23-Ind-B1 Reliability and Maintainability · December 2014
Question 8 of 9: Two-Factor ANOVA — Line Speed and Repair Policy on Throughput
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2014 — 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: four sections — Section A: do 2 of 3 questions (20 marks); Section B: do 1 of 2 (25 marks); Section C: do 1 of 2 (25 marks); Section D: do 1 of 2 (30 marks) — a 5-question, 100-mark paper as printed. All nine questions across the four sections are solved below for completeness. Page 1's own NOTES list is mis-numbered (two items both labelled "4."), transcribed as printed.
Reference texts: Montgomery & Runger, Applied Statistics and Probability for Engineers (7th ed., Wiley) — point/interval estimation (ch. 8–9), hypothesis testing (ch. 9–10), simple/multiple linear regression (ch. 11–12), single- and two-factor ANOVA (ch. 13). Montgomery, Peck & Vining, Introduction to Linear Regression Analysis (6th ed., Wiley) — multiple regression by matrices, confidence/prediction intervals (ch. 2–3).
Question 8 (Section D.1): Two-Factor ANOVA — Line Speed and Repair Policy on Throughput (30 marks)
Given. A $4\times2$ factorial (line speed $\times$ repair policy), $n=4$ replicates per cell (32 observations total):
Policy
0.5 m/s
0.6 m/s
0.7 m/s
0.8 m/s
Individual
2.3, 2.9, 3.1, 3.2
3.4, 3.7, 3.6, 3.2
3.8, 3.9, 4.1, 3.8
3.9, 3.2, 3.0, 2.7
Dedicated
4.3, 3.9, 3.9, 4.2
3.8, 3.8, 3.9, 3.5
3.9, 4.0, 3.7, 3.6
3.5, 3.6, 3.8, 3.9
Find. (a) significance of speed; (b) significance of repair policy; (c) interaction significance; (d) interpretation and optimal setting; (e) $R^2$; (f) model adequacy; (g) assumptions.
Approach. Compute the speed-factor and policy-factor sums of squares directly from the cell/marginal totals, then use $SST=SS(A)+SS(B)+SS(AB)+SSE$ (with $SST$ and $SS(AB)$ supplied) to isolate $SSE$, build the ANOVA table, and $F$-test each source against $MSE$.
Setup — marginal totals and $SS(A)$, $SS(B)$. The four speed totals (summed over both policies, $n=8$ each) are $27.8,\ 28.9,\ 30.8,\ 27.6$ (summing to the given grand sum $115.1$); the two policy totals (summed over all four speeds, $n=16$ each) are Individual $=53.8$, Dedicated $=61.3$. With correction factor $CF=(115.1)^2/32=414.00$:
$$SS(A)=\frac{27.8^2+28.9^2+30.8^2+27.6^2}{8}-CF=\boxed{0.806},\qquad SS(B)=\frac{53.8^2+61.3^2}{16}-CF=\boxed{1.758}.$$
Setup — complete the ANOVA table. Using the supplied $SST=6.25$ and $SS(AB)=1.893$:
$$SSE = SST-SS(A)-SS(B)-SS(AB) = 6.25-0.806-1.758-1.893 = \boxed{1.793}.$$
Degrees of freedom: $df_A=3$, $df_B=1$, $df_{AB}=3$, $df_E=32-8=24$ (8 cells $\times$ 3 each). Mean squares: $MS(A)=0.806/3=0.2686$, $MS(B)=1.758/1=1.758$, $MS(AB)=1.893/3=0.631$, $MSE=1.793/24=0.0747$.
Part (a) — line speed.
$$F_A=\frac{MS(A)}{MSE}=\frac{0.2686}{0.0747}=\boxed{3.60}.$$
Against $F_{0.05,3,24}=3.01$: $3.60>3.01$, so line speed has a significant effect on throughput at $\alpha=0.05$.
Part (b) — repair policy.
$$F_B=\frac{MS(B)}{MSE}=\frac{1.758}{0.0747}=\boxed{23.53}.$$
Against $F_{0.05,1,24}=4.26$: $23.53\gg4.26$, so repair policy has a highly significant effect.
Part (c) — interaction.
$$F_{AB}=\frac{MS(AB)}{MSE}=\frac{0.631}{0.0747}=\boxed{8.45}.$$
Against $F_{0.05,3,24}=3.01$: $8.45>3.01$, so the speed×policy interaction is significant — the effect of line speed on throughput genuinely depends on which repair policy is in use (and vice versa).
Part (d) — interpreting the interaction and finding the optimum. Because the interaction is significant, the main effects in (a)/(b) cannot be read in isolation to pick a "best" setting — a significant interaction means the two main-effect conclusions do not simply add. The individual cell means make this concrete: under the Individual-repair policy throughput rises with speed up to 0.7 m/s (peak mean 3.90) then falls at 0.8 m/s, while under Dedicated repair throughput is highest at the slowest speed, 0.5 m/s (mean 4.075), and generally declines as speed increases. The global maximum across all eight cells is Dedicated repair at 0.5 m/s, mean throughput 4.075 parts/min — the IE should read the optimal setting directly off the cell-mean table (or a full interaction plot) rather than optimizing each factor separately, since the "best repair policy" is not the same at every speed and the "best speed" is not the same under every policy.
Part (f) — is this a good model? With $R^2=0.713$, the two factors and their interaction jointly explain about 71% of the observed variability in throughput, and all three $F$-tests (speed, policy, interaction) are significant at $\alpha=0.05$ — so the model captures real, structured effects rather than noise. It is not a complete description (29% of the variance remains unexplained, likely operator-to-operator and run-to-run variation not captured by the two controlled factors), but for a designed engineering experiment identifying which settings drive throughput, $R^2=0.71$ with three clearly significant sources is a good, actionable model.
Part (g) — ANOVA assumptions. The two-factor fixed-effects model assumes: (i) the response within each of the 8 speed×policy cells is normally distributed; (ii) the 8 cell variances are homogeneous (equal error variance, $\sigma^2$, across all cells); (iii) observations are independent, both within and across cells (no serial correlation between replicate runs); and (iv) the two factor levels tested (4 speeds, 2 policies) are the specific, fixed levels of interest, not a random sample from a larger population of possible speeds/policies.
