23-Ind-B1 Reliability and Maintainability · May 2015
Question 10 of 10: 2³ Factorial Design — Design Matrix, Effects, Sums of Squares, and ANOVA
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2015 — 98-Ind-B1 Applied Probability & Statistics. Three-hour, closed-book exam; one of two permitted calculators (Sharp or Casio), one 8.5″×11.0″ aid sheet (both sides), statistical tables supplied. Format: three sections — Section A: do 2 of 4 questions (10 marks); Section B: do 2 of 3 (20 marks); Section C: do 1 of 3 (20 marks) — a 5-question, 50-mark paper as printed. All ten questions across the three sections are solved below for completeness.
Reference texts: Montgomery & Runger, Applied Statistics and Probability for Engineers (7th ed., Wiley) — joint distributions (ch. 5), binomial/geometric probability (ch. 3), normal distribution and sampling distributions (ch. 4–7), point/interval estimation (ch. 8–9), hypothesis testing incl. sample-size design (ch. 9–10), simple linear regression and ANOVA (ch. 11–13), Bartlett's and Tukey's tests, single-degree-of-freedom contrasts and 23 factorial designs (ch. 13–14). Montgomery, Peck & Vining, Introduction to Linear Regression Analysis (6th ed., Wiley) — multiple regression by matrices, confidence/prediction intervals (ch. 2–3).
Question 10 (Section C.3): 2³ Factorial Design — Design Matrix, Effects, Sums of Squares, and ANOVA (20 marks)
Find. (a) AB/AC/BC/ABC sign columns; (b)–(c) contrasts, effects, and SS for A, AB, ABC; (d) full ANOVA table; (e) reduced regression model.
Approach. Interaction columns are the sign-products of the main-effect columns; each contrast is $\sum(\pm)(\text{run total})$, giving effect $=$ contrast$/(n_{rep}2^{k-1})$ and $SS=$ contrast$^2/(n_{rep}2^k)$; the ANOVA error term comes from the within-run replication (8 runs $\times$ 1 df each $=8$ error df).
(a) Design matrix for the interactions (signs multiply).
Run
1
2
3
4
5
6
7
8
AB
1
−1
−1
1
1
−1
−1
1
AC
1
−1
1
−1
−1
1
−1
1
BC
1
1
−1
−1
−1
−1
1
1
ABC
−1
1
1
−1
1
−1
−1
1
Total ($y_1+y_2$)
74
54
70
81
84
82
99
91
(b) Contrast and mean effect for A, AB, ABC. Contrast $=\sum(\text{sign})(\text{run total})$; effect $=$ contrast$/(2\cdot4)=$ contrast$/8$. $$\text{Contrast}_A=-74+54-70+81-84+82-99+91=-19,\qquad \text{Effect}_A=-19/8=\boxed{-2.375}$$ $$\text{Contrast}_{AB}=25\ (\text{matches given}),\qquad \text{Effect}_{AB}=25/8=\boxed{3.125}$$ $$\text{Contrast}_{ABC}=-37\ (\text{matches given}),\qquad \text{Effect}_{ABC}=-37/8=\boxed{-4.625}$$
(c) Sums of squares, $SS=\text{contrast}^2/(n_{rep}2^k)=\text{contrast}^2/16$. $$SS_A=(-19)^2/16=\boxed{22.5625},\qquad SS_{AB}=25^2/16=\boxed{39.0625}\ (\text{matches given}),\qquad SS_{ABC}=(-37)^2/16=\boxed{85.5625}$$
(d) Full ANOVA table. Filling in the exam's starred entries with the values computed the same way as (b)/(c) above (contrast$_A=-19$, contrast$_C=77$, contrast$_{AC}=-1$; $SS_A=22.5625$, $SS_B=138.0625$, $SS_{ABC}=85.5625$), and computing the error term from the 8 runs' own replication: $SSE=\sum_{run}(y_1-y_2)^2/2=507.5$ with $df_E=8$, $MSE=63.4375$. $F_{0.05,1,8}=5.318$.
ANOVA table
Source
A
B
C
AB
AC
BC
ABC
Error
Total
SS
22.56
138.06
370.56
39.06
0.06
0.06
85.56
507.50
1163.44
df
1
1
1
1
1
1
1
8
15
F
0.36
2.18
5.84
0.62
0.001
0.001
1.35
—
—
Sig. (α=0.05)?
no
no
yes
no
no
no
no
—
—
(e) Reduced regression model. In coded $\pm1$ units, each regression coefficient is half its effect: $b_0=\bar y=39.6875$, $b_A=-1.1875$, $b_B=2.9375$, $b_C=4.8125$, $b_{AB}=1.5625$, $b_{AC}=-0.0625$, $b_{BC}=0.0625$, $b_{ABC}=-2.3125$. Only $C$ clears $F_{0.05,1,8}=5.318$ (part d), so the significant reduced model is $$\boxed{\hat y = 39.6875 + 4.8125\,x_C}$$ — cutting angle (C) is the only factor (main effect or interaction) with a statistically detectable effect on tool life at $\alpha=0.05$; speed (A), tool geometry (B), and all interactions are not distinguishable from noise at this replication level.
Check — the exam's own printed "SST = 22.5625" is mislabeled Summing the squared deviations of all 16 raw observations from the grand mean gives the true $SST=1163.4375$ — exactly equal to the sum of all 7 effect sums of squares (655.9375) plus the replication error SS (507.5), confirming this is correct. The value $22.5625$ printed under "SST" in the exam's own given-Sums-of-Squares table is, numerically, exactly $SS_A$ as independently computed here — almost certainly a column-labelling slip in the original exam (an "SST" heading printed one column over from an "SS(A)" entry), since $22.5625$ is far too small to be a total that must exceed each individual component (e.g. $SS_C=370.5625$ alone). The ANOVA table above uses the correctly-derived $SST=1163.4375$.
