12_Equivalence

# 12_Equivalence - Equivalence of model comparison and Cb F...

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Unformatted text preview: Equivalence of model comparison and Cb F tests Continuation of the Storage time example from Part 9. Data: Storage Temperature Storage Time 20 ◦ C 30 ◦ C 3 months 2 5 9 12 15 6 months 6 6 7 7 16 Copyright c 2011 Dept. of Statistics (Iowa State University) Statistics 511 1 / 1 Summary of results from Cb estimates and tests Quantity Estimate F stat p value Time main effect 3.5 6.00 0.049 Temp main effect 9.0 39.71 0.00074 Interaction 1.0 0.12 0.74 These correspond to tests of: Time Ho: μ 3 , 20 + μ 3 , 30 = μ 6 , 20 + μ 6 , 30 Temp Ho: μ 3 , 20 + μ 6 , 20 = μ 3 , 30 + μ 6 , 30 Interaction Ho: μ 3 , 20- μ 3 , 30 = μ 6 , 20- μ 6 , 30 No question about these Ho. Copyright c 2011 Dept. of Statistics (Iowa State University) Statistics 511 2 / 1 Model comparisons If you fit two specific models and compare residual SS, no question what models being compared. Interaction: In R: lm(y ∼ time+temp+time:temp) SSE = 23.50 with 6 d.f. lm(y ∼ time+temp) SSE = 23.98 with 7 d.f. In SAS: model y=time temp time*temp; SSE = 23.50 with 6 d.f. model y=time temp; SSE = 23.98 with 7 d.f. F = ( 23 . 98- 23 . 50 ) / ( 7- 6 ) 23 . 50 / 6 = . 48 / 3 . 917 = . 122 Same as Cb test Copyright c 2011 Dept. of Statistics (Iowa State University) Statistics 511 3 / 1 Model comparisons I Time main effect: In R: lm(y ∼ time+temp+time:temp) SSE = 23.50 with 6 d.f. lm(y ∼ temp+time:temp) SSE = 23.50 with 6 d.f. In SAS: model y=time temp time*temp; SSE = 23.50 with 6 d.f....
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12_Equivalence - Equivalence of model comparison and Cb F...

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