Question

A researcher would like to determine if meditation training will affect anxiety-related distress among a group of participants. For a week prior to training, each participant records the number of times they feel anxious. Participants then receive meditation training and for the week following training the number of times they feel anxious is again measured. The data are as follows:

Before      After

8             5

7             2

5             4

5             2

8             3

4             2

6             4

6             3

a. Compute the mean and sum of the squared deviations for the sample of difference scores.

b. Do the results indicate a significant difference? Use a two-tailed test with α = .05.

c. Compute Cohen’s d to measure the size of the effect.

Answer 1

From the sample data, it is found that the corresponding sample means are:

Xbar after= 6.125

Xbar before = 3.125

Also, the provided sample standard deviations are:

s_1 = 1.458 and s_2 = 1.126

and the sample size is n = 8. For the score differences we have

Dbar=3 and s D=1.414
(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μD= 0

Ha: μ D ≠ 0

This corresponds to a two-tailed test, for which a t-test for two paired samples be used.

(2) Rejection Region

Based on the information provided, the significance level is alpha = 0.05α=0.05, and the degrees of freedom are df = 7df=7.

Hence, it is found that the critical value for this two-tailed test is t c=2.365, for α=0.05 and df=7.

The rejection region for this two-tailed test is R={t:∣t∣>2.365}.

(3) Test Statistics

The t-statistic is computed as shown in the following formula:

t =Dbar/sd/sqrt(n)

t= 3/1.414/sqrt(8)

t=6

(4) Decision about the null hypothesis

Since it is observed that ∣t∣=6>t c =2.365, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0.0005, and since p = 0.0005 & p=0.0005<0.05, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population mean μ1 is different than μ 2 at the 0.05 significance level.

c) Cohen's d = (3.125 - 6.125) ⁄ 1.30262 = 2.30305.