1. Problem (Q1): Test at level $\alpha=0.05$ whether the three Solutions have different mean responses across Days. 2. Data and grouping for Q1: group 1 = $[13,22,18,39]$, group 2 = $[16,24,17,44]$, group 3 = $[5,4,1,22]$. 3. Compute group sums and means for Q1. $$n_1=4,\ S_1=92,\ \bar x_1=23.0$$ $$n_2=4,\ S_2=101,\ \bar x_2=25.25$$ $$n_3=4,\ S_3=32,\ \bar x_3=8.0$$ 4. Grand total and grand mean for Q1. $$N=12,\ T=225,\ \bar x = \dfrac{225}{12}=18.75$$ 5. Between-group sum of squares (SSB) for Q1. $$SSB=\sum n_i(\bar x_i-\bar x)^2=4(23-18.75)^2+4(25.25-18.75)^2+4(8-18.75)^2=703.5$$ 6. Total sum of squares (SST) for Q1 computed from deviations. $$SST=\sum (x-\bar x)^2=1862.25$$ 7. Within-group sum of squares (SSW) for Q1. $$SSW=SST-SSB=1862.25-703.5=1158.75$$ 8. ANOVA table entries and F for Q1. $$df_{B}=2,\ df_{W}=9$$ $$MSB=\dfrac{703.5}{2}=351.75,\ MSW=\dfrac{1158.75}{9}=128.75$$ $$F=\dfrac{MSB}{MSW}=\dfrac{351.75}{128.75}\approx2.733$$ 9. Decision for Q1: The critical value $F_{0.05,2,9}\approx4.256$ and $F<4.256$, so do not reject $H_0$. Conclusion Q1: At $\alpha=0.05$ there is insufficient evidence to conclude the three Solutions differ in mean response. 1. Problem (Q2): Latin-square design ($t=5$) test at $\alpha=0.05$ whether treatment (letter) effects are significant, and assess row and column effects. 2. Numeric $5\times5$ data matrix for Q2 (rows = batches, cols = days): Row1 $[8,7,1,7,3]$, Row2 $[11,2,7,3,8]$, Row3 $[4,9,10,1,5]$, Row4 $[6,8,6,6,10]$, Row5 $[4,2,3,8,8]$. 3. Extract treatment totals (by label positions given) for Q2. $$S_A=42,\ S_B=28,\ S_C=44,\ S_D=17,\ S_E=16$$ 4. Grand total and grand mean for Q2. $$T=147,\ N=25,\ \bar x=\dfrac{147}{25}=5.88$$ 5. Total sum of squares for Q2 (sum of squared deviations). $$SST=206.64$$ 6. Row, column and treatment sums of squares for Q2 (Latin-square formulas). $$SS_{rows}=\frac{1}{t}\sum R_i^2-\frac{T^2}{N}=15.44$$ $$SS_{cols}=\frac{1}{t}\sum C_j^2-\frac{T^2}{N}=12.24$$ $$SS_{trt}=\frac{1}{t}\sum S_k^2-\frac{T^2}{N}=141.44$$ 7. Error sum of squares for Q2. $$SS_{E}=SST-(SS_{rows}+SS_{cols}+SS_{trt})=206.64-169.12=37.52$$ 8. Degrees of freedom and mean squares for Q2. $$df_{rows}=4,\ df_{cols}=4,\ df_{trt}=4,\ df_{E}=12$$ $$MS_{trt}=\dfrac{141.44}{4}=35.36,\ MS_E=\dfrac{37.52}{12}=3.1267$$ 9. F statistic and decision for Q2. $$F_{trt}=\dfrac{35.36}{3.1267}\approx11.31$$ Critical $F_{0.05,4,12}\approx3.26$, and $F_{trt}>3.26$, so reject $H_0$ for treatments. Conclusion Q2: At $\alpha=0.05$ treatment (letter) effects are significant; row and column effects have $F_{rows}\approx1.234$ and $F_{cols}\approx0.979$ and are not significant. 1. Problem (Q3): Two-way ANOVA without replication on a $4\times4$ table to test row (assembly order) and column (operator) effects at $\alpha=0.05$. 