2 . 0690 . 1776 . 0005 . 0653 . 0251 . 0995 b. r = . 988 c. Y ij = µ . + τ 1 I ij 1 + τ 2 I ij 2 + γx ij + β 1 I ij 1 x ij + β 2 I ij 2 x ij + ε ij H 0 : β 1 = β 2 = 0, H a : not both β 1 and β 2 equal zero. SSE ( F ) = . 9572, SSE ( R ) = 1 . 3175, F = ( . 3603 / 2) ÷ ( . 9572 / 21) = 3 . 95, F ( . 99; 2 , 21) = 5 . 78. If F 5 . 78 conclude H 0 , otherwise H a . Conclude H 0 . P -value = . 035 d. Yes, 5 22.8. b. Full model: Y ij = µ . + τ 1 I ij 1 + τ 2 I ij 2 + γx ij + ε ij , ( ¯ X .. = 9 . 4) . Reduced model: Y ij = µ . + γx ij + ε ij . c. Full model: ˆ Y = 7 . 80627 + 1 . 65885 I 1 . 17431 I 2 + 1 . 11417 x , SSE ( F ) = 1 . 3175 Reduced model: ˆ Y = 7 . 95185 + . 54124 x , SSE ( R ) = 5 . 5134 H 0 : τ 1 = τ 2 = 0, H a : not both τ 1 and τ 2 equal zero. F = (4 . 1959 / 2) ÷ (1 . 3175 / 23) = 36 . 625, F ( . 95; 2 , 23) = 3
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Unformatted text preview: . 42. If F ∗ ≤ 3 . 42 conclude H , otherwise H a . Conclude H a . P-value = 0+ d. MSE ( F ) = . 0573, MSE = . 6401 e. ˆ Y = ˆ µ . + ˆ τ 2 − . 4ˆ γ = 7 . 18629, s 2 { ˆ µ . } = . 00258, s 2 { ˆ τ 2 } = . 00412 , s 2 { ˆ γ } = . 00506, s { ˆ µ . , ˆ τ 2 } = − . 00045, s { ˆ τ 2 , ˆ γ } = − . 00108 , s { ˆ µ . , ˆ γ } = − . 00120, s { ˆ Y } = . 09183, t ( . 975; 23) = 2 . 069 , 7 . 18629 + 2 . 069( . 09183), 6 . 996 ≤ µ . + τ 2 − . 4 γ ≤ 7 . 376 22-1...
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