A researcher is studying treatments for back pain. 120 patients were randomly di
ID: 3226942 • Letter: A
Question
A researcher is studying treatments for back pain. 120 patients were randomly divided into four groups of 30 each. One group was assigned to placebo and the other three groups were assigned to one of the other treatments. After 5 weeks on treatment a measure of pain was evaluated (high score indicating a higher level of pain). Assume the data from the four groups are independent and the responses are approximately normal.
The researchers did an ANOVA test of the data and obtained the following results.
Source
DF
SS
MS
F
P-value
Groups
3
3727.8
1242.6
3.997
0.0094
Error
116
36,060.92
310.87
Total
119
39,788.72
1. What is the value of the estimate for the pooled variance within each group?
2. At a = 0.05, do we reject the null hypothesis?
3. What percent of the overall variation in the outcome is explained by group membership?
4. If we wish to make all pair-wise comparisons using a t-test, what is the new value we need to compare our p-value to for conclusions in order keep our overall type I error rate at 0.05 using the Boneferroni technique?
Source
DF
SS
MS
F
P-value
Groups
3
3727.8
1242.6
3.997
0.0094
Error
116
36,060.92
310.87
Total
119
39,788.72
Explanation / Answer
'Pooled Variance' is nothing but the sum the 'sum of squares' of all your individual treatments, and dividing that by the "within" degrees of freedom (df within).
sp = (SS1 + SS2 + SS3 +SS4 ) / 116
The numerator is nothing but error sum of square =(SS1 + SS2 + SS3 +SS4 ) =SSE = 36060.92
So that pooled variance of each group = 310.87
2) Since p-value corresponding to the F statistics =0.0094
Decision rule:
1) If p-value < level of significance then we reject null hypothesis
2) If p-value > level of significance then we fail to reject null hypothesis
Here p-value < 0.05 (the given level of significance)
So use first decision rule
Therefore at a = 0.05, do we reject the null hypothesis
3) Formula of coefficient of determination (R2) = SSR/SST = 3727.8/39788.72 = 0.09369 = 9.37%
Therefore 9.37% of the overall variation in the outcome is explained by group membership.
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