1. **Stating the problem:** We have typing speeds (words per minute) from four word processors (A, B, C, D), with multiple measurements each. We want to test if the observed differences in the sample means are statistically significant or if they can be attributed to random chance at the $0.01$ significance level. 2. **Step 1: Organize the data.** - Processor A: 68, 75, 65, 77, 61, 72, 78, 71 - Processor B: 68, 71, 74, 65, 70, 67, 62, 70 - Processor C: 75, 72, 82, 73, 78, 72 - Processor D: 62, 59, 71, 68, 65 3. **Step 2: State hypotheses.** - Null hypothesis $H_0$: The means of the four groups are equal. - Alternative hypothesis $H_a$: At least one mean is different. 4. **Step 3: Calculate sample means.** - $\bar{x}_A = \frac{68+75+65+77+61+72+78+71}{8} = \frac{567}{8} = 70.875$ - $\bar{x}_B = \frac{68+71+74+65+70+67+62+70}{8} = \frac{547}{8} = 68.375$ - $\bar{x}_C = \frac{75+72+82+73+78+72}{6} = \frac{452}{6} = 75.333$ - $\bar{x}_D = \frac{62+59+71+68+65}{5} = \frac{325}{5} = 65$ 5. **Step 4: Calculate overall mean.** - Total observations $N = 8+8+6+5 = 27$ - Sum of all observations $= 567 + 547 + 452 + 325 = 1891$ - Overall mean $\bar{x} = \frac{1891}{27} \approx 70.037$ 6. **Step 5: Calculate Sum of Squares Between (SSB).** $$ \text{SSB} = \sum n_i (\bar{x}_i - \bar{x})^2 = 8(70.875 - 70.037)^2 + 8(68.375 - 70.037)^2 + 6(75.333 - 70.037)^2 + 5(65 - 70.037)^2 $$ Calculating each term: - $8 (0.838)^2 = 8 \times 0.702 = 5.616$ - $8 (-1.662)^2 = 8 \times 2.763 = 22.10$ - $6 (5.296)^2 = 6 \times 28.05 = 168.3$ - $5 (-5.037)^2 = 5 \times 25.38 = 126.90$ Sum: $$ \text{SSB} = 5.616 + 22.10 + 168.3 + 126.90 = 322.92 $$ 7. **Step 6: Calculate Sum of Squares Within (SSW).** Calculate variance within each group: - Processor A variance sums: $\sum (x_i - \bar{x}_A)^2 = (68-70.875)^2 + (75-70.875)^2 + ... + (71-70.875)^2 = 166.875$ - Processor B variance sums: 114.875 - Processor C variance sums: 107.33 - Processor D variance sums: 86.80 Sum of all: $$ \text{SSW} = 166.875 + 114.875 + 107.33 + 86.80 = 475.88 $$ 8. **Step 7: Degrees of freedom.** - Between groups: $df_b = k-1 = 4-1 = 3$ - Within groups: $df_w = N - k = 27 - 4 = 23$ 9. **Step 8: Calculate Mean Squares.** $$ \text{MSB} = \frac{\text{SSB}}{df_b} = \frac{322.92}{3} = 107.64 $$ $$ \text{MSW} = \frac{\text{SSW}}{df_w} = \frac{475.88}{23} = 20.69 $$ 10. **Step 9: Calculate F statistic.** $$ F = \frac{MSB}{MSW} = \frac{107.64}{20.69} = 5.20 $$ 11. **Step 10: Determine critical value and compare.** For $\alpha=0.01$ with $df_b=3$, $df_w=23$, the critical $F$ value from F-distribution tables is approximately $4.35$. Since $5.20 > 4.35$, reject the null hypothesis. 12. **Conclusion:** At the 0.01 significance level, there is sufficient evidence to conclude that the differences among the four sample means are statistically significant and cannot be attributed to chance alone.