In: Statistics and Probability
A clinical psychologist is interested in comparing the effectiveness of short term relaxation and cognitive-behavioral therapy in treating mild depression. An experiment is conducted in which 15 patients with mild depression are randomly selected and assigned 5 each to a relaxation therapy group, a cognitive/behavioral therapy group, and an attention placebo group. Therapy is administered until the patient is judged no longer depressed or until 10 treatment sessions have elapsed. The following data is obtained. Scores are the number of sessions for each patient. Using an alpha of .05, test the hypothesis that there is a difference in effectiveness among these therapies.
Cognitive/Behavioral
Therapy |
Relaxation |
Attention-Placebo |
5 |
6 |
5 8 9 7 6 |
a) What is the appropriate test?
b) What is the Null Hypothesis? What is the Alternative Hypothesis? (include the means)
c) Using a = 0.05, Fcrit = _________.
d) Calculate
a)
We Can One way ANOVA test
b)
Ho: 1 = 2 = 3
Ha: At least one of the mean is different
C)
CALCULATION:
Number of Treatment, t = 3 n = 15
T1 (sum of Cognitive/Behavioral therapy) = 28, T2(Sum of Relaxation therapy) = 37, T3(Sum of Attention-Placebo group) = 35
G = Grand Total = 100
CF = Correction Factor = G2/N = 1002 / 15 = 666.67
= (5)2 + (6)2 + ......................+ (6)2 - 666.67
TSS = 29.33
SSTR = (1 / 5) * [(28)2 + (37)2 + (35)2] - 666*.67
SSTR = 8.93
SSE = TSS - SSTR
= 29.33 - 8.93
SSE = 20.40
MSSTR = SSTR/t-1 = 8.93 / 3-1 = 4.47
MSSE = SSE / n-t = 20.40 /15-3 = 1.70
F = MSSTR / MSSE = 4.47 / 1.70 = 2.63
Ftabulate = F,(t-1,n-t) = F0.05,(2,12) = 3.8853 ..............Using F table
ANOVA TABLE:
Conclusion:
Test statistic < Critical value, i.e. 2.63 < 3.8853, That is Fail to Reject Ho at 5% level of significance.
Therefore, there is a difference in effectiveness among these therapies