HW Solutions Stat 75

# HW Solutions Stat 75 - 777 F 95 2 17 = 3 59 If F ∗ ≤ 3...

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1 if experimental unit from block 1 I ij 1 = 1 if experimental unit from block 10 0 otherwise I ij 2 ,...,I ij 9 are deFned similarly 1 if experimental unit received treatment 1 I ij 10 = 1 if experimental unit received treatment 3 0 otherwise 1 if experimental unit received treatment 2 I ij 11 = 1 if experimental unit received treatment 3 0 otherwise x ij = X ij ¯ X . . ( ¯ X .. =80 . 033333) c. ˆ Y =77 . 10000 + 4 . 87199 I 1 +3 . 87266 I 2 +2 . 21201 I 3 +3 . 22003 I 4 +1 . 23474 I 5 + . 90876 I 6 1 . 09124 I 7 3 . 74253 I 8 4 . 08322 I 9 6 . 50033 I 10 2 . 49993 I 11 + . 00201 x SSE ( F ) = 112 . 3327 d. Y ij = µ .. + ρ 1 I ij 1 + ρ 2 I ij 2 + ρ 3 I ij 3 + ρ 4 I ij 4 + ρ 5 I ij 5 + ρ 6 I ij 6 + ρ 7 I ij 7 + ρ 8 I ij 8 + ρ 9 I ij 9 + γx ij + ± ij ˆ Y =77 . 10000 + 6 . 71567 I 1 +5 . 67233 I 2 +3 . 61567 I 3 +4 . 09567 I 4 +1 . 14233 I 5 + . 33233 I 6 1 . 66767 I 7 5 . 33100 I 8 5 . 18767 I 9 . 13000 x SSE ( R )=1 , 404 . 5167 e. H 0 : τ 1 = τ 2 =0 , H a : not both τ 1 and τ 2 equal zero. F =(1 , 292 . 18 / 2) ÷ (112 . 3327 /
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Unformatted text preview: . 777, F ( . 95; 2 , 17) = 3 . 59. If F ∗ ≤ 3 . 59 conclude H , otherwise H a . Conclude H a . P-value = 0+ f. ˆ τ 1 = − 6 . 50033, ˆ τ 2 = − 2 . 49993, ˆ L = − 4 . 0004, L 2 { ˆ τ 1 } = . 44162, s 2 { ˆ τ 2 } = . 44056, s { ˆ τ 1 , ˆ τ 2 } = − . 22048, s { ˆ L } = 1 . 1503, t ( . 975; 17) = 2 . 11, − 4 . 0004 ± 2 . 11(1 . 1503), − 6 . 43 ≤ L ≤ − 1 . 57 22.21. a. Source SS df MS Between treatments 25 . 5824 2 12 . 7912 Error 1 . 4650 24 . 0610 Total 27 . 0474 26 b. Covariance: MSE = . 0573, ˆ γ = 1 . 11417 22-4...
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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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