Source Between treatments Within treatments Between subjects Error T= 12.59 SS = 2.0539 Total The three treatment evaluations define three populations of interest. Use analysis of variance (ANOVA) to test the hypothesis that the three popula means are equal. SS 2.7352 2.3113 1.0332 1.2781 5.0464 T = 8.04 Select a Distribution SS = 0.0484 violence. ANOVA Table df 2 15 5 10 17 Fill in the missing value for the F test statistic in the ANOVA table. MS 0 1.3676 0.1541 0.2066 0.1278 To measure the effect size, calculate n². The n² is 1 T = 7.30 SS = 0.2090 2 F Distributions 3 Use the F distribution in the Distributions tool to find the critical value of F for a = 0.05. The critical value is F = At a significance level of a = 0.05, evaluate the null hypothesis that the population means for all treatments are equal. The null hypothesis is You conclude that parent-child therapy has an effect on oppositional defiance in children who have witnessed family The percentage of variance accounted for by the treatment effect if
Source Between treatments Within treatments Between subjects Error T= 12.59 SS = 2.0539 Total The three treatment evaluations define three populations of interest. Use analysis of variance (ANOVA) to test the hypothesis that the three popula means are equal. SS 2.7352 2.3113 1.0332 1.2781 5.0464 T = 8.04 Select a Distribution SS = 0.0484 violence. ANOVA Table df 2 15 5 10 17 Fill in the missing value for the F test statistic in the ANOVA table. MS 0 1.3676 0.1541 0.2066 0.1278 To measure the effect size, calculate n². The n² is 1 T = 7.30 SS = 0.2090 2 F Distributions 3 Use the F distribution in the Distributions tool to find the critical value of F for a = 0.05. The critical value is F = At a significance level of a = 0.05, evaluate the null hypothesis that the population means for all treatments are equal. The null hypothesis is You conclude that parent-child therapy has an effect on oppositional defiance in children who have witnessed family The percentage of variance accounted for by the treatment effect if
Transcribed Image Text:T = 8.04
T = 7.30
T = 12.59
SS = 0.0484
SS = 0.2090
SS = 2.0539
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.
ANOVA Table
Source
df
MS
F
Between treatments
2.7352
1.3676
Within treatments
2.3113
15
0.1541
Between subjects
1.0332
15
0.2066
Error
1.2781
10
0.1278
Total
5.0464
17
Fill in the missing value for the F test statistic in the ANOVA table.
Select a Distribution
Distributions
Use the F distribution in the Distributions tool to find the critical value of F for q = 0.05. The critical value is F =
At a significance level of a = 0.05, evaluate the null hypothesis that the population means for all treatments are equal. The null hypothesis is
▼ , You
condude that parent-child therapy has an effect on oppositional defiance in children who have witnessed family
violence.
To measure the effect size, calculate n2. The n? is
The percentage of variance accounted for by the treatment effect is
Transcribed Image Text: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 oppositional defiance.
In the experiment above, the null hypothesis is:
O At least one oppositional defiance mean is different from another
O There are no differences among the oppositional defiance means across the three time points
O There are no individual differences in the oppositional defiance means
The results of the study are presented in the following data table. All scores are from the Achenbach Child Behavior Checklist.
Defiance Score
Child
Before Treatment
After Treatment
12-Month Follow-Up
Participant Totals
A
2.80
1.32
1.50
P = 5.62
n = 6
B
1.53
1.48
1.12
P = 4.13
k= 3
2.67
1.29
1.44
P = 5.40
N= 18
D
1.42
1.43
0.99
P = 3.84
G= 27.93
1.61
1.21
1.10
P = 3.92
EX2 = 48.3845
2.56
1.31
1.15
P = 5.02
T = 12.59
T = 8.04
T= 7.30
SS = 2.0539
SS = 0.0484
SS =
090
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.
ANOVA Table
Source
SS
df
MS
Between treatments
2.7352
1.3676
Within treatments
2.3113
15
0.1541
Between subjects
1.0332
0.2066
Error
1.2781
10
0.1278
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