23-Ind-B1 Reliability and Maintainability · December 2017

Question 10 of 10: $2^2$ Factorial Experiment — Design Matrix, Contrasts, Effects, ANOVA, Regression

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 10 (Section C.3): $2^2$ Factorial Experiment — Design Matrix, Contrasts, Effects, ANOVA, Regression (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Two factors $A,B$ at two levels each, 3 replicates/treatment ($N=12$):

Find. (a) coded design matrix; (b) contrasts and effects for $A$, $B$, $AB$; (c) ANOVA significance; (d) regression model and coefficient significance.

Approach. Code each factor $\pm1$, form the interaction column as the sign product, compute each effect's contrast from treatment totals, convert to single-df sums of squares ($SS=\text{contrast}^2/(n\,2^k)$), and confirm that the least-squares regression on the coded columns reproduces the same sums of squares and significance conclusions.

1. (a) Design matrix. With $A,B\in\{-1,+1\}$ and $AB=A\times B$: $$\begin{array}{c|ccc}\text{Treatment} & A & B & AB\\\hline (1) & -1 & -1 & +1\\ a & +1 & -1 & -1\\ b & -1 & +1 & -1\\ ab & +1 & +1 & +1\end{array}$$
2. Treatment totals. $(1)=41$, $a=51$, $b=57$, $ab=67$ (each the sum of 3 replicates).
3. (b) Contrasts and mean effects. $$\text{Contrast}_A=a+ab-b-(1)=51+67-57-41=\boxed{20},\qquad \bar A=\frac{\text{Contrast}_A}{n\cdot2^{k-1}}=\frac{20}{3(2)}=\boxed{3.333}$$ $$\text{Contrast}_B=b+ab-a-(1)=57+67-51-41=\boxed{32},\qquad \bar B=\frac{32}{6}=\boxed{5.333}$$ $$\text{Contrast}_{AB}=ab+(1)-a-b=67+41-51-57=\boxed{0},\qquad \overline{AB}=\frac{0}{6}=\boxed{0}$$
4. (c) ANOVA. $SS=\text{contrast}^2/(n\,2^k)=\text{contrast}^2/12$: $$SS_A=\frac{20^2}{12}=\boxed{33.33},\qquad SS_B=\frac{32^2}{12}=\boxed{85.33},\qquad SS_{AB}=\frac{0^2}{12}=\boxed{0}$$ Using $SST=266$ (given) and $SS_{Treatments}=SS_A+SS_B+SS_{AB}=118.67$: $$SSE=SST-SS_{Treatments}=266-118.67=\boxed{147.33}$$ at $df_E=N-4=8$, so $MSE=147.33/8=18.42$; $F_{0.05,1,8}=5.318$. $$F_A=\frac{33.33}{18.42}=1.81\ (\text{n.s.}),\qquad F_B=\frac{85.33}{18.42}=\boxed{4.63}\ (\text{n.s.},\ p=0.064),\qquad F_{AB}=0\ (\text{n.s.})$$ None of the three effects clears $F_{crit}=5.318$ at $\alpha=0.05$, though $B$ comes closest ($p=0.064$) and would be significant at $\alpha=0.10$.
5. (d) Regression model. Fitting $\hat y=b_0+b_A A+b_B B+b_{AB}AB$ by least squares on the coded $\pm1$ columns reproduces the effects directly (coefficient $=$ half the effect): $$\hat y=18.00+1.667A+2.667B+0\cdot AB$$ The residual sum of squares from this regression equals $SSE=147.33$ exactly, confirming consistency with (c). Standard errors are identical for all three coefficients by the design's orthogonality, $se=\sqrt{MSE/N}=\sqrt{18.42/12}=1.239$: $$t_A=\frac{1.667}{1.239}=1.35\ (p=0.215),\qquad t_B=\frac{2.667}{1.239}=\boxed{2.15}\ (p=0.064),\qquad t_{AB}=0$$ Each $t_j^2$ equals its corresponding single-df $F$ from (c), as expected. $$\boxed{\text{No coefficient is significant at }\alpha=0.05\text{; }B\text{ is the strongest candidate but falls short}}$$