Scheffe Test 854Db3
1. **State the problem:** We want to use the Scheffe test to compare pairs of group means from three groups (A: no sleep apnea, B: untreated sleep apnea, C: treated sleep apnea) after an ANOVA showed significant differences.
2. **Recall formulas:**
- Total sample size per group: $n=11$
- Treatment totals: $T_i = n \times \text{mean}_i$
- Sum of squares between two groups $i$ and $j$:
$$SS_{between\,ij} = \frac{(T_i - T_j)^2}{n_i + n_j}$$
- Scheffe test statistic for groups $i$ and $j$:
$$F_{i\,vs\,j} = \frac{SS_{between\,ij}}{MS_{within}} \times \frac{k-1}{1}$$
where $MS_{within} = 0.0330$ and $k=3$ groups.
3. **Calculate treatment totals:**
- $T_A = 11 \times 0.42 = 4.62$
- $T_B = 11 \times 0.56 = 6.16$
- $T_C = 11 \times 0.30 = 3.30$
4. **Calculate $SS_{between}$ and $F$ for pairs:**
- For A vs B:
$$SS_{between\,AB} = \frac{(4.62 - 6.16)^2}{11 + 11} = \frac{(-1.54)^2}{22} = \frac{2.3716}{22} = 0.1078$$
$$F_{A\,vs\,B} = \frac{0.1078}{0.0330} \times (3-1) = 3.266 \times 2 = 6.532$$
- For A vs C:
$$SS_{between\,AC} = \frac{(4.62 - 3.30)^2}{22} = \frac{1.32^2}{22} = \frac{1.7424}{22} = 0.0792$$
$$F_{A\,vs\,C} = \frac{0.0792}{0.0330} \times 2 = 2.4 \times 2 = 4.8$$
- For B vs C:
$$SS_{between\,BC} = \frac{(6.16 - 3.30)^2}{22} = \frac{2.86^2}{22} = \frac{8.1796}{22} = 0.3718$$
$$F_{B\,vs\,C} = \frac{0.3718}{0.0330} \times 2 = 11.27 \times 2 = 22.54$$
5. **Interpret results at $\alpha=0.05$:**
- Critical $F$ for Scheffe test is the same as ANOVA critical value $3.316$.
- Since $F_{A\,vs\,B} = 6.532 > 3.316$, conclude means differ between no sleep apnea and untreated sleep apnea.
- Since $F_{A\,vs\,C} = 4.8 > 3.316$, conclude means differ between no sleep apnea and treated sleep apnea.
- Since $F_{B\,vs\,C} = 22.54 > 3.316$, conclude means differ between untreated and treated sleep apnea.
**Final answers:**
- $SS_{between\,AB} = 0.1078$, $F_{A\,vs\,B} = 6.532$
- $SS_{between\,AC} = 0.0792$, $F_{A\,vs\,C} = 4.8$
- $SS_{between\,BC} = 0.3718$, $F_{B\,vs\,C} = 22.54$
All pairs show significant differences at $\alpha=0.05$.