Question

The researcher from Q2 decided that they want to follow up with the 6 children one year after they have completed the intervention program to see how they are now coping with negative emotions. The same method of assessment was used to examine coping responses. The data for the one-year follow-up list is listed below (along with results from Q2).

Before One-Year Person Follow-U After Treatment Child Treatment Total Y 297 n=6 k 3 N 18 G 65 17 10 14 4 4 12 SSbefore 11.33 SSafter 14.83 SSone year 21.33 7before = 14 | Tatter-25 | Tone vear = 26 a. Use a repeated-measures ANOVA with a 0.05 to determine whether there are any significant mean differences among the three conditions. (32 points) Calculate η2 to determine the effect size. (3 points) b.

A. Use a repeated measures ANOVA with a=0.05 to determine whether there are any significant mean differences among the three conditions.

B. Calculate n2 to determine the effect size.

Answer 1

Date: 04/05/2019 Answer To test the hypothesis is that there any significant differences among the three treatments means at 5% level of significance. a) The null and alternative hypothesis is, H, : at least one μ,is unequal. The F-test statistics is, First, find sum, average and sum of square for all groups for finding sum of squares between groups The total sample size is 18; sample size is 6 and 3 levels. The sum, average and sum of square (Sum sq) table for Treatment 1 to Treatment 3 is shown below.

Before After Follow-up 4 ty 4 4 3 25 4.17 119 14 26 4.33 134 Sum Average2.33 Sum sq Grand Total Sum sq is 297 Sum is 65 The sum of squares between group s is,

142 252 262 (65)2 18 = 249.50-234.72 -14.78 The sum of square totalis. (GT) SST -Sumof squareGT - 2 = 297-(45) = 297-234 .72 = 62.28 12 The sumof square error is, SSE = SST-SSBG 62.28-14.78 = 47.50

The degree of freedom for between groups is, =3-1 The degree of freedom error is =18-3 -15 The degree of freedom total is, d, = N-1 - 18-1 17 The mean square between groups is,

BG dfsG 14.78 7.39 The mean square error is, MSE 47.50 15 3.17 The F-test statistics is, MSBG obt MSE 7.39 3.17 - 2.33

The F-test statistics is 2.33 The complete ANOVA table is, Source of VariationSS MS 7.39 2.33 14.78 2 Treatment Error Total 47.50 153.17 62.28 17 The F-critical value is, From the F distribution at 5% level of significance table, show degree of freedom treatment 2 and degree of freedom error 17 as result the F-critical value is 3.68. The F-critical value is 3.59 Decision

The F-test statistics is 2.33 is less than F-critical value which is 3.68, the null hypothesis is not rejected at 5% level of significance. There is insufficient evidence that there are significant differences among the three treatments means. The result is not statistically significant. b) The eta-squared is, 2 SS Treatment Total 14.78 62.28 = 0.24 The eta-squared is [0.24