1. **Problem Statement:** A researcher measures hormone X levels in three groups: acute schizophrenics (n1=5), chronic schizophrenics (n2=6), and normal controls (n3=7). We want to calculate the F ratio to test if there is a significant difference in mean hormone levels among the groups using one-way ANOVA. 2. **Hypotheses:** - Null hypothesis $H_0$: $\mu_1 = \mu_2 = \mu_3$ (no difference in means) - Alternative hypothesis $H_a$: At least one mean differs 3. **ANOVA formula:** - Total number of observations $N_t = n_1 + n_2 + n_3 = 18$ - Number of groups $k=3$ - Degrees of freedom between groups $df_{bet} = k-1 = 2$ - Degrees of freedom within groups $df_w = N_t - k = 15$ 4. **Calculate group means:** - Acute mean $\bar{x}_1 = \frac{86 + 23 + 47 + 51 + 63}{5} = 54$ - Chronic mean $\bar{x}_2 = \frac{75 + 42 + 35 + 56 + 70 + 40}{6} = 53.67$ (approx) - Normal mean $\bar{x}_3 = \frac{49 + 28 + 68 + 52 + 63 + 82 + 36}{7} = 54.29$ (approx) 5. **Calculate grand mean:** $$\bar{x}_G = \frac{n_1 \bar{x}_1 + n_2 \bar{x}_2 + n_3 \bar{x}_3}{N_t} = \frac{5(54) + 6(53.67) + 7(54.29)}{18} = 54$$ 6. **Calculate sum of squares between groups (SSB):** $$SSB = \sum n_i (\bar{x}_i - \bar{x}_G)^2 = 5(54 - 54)^2 + 6(53.67 - 54)^2 + 7(54.29 - 54)^2 = 0 + 0.66 + 0.58 = 1.24$$ 7. **Calculate sum of squares within groups (SSW):** Calculate variance for each group: - Acute variance $s_1^2 = 531$ - Chronic variance $s_2^2 = 254$ - Normal variance $s_3^2 = 349.33$ Then, $$SSW = (n_1 - 1)s_1^2 + (n_2 - 1)s_2^2 + (n_3 - 1)s_3^2 = 4(531) + 5(254) + 6(349.33) = 2124 + 1270 + 2096 = 5490$$ 8. **Calculate mean squares:** - Mean square between groups (MSB): $$MSB = \frac{SSB}{df_{bet}} = \frac{1.24}{2} = 0.62$$ - Mean square within groups (MSW): $$MSW = \frac{SSW}{df_w} = \frac{5490}{15} = 366$$ 9. **Calculate F ratio:** $$F = \frac{MSB}{MSW} = \frac{0.62}{366} \approx 0.0017$$ 10. **Decision:** Critical F value at $\alpha=0.05$ with $df_{bet}=2$ and $df_w=15$ is approximately 3.68. Since $F_{comp} = 0.0017 < 3.68$, we fail to reject $H_0$. 11. **Interpretation:** There is no significant difference in mean hormone X levels among the three groups. 12. **How inspecting means could save effort:** The group means are very close (54, 53.67, 54.29), indicating little difference. This suggests the F ratio will be very small, so extensive calculations could be avoided by first checking means.