5. Use the following table to complete the ANOVA table below for a within subjects one-way ANOVA with 6 participants who were measured on the same variable (x) following 4 different treatments: | X1 | X1^2 | X2 | X2^2 | X3 | X3^2 | X4 | X4^2 | ?P | |---|---|---|---|---|---|---|---|---| | 3 | | 4 | | 2 | | 6 | | | | 2 | | 3 | | 3 | | 7 | | | | 3 | | 3 | | 2 | | 4 | | | | 4 | | 3 | | 3 | | 5 | | | | 3 | | 4 | | 2 | | 5 | | | | 3 | | 4 | | 2 | | 4 | | | | ?x1 = | | ?x2 = | | ?x3 = | | ?x4 = | | ?xT = | | | ?x1^2 = | | ?x2^2 = | | ?x3^2 = | | ?x4^2 = | ?xT^2 = | ANOVA Table | | SS | df | MS | F | |---|---|---|---|---| | Between Groups | | | | | | Between Persons | | | | | | Within Groups | | | | -- | | Total | | | -- | -- | Do we retain or reject the null hypothesis? Substantively interpret this decision. If significant, calculate the effect size and interpret. What would be our next step if we wanted to determine which pairs of means were statistically significantly different?
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This involves calculating the grand mean (the mean of all scores regardless of group), then finding the sum of the squared deviations of each score from the grand mean. - First, calculate the grand mean. Add all the scores together and divide by the total number Show moreβ¦
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The ANOVA summary table for an experiment with groups, with five values in each group, is shown below. Complete parts (a) through (d) below. Source | Degrees of Freedom | Sum of Squares | Mean Square (Variance) | F Among groups | c - 1 = 5 | SSA = 200 | MSA = 40 | F_STAT = 4.00 Within groups | n - c = 24 | SSW = 240 | MSW = 10 | Total | n - 1 = 29 | SST = 440 | | At the 0.05 level of significance, state the decision rule for testing the null hypothesis that all six groups have equal population means. (Based on the statement, finish the following 4 questions) I. Determine the hypotheses. Choose the correct answer below. A. H0: ΞΌ1 = ΞΌ2 = ... = ΞΌ5; H1: ΞΌ1 β ΞΌ2 β ... β ΞΌ5 B. H0: ΞΌ1 = ΞΌ2 = ... = ΞΌ5; H1: Not all the means are equal. C. H0: ΞΌ1 = ΞΌ2 = ... = ΞΌ6; H1: Not all the means are equal. D. H0: ΞΌ1 = ΞΌ2 = ... = ΞΌ6; H1: ΞΌ1 β ΞΌ2 β ... β ΞΌ6 II. Compute the test statistic (Keep 2 decimal places) III. Determine the critical value. (Keep 2 decimal places) IV. (T/F) There is sufficient evidence to conclude that there is a difference in the population means of the groups. (Put "T" for True; "F" for False)
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A one-way between subjectsβ ANOVA was conducted to compare the weight between three professional football teams. Each group had seventeen players for a total of 51 participants. The null hypothesis indicates there is not any significant difference between the three population means. The research hypothesis states there is a significant difference between at least two population means. Results from One-Way ANOVA indicated that the means of the three groups were not significantly different, F(2,48) = 0.819 , p-value = 0.477 SSE b/w Groups is 431.412, df = 2, SSE within Groups is 12648, df = 48, F Statistics of 0.818, p = 0.44, is not significant. Is this correctly reported for a one way ANOVA report in APA?
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