Question

A researcher conducts an independent-measures study examining the effectiveness of a group exercise program at an assisted living facility for elderly adults. One group of residents is selected to participate in the program, and a second group serves as a control. After 6 weeks, the researcher assessed physical fitness of each participant.

The data are as follows: Exercise Group: Control Group: n = 15 n = 15 M = 37 M = 34 SS = 230 SS = 160 A. Does the exercise program have a significant effect on physical fitness? Use p < .05, 2-tails test.

B. If there is a significant effect, compute the estimated Cohen’s d to measure the size of the treatment effect.

C. Write a conclusion demonstrating how the outcome of the hypothesis test and the measure of effect size would appear in a research report in APA style.

Note: For full credit show all steps of hypothesis testing (i.e., write hypotheses, all computations, the tcritical used for the decision about H0 and the conclusion in APA reporting format).

Answer 1

Solution:

Part A

Here, we have to use two sample t test for the population mean. The null and alternative hypotheses for this test are given as below:

Null hypothesis: H₀: Exercise program don’t have any significant effect on physical fitness.

Alternative hypothesis: H_a: Exercise program have a significant effect on physical fitness.

H₀: µ₁ = µ₂ versus H_a: µ₁ ≠ µ₂

This is a two tailed test.

Test statistic formula for pooled variance t test is given as below:

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

Where S_p² is pooled variance

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

We are given

X1bar = 37

X2bar = 34

n1 = 15

n2 = 15

SS1 = 230, VAR = SS1/(n – 1) = 230/(15 – 1) = 16.42857, S1 = sqrt(16.42857) = 4.053217

S1 = 4.05

SS2 = 160, VAR = SS2/(n2 – 1) = 160/(15 – 1) = 11.42857, S2 = sqrt(11.42857) = 3.380617

S2 = 3.38

DF = n1 + n2 – 2 = 15 + 15 – 2 = 28

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

S_p² = [(15 – 1)*4.05^2 + (15 – 1)*3.38^2]/(15 + 15 – 2)

S_p² = 13.9135

Test statistic is given as below:

t = (37 – 34)/sqrt[13.9135*((1/15)+(1/15))]

t = 3/ 1.3620

t = 2.2026

Critical value = ±2.0484

(by using t-table or excel)

P-value = 0.0360

(by using t-table or excel)

α =0.05

P-value < α

So, we reject the null hypothesis that Exercise program don’t have any significant effect on physical fitness.

There is sufficient evidence to conclude that Exercise program have a significant effect on physical fitness.

Part B

Cohen’s d = (M1 – M2) / SD_pooled

We have

M1 = 37

M2 = 34

Pooled variance = S_p² = 13.9135

Pooled SD = sqrt(13.9135) = 3.730073726

Cohen’s d = (37 – 34) / 3.730073726

Cohen’s d = 0.804273647

Cohen’s d > 0.8

So, it is a large effect size.

The difference between two groups is large enough.

Part C

The independent samples t-test assuming equal population variances conclude that there is sufficient evidence (t=2.2, p = 0.036) to conclude that Exercise program has a significant effect on physical fitness. The effect size (Cohen’s d = 0.8) suggests that the difference between two groups is large enough.