Question

A student ran a between-subjects experiment comparing three treatments for depression: cognitive-behavioral therapy (CBT), client-centered therapy (CCT), and a no-treatment condition. Subjects were randomly assigned to the experimental condi- tion. After a sufficient duration on these treatments, the difference in subject’s depression scores pre/post study were measured using a valid depression assessment instrument. The data are summarized in the tables below. Conduct a Oneway ANOVA with α = 0.05 by first completing the values in the lower table, then using those values to demonstrate how the F test statistic is computed.

The following is a summary of the results:

	Sample Size	Sample Means	Standard Deviations
No Treatment	22	20.6	4.38
CBT	21	17.6	3.90
CCT	19	17.1	4.46
Overall	62	18.44	N/A

Suppose we have a null hypothesis that these three groups are drawn from the same distribution and an alternative hypothesis that at least one differs. We want to test these at 95% confidence.

To compute F-score, we need degrees of freedom, sum of squares results, and mean squared error results. The table below has most of this information, please complete the missing items, rounding to the nearest hundredth place in the case of Sum of Squares.

	Degrees of Freedom	Sum of Squares	Mean Squares
Total	61	1218.92	N/A
Group		145.28	72.64
Error			18.20

What is the F-Score, to the nearest hundredth place?  

What is the conclusion? If you think we reject the null hypothesis, type "reject". If you think we do not reject, type "none".  

Answer 1

Solution:
ANOVA table is given in the question so F-score can be calculated as
F Score = Mean Square for Group/ Mean Square for Error = 72.64/18.20 = 3.99
And Degree of freedom for Group = Sum of the square for Group/Mean square group = 145.28/72.64 = 2
Degree of freedom for Error = Total Degree of freedom - DF for Group = 61-2 = 59
So Complete ANOVA table is

	DF	Sum of Square	Mean Square	Fstat	P-value
Group	2	145.28	72.64	3.99	0.0237
Error	59	1073.64	18.2
Total	61	1218.92

So P-value from F table at Fstat = 3.99, DF numerator = 2, DF Denominator = 59 is 0.0237
At 95% confidence level, Alpha = 0.05, So we can say that we can reject the null hypothesis as p-value is less than alpha value i.e. (0.0237<0.05) and we have significant evidence to support the claim that at least one differs.