1. The problem involves analyzing an ANOVA table to understand the variance between and within groups. 2. ANOVA (Analysis of Variance) tests whether there are statistically significant differences between the means of three or more groups. 3. The key formulas are: - Mean Square Between Groups: $MS_{between} = \frac{SS_{between}}{df_{between}}$ - Mean Square Within Groups: $MS_{within} = \frac{SS_{within}}{df_{within}}$ - F-statistic: $F = \frac{MS_{between}}{MS_{within}}$ 4. From the table: - $SS_{between} = 56.66$, $df_{between} = 4$, so $MS_{between} = \frac{56.66}{4} = 14.165$ - $SS_{within} = 85.05$, $df_{within} = 95$, so $MS_{within} = \frac{85.05}{95} = 0.895263$ - The calculated F-statistic is $15.822$ 5. The P-value is $5.87302 \times 10^{-10}$, which is much smaller than common significance levels (e.g., 0.05), indicating strong evidence against the null hypothesis. 6. The critical F-value at the given degrees of freedom is $2.4675$. 7. Since the calculated F ($15.822$) is greater than the critical F ($2.4675$), we reject the null hypothesis and conclude that there are significant differences between group means. Final answer: There is strong statistical evidence to conclude that at least one group mean is different from the others based on the ANOVA test.