Question

A group of biochemistry researchers developed a new medication for treating depression. To assess if the new medication is effective at treating depression, the researchers obtained a sample of depressive patients from a local clinic. The patients were then randomly assigned into taking the medication for a month or not. In addition, the patients were also randomly assigned into receiving psychotherapy for a month or not. At the end of the study, the patients were asked to fill out a depression inventory in which a higher score indicates more depression. The data are below. What can be concluded with α = 0.05?

                Psychotherapy

Medication	no	yes
no	10 11 9 13	12 10 13 8
yes	11 13 8 9	4 3 7 6

a) What is the appropriate test statistic?
---Select--- na one-way ANOVA within-subjects ANOVA two-way ANOVA

b) Compute the appropriate test statistic(s) to make a decision about H₀.
Medication: p-value =  ; Decision:  ---Select--- Reject H0 Fail to reject H0
Psychotherapy: p-value =  ; Decision:  ---Select--- Reject H0 Fail to reject H0
Interaction: p-value =  ; Decision:  ---Select--- Reject H0 Fail to reject H0


c) Compute the corresponding effect size(s) and indicate magnitude(s).
Medication: η² =  ;  ---Select--- na trivial effect small effect medium effect large effect
Psychotherapy: η² =  ;  ---Select--- na trivial effect small effect medium effect large effect
Interaction: η² =  ;  ---Select--- na trivial effect small effect medium effect large effect


d) Make an interpretation based on the results.

There is a medication difference in depression.There is no medication difference in depression.    

There is a psychotherapy difference in depression.There is no psychotherapy difference in depression.    

There is a medication by psychotherapy interaction in depression.There is no medication by psychotherapy interaction in depression.    

Answer 1

(a) two way anova

(b)

medication - decision: reject hO, as the p-value is less than typical alpha=0.05

psychotherapy: decision: reject hO, as the p-value is less than typical alpha=0.05

interaction: decision: reject hO , as the p-value is less than typical alpha=0.05

here we want to test the null hypothesis H0 against the alternate hypothesis H1

H0: 1. medication has no effect, against H1: medication has effect

H0: 2. psychotherapy has no effect against H1: psychotherapy has effect

H0: 3. there is no interaction agaist H1: there is interaction

(c)=SS(effect)/Total SS

Source of Variation
medication	49/164=0.2988, large effect
psyschotherapy	36/164=0.2195, large effect
Interaction	36/164=0.2195, large effect

Cohen’s (1988) guidelines. According to him:

• Small: 0.01
• Medium: 0.059
• Large: 0.138

(d) since we reject the null hypothesis in part (b) so right choices are

following information has been generated using ms-excel

Anova: Two-Factor With Replication
SUMMARY	pyed	pno	Total
myes
Count	4	4	8
Sum	43	43	86
Average	10.75	10.75	10.75
Variance	2.916667	4.916667	3.357143
mno
Count	4	4	8
Sum	41	17	58
Average	10.25	4.25	7.25
Variance	4.916667	1.583333	13.07143
Total
Count	8	8
Sum	84	60
Average	10.5	7.5
Variance	3.428571	14.85714
ANOVA
Source of Variation	SS	df	MS	F	P-value	F crit
medication	49	1	49	13.67442	0.003047	4.747225
psyschotherapy	36	1	36	10.04651	0.008075	4.747225
Interaction	36	1	36	10.04651	0.008075	4.747225
Within	43	12	3.583333
Total	164	15