1. **State the problem:** We want to test if there is a significant difference in the average life of four brands of fluorescent bulbs using one-way ANOVA at the 0.05 significance level. 2. **Data:** Brand A: 12, 13, 14, 11, 15 Brand B: 11, 10, 13, 15, 14 Brand C: 12, 11, 15, 10, 16 Brand D: 12, 15, 15, 12, 11 3. **Calculate group means:** $$\bar{X}_A = \frac{12+13+14+11+15}{5} = \frac{65}{5} = 13$$ $$\bar{X}_B = \frac{11+10+13+15+14}{5} = \frac{63}{5} = 12.6$$ $$\bar{X}_C = \frac{12+11+15+10+16}{5} = \frac{64}{5} = 12.8$$ $$\bar{X}_D = \frac{12+15+15+12+11}{5} = \frac{65}{5} = 13$$ 4. **Calculate overall mean:** $$\bar{X} = \frac{65 + 63 + 64 + 65}{20} = \frac{257}{20} = 12.85$$ 5. **Calculate Sum of Squares Between Groups (SSB):** $$SSB = 5[(13 - 12.85)^2 + (12.6 - 12.85)^2 + (12.8 - 12.85)^2 + (13 - 12.85)^2]$$ $$= 5[(0.15)^2 + (-0.25)^2 + (-0.05)^2 + (0.15)^2]$$ $$= 5[0.0225 + 0.0625 + 0.0025 + 0.0225] = 5 \times 0.11 = 0.55$$ 6. **Calculate Sum of Squares Within Groups (SSW):** Calculate variance within each group and sum: For Brand A: $$\sum (X_i - \bar{X}_A)^2 = (12-13)^2 + (13-13)^2 + (14-13)^2 + (11-13)^2 + (15-13)^2 = 1 + 0 + 1 + 4 + 4 = 10$$ For Brand B: $$= (11-12.6)^2 + (10-12.6)^2 + (13-12.6)^2 + (15-12.6)^2 + (14-12.6)^2 = 2.56 + 6.76 + 0.16 + 5.76 + 1.96 = 17.2$$ For Brand C: $$= (12-12.8)^2 + (11-12.8)^2 + (15-12.8)^2 + (10-12.8)^2 + (16-12.8)^2 = 0.64 + 3.24 + 4.84 + 7.84 + 10.24 = 26.8$$ For Brand D: $$= (12-13)^2 + (15-13)^2 + (15-13)^2 + (12-13)^2 + (11-13)^2 = 1 + 4 + 4 + 1 + 4 = 14$$ Total SSW: $$10 + 17.2 + 26.8 + 14 = 68$$ 7. **Degrees of freedom:** Between groups: $$df_{between} = k - 1 = 4 - 1 = 3$$ Within groups: $$df_{within} = N - k = 20 - 4 = 16$$ 8. **Calculate Mean Squares:** $$MSB = \frac{SSB}{df_{between}} = \frac{0.55}{3} \approx 0.1833$$ $$MSW = \frac{SSW}{df_{within}} = \frac{68}{16} = 4.25$$ 9. **Calculate F-statistic:** $$F = \frac{MSB}{MSW} = \frac{0.1833}{4.25} \approx 0.0431$$ 10. **Decision:** At $$\alpha = 0.05$$, the critical value $$F_{critical}$$ for $$df_1=3$$ and $$df_2=16$$ is approximately 3.24. Since $$F = 0.0431 < 3.24$$, we fail to reject the null hypothesis. 11. **Conclusion:** There is no significant evidence at the 0.05 level to conclude that the average life of the four brands of fluorescent bulbs differs.