1. The problem involves analyzing data from a 4x4 Latin square design with 4 subjects, 4 periods, and 4 treatments (A-no treatment, B-drug X, C-drug Y, D-both drugs) to perform an analysis of variance (ANOVA) to determine the significance of treatment effects on growth rate. 2. First, we compile the observed data into a matrix: $$\begin{array}{c|cccc} \text{Subject} & k=1 & k=2 & k=3 & k=4 \\ \hline 1 & 0.02(A) & 0.15(B) & 0.45(D) & 0.18(C) \\ 2 & 0.27(B) & 0.24(C) & -0.01(A) & 0.58(D) \\ 3 & 0.11(C) & 0.35(D) & 0.14(B) & -0.03(A) \\ 4 & 0.48(D) & 0.04(A) & 0.18(C) & 0.22(B) \end{array}$$ 3. Calculate the grand total $G$ and grand mean $\bar{Y}_{..}$: $$G = \sum_{i=1}^4 \sum_{k=1}^4 Y_{ik} = 0.02 + 0.15 + 0.45 + 0.18 + 0.27 + 0.24 - 0.01 + 0.58 + 0.11 + 0.35 + 0.14 - 0.03 + 0.48 + 0.04 + 0.18 + 0.22 = 3.33$$ $$\bar{Y}_{..} = \frac{G}{16} = \frac{3.33}{16} = 0.2081$$ 4. Calculate sums of squares total (SST): $$SST = \sum (Y_{ik} - \bar{Y}_{..})^2$$ Calculate each squared difference and sum: For example,\n$(0.02-0.2081)^2 = 0.03536$, and similarly for all others. Summing gives $$SST = 0.6777$$ 5. Calculate row (subjects) sums and their means: $$\text{Subject 1 mean} = \frac{0.02+0.15+0.45+0.18}{4} = 0.2$$ $$\text{Subject 2 mean} = \frac{0.27+0.24-0.01+0.58}{4}=0.27$$ $$\text{Subject 3 mean} = \frac{0.11+0.35+0.14-0.03}{4} = 0.1425$$ $$\text{Subject 4 mean} = \frac{0.48+0.04+0.18+0.22}{4} = 0.23$$ Subjects sum of squares (SSR): $$SSR = 4\sum(\bar{Y}_{i.} - \bar{Y}_{..})^2$$ $$=4[(0.2-0.2081)^2 + (0.27-0.2081)^2 + (0.1425-0.2081)^2 + (0.23-0.2081)^2] = 0.0172$$ 6. Calculate column (period) means and sum of squares similarly: Period means: $$k=1: \frac{0.02+0.27+0.11+0.48}{4}=0.22$$ $$k=2: \frac{0.15+0.24+0.35+0.04}{4}=0.195$$ $$k=3: \frac{0.45-0.01+0.14+0.18}{4}=0.19$$ $$k=4: \frac{0.18+0.58-0.03+0.22}{4}=0.2375$$ Column sum of squares (SSC): $$SSC=4\sum(\bar{Y}_{.k} - \bar{Y}_{..})^2=0.0179$$ 7. Calculate treatment means: Treatment means are averages of the respective treatment's values from data: A: (0.02 -0.01 -0.03 +0.04)/4 = 0.005 B: (0.15 +0.27 +0.14 +0.22)/4 = 0.195 C: (0.18 +0.24 +0.11 +0.18)/4 = 0.1775 D: (0.45 +0.58 +0.35 +0.48)/4 = 0.465 Treatment sum of squares (SSTr): $$SSTr=4\sum(\bar{Y}_{.j} - \bar{Y}_{..})^2=4[(0.005-0.2081)^2 + (0.195-0.2081)^2 + (0.1775-0.2081)^2 + (0.465-0.2081)^2]=0.3462$$ 8. Calculate sum of squares error (SSE): $$SSE= SST - SSR - SSC - SSTr = 0.6777 - 0.0172 - 0.0179 - 0.3462 = 0.2964$$ 9. Degrees of freedom: - Total: 16 -1 = 15 - Rows (Subjects): 4 -1 = 3 - Columns (Periods): 4 -1 = 3 - Treatments: 4 -1 = 3 - Error: 15 - (3+3+3) = 6 10. Mean squares: $$MSR = SSR/3 = 0.0172/3 = 0.0057$$ $$MSC = SSC/3 = 0.0179/3 = 0.0060$$ $$MSTr = SSTr/3 = 0.3462/3 = 0.1154$$ $$MSE = SSE/6 = 0.2964/6 = 0.0494$$ 11. Calculate F-statistics: $$F_{treatment} = MSTr/MSE = 0.1154/0.0494 = 2.34$$ $$F_{row} = MSR/MSE = 0.0057/0.0494 = 0.115$$ $$F_{column} = MSC/MSE = 0.0060/0.0494 = 0.121$$ 12. Compare $F$ values to critical $F_{3,6}(0.05) \approx 4.76$: - $F_{treatment}=2.34 < 4.76$: Not significant - $F_{row}=0.115 <4.76$: Not significant - $F_{column}=0.121 <4.76$: Not significant 13. Interpretation: None of the factors (treatments, subjects, periods) show statistically significant effect on growth rate at 5% significance level in this Latin square design. Final ANOVA Table: $$\begin{array}{lcccc} \text{Source} & \text{SS} & \text{df} & \text{MS} & F \\ \hline \text{Subjects} & 0.0172 & 3 & 0.0057 & 0.115 \\ \text{Periods} & 0.0179 & 3 & 0.0060 & 0.121 \\ \text{Treatments} & 0.3462 & 3 & 0.1154 & 2.34 \\ \text{Error} & 0.2964 & 6 & 0.0494 & - \\ \text{Total} & 0.6777 & 15 & - & - \end{array}$$