hw33soluf08

# hw33soluf08 - Assignment 3 Answer Keys 1 In the experiment...

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Unformatted text preview: Assignment 3 Answer Keys 1. In the experiment, a = 5 and n 1 = n 2 = . . . = n a = 13. Only the summary statistics, sample means (¯ y i. ) and sample standard deviatiuons ( s i ), are given. Although the original observations are not reported, the summary statistics are enough for calculting various SS’s of ANOVA. Recall that SS treatment = X i n i (¯y i .- ¯y .. ) 2 and SS E = X i X j (y ij- ¯y i . ) 2 . The grand mean ¯ y .. is in fact the average of the sample means, i.e., ¯ y .. = ¯ y 1 . + ¯ y 2 . + ¯ y 3 . + ¯ y 4 . + ¯ y 5 . 5 = 11 . 336 So SS treatment = 2 . 28176. To calculate SS E , note that, for any fixed i , ∑ j ( y ij- ¯ y i. ) 2 = ( n i- 1) s 2 i , hence SS E = ( n 1- 1) s 2 1 + ( n 2- 1) s 2 2 + . . . + ( n 5- 1) s 2 5 = 7 . 9488 Based on SS treatment and SS E , the ANOVA table can be constructed, and F = 4 . 036, and P-value=.004. The conclusion is that there is sufficient evidence to suggest that the treatments are not all equal. 2. a) The ANOVA table from SAS is Sum of Source DF Squares Mean Square FValue Pr > F Model 2 3.28516000 1.64258000 74.11 <.0001 1 Error 57 1.26340000 0.02216491 CorTotal 59...
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hw33soluf08 - Assignment 3 Answer Keys 1 In the experiment...

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