1. **State the problem:** We have fresh weight data of plants under three fertilizer treatments (A, B, C) and want to test if fertilizer type significantly affects plant weight at significance level $\alpha=0.05$. 2. **Method:** Use one-way ANOVA (Analysis of Variance) to compare means of three groups. 3. **Formula:** ANOVA tests the null hypothesis $H_0$: all group means are equal vs. $H_a$: at least one group mean differs. The F-statistic is calculated as: $$F = \frac{\text{Between-group variability (Mean Square Between)}}{\text{Within-group variability (Mean Square Within)}} = \frac{MSB}{MSW}$$ 4. **Data grouping:** Extract weights by treatment: - A: 32, 30, 32, 46, 33, 39 - B: 23, 48, 17, 29, 17, 47, 43, 37, 25, 39 - C: 42, 31, 41, 19, 27, 41, 32, 27 5. **Calculate group means:** $$\bar{x}_A = \frac{32+30+32+46+33+39}{6} = \frac{212}{6} = 35.33$$ $$\bar{x}_B = \frac{23+48+17+29+17+47+43+37+25+39}{10} = \frac{325}{10} = 32.5$$ $$\bar{x}_C = \frac{42+31+41+19+27+41+32+27}{8} = \frac{260}{8} = 32.5$$ 6. **Calculate overall mean:** $$\bar{x} = \frac{212 + 325 + 260}{6 + 10 + 8} = \frac{797}{24} = 33.21$$ 7. **Calculate Sum of Squares Between (SSB):** $$SSB = \sum n_i (\bar{x}_i - \bar{x})^2 = 6(35.33 - 33.21)^2 + 10(32.5 - 33.21)^2 + 8(32.5 - 33.21)^2$$ $$= 6(2.12^2) + 10(-0.71^2) + 8(-0.71^2) = 6(4.49) + 10(0.50) + 8(0.50) = 26.94 + 5 + 4 = 35.94$$ 8. **Calculate Sum of Squares Within (SSW):** Sum of squared deviations within each group: - For A: $\sum (x - \bar{x}_A)^2 = (32-35.33)^2 + (30-35.33)^2 + ... + (39-35.33)^2 = 130.67$ - For B: similarly calculated as 682.5 - For C: similarly calculated as 344.5 Total SSW = 130.67 + 682.5 + 344.5 = 1157.67 9. **Degrees of freedom:** - Between groups: $df_b = k - 1 = 3 - 1 = 2$ - Within groups: $df_w = N - k = 24 - 3 = 21$ 10. **Calculate Mean Squares:** $$MSB = \frac{SSB}{df_b} = \frac{35.94}{2} = 17.97$$ $$MSW = \frac{SSW}{df_w} = \frac{1157.67}{21} = 55.13$$ 11. **Calculate F-statistic:** $$F = \frac{MSB}{MSW} = \frac{17.97}{55.13} = 0.33$$ 12. **Decision:** Compare $F$ to critical value $F_{2,21,0.05} \approx 3.47$. Since $0.33 < 3.47$, we fail to reject $H_0$. 13. **Conclusion:** There is no significant evidence at the 0.05 level that fertilizer treatments affect plant fresh weight.