Question: make this shorter, even though the ANOVA table is included in the appendix of this assignment: The ANOVA test was conducted to analyze the MAAP

make this shorter, even though the ANOVA table is included in the appendix of this assignment: The ANOVA test was conducted to analyze the MAAP ELA scores across students in the third through sixth grades (see Table 1). The null hypothesis for this analysis was there was no significant difference in the MAAP ELA scores as measured across different grade levels. The results indicated varying mean scores across the third, fourth, fifth, and sixth grades. Specifically, the mean scores were 73.4 (n = 10, sum = 734), 68 (n = 8, sum = 544), 55.12 (n = 17, sum = 937), and 52.41 (n = 22, sum = 1153) for the third, fourth, fifth, and sixth grades, respectively. The degrees of freedom (Df) were calculated between and within groups. The between-groups Df was 3, derived by subtracting one from the total number of groups (4-1). Conversely, the within-groups Df was 53, calculated by subtracting the number of groups from the total observations (57-4). The total Df, the sum between and within groups Df, was 56. The F statistic was 5.89, which measures the variance ratio between the groups to the variance within the groups. The P-value was 0.002, less than the significance level of 0.05. This indicates a statistically significant difference in the mean scores of the four grades. The F critical value was 2.78, which is the value the F statistic must exceed for the results to be statistically significant. We reject the null hypothesis since the F-value is greater than the critical F-value, and

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