1. **Stating the problem:** We have four word processors (A, B, C, and D) with observations of words typed per minute. We want to test if the differences among the means of these four samples are statistically significant at the 0.01 significance level. 2. **Organizing the data:** Word Processor A: 68, 75, 65, 77, 61, 72, 78, 71 Word Processor B: 68, 71, 74, 65, 70, 67, 62, 70 Word Processor C: 75, 72, 82, 73, 78, 72 Word Processor D: 62, 59, 71, 68, 65 3. **Calculate sample sizes $(n_i)$:** $ n_A = 8, n_B = 8, n_C = 6, n_D = 5 $ 4. **Calculate sample means $(\bar{x}_i)$:** $ \bar{x}_A = \frac{68 + 75 + 65 + 77 + 61 + 72 + 78 + 71}{8} = \frac{567}{8} = 70.875 $ $ \bar{x}_B = \frac{68 + 71 + 74 + 65 + 70 + 67 + 62 + 70}{8} = \frac{547}{8} = 68.375 $ $ \bar{x}_C = \frac{75 + 72 + 82 + 73 + 78 + 72}{6} = \frac{452}{6} = 75.333 $ $ \bar{x}_D = \frac{62 + 59 + 71 + 68 + 65}{5} = \frac{325}{5} = 65.0 $ 5. **Calculate the overall mean $(\bar{x})$:** $ \bar{x} = \frac{567 + 547 + 452 + 325}{8 + 8 + 6 + 5} = \frac{1891}{27} \approx 70.037 $ 6. **Calculate the Sum of Squares Between Groups (SSB):** $ SSB = \sum n_i (\bar{x}_i - \bar{x})^2 $ $ = 8(70.875 - 70.037)^2 + 8(68.375 - 70.037)^2 + 6(75.333 - 70.037)^2 + 5(65-70.037)^2 $ $ = 8(0.838^2) + 8(-1.662^2) + 6(5.296^2) + 5(-5.037^2) $ $ = 8(0.702) + 8(2.763) + 6(28.047) + 5(25.373) $ $ = 5.619 + 22.103 + 168.282 + 126.865 = 322.869 $ 7. **Calculate the Sum of Squares Within Groups (SSW):** Calculate variance within each group and sum the squared deviations: For brevity, compute sum of squared differences: A: $\sum (x - \bar{x}_A)^2 = (68-70.875)^2 + ... + (71-70.875)^2 = 127.875$ B: similarly, $\sum (x - \bar{x}_B)^2 = 105.875$ C: $\sum (x - \bar{x}_C)^2 = 63.333$ D: $\sum (x - \bar{x}_D)^2 = 61.0$ $ SSW = 127.875 + 105.875 + 63.333 + 61.0 = 358.083 $ 8. **Degrees of freedom:** Between groups: $df_{between} = k - 1 = 4 - 1 = 3$ Within groups: $df_{within} = N - k = 27 - 4 = 23$ 9. **Calculate Mean Squares:** $ MSB = \frac{SSB}{df_{between}} = \frac{322.869}{3} = 107.623 $ $ MSW = \frac{SSW}{df_{within}} = \frac{358.083}{23} = 15.57 $ 10. **Calculate the F-statistic:** $ F = \frac{MSB}{MSW} = \frac{107.623}{15.57} \approx 6.91 $ 11. **Decision rule:** For $df_1=3$, $df_2=23$, and $\alpha=0.01$, the critical F value $\approx 4.35$ (from F-table). Since $6.91 > 4.35$, we reject the null hypothesis. 12. **Conclusion:** There is sufficient evidence at the 0.01 significance level to conclude that the differences among the word processors' mean typing speeds are statistically significant and not likely due to chance.