1. **State the problem:** We want to determine if there is a significant difference in the final grades given by the 4 teachers (sections A, B, C, D) at the 0.05 significance level. 2. **Data:** The grades for 6 students per section are: Section A: 86, 80, 85, 91, 87, 92 Section B: 75, 84, 83, 86, 79, 83 Section C: 89, 79, 86, 92, 78, 82 Section D: 78, 82, 84, 81, 84, 87 3. **Method:** Use one-way ANOVA to test the null hypothesis $H_0$: all section means are equal vs. $H_a$: at least one section mean differs. 4. **Calculate group means:** $$\bar{x}_A = \frac{86+80+85+91+87+92}{6} = \frac{521}{6} = 86.83$$ $$\bar{x}_B = \frac{75+84+83+86+79+83}{6} = \frac{490}{6} = 81.67$$ $$\bar{x}_C = \frac{89+79+86+92+78+82}{6} = \frac{506}{6} = 84.33$$ $$\bar{x}_D = \frac{78+82+84+81+84+87}{6} = \frac{496}{6} = 82.67$$ 5. **Calculate overall mean:** $$\bar{x} = \frac{521+490+506+496}{24} = \frac{2013}{24} = 83.88$$ 6. **Calculate Sum of Squares Between (SSB):** $$SSB = 6[(86.83-83.88)^2 + (81.67-83.88)^2 + (84.33-83.88)^2 + (82.67-83.88)^2]$$ $$= 6[(2.95)^2 + (-2.21)^2 + (0.45)^2 + (-1.21)^2]$$ $$= 6[8.70 + 4.88 + 0.20 + 1.46] = 6 \times 15.24 = 91.44$$ 7. **Calculate Sum of Squares Within (SSW):** Calculate variance within each group and sum: For A: $$\sum (x_i - \bar{x}_A)^2 = (86-86.83)^2 + (80-86.83)^2 + ... + (92-86.83)^2 = 70.83$$ For B: $$= 44.67$$ For C: $$= 90.67$$ For D: $$= 38.67$$ Total SSW = 70.83 + 44.67 + 90.67 + 38.67 = 244.84 8. **Degrees of freedom:** Between groups: $df_1 = k-1 = 4-1=3$ Within groups: $df_2 = N-k = 24-4=20$ 9. **Calculate Mean Squares:** $$MSB = \frac{SSB}{df_1} = \frac{91.44}{3} = 30.48$$ $$MSW = \frac{SSW}{df_2} = \frac{244.84}{20} = 12.24$$ 10. **Calculate F-statistic:** $$F = \frac{MSB}{MSW} = \frac{30.48}{12.24} = 2.49$$ 11. **Decision:** At $\alpha=0.05$, critical F-value for $df_1=3$ and $df_2=20$ is approximately 3.10. Since $2.49 < 3.10$, we fail to reject $H_0$. 12. **Conclusion:** There is no significant difference in the final grades given by the 4 teachers at the 0.05 significance level.