1. **State the problem:** We need to compute the F ratio for three groups of scores. 2. **Recall the formula:** The F ratio in ANOVA is given by $$F = \frac{\text{Mean Square Between Groups (MSB)}}{\text{Mean Square Within Groups (MSW)}}$$ where - MSB = $\frac{\text{Sum of Squares Between Groups (SSB)}}{\text{df between}}$ - MSW = $\frac{\text{Sum of Squares Within Groups (SSW)}}{\text{df within}}$ 3. **Calculate group means:** Group 1: $\frac{10+10+9+8+11}{5} = \frac{48}{5} = 9.6$ Group 2: $\frac{9+8+7+7+6}{5} = \frac{37}{5} = 7.4$ Group 3: $\frac{8+10+6+8+8}{5} = \frac{40}{5} = 8.0$ 4. **Calculate overall mean:** $$\text{Grand Mean} = \frac{48 + 37 + 40}{15} = \frac{125}{15} = 8.3333$$ 5. **Calculate SSB:** $$SSB = 5 \times ((9.6 - 8.3333)^2 + (7.4 - 8.3333)^2 + (8.0 - 8.3333)^2)$$ $$= 5 \times (1.6044 + 0.8711 + 0.1111) = 5 \times 2.5866 = 12.933$$ 6. **Calculate SSW:** Sum of squared deviations within each group: Group 1: $$(10-9.6)^2 + (10-9.6)^2 + (9-9.6)^2 + (8-9.6)^2 + (11-9.6)^2 = 0.16 + 0.16 + 0.36 + 2.56 + 1.96 = 5.2$$ Group 2: $$(9-7.4)^2 + (8-7.4)^2 + (7-7.4)^2 + (7-7.4)^2 + (6-7.4)^2 = 2.56 + 0.36 + 0.16 + 0.16 + 1.96 = 5.2$$ Group 3: $$(8-8)^2 + (10-8)^2 + (6-8)^2 + (8-8)^2 + (8-8)^2 = 0 + 4 + 4 + 0 + 0 = 8$$ Total SSW = 5.2 + 5.2 + 8 = 18.4 7. **Degrees of freedom:** Between groups: $df_b = k - 1 = 3 - 1 = 2$ Within groups: $df_w = N - k = 15 - 3 = 12$ 8. **Calculate MSB and MSW:** $$MSB = \frac{12.933}{2} = 6.4665$$ $$MSW = \frac{18.4}{12} = 1.5333$$ 9. **Calculate F ratio:** $$F = \frac{6.4665}{1.5333} = 4.22$$ **Final answer:** The F ratio is approximately **4.22**. --- **Excel format for the calculations:** | Group | Scores | Mean | Sum of Squares Within | |-------|------------------|------|-----------------------| | 1 | 10,10,9,8,11 | 9.6 | 5.2 | | 2 | 9,8,7,7,6 | 7.4 | 5.2 | | 3 | 8,10,6,8,8 | 8.0 | 8.0 | | Overall Mean | 8.3333 | | SSB | 12.933 | | SSW | 18.4 | | df Between | 2 | | df Within | 12 | | MSB | 6.4665 | | MSW | 1.5333 | | F Ratio | 4.22 |