1. **Problem Statement:** A teacher wants to test if the means of three teaching strategies (A, B, and C) are equal based on student scores. 2. **Hypotheses:** - Null Hypothesis ($H_0$): $\mu_A = \mu_B = \mu_C$ - Alternative Hypothesis ($H_a$): At least one mean differs. - Significance level: $\alpha = 0.05$ 3. **ANOVA Test Formula:** The F-statistic is calculated as: $$F = \frac{MS_{Between}}{MS_{Within}}$$ where $MS_{Between} = \frac{SS_{Between}}{df_{Between}}$ and $MS_{Within} = \frac{SS_{Within}}{df_{Within}}$. 4. **Given Data:** - $SS_{Between} = 178.47$, $df_{Between} = 2$ - $SS_{Within} = 409.40$, $df_{Within} = 27$ - $MS_{Between} = 89.235$, $MS_{Within} = 15.163$ - Calculated $F = 5.88$ - p-value = 0.0075 5. **Decision Rule:** Compare calculated $F$ to critical $F_{crit}$ (approx. 3.35 for $df_1=2$, $df_2=27$ at $\alpha=0.05$). 6. **Evaluation:** Since $5.88 > 3.35$ and $p = 0.0075 < 0.05$, reject $H_0$. 7. **Effect Size:** Eta-squared $= \frac{SS_{Between}}{SS_{Total}} = \frac{178.47}{587.87} \approx 0.304$ (large effect). 8. **Post Hoc Analysis:** - Significant differences between A vs B and A vs C. - No significant difference between B vs C. 9. **Conclusion:** The null hypothesis is rejected. Teaching strategies B and C are significantly more effective than A. Your answer is correct and well supported by the ANOVA results.