23-Ind-B1 Reliability and Maintainability · December 2016

Question 11 of 11: Machine-Tool Life — $2^3$ Factorial Design

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

Notes on this paper

National Exams — December 2016 — 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 (30 marks); Section B: do 2 of 3 (30 marks); Section C: do 2 of 4 (40 marks) — a 6-question, 100-mark paper as printed. All eleven 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 and covariance (ch. 5), point/interval estimation (ch. 8), hypothesis testing incl. two-sample and goodness-of-fit tests (ch. 9–10), simple linear regression (ch. 11), design and analysis of single-factor and factorial experiments (ch. 13–14). Montgomery, Peck & Vining, Introduction to Linear Regression Analysis (6th ed., Wiley) — multiple regression by matrices, confidence/prediction intervals (ch. 2–3). Montgomery, Design and Analysis of Experiments (9th ed., Wiley) — two-way factorial ANOVA and $2^k$ designs (ch. 5, 6–7).

Question 11 (Section C.4): Machine-Tool Life — $2^3$ Factorial Design (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. 8 design points, 2 replicates each ($N=16$):

Run	A	B	C	Rep 1	Rep 2
1	−1	−1	−1	32	31
2	1	−1	−1	15	43
3	−1	1	−1	35	34
4	1	1	−1	35	47
5	−1	−1	1	44	45
6	1	−1	1	40	37
7	−1	1	1	60	50
8	1	1	1	39	41

Given Contrasts: A$=-34$, B$=*$, C$=84$, AB$=**$, AC$=-50$, BC$=-6$, ABC$=***$. Given SS: SS(A)$=*$, SS(B)$=182.25$, SS(C)$=**$, SS(AB)$=0$, SS(AC)$=***$, SS(BC)$=2.25$, SS(ABC)$=81$, SST$=1457$.

Find. (a) interaction columns; (b)/(c) contrasts, mean effects and SS for A/AB/ABC; (d) full ANOVA table; (e) significant terms in a fitted regression model.

Approach. Each interaction column is the sign-product of its component columns; a contrast is $\sum(\text{sign})\times(\text{run total})$; $SS(\text{effect})=\text{contrast}^2/(n\,2^k)$ with $n=2$ reps, $k=3$ factors; the two-replicate design gives a pure-error $SSE$ with $df=N-2^k=8$.

(a) Interaction columns (sign products). AB, AC, BC, ABC are computed row-by-row as products of the A, B, C signs, e.g. run 1: AB$=(-1)(-1)=+1$, AC$=(-1)(-1)=+1$, BC$=(-1)(-1)=+1$, ABC$=(-1)(-1)(-1)=-1$; run 2: AB$=-1$, AC$=-1$, BC$=+1$, ABC$=+1$; and so on for all 8 runs following the same rule.
(b) Contrasts and mean effects for A, AB, ABC. Using run totals ($=$Rep1$+$Rep2) and summing with each column's signs: $$\text{Contrast}_A=-34,\qquad \text{Contrast}_{AB}=0,\qquad \text{Contrast}_{ABC}=-36$$ Mean effect $=\text{contrast}/(n\,2^{k-1})=\text{contrast}/8$: $$\bar A=-4.25,\qquad \overline{AB}=0,\qquad \overline{ABC}=-4.50$$
(c) Sums of squares for A, AB, ABC. $SS=\text{contrast}^2/(n\,2^k)=\text{contrast}^2/16$: $$SS(A)=(-34)^2/16=\boxed{72.25}\qquad SS(AB)=0^2/16=\boxed{0}\qquad SS(ABC)=(-36)^2/16=\boxed{81}$$
Filling the remaining blanks. Since $SS(BC)=2.25=(-6)^2/16$ confirms the $SS=\text{contrast}^2/16$ relation, the same relation gives the rest directly from the raw run totals: $$B=\boxed{54}\ (\Rightarrow SS(B)=54^2/16=182.25\ \checkmark),\qquad AB=\boxed{0},\qquad ABC=\boxed{-36}$$ $$SS(A)=\boxed{72.25},\qquad SS(C)=84^2/16=\boxed{441},\qquad SS(AC)=(-50)^2/16=\boxed{156.25}$$
(d) ANOVA table. Sum of the seven effect SS $=72.25+182.25+441+0+156.25+2.25+81=935$. With pure-error $SSE=SST-935=1457-935=\boxed{522}$ at $df_{Err}=N-2^k=16-8=8$, $MSE=522/8=65.25$. $F_{0.05,1,8}=5.318$.

Effect A B C AB AC BC ABC

$SS$ 72.25 182.25 441.0 0 156.25 2.25 81.0

$F$ 1.107 2.793 6.759 0 2.395 0.034 1.241

Only $C$ (cutting angle) exceeds $F_{crit}=5.318$: $$\boxed{\text{only the C main effect is statistically significant}}$$
(e) Regression model and significant terms. Regression coefficients $=\text{effect}/2=\text{contrast}/(n\,2^k)$: $$\hat y=39.25-2.125A+3.375B+5.25C+0\cdot AB-3.125AC-0.375BC-2.25ABC$$ Since each single-df effect's $F$-statistic equals its coefficient's $t^2$, the same conclusion carries over: only the $C$ (cutting angle) term is significant; cutting speed, tool geometry, and every interaction are statistically indistinguishable from noise at $\alpha=0.05$.

Effect	A	B	C	AB	AC	BC	ABC
$SS$	72.25	182.25	441.0	0	156.25	2.25	81.0
$F$	1.107	2.793	6.759	0	2.395	0.034	1.241

Summary
Quantity	Value
$B$, $AB$, $ABC$ contrasts	54, 0, −36
$SS(A)$, $SS(C)$, $SS(AC)$	72.25, 441.0, 156.25
$SSE$, $MSE$ ($df=8$)	522.0, 65.25
Significant effect(s)	C only ($F=6.76\gt5.32$)

Topic: $2^k$ factorial designs: contrasts, effects, sums of squares, and single-degree-of-freedom $F$-tests.

Key relations: Contrast $=\sum(\text{sign})\times(\text{run total})$; effect $=\text{contrast}/(n\,2^{k-1})$; $SS=\text{contrast}^2/(n\,2^k)$; with replicated runs, pure-error $SSE=SST-\sum SS_{effects}$ at $df=N-2^k$.

Why this works: Because the design is orthogonal (every column has equal $+1/-1$ counts and every pairwise column-product also balances), each effect's sum of squares is estimated completely independently of every other effect — that is what lets a single ANOVA table cleanly test all seven effects at once.

Common pitfall: Using $\text{contrast}^2/(n\,2^{k-1})$ (the mean-effect denominator) instead of $/(n\,2^k)$ for the sum of squares — the two formulas look similar but differ by a factor of 2 and will double every reported $SS$ and $F$.

Source: Montgomery, Design and Analysis of Experiments (9th ed.), ch. 6–7 ($2^k$ factorial designs).

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