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Homework10solution[1]

# Homework10solution[1] - Homework 10 solution 1 a H0 1 = 2 =...

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Homework 10 solution 1. a) H 0 : μ 1 = μ 2 = μ 3 = μ 4 H 1 : Not all means are equal. α = 0.05 Control T15 T40 T50 Overall X .j 16.2 20.4 17.2 12.8 X ..=16.7 SS b = Σ n j ( X .j - X ..) 2 = 5{(16.2-16.7) 2 + (20.4 - 16.7) 2 + (17.2 - 16.7) 2 + (12.8 - 16.7) 2 } = 5(0.25 + 13.69 + 0.25 + 15.21) = 147.0 SS w = Σ Σ (X ij - X .j ) 2 = 36.8 + 5.2 + 12.8 + 6.8 = 61.6 Source SS df MS F Between 147.0 3 49.0 12.72 Within 61.6 16 3.85 Total 208.6 19 Reject H 0 if F > F 0.05 (3,16) = 3.24 Reject H 0 since 12.72 > 3.24. We have significant evidence, α = 0.05, to show that the means are not all equal. b) H 0 : μ C = μ T15 H 1 : μ C μ T15 α = 0.05 Reject H 0 if F > (k-1) F 0.05 (3,16) = 3(3.24) = 9.72 ) 5 1 + 5 1 3.85( ) 20.4 - (16.2 = ) n 1 + n 1 MSE( ) X - X ( = F 2 T15 C 2 T15 C = 17.64/1.54 = 11.5 Reject H 0 since 11.5 > 9.72. We have significant evidence, α = 0.05, to show that μ C μ T15 . H 0 : μ C = μ T40 H 1 : μ C μ T40 α = 0.05 Reject H 0 if F > (k-1) F 0.05 (3,16) = 3(3.24) = 9.72 ) 5 1 + 5 1 3.85( ) 17.2 - (16.2 = ) n 1 + n 1 MSE( ) X - X ( = F 2 T40 C 2 T40 C = 0.65 Do not Reject H 0 since 0.65 < 9.72. We do not have significant evidence, α = 0.05, to show that μ C μ T40 .

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H 0 : μ C = μ T50 H 1 : μ C μ T50 α = 0.05 Reject H 0 if F > (k-1) F 0.05 (3,16) = 3(3.24) = 9.72 ) 5 1 + 5 1 3.85( ) 12.8 - (16.2 = ) n 1 + n 1 MSE( ) X - X ( = F 2 T50 C 2 T50 C = 7.51 Do not Reject H 0 since 7.51 < 9.72. We do not have significant evidence, α = 0.05, to show that μ C μ T50 .
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Homework10solution[1] - Homework 10 solution 1 a H0 1 = 2 =...

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