1. **State the problem:** We want to test if there is a significant difference among the means of typing speeds (words per minute) of four different word processors (A, B, C, D). 2. **Data and Hypotheses:** We have four samples corresponding to each word processor. - Null hypothesis $H_0$: The means are equal, i.e., $\mu_A = \mu_B = \mu_C = \mu_D$. - Alternative hypothesis $H_a$: At least one mean is different. We test at significance level $\alpha = 0.01$. 3. **Organize 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 4. **Calculate sample means ($\bar{x}$):** $\bar{x}_A = \frac{68+75+65+77+61+72+78+71}{8} = \frac{567}{8} = 70.875$ $\bar{x}_B = \frac{68+71+74+65+70+67+62+70}{8} = \frac{547}{8} = 68.375$ $\bar{x}_C = \frac{75+72+82+73+78+72}{6} = \frac{452}{6} = 75.333$ $\bar{x}_D = \frac{62+59+71+68+65}{5} = \frac{325}{5} = 65$ 5. **Calculate the overall mean ($\bar{x}_{\text{total}}$):** Total sum = 567 + 547 + 452 + 325 = 1891 Total count = 8 + 8 + 6 + 5 = 27 $\bar{x}_{\text{total}} = \frac{1891}{27} \approx 70.04$ 6. **Calculate Sum of Squares Between (SSB):** $$SSB = \sum n_i (\bar{x}_i - \bar{x}_{\text{total}})^2$$ where $n_i$ and $\bar{x}_i$ are the sample size and mean of group $i$. $$SSB = 8(70.875 - 70.04)^2 + 8(68.375 - 70.04)^2 + 6(75.333 - 70.04)^2 + 5(65 - 70.04)^2$$ Calculate each term: $8(0.835)^2 = 8 \times 0.697 = 5.576$ $8(-1.665)^2 = 8 \times 2.772 = 22.176$ $6(5.293)^2 = 6 \times 28.016 = 168.096$ $5(-5.04)^2 = 5 \times 25.402 = 127.01$ Sum: $$SSB = 5.576 + 22.176 + 168.096 + 127.01 = 322.858$$ 7. **Calculate Sum of Squares Within (SSW):** Calculate variance within each group: For each group, sum the squared deviations from the group mean: Processor A: $(68 - 70.875)^2 + (75 - 70.875)^2 + \dots + (71 - 70.875)^2$ $= 8.266 + 17.016 + 34.516 + 37.516 + 96.016 + 1.266 + 51.516 + 0.016 = 246.13$ Processor B: $(-0.375)^2 + (2.625)^2 + (5.625)^2 + (-3.375)^2 + (1.625)^2 + (-1.375)^2 + (-6.375)^2 + (1.625)^2$ $= 0.141 + 6.891 + 31.641 + 11.391 + 2.641 + 1.891 + 40.641 + 2.641 = 97.879$ Processor C: $(-0.333)^2 + (-3.333)^2 + (6.667)^2 + (-2.333)^2 + (2.667)^2 + (-3.333)^2$ $= 0.111 + 11.111 + 44.445 + 5.444 + 7.111 + 11.111 = 79.333$ Processor D: $(-3)^2 + (-6)^2 + (6)^2 + (3)^2 + (0)^2 = 9 + 36 + 36 + 9 + 0 = 90$ Sum: $$SSW = 246.13 + 97.879 + 79.333 + 90 = 513.342$$ 8. **Degrees of freedom:** Between groups: $df_b = k - 1 = 4 - 1 = 3$ Within groups: $df_w = N - k = 27 - 4 = 23$ 9. **Calculate Mean Squares:** $$MSB = \frac{SSB}{df_b} = \frac{322.858}{3} = 107.619$$ $$MSW = \frac{SSW}{df_w} = \frac{513.342}{23} = 22.322$$ 10. **Calculate F-statistic:** $$F = \frac{MSB}{MSW} = \frac{107.619}{22.322} = 4.82$$ 11. **Compare with critical value:** For $df_b=3$ and $df_w=23$ at $\alpha = 0.01$, critical value $F_{crit} \approx 4.35$ (from F-distribution tables). Since $4.82 > 4.35$, reject the null hypothesis. 12. **Conclusion:** There is sufficient evidence at the 0.01 level of significance to conclude that the mean typing speeds differ among the four word processors and the differences are not due to chance.