1. **State the problem:** We want to use the Scheffe test to compare pairs of group means from an ANOVA with three groups: A (No sleep apnea), B (Untreated sleep apnea), and C (Treated sleep apnea). 2. **Given data:** - Sample size per group: $n=11$ - Means: $\bar{X}_A=0.42$, $\bar{X}_B=0.56$, $\bar{X}_C=0.30$ - Mean square within treatments: $MS_{within}=0.0330$ - Degrees of freedom within treatments: $df_{within}=30$ - Number of groups: $k=3$ 3. **Formula for Scheffe test:** The Scheffe test statistic for comparing groups $i$ and $j$ is $$ F_{i,j} = \frac{(\bar{X}_i - \bar{X}_j)^2}{MS_{within} \times \frac{2}{n}} $$ This follows from the sum of squares between groups for the pair: $$ SS_{between\,ij} = \frac{n}{2} (\bar{X}_i - \bar{X}_j)^2 $$ 4. **Calculate $SS_{between\,AB}$:** $$ SS_{between\,AB} = \frac{11}{2} (0.42 - 0.56)^2 = 5.5 \times ( -0.14 )^2 = 5.5 \times 0.0196 = 0.1078 $$ 5. **Calculate $F_{A\,vs\,B}$:** $$ F_{A\,vs\,B} = \frac{(0.42 - 0.56)^2}{0.0330 \times \frac{2}{11}} = \frac{0.0196}{0.0330 \times 0.1818} = \frac{0.0196}{0.006} = 3.27 $$ 6. **Decision for A vs B:** Critical value for Scheffe test at $\alpha=0.05$ with $k=3$ groups and $df_{within}=30$ is $$ F_{critical} = (k-1) \times F_{\alpha, k-1, df_{within}} = 2 \times 3.316 = 6.632 $$ Since $F_{A\,vs\,B} = 3.27 < 6.632$, we **fail to reject** the null hypothesis and conclude the means for groups A and B do not differ significantly. 7. **Calculate $SS_{between\,AC}$:** $$ SS_{between\,AC} = \frac{11}{2} (0.42 - 0.30)^2 = 5.5 \times 0.12^2 = 5.5 \times 0.0144 = 0.0792 $$ 8. **Calculate $F_{A\,vs\,C}$:** $$ F_{A\,vs\,C} = \frac{(0.42 - 0.30)^2}{0.0330 \times \frac{2}{11}} = \frac{0.0144}{0.006} = 2.4 $$ 9. **Decision for A vs C:** Since $F_{A\,vs\,C} = 2.4 < 6.632$, we **fail to reject** the null hypothesis and conclude the means for groups A and C do not differ significantly. 10. **Summary:** - $SS_{between\,AB} = 0.1078$, $F_{A\,vs\,B} = 3.27$, no significant difference. - $SS_{between\,AC} = 0.0792$, $F_{A\,vs\,C} = 2.4$, no significant difference. The psychologist cannot conclude that the population means differ between these pairs at $\alpha=0.05$.