1. The first question asks why blocking is done to maximize differences between blocks. Blocking is used to control for variability among experimental units by grouping similar units (e.g., patients) into blocks. By maximizing the differences between blocks, we isolate variability due to blocks and reduce error variance within blocks, increasing the precision of treatment comparisons. 2. The second question is whether dosage levels significantly affect blood pressure when accounting for individual patient differences at α = 0.05. We use a One-Way Repeated Measures ANOVA because the same patients receive all three dosages, and patient serves as the blocking factor. 3. Define hypotheses: - Null hypothesis $H_0$: mean systolic BP is the same across Low, Medium, and High dosages. - Alternative hypothesis $H_a$: at least one dosage differs in mean systolic BP. 4. Organize the data: \begin{array}{c|ccc} \text{Patient} & \text{Low} & \text{Medium} & \text{High} \\ \hline 1 & 130 & 125 & 120 \\ 2 & 140 & 138 & 132 \\ 3 & 135 & 130 & 125 \\ \end{array} 5. Calculate the means: - Mean Low: $\frac{130 + 140 + 135}{3} = 135$ - Mean Medium: $\frac{125 + 138 + 130}{3} \approx 131$ - Mean High: $\frac{120 + 132 + 125}{3} \approx 125.67$ - Grand mean: $\frac{130+125+120+140+138+132+135+130+125}{9} = \frac{1175}{9} \approx 130.56$ 6. Compute sums of squares: - Total Sum of Squares ($SS_{Total}$): sum of squared differences of every observation from the grand mean - Between Treatments Sum of Squares ($SS_{Treatment}$): involves squared differences between treatment means and grand mean - Between Subjects Sum of Squares ($SS_{Subjects}$): accounts for differences between patients' average measurements and grand mean - Error Sum of Squares ($SS_{Error}$): $SS_{Total} - SS_{Treatment} - SS_{Subjects}$ Calculations: - Patient means: - Patient 1: $\frac{130 + 125 + 120}{3} = 125$ - Patient 2: $\frac{140 + 138 + 132}{3} = 136.67$ - Patient 3: $\frac{135 + 130 + 125}{3} = 130$ - $SS_{Total} = \sum (x_{ij} - \text{grand mean})^2$ = (130 - 130.56)^2 + (125 - 130.56)^2 + (120 - 130.56)^2 + (140 - 130.56)^2 + (138 - 130.56)^2 + (132 - 130.56)^2 + (135 - 130.56)^2 + (130 - 130.56)^2 + (125 - 130.56)^2 $= 0.31 + 30.94 + 111.14 + 88.61 + 55.54 + 2.07 + 19.64 + 0.31 + 30.94 = 339.5$ - $SS_{Treatment} = n \sum (\text{treatment mean} - \text{grand mean})^2$ where $n=3$ patients = $3[(135 - 130.56)^2 + (131 - 130.56)^2 + (125.67 - 130.56)^2]$ = $3[19.72 + 0.19 + 24.04] = 3 * 43.95 = 131.85$ - $SS_{Subjects} = k \sum (\text{subject mean} - \text{grand mean})^2$ where $k=3$ treatments = $3[(125 - 130.56)^2 + (136.67 - 130.56)^2 + (130 - 130.56)^2]$ = $3[30.94 + 37.31 + 0.31] = 3 * 68.56 = 205.68$ - $SS_{Error} = SS_{Total} - SS_{Treatment} - SS_{Subjects} = 339.5 - 131.85 - 205.68 = 2.0$ 7. Degrees of freedom: - $df_{Treatment} = k - 1 = 2$ - $df_{Subjects} = n - 1 = 2$ - $df_{Error} = (k - 1)(n - 1) = 4$ 8. Mean Squares: - $MS_{Treatment} = \frac{SS_{Treatment}}{df_{Treatment}} = \frac{131.85}{2} = 65.92$ - $MS_{Error} = \frac{SS_{Error}}{df_{Error}} = \frac{2.0}{4} = 0.5$ 9. Calculate F-ratio: $$F = \frac{MS_{Treatment}}{MS_{Error}} = \frac{65.92}{0.5} = 131.84$$ 10. Look up critical F-value for $df_1=2$, $df_2=4$ at $\alpha=0.05$: approximately 6.94. Since $131.84 >> 6.94$, we reject $H_0$. **Conclusion:** There is a statistically significant effect of dosage level on systolic blood pressure when accounting for individual differences.