Question

The data below are from an independent-measures experiment comparing three different treatment conditions on number of panic attacks. (10 POINTS)

CBT	Psychotherapy	Placebo
1	2	4
1	3	2	G = 42
2	1	4	SX2 = 122
3	3	3
1	2	4
1	1	4
T = 9	T = 12	T = 21
SS = 3.5	SS = 4	SS = 3.5

1) State the null hypothesis

2) calculate necessary statistics using an alpha of .05

3) make a decision about your null hypothesis and explain what that decision means

4) Make a conclusion about your IV and DV that includes an APA format summary of your statistical analysis.

5) Indicate what you would do next

Answer 1

1) Ho: µ1=µ2=µ3  

2)

one way anova

treatment	CBT	psychotherepy	Placebo
count, ni =	6	6	6
mean , x̅ i =	1.500	2.00	3.50
std. dev., si =	0.8	0.9	0.8
sample variances, si^2 =	0.700	0.800	0.700
total sum	9	12	21				42	(grand sum)
grand mean , x̅̅ =	Σni*x̅i/Σni =	2.33
( x̅ - x̅̅ )²	0.694	0.111	1.361
							TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	4.167	0.667	8.167				13
SS(within ) = SSW = Σ(n-1)s² =	3.500	4.000	3.500				11.0000

no. of treatment , k =   3  
df between = k-1 =    2  
N = Σn =   18  
df within = N-k =   15  
      
mean square between groups , MSB = SSB/k-1 =    13/2=   6.5000
mean square within groups , MSW = SSW/N-k =    11/15=   0.7333
      
F-stat = MSB/MSW =    6.5/0.7333=   8.86

ANOVA
	SS	df	MS	F	p-value	F-critical
Between:	13.00	2	6.50	8.864	0.0029	3.6823
Within:	11.00	15	0.73
Total:	24.00	17
					α =	0.05

3)

   Decision:   p-value<α , reject null hypothesis    
Test is statiscally significant

4) conclusion :    there is enough evidence of significant mean difference among three treatments  

This test was found to be statistically significant, F(2,15) = 8.864, p < .05;

5)

since, null hypothesis is rejected, we will do post hoc test ,tukey Kramer test to test which pair of means are significantly different