Question 14 of 14: Question 14 (Part D, Set 4 — ANOVA on the Screening Design)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2013 — 98-Ind-A6 Systems Simulation. Three-hour, closed-book exam; one of two permitted calculators (Sharp or Casio), one 8.5″×11.0″ aid sheet (both sides). Format: 14 sub-questions across four parts — Part A (do 1 of 2, 25 marks), Part B (do 3 of 5, 15 marks), Part C (do 1 of 3, 10 marks), Part D (do 2 of 4, 20 marks); 7 questions, 70 marks constitute a complete paper. All fourteen sub-questions are solved below for completeness (the source restarts its own numbering at 1 within each Part). Statistical tables (Normal, t, chi-square, F) were supplied with the exam; the values below are the same table values obtained by direct computation.
Reference texts: Banks, Carson, Nelson & Nicol, Discrete-Event System Simulation (5th ed., Pearson) — simulation study design, random-number generation, input/output data analysis, variance reduction, verification & validation, queueing simulation; Montgomery, Design and Analysis of Experiments (9th ed., Wiley) — factorial designs and ANOVA.
Question 14 (Part D, Set 4 — ANOVA on the Screening Design) (10 marks)
Check — source discrepancy: the question's own prose names the two screened factors "transport type" and "order lead time," but the printed design-matrix table headers read "Line Speed" and "Rework" — a genuine inconsistency in the printed paper. The numeric $(+/-)$ design matrix and response table are tied to the printed column headers, so this solution analyzes the data exactly as tabulated, labelling the two factors by their printed table names (A = Line Speed, B = Rework); the ANOVA mechanics and conclusions are unaffected by which cover-story label is attached to each factor.
Given. $2^2$ full factorial, $n=5$ replications per cell:
Run
A (Line Speed)
B (Rework)
Rep 1
Rep 2
Rep 3
Rep 4
Rep 5
Total
(1)
−
−
86
93
86
84
94
443
a
+
−
141
156
143
141
149
730
b
−
+
140
142
133
127
129
671
ab
+
+
154
170
162
161
163
810
Grand mean $\bar y=132.7$, sample variance $s^2=814.12$ over all $N=20$ runs (both reproduced exactly from the raw data above, confirming the stated summary figures).
Find. Whether the main effects (A, B) and the interaction (AB) are statistically significant.
Approach. Use the standard $2^2$ contrast method to get $SS_A$, $SS_B$, $SS_{AB}$ directly from the treatment totals, get $SS_{total}$ from the given/derived grand variance, obtain $SS_{error}$ by subtraction, and form three $F$-ratios against $F_{1,16}$ at $\alpha=0.05$.
Total and error sums of squares. $SS_{total}=(N-1)s^2=19(814.12)=15{,}468.2$:
$$SS_{error} = SS_{total}-SS_A-SS_B-SS_{AB} = 15468.2-9073.8-4743.2-1095.2 = \boxed{556.0},\quad df_{error}=N-4=16.$$
Mean squares and F-ratios. $MS_{error}=556.0/16=34.75$; each effect has 1 df so $MS=SS$:
$$F_A=\frac{9073.8}{34.75}=261.1,\qquad F_B=\frac{4743.2}{34.75}=136.5,\qquad F_{AB}=\frac{1095.2}{34.75}=31.5.$$
ANOVA table and decision ($\alpha=0.05$). Critical value $F_{0.05,1,16}=4.494$:
Source
SS
df
MS
F
Significant at $\alpha=0.05$?
A (Line Speed)
9073.8
1
9073.8
261.1
Yes
B (Rework)
4743.2
1
4743.2
136.5
Yes
AB (interaction)
1095.2
1
1095.2
31.5
Yes
Error
556.0
16
34.75
—
—
Total
15,468.2
19
—
—
—
All three F-ratios exceed $F_{crit}=4.494$ by a wide margin ($p<0.0001$ for all three), so both main effects AND the interaction are statistically significant.
