A health psychologist tests a new intervention to determine if it can change healthy behaviors among children. To conduct this test using a matched-pairs design, the researcher gives one group of children an intervention, and a second group is given a control task without the intervention. The number of healthy behaviors observed in the children during a 5-minuteobservation are given in the table.
Intervention | |
---|---|
Yes | No |
6 | 4 |
3 | 5 |
6 | 4 |
6 | 5 |
6 | 4 |
5 | 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.)
You may need to use the appropriate table in Appendix C to answer
this question.
Yes No d
6 4 2
3 5 -2
6 4 2
6 5 1
6 4 2
5 4 -1
d-bar= 4
a) Given that
n1= 6 df= n- 1 = 5
x̄1 = Sample Mean = 5.33
s1^2 = Sample Variance = 1.47
s1 = Sample Standard Deviation = 1.21
n2= 6 df= n- 1 = 5
x̄2 = Sample Mean = 4.33
s2^2 = Sample Variance = 0.27
s2 = Sample Standard Deviation = 0.52
Test Statistic = 1.859
α = 0.05
Left-tailed area corresponding to z-score of -1.859895 = 0.03145
Right-tailed area corresponding to z-score of 1.859895 = 0.03145
P-value = Left-Tailed Area + Right-Tailed Area = 0.0629
p-value is greater than α then do not reject the null hypothesis.
b) Effect size
cohens d =(4.33 - 5.33) ⁄ 0.931263 = 1.07
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