23-Ind-A6 Systems Simulation · December 2013

Question 6 of 14: Question 6 (Part B, Q4 — Goodness-of-Fit Test)

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 6 (Part B, Q4 — Goodness-of-Fit Test) (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. 100 Firm B transit times, binned:

Shipping time (days)	Observed count
0–4.999	10
5–9.999	40
10–14.999	30
15–19.999	8
20+	12

Hypothesized distribution: Normal with mean $\mu=8$ and variance $\sigma^2=4$ (so $\sigma=2$).

Find. Whether a chi-square goodness-of-fit test supports $H_0:$ data $\sim$ Normal(8, 4).

Approach. Compute the Normal(8,2) expected count in each bin; because the two upper bins have expected counts far below the rule-of-thumb minimum of 5, pool the top three bins into a single "10+" tail bin before forming the chi-square statistic.

1. Expected counts under Normal(8,2), original 5 bins. Using $\Phi\!\left(\frac{x-8}{2}\right)$ at each boundary: $E=(6.68,\ 77.45,\ 15.84,\ 0.023,\ \approx0)$. The last two bins are far under 5 — combine bins 3–5 into one "$\ge10$" bin.
2. Pooled 3-bin table.

   Bin	Observed $O$	Expected $E$ (Normal 8, 2)
   0–4.999	10	6.68
   5–9.999	40	77.45
   ≥10	50	15.87

3. Chi-square statistic. $$\chi^2 = \sum \frac{(O-E)^2}{E} = \frac{(10-6.68)^2}{6.68}+\frac{(40-77.45)^2}{77.45}+\frac{(50-15.87)^2}{15.87} = \boxed{93.2}.$$
4. Critical value and decision. $k=3$ bins, both parameters GIVEN (not estimated from these 100 points), so $df=k-1=2$: $$\chi^2_{0.05,2} = 5.991.$$ Since $\chi^2=93.2 \gg \chi^2_{crit}=5.991$, reject $H_0$ — the data is not consistent with Normal(8, 4). The empirical tail is far heavier than a $\sigma=2$ Normal allows (50 of 100 observations at 10+ days, vs. ≈16 expected).

Quantity	Result
Test statistic $\chi^2$ (3 bins, pooled)	93.2
Critical value ($df=2$, $\alpha=0.05$)	5.991
Conclusion	Reject Normal(8,4); data is right-skewed / heavy-tailed

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