1. **State the problem:** We have three groups (Section 1, Section 2, Section 3) with sample data and want to perform a one-way ANOVA test at significance level $\alpha=0.05$ to determine if there is a significant difference among the group means. 2. **Recall the ANOVA formulas:** - Total number of observations: $N=27$ (9 per group) - Number of groups: $k=3$ - Calculate group means $\bar{x}_i$ and overall mean $\bar{x}$ - Sum of Squares Between (SSB): $$SSB=\sum_{i=1}^k n_i (\bar{x}_i - \bar{x})^2$$ - Sum of Squares Within (SSW): $$SSW=\sum_{i=1}^k \sum_{j=1}^{n_i} (x_{ij} - \bar{x}_i)^2$$ - Degrees of freedom: $df_{between}=k-1=2$, $df_{within}=N-k=24$ - Mean Squares: $$MSB=\frac{SSB}{df_{between}},\quad MSW=\frac{SSW}{df_{within}}$$ - F-ratio: $$F=\frac{MSB}{MSW}$$ 3. **Calculate group means:** - Section 1 mean: $\bar{x}_1=\frac{56.8+47.9+42.9+51+49+59.6+61.6+65.3+55.9}{9}=54.67$ - Section 2 mean: $\bar{x}_2=\frac{60.9+38.9+38+22.2+43.8+24.8+46.2+49+64.5}{9}=43.7$ - Section 3 mean: $\bar{x}_3=\frac{48.2+61.9+55.3+59.4+54.5+46.2+55.9+51.4+44.5}{9}=53.7$ 4. **Calculate overall mean:** $$\bar{x}=\frac{54.67\times9 + 43.7\times9 + 53.7\times9}{27} = \frac{492 + 393.3 + 483.3}{27} = \frac{1368.6}{27} = 50.69$$ 5. **Calculate SSB:** $$SSB = 9[(54.67 - 50.69)^2 + (43.7 - 50.69)^2 + (53.7 - 50.69)^2]$$ $$=9[(3.98)^2 + (-6.99)^2 + (3.01)^2] = 9[15.84 + 48.86 + 9.06] = 9 \times 73.76 = 663.84$$ 6. **Calculate SSW:** Sum of squared deviations within each group: - Section 1: $\sum (x_{1j} - 54.67)^2 = 334.89$ - Section 2: $\sum (x_{2j} - 43.7)^2 = 1021.01$ - Section 3: $\sum (x_{3j} - 53.7)^2 = 234.01$ $$SSW = 334.89 + 1021.01 + 234.01 = 1589.91$$ 7. **Calculate mean squares:** $$MSB = \frac{663.84}{2} = 331.92$$ $$MSW = \frac{1589.91}{24} = 66.25$$ 8. **Calculate F-ratio:** $$F = \frac{331.92}{66.25} = 5.009$$ 9. **Find p-value:** Using F-distribution with $df_1=2$ and $df_2=24$, $p \approx 0.0141$ (rounded to four decimals). 10. **Conclusion:** Since $p=0.0141 < \alpha=0.05$, we reject the null hypothesis. **Final answers:** $$F = 5.009$$ $$p = 0.0141$$ **Conclusion:** Reject the null hypothesis: at least one of the group means is different.