Question

"Three Sample ANOVA QA"

Researcher wants to compare three Types of Therapy [Behavior Therapy (BT), Cognitive Therapy (CT), and Cognitive-Behavior Therapy (CBT)] to determine which therapy works best (lowers) for treating clinical depression. The researcher selects a sample of individuals who have been diagnosed with clinical depression of size 33 (N = 33) to test the research hypothesis (alternative hypothesis) that there is a difference in the ability of the three therapies to treat clinical depression. The 33 participants are randomly assigned to each type of therapy (BT, CT, and CBT). Each participant receives their assigned therapy for a 6-month period. At the end of the 6-month period, the researcher administers a standardized test of depression to each participant (low numbers indicate less depression). The data are listed below. The researcher wants to maximize the power of this statistical test.

   BT CT CBT   
90 88 86
89 87 85
88 86 84
87 85 83
86 84 82
85 83 81
84 82 80
83 81 76
82 80 78
81 79 77
    80 78 79   

Means 85 83 81 Grand Mean = 83

2)What is the correct observed test statistic value for this statistical test of the means (Round to two decimal places)?

3)What is the correct critical value for this statistical test of the means?

6)If appropriate, conduct post-hoc analysis using Tukey’s HSD. What is the critical value?

7)If appropriate, what is the value of Tukey’s HSD for comparing BT to CT (Round to two decimal places)?

8)If appropriate, what is the value of Tukey’s HSD for comparing BT to CBT (Round to two decimal places)?

9)If appropropriate, what is the value of Tukey’s HSD for comparing CT to CBT (Round to two decimal places)?

Answer 1

	A	B	C	D
count, ni =	11	11	11
mean , x̅ i =	85.000	83.00	81.00
std. dev., si =	3.317	3.317	3.317
sample variances, si^2 =	11.000	11.000	11.000
total sum	935	913	891		2739	(grand sum)
grand mean , x̅̅ =	Σni*x̅i/Σni =				83.00
square of deviation of sample mean from grand mean,( x̅ - x̅̅)²	4.000	0.000	4.000
					TOTAL
SS(between)= SSB = Σn( x̅ - x̅̅)² =	44.000	0.000	44.000		88
SS(within ) = SSW = Σ(n-1)s² =	110.000	110.000	110.000		330.0000

no. of treatment , k =   3
df between = k-1 =    2
N = Σn =   33
df within = N-k =   30
  
mean square between groups , MSB = SSB/k-1 =    44.0000
  
mean square within groups , MSW = SSW/N-k =    11.0000
  
F-stat = MSB/MSW =    4.0000

anova table
	SS	df	MS	F	p-value	F-critical
Between:	88.00	2	44.00	4.00	0.029	3.32
Within:	330.00	30	11.00
Total:	418.00	32

       α =    0.05  

F stat = 4.00

critical value = 3.32

f stat > critical, reject null hypothesis    

  
              
...........

Level of significance	0.05
no. of treatments,k	3
DF error =N-k=	30
MSE	11.000
q-statistic value(α,k,N-k)	3.4900

tukey HSD = 3.49

critical value = q*√(MSE/n)

if absolute difference of means > critical value,means are significnantly different ,otherwise not

			confidence interval
population mean difference		critical value	lower limit	upper limit	result
µ1-µ2	2.00	3.49	-1.49	5.49	means are not different
µ1-µ3	4.00	3.49	0.51	7.49	means are different
µ2-µ3	2.00	3.49	-1.49	5.49	means are not different

...................

THANKS

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