Question

a= .05 A therapist would like to evaluate the effectiveness of a procedure for treating depression. A sample of n = 16 depressed clients is obtained and each person’s level of depression is measured using a standardized questionnaire before they begin the therapy program. Two weeks after therapy, each client’s level of depression is measured again. The average level of depression dropped by 6.8 points following therapy. The difference scores had a variance of 64. a. State the null hypothesis. b. Which statistical test would you use to test this hypothesis? c. Calculate the appropriate test statistic and state your conclusions. Show your work. d. Compute an effect size. e. Find critical value

Answer 1

a) Ho: There is no significant drop in the level of depression (mu=0)

Ha: There is significant drop in the level of depression (mu>0)

b) Here we have sample std deviation known and sample size is less than 30 and hence we have to use the t test

c,d,e)t=(Xbar-mu)/(sd/sqrt(n))

t=(6.8-0)/(8/sqrt(16))

t=3.4

Test stat is 3.4

critical values for alpha =0.05 at df=n-1=15 is from the t table check for area under the curve = 95% i.e. 0.95

(Done this in R however you can find the same fom t table as shown above)

> qt(.95, df = 15) [1] 1.75305

t_critical = 1.75 and here t>t_critical we have to reject Ho

Hence we can conclude that There is significant drop in the level of depression after the treatment

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