Question

• Please state the null and alternative hypotheses.
• Using an alpha level of .05, please test the null hypothesis (what is the appropriate test, what is the critical value?). Draw the rejection region.
• Calculate the test statistic
• State the conclusion you are entitled to draw as a result of this test.
• Calculate the effect size as appropriate for the hypothesis test used.

1. A researcher has developed a new therapy for the treatment of depression. The researcher has selected a small sample of four participants, and assesses them on the Beck Depression Inventory (scores of 25 or higher indicate depression). After the initial measurement, the four participants engage in the new therapy for two weeks, and are re-assessed on the Beck Depression Inventory. Did the therapy make a difference? What is the strength of this difference?

Pre-therapy	Post-therapy
26 31 29 24	23 25 28 19

Answer 1

Pre	Post	Difference
26	23	3
31	25	6
29	28	1
24	19	5

∑d = 15

∑d² = 71

n = 4

Mean , x̅d = Ʃd/n = 15/4 = 3.7500

Standard deviation, sd = √[(Ʃd² - (Ʃd)²/n)/(n-1)] = √[(71-(15)²/4)/(4-1)] = 2.2174

Null and Alternative hypothesis:

Ho : µd = 0

H1 : µd ≠ 0

df = n-1 = 3

Critical value :

Two tailed critical value, t-crit = T.INV.2T(0.05, 3) = 3.182

Rejection region: Reject Ho if t < -3.182 or if t > 3.182

Test statistic:

t = (x̅d)/(sd/√n) = (3.75)/(2.2174/√4) = 3.3824

Decision:

Reject the null hypothesis

Conclusion:

There is enough evidence to conclude that the therapy make a difference at 0.05 significance level.

Effect size : Cohen's d = (x̅d - µd)/sd = 1.6912