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A health psychologist tests a new intervention to determine if it can change healthy behaviors among...

A health psychologist tests a new intervention to determine if it can change healthy behaviors among siblings. To conduct the this test using a matched-pairs design, the researcher gives one sibling an intervention, and the other sibling is given a control task without the intervention. The number of healthy behaviors observed in the siblings during a 5-minute observation were then recorded. Intervention Yes 6, 2, 5,6,6,5 No 4,5,4,5,4,4

(a) Test whether or not the number of healthy behaviors differ at a 0.05 level of significance. State the value of the test statistic. (Round your answer to three decimal places.)

State the decision to retain or reject the null hypothesis. Retain the null hypothesis. Reject the null hypothesis. (

b) Compute effect size using eta-squared. (Round your answer to two decimal places.)

Solutions

Expert Solution

a) (1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μD​ = 0

Ha:μD​ ≠ 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 α=0.05, and the degrees of freedom are df=5.

Hence, it is found that the critical value for this two-tailed test is tc​=2.571, for α=0.05 and df=5.

The rejection region for this two-tailed test is R={t:∣t∣>2.571}.

(3) Test Statistics:

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

(4) Decision about the null hypothesis

Since it is observed that |t| = 0.877 ∣t∣=0.877≤tc​=2.571, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.4206, and since p=0.4206≥0.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ1​ is different than μ2​, at the 0.05 significance level.

b) eta squared= r^2= t^2/(t^2+df)= (0.877)^2/((0.877)^2+5)= 0.7691/0.7691+5= 0.13

Effect size is SMALL.


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