Question

Scenario 3 (Mental Health) The efficacy of two kinds of cognitive-behavioral therapy (CBT) in relation to a clinical population were compared. The therapies differed on the dimension of how wedded they were to the rational-emotive behavioral therapy (REBT), a subtype of CBT that emphasizes a directive, confrontational approach to encourage a patient to recognize the irrationality of specific thought patterns. Forty adolescents at an inpatient clinic for treatment of self-destructive behaviors were randomly divided into two groups of equal size, one of which received the less challenging type of CBT (Treatment A) and one of which received the more challenging kind of CBT (Treatment B). All patients were treated by trained therapists in one-on-one sessions for 1.5 hours per day (broken down into 45-minute sessions) for six weeks. All participants were apprised that they were part of a study, all participants signed consent forms, and all were told they would be informed of the results at its conclusion; participants exhibiting any behaviors that required critical intervention were promptly treated outside the plan of the study. Outcome data on the Revised Behavior Problem Checklist (RBPC)-PAR Edition* collected at the conclusion of six weeks as shown below (also found in the Data Set Scenario 3 Excel file).

Treatment A
74
50
70
60
30
37
34
40
39
70
19
43
25
15
20
55
48
42
60
27

Treatment B
80
82
49
44
60
65
77
45
51
70
53
16
38
44
29
51
54
46
18
61

1. Prepare an appropriately labeled histogram for each set of data.

2. Evaluate the shape of each distribution using your created histograms. In other words, what does the shape of each distribution tell us about the data

3. resent properly labeled graphs representing the data analysis results detailed clearly for ease of stakeholder interpretation.

Answer 1

From the above histograms, we observed that the two distributions are bell-shaped.

From above two box plots, both distributions are positively skewed since for both third quartile-2nd quartile>2nd quartile-1st quartile. Median for Treatment A<median for treatment B and moreover 75% of data for treatment B above the 50% of data for treatment A.

Hence treatment B is more effective than treatment A.

From the above probability plots, normality assumption holds for two data sets.

Now we use 2 sample t test to justify the statement "treatment B is more effective than treatment A" (which is obtained from box plots).

Step 1:

Test and CI for Two Variances: Treatment A, Treatment B

Method

Null hypothesis Sigma(Treatment A) / Sigma(Treatment B) = 1
Alternative hypothesis Sigma(Treatment A) / Sigma(Treatment B) not = 1
Significance level Alpha = 0.05

Statistics

Variable N StDev Variance
Treatment A 20 17.764 315.568
Treatment B 20 18.282 334.239

Ratio of standard deviations = 0.972
Ratio of variances = 0.944

95% Confidence Intervals

CI for
Distribution CI for StDev Variance
of Data Ratio Ratio
Normal (0.611, 1.544) (0.374, 2.385)
Continuous (0.629, 1.842) (0.396, 3.395)

Tests

Test
Method DF1 DF2 Statistic P-Value
F Test (normal) 19 19 0.94 0.902
Levene's Test (any continuous) 1 38 0.03 0.853

Since p-value>0.05 so we can assume the population variances are same.

Step 2:

Two-Sample T-Test and CI: Treatment A, Treatment B

Two-sample T for Treatment A vs Treatment B

N Mean StDev SE Mean
Treatment A 20 42.9 17.8 4.0
Treatment B 20 51.6 18.3 4.1

Difference = mu (Treatment A) - mu (Treatment B)
Estimate for difference: -8.75
95% upper bound for difference: 0.86
T-Test of difference = 0 (vs <): T-Value = -1.54 P-Value = 0.067 DF = 38
Both use Pooled StDev = 18.0251

Since 0.05<p-value<0.1 hence treatment B is more effective than treatment A at 10% level of significance whereas this difference is insignificant at 5% level of significant.