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 records a combined score measuring balance and strength for each individual. The data are as follows:

                                                Control               Exercise

                                                 n₁ = 10                 n₂ = 15

                                                M₁ = 12               M₂ = 15.5

                                               SS₁ = 120.5          SS₂ = 190.0

Conduct an independent samples t-test to evaluate the significance of the treatment effect and calculate Cohen's d to measure the size of the treatment effect. Use a two-tailed test with α = .05. In your response, make sure to state:

1. The null and alternative hypotheses (.5 points per hypothesis)

2. The t statistic representing the critical region cutoff point (i.e., the t value you find using the t distribution table and df). (.5 points)

3. The value of the t test statistic and the value of Cohen's d. (.5 points for t and .5 points for d)

4. Your decision about whether to reject the null hypothesis (you do not need to write an APA format statement). (.5 points)

Answer 1

Null hypothesis:

H0 : μ1 = μ2

That is, there is no significant difference between the mean number of aggression responses of treatment group and control group.

Alternative hypothesis:

H1 : μ1 ≠ μ2

That is, there is a significant difference between the mean number of aggression responses of treatment group and control group.

In order to test the hypothesis regarding the significant difference between the means of two independent samples, when the population standard deviations are unknown an independent sample t-test is appropriate.

Degrees of freedom:

The size of the treatment group is n1 = 15.

The size of the control group is n2 = 10.

Assume that the population variances are equal.

The degrees of freedom is obtained as 23 from the calculations given below:

Critical value:

The level of significance is α = 0.05.

The critical value is obtained as ±2.0687 from the calculation given below:

Since, the t-distribution is symmetric, the two critical values are –t(α/2) = -2.0687 and +t(α/2) = +2.0687.

Pooled sample variance:

The sum of squares for treatment group is SS1 = 120.5

The sum of squares for control  group is SS2 = 190

The pooled sample variance is obtained as 13.5 from the calculation given below:

Standard error:

The standard error of the difference of means is obtained as 1.5 from the calculation given below:

Test statistic:

The test statistic is obtained as 2.333 from the calculations given below:

=2.333

Decision rule:

Denote t as test statistic value and t(α/2) as the critical value.

Decision rule based on critical approach:

If t ≤ –t(α/2) (or) t ≥ t(α/2), then reject the null hypothesis H0.

If –t(α/2) < t < t(α/2), then fail to reject the null hypothesis H0.

Conclusion:

Conclusion based on critical value approach:

The test statistic value is 2.333 and critical value is ±2.0687.

Here, t ≥ t(α/2). That is, 2.333 (=t) > 2.0687 (= t(α/2)).

By the rejection rule, reject the null hypothesis H0.

Therefore, there is a significant difference between the mean number of aggression responses of treatment group and control group.

Hence, at the 0.05 level of significance, there is enough evidence to conclude that the drug have a significant effect on aggression.

The null hypothesis: H0 = μ1 = μ2

The degrees of freedom is df = 23

t critical value tcritical : –t(α/2) = -2.0687 and +t(α/2) = +2.0687.

The pooled sample variance sp2 = 13.5.

The standard error S(M1 – M2) = 1.5

cohen's d effect size =