A study of 8 different weight loss programs involved 200 subjects. Each of the 8
ID: 3310239 • Letter: A
Question
A study of 8 different weight loss programs involved 200 subjects. Each of the 8 programs had 25 subjects in it. The subjects were followed for 12 months. Weight change for each subject was recorded. We want to test the claim that the mean weight loss is the same for the 8 programs. 20. (a) Complete the following ANOVA table with sum of squares, degrees of freedom, and mean square (Show all work) Source of Variation Sum of Squares Degrees of Freedom Mean Square (MS) Factor (Between) Error (Within) Total 50.8 N/A 620.05 199 (b) Determine the test statistic. Show all work; writing the correct test statistic, without supporting (c) Determine the P-value. Show all work wrting the correct P-value, without supporting work (d) Is there sufficient evidence to support the claim that the mean weright loss is the same for the s work, will receive no credit will receive no credit programs at the significance level of 0.05? Explain.Explanation / Answer
We know that, one way ANOVA is
Source of Variation
Sum of Squares (SS)
Degree of Freedom (df)
Mean Square (MS)
F
p
Factor (Between)
SSB
K-1
MSB=SSB/(K-1)
=MSB/MSE
Error (Within)
SSE
N-K
MSE=SSE/(N-K)
Total
SST
N-1
Where, SST = SSB +SSE and given that, K=8 , N=200
(A)
From above, the complete one way ANOVA table is
Source of Variation
Sum of Squares (SS)
Degree of Freedom (df)
Mean Square (MS)
F
p
Factor (Between)
50.8
7
7.257142857
2.447731978
0.019993
Error (Within)
569.25
192
2.96484375
Total
620.05
199
N/A
(B)
From part (A),
The test statistic F = 2.447731978
(C)
From F distribution table at df(7,192), for F = 2.447731978
p-value = 0.019993
(D)
The p-value is 0 .019993. The result is significant at p < 0.05. Therefore, test is fail to reject null hypothesis. Hence, There is sufficient evidence to support the claim that the mean weight loss is the same for the 8 programs at the significance level of 0.05.
Source of Variation
Sum of Squares (SS)
Degree of Freedom (df)
Mean Square (MS)
F
p
Factor (Between)
SSB
K-1
MSB=SSB/(K-1)
=MSB/MSE
Error (Within)
SSE
N-K
MSE=SSE/(N-K)
Total
SST
N-1
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