Question

5. Measuring effect size for the repeated-measures ANOVA Alicia F. Lieberman and Patricia Van Horn have...

5. Measuring effect size for the repeated-measures ANOVA

Alicia F. Lieberman and Patricia Van Horn have created a psychotherapy model for young children who have witnessed family violence. The therapy focuses on building the parent’s capacity to nurture and protect the child, thereby promoting the child’s emotional health and repairing the parent-child relationship, which has been disrupted by the stress and trauma of family violence.

As a clinical psychology intern, you are learning parent-child therapy with Drs. Lieberman and Van Horn. You see six parent-child dyads for a year and then evaluate them before therapy, after therapy, and again one year after therapy is finished. You are interested in assessing whether the treatment had an effect on the child’s anxiety.

In the experiment above, the null hypothesis is:

There are no differences among the anxiety means across the three time points

At least one anxiety mean is different from another

There are no individual differences in the anxiety means

The results of the study are presented in the following data table. All scores are from the Achenbach Child Behavior Checklist.

Anxiety Score

Child

Before Treatment

After Treatment

12-Month Follow-Up

Participant Totals

A 2.23 1.91 1.71 P = 5.85 n = 6
B 1.75 1.52 1.38 P = 4.65 k = 3
C 0.96 1.09 1.11 P = 3.16 N = 18
D 1.35 1.26 1.16 P = 3.77 G = 25.68
E 0.77 0.51 0.52 P = 1.80 ΣX² = 41.9396
F 2.27 2.15 2.03 P = 6.45
T = 9.33 T = 8.44 T = 7.91
SS = 2.0173 SS = 1.7446 SS = 1.3695

The three treatment evaluations define three populations of interest. Use analysis of variance (ANOVA) to test the hypothesis that the three population means are equal.

Source

SS

ANOVA Table

MS

F

df

Between treatments 0.1716 2 0.0858   
Within treatments 5.1314 15 0.3421
Between subjects 4.9919 5 0.9984
Error 0.1395 10 0.0140
Total 5.3028 17

Fill in the missing value for the F test statistic in the ANOVA table.

1. Use the F distribution in the Distributions tool to find the critical value of F for α = 0.05. The critical value is F = _______ .

2. At a significance level of α = 0.05, evaluate the null hypothesis that the population means for all treatments are equal. The null hypothesis is _____________ (rejected/not rejected? . You ____________ (can/cant) conclude that parent-child therapy has an effect on anxiety in children who have witnessed family violence.

3. To measure the effect size, calculate η². The η² is __________ . The percentage of variance accounted for by the treatment effect is ___________ .

Homework Answers

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
12. Measuring effect size for two-factor ANOVA It is projected that approximately 580,000 veterans will take...
12. Measuring effect size for two-factor ANOVA It is projected that approximately 580,000 veterans will take advantage of the GI Bill for the 21st Century. Boots to Books is a course for all veterans, current military members, and their family members, friends, and supporters. The goal of Boots to Books is to assist deployed, postdeployed, and veteran students in making a positive transition from military to civilian life or from deployment to postdeployment life, including the acquisition of college survival...
Consider the partial ANOVA table shown below. Let a = .01 Source of Variation DF SS...
Consider the partial ANOVA table shown below. Let a = .01 Source of Variation DF SS MS F Between Treatments 3 180 Within Treatments (Error) Total 19 380 If all the samples have five observations each: there are 10 possible pairs of sample means. the only appropriate comparison test is the Tukey-Kramer. all of the absolute differences will likely exceed their corresponding critical values. there is no need for a comparison test – the null hypothesis is not rejected. 2...
The results of a one-way ANOVA are reported below. Source of variation SS df MS F...
The results of a one-way ANOVA are reported below. Source of variation SS df MS F Between groups 7.18 3 2.39 21.73 Within groups 48.07 453 0.11 Total 55.25 456 How many treatments are there in the study? What is the total sample size? What is the critical value of F if p = 0.05? Write out the null hypothesis and the alternate hypothesis. What is your decision regarding the null hypothesis? Can we conclude any of the treatment means...
The following data were obtained from an independent-measures research study comparing three treatment conditions. I II...
The following data were obtained from an independent-measures research study comparing three treatment conditions. I II III n = 6 n = 4 n = 4                    M = 2 M = 2.5 M = 5 N = 14 T = 12 T = 10 T = 20 G = 42 SS = 14 SS = 9 SS = 10 ΣX2tot = 182 Use an ANOVA with α = .05 to determine whether there are any significant mean differences among the...
The following data summarize the results from an independent measures study comparing three treatment conditions. I...
The following data summarize the results from an independent measures study comparing three treatment conditions. I II III n = 6 n = 6 n = 6                    M = 4 M = 5 M = 6 N = 18 T = 24 T = 30 T = 36 G = 90 SS = 30 SS = 35 SS = 40 ΣX2tot = 567 Use an ANOVA with α = .05 to determine whether there are any significant differences among the...
The following data are given for a two-factor ANOVA with two treatments and three blocks.   ...
The following data are given for a two-factor ANOVA with two treatments and three blocks.    Treatment Block 1 2 A 43 31 B 33 22 C 46 36    Using the 0.05 significance level conduct a test of hypothesis to determine whether the block or the treatment means differ. State the null and alternate hypotheses for treatments. State the decision rule for treatments. (Round your answer to 1 decimal place.)    State the null and alternate hypotheses for blocks....
1a The following data were obtained from an independent-measures research study comparing three treatment conditions. I...
1a The following data were obtained from an independent-measures research study comparing three treatment conditions. I II III 2 5 7 5 2 3 0 1 6 1 2 4 2 2 T =12 T =10 T =20 G = 42 SS =14 SS =9 SS =10 ΣX2= 182 Use an ANOVA with α = .05 to determine whether there are any significant mean differences among the treatments. The null hypothesis in words is Group of answer choices a. There...
The following data were obtained from an independent-measures research study comparing three treatment conditions. I II...
The following data were obtained from an independent-measures research study comparing three treatment conditions. I II III n = 6 n = 4 n = 4                    M = 2 M = 2.5 M = 5 N = 14 T = 12 T = 10 T = 20 G = 42 SS = 14 SS = 9 SS = 10 ΣX2tot = 182 Use an ANOVA with α = .05 to determine whether there are any significant mean differences among the...
The following data are given for a two-factor ANOVA with two treatments and three blocks. Using...
The following data are given for a two-factor ANOVA with two treatments and three blocks. Using the 0.05 significance level conduct a test of hypothesis to determine whether the block or the treatment means differ. A 48 36 B 36 22 C 45 37 1. State the null and alternate hypotheses for treatments. 2. State the decision rule for treatments. (Round your answer to 1 decimal place.) 3. State the null and alternate hypotheses for blocks. (Round your answer to...
The following data are given for a two-factor ANOVA with two treatments and three blocks.   ...
The following data are given for a two-factor ANOVA with two treatments and three blocks.       Treatment Block 1 2 A 46 31 B 37 26 C 44 35    Using the 0.05 significance level conduct a test of hypothesis to determine whether the block or the treatment means differ.     a. State the null and alternate hypotheses for treatments.   H0 (Click to select)The means are the same.The standard deviations are the same.The standard deviations are different.The means are...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT