23-Ind-A6 Systems Simulation · December 2018

Question 6 of 8: One-Way ANOVA Across Three Routing Scenarios

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

Notes on this paper

National Exams — December 2018 — 17-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: three sections — Part A (Input Modelling): do 2 of 3 questions, 30 marks total; Part B (Modelling Concepts): do 1 of 2, 20 marks; Part C (Output Analysis): do 1 of 2, 20 marks — plus a 1-mark trivia bonus. All seven graded questions and the bonus are solved below for completeness. The exam's own front matter has two internal quirks, transcribed as printed: NOTES item "4." appears twice on page 1, and every page footer reads "17-Ind-A6/Dec. 2019" against a "December 2018" masthead.

Reference texts: Banks, Carson, Nelson & Nicol, Discrete-Event System Simulation (5th ed., Pearson) — input data analysis (ch. 9), random-variate generation, output analysis for a single system and comparing alternative systems (ch. 11–12), verification and validation (ch. 10).

Question 6 (Part C.1): One-Way ANOVA Across Three Routing Scenarios (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. Five replications of average wait time (min) under each of 3 scenarios:

Rep	Scenario 1	Scenario 2	Scenario 3
1	47.6	40.6	37.5
2	46.6	42.5	38.8
3	47.2	46.4	33.3
4	44.9	39.4	36.0
5	47.7	41.0	36.5
Average	46.8	42.0	36.4

Find. Whether the three scenarios' mean wait times are statistically different, and specifically which pair(s).

Approach. A one-way, single-factor ANOVA is the appropriate method for comparing more than two group means at once (a series of pairwise $t$-tests would inflate the overall Type I error rate); follow a significant overall $F$-test with a pairwise (Fisher LSD) comparison to identify which scenario(s) differ.

1. Sums of squares. With $k=3$ scenarios, $n_i=5$ reps each, $N=15$, grand mean $\bar{\bar x}=41.73$ (matches the stated overall mean of 41.7): $$SS_B = n\sum_i(\bar x_i-\bar{\bar x})^2 = \boxed{269.82},\qquad SS_W=\sum_i\sum_j(x_{ij}-\bar x_i)^2=\boxed{51.34}.$$ $$SS_T = SS_B+SS_W = 321.15\ \ \left(\text{cross-checks the given grand variance: } (N-1)\times22.94=321.16\right).$$
2. $F$-test. $df_B=2$, $df_W=12$: $$MS_B=\frac{269.82}{2}=134.91,\qquad MS_W=\frac{51.34}{12}=4.28,\qquad F=\frac{MS_B}{MS_W}=\boxed{31.54}.$$ $$F_{0.05,2,12}=3.89.$$ Since $31.54 \gg 3.89$, reject $H_0$: at least one scenario's mean wait time differs from the others.
3. Which scenario(s) differ — Fisher's LSD. $$LSD = t_{0.025,12}\sqrt{\frac{2\,MS_W}{n}} = 2.179\sqrt{\frac{2(4.28)}{5}} = \boxed{2.85\ \text{min}}.$$

Pairwise comparison	$|\bar x_i-\bar x_j|$	vs. LSD $=2.85$	Conclusion
Scenario 1 vs. 2	4.82	$>2.85$	Significant
Scenario 1 vs. 3	10.38	$>2.85$	Significant
Scenario 2 vs. 3	5.56	$>2.85$	Significant

All three pairwise differences exceed the LSD threshold, so every scenario is statistically distinguishable from every other scenario at the 5% level — not just "some" pair. Ranked by mean wait time, Scenario 3 (36.4 min) is fastest, Scenario 2 (42.0 min) is intermediate, and Scenario 1 (46.8 min) is slowest; on the wait-time criterion alone, Scenario 3 is the statistically best-supported choice (a full recommendation would also weigh each scenario's bus/routing cost, not analyzed here).

Quantity	Result
$F$ statistic (2,12 df)	31.54 vs. $F_{crit}=3.89$ — reject $H_0$
Fisher LSD	2.85 min
Pairwise result	All three scenarios differ significantly from each other
Best scenario (lowest wait)	Scenario 3 (36.4 min)

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