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23-Ind-A6 Systems Simulation · December 2013

Question 5 of 14: Question 5 (Part B, Q3 — Firm Selection)

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 5 (Part B, Q3 — Firm Selection) (5 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. Firm A (outlier removed per Question 3): mean 9.333 days, SD 1.732 days, $n=9$. Firm B: mean 8.8 days, SD 4.077 days, $n=10$.

Find. A recommendation, with the statistical basis stated.

Approach. Compare the two firms on both central tendency (two-sample $t$-test) and variability (F-test on the variance ratio), since a supply chain that must hold buffer stock cares about predictability as much as the average.

  1. Mean comparison (Welch's $t$-test). $H_0:\mu_A=\mu_B$: $$t = \frac{9.333-8.8}{\sqrt{1.732^2/9+4.077^2/10}} = 0.377,\qquad p=0.71\ (\text{2-sided}).$$ No evidence the average transit times differ — Firm A's slightly slower mean is not statistically significant.
  2. Variance comparison (F-test). $H_0:\sigma_A^2=\sigma_B^2$, using the larger sample variance in the numerator: $$F = \frac{s_B^2}{s_A^2} = \frac{16.622}{3.000} = \boxed{5.541},\qquad F_{0.025,\,9,8} = 4.357.$$ Since $F=5.541 > F_{crit}=4.357$, Firm B is significantly MORE VARIABLE than Firm A.
QuantityResult
Mean difference test$t=0.377$, $p=0.71$ — not significant
Variance ratio test$F=5.541 > F_{crit}=4.357$ — Firm B significantly more variable
RecommendationFirm A

With average delivery time statistically indistinguishable between the firms but Firm B's variability more than five times Firm A's (and significantly so), Firm A is the better choice: for a supply chain that must size a fixed yard buffer against transit-time uncertainty, a firm that is consistently on time is more valuable than one that is occasionally faster but occasionally much slower (Firm B's own worst case, 18 days, is more than twice its mean).

Check: assumes the anomalous 2-day Firm A entry (Question 3) is a genuine data-quality issue and is excluded before comparing firms; including it would make Firm A's SD (2.836) still smaller than Firm B's (4.077), so the recommendation is unchanged either way.