2. Data matrix for Q3: rows = [10,14,7,8], [7,18,11,8], [5,10,11,9], [10,10,12,14]. 3. Row and column sums for Q3. $$R=(39,44,35,46),\ C=(32,52,41,39),\ T=164,\ N=16,\ \bar x=10.25$$ 4. Total sum of squares for Q3. $$SST=153.0$$ 5. Row and column sums of squares and error for Q3. $$SS_{rows}=\frac{\sum R_i^2}{4}-\frac{T^2}{N}=18.5$$ $$SS_{cols}=\frac{\sum C_j^2}{4}-\frac{T^2}{N}=51.5$$ $$SS_E=SST-SS_{rows}-SS_{cols}=153.0-18.5-51.5=83.0$$ 6. Degrees of freedom and mean squares for Q3. $$df_{rows}=3,\ df_{cols}=3,\ df_E=9$$ $$MS_{rows}=\dfrac{18.5}{3}=6.1667,\ MS_{cols}=\dfrac{51.5}{3}=17.1667,\ MS_E=\dfrac{83}{9}=9.2222$$ 7. F statistics and decisions for Q3. $$F_{rows}=\dfrac{6.1667}{9.2222}\approx0.669,\ F_{cols}=\dfrac{17.1667}{9.2222}\approx1.861$$ Critical $F_{0.05,3,9}\approx3.29$, and both $F$ values are less than the critical value, so do not reject the nulls. Conclusion Q3: At $\alpha=0.05$ there is no significant evidence of differences among assembly orders or operators. 1. Problem (Q4): Graeco-Latin $4\times4$ square test at $\alpha=0.05$ to assess row, column, Latin (uppercase) and Greek effects. 2. Numeric response matrix for Q4: rows = [11,10,14,8], [8,12,10,12], [9,11,7,15], [9,8,18,6]. 3. Totals and grand mean for Q4. $$T=168,\ N=16,\ \bar x=10.5$$ 4. Treatment totals separated by uppercase (Latin) and Greek factor for Q4. Uppercase sums: $$S_A=35,\ S_B=31,\ S_C=56,\ S_D=46$$ Greek sums: $$S_{\alpha}=45,\ S_{\beta}=38,\ S_{\gamma}=44,\ S_{\delta}=41$$ 5. Total sum of squares for Q4. $$SST=150$$ 6. Row and column sums of squares and treatment sums of squares for Q4. $$SS_{rows}=0.5,\ SS_{cols}=19,\ SS_{Latin}=95.5,\ SS_{Greek}=7.5$$ 7. Error sum of squares for Q4 using $df_E=(t-1)(t-3)$ with $t=4$. $$SS_E=SST-(SS_{rows}+SS_{cols}+SS_{Latin}+SS_{Greek})=150-122.5=27.5$$ 8. Degrees of freedom and mean squares for Q4. $$df_{rows}=3,\ df_{cols}=3,\ df_{Latin}=3,\ df_{Greek}=3,\ df_E=3$$ $$MS_{Latin}=\dfrac{95.5}{3}=31.8333,\ MS_E=\dfrac{27.5}{3}=9.1667$$ 9. F statistics and decisions for Q4. $$F_{Latin}=\dfrac{31.8333}{9.1667}\approx3.472,\ F_{Greek}=\dfrac{7.5}{9.1667}\approx0.818$$ Critical $F_{0.05,3,3}\approx9.28$, and both $F$ values are less than the critical value, so do not reject the nulls. Conclusion Q4: At $\alpha=0.05$ there is no significant evidence of effects of rows, columns, Latin-treatment, or Greek-treatment in this Graeco-Latin square. Overall final conclusions at $\alpha=0.05$: - Q1: No significant difference among the three Solutions. - Q2: Treatment (letter) effect is significant; rows and columns are not. - Q3: Neither assembly order nor operator shows a significant effect. - Q4: No significant effects detected for any factor in the Graeco-Latin square.