23-Ind-A6 Systems Simulation · Undated paper

Question 3 of 10: Project-Estimate Accuracy — Distribution, Confidence Interval, Anomaly Test

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

Notes on this paper

National Exams May 2019 — 17-Ind-A6, Systems Simulation, 3 hours duration. Section A: do 3 of 5 (30 marks); Section B: do 1 of 2 per the front-page summary table — the Part B section instructions on the page itself read “do 2 of 3” instead, a genuine inconsistency in the source paper's own front matter (flagged below); Section C: do 1 of 2 (20 marks). This study resource answers all ten questions across the three parts.

Check — front-page/section instructions disagree. The cover page's summary table states “Section B: Do 1 of 2 Questions, Total marks: 20,” but Part B's own section banner (page 4) reads “Complete two of the following three sets of questions,” and Part B in fact contains three questions worth 15 marks each. The cover page's own overall total (70 marks across “5 Questions”) is likewise only consistent with the “2 of 3” reading (30+30+... does not match 70 either way exactly, since 30(A)+2×15(B)+20(C)=80, one section choice short of 70); this is an internal inconsistency in the exam's own printed materials. All three Part B questions are answered in full below regardless.

Reference texts. Banks, Carson, Nelson & Nicol, Discrete-Event System Simulation (5th ed.) — primary text for random-variate generation, input/output analysis, variance reduction, verification & validation, and design of simulation experiments.

Question 3 (Part A.3): Project-Estimate Accuracy — Distribution, Confidence Interval, Anomaly Test (10 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. Sample of $n=250$ (Actual−Estimated) differences: mean $=1.001$, sample variance $=0.950$ (so $s=\sqrt{0.950}=0.9747$), median $=0.991$, skewness $=0.004$, range $=5.136$, min $=-5.081$, max $=6.860$.

Find. (a) a plausible distribution and rationale; (b) a 95% CI for the mean difference; (c) whether a project finishing 3.0 person-months ahead of schedule (difference $=-3.0$) is an anomaly.

Approach. Compare mean/median/skewness to diagnose shape and symmetry for (a); build the standard $t$-interval for (b); express the $-3.0$ observation as a $z$-score against the fitted model for (c).

(a) Suggest a distribution. Three clues point the same way: (i) the quantity is a difference that can be positive or negative (min $=-5.081$, max $=6.860$), ruling out any distribution confined to non-negative values; (ii) mean ($1.001$) and median ($0.991$) are nearly identical, the signature of a symmetric (not skewed) distribution; (iii) the reported skewness, $0.004$, is essentially zero, directly confirming symmetry. A difference-of-two-estimates quantity that is symmetric about a small positive bias is the classic setting for a $$\boxed{\text{Normal distribution}, \; N(\mu\approx1.00,\ \sigma^2\approx0.95)}$$ — consistent with a Central-Limit-style accumulation of many small, roughly independent estimation errors around a mean bias of about $+1$ person-month.
(b) 95% CI for the mean difference. With $n=250$ (large, so $t\approx z$, but $t$ is used for rigor), $s=\sqrt{0.950}=0.9747$: $$SE=\frac{s}{\sqrt n}=\frac{0.9747}{\sqrt{250}}=0.06164,\qquad t_{0.025,249}=1.9695$$ $$\bar x\pm t_{0.025,249}\,SE = 1.001\pm(1.9695)(0.06164)=1.001\pm0.1214$$ $$\boxed{95\%\ \text{CI} = (0.880,\ 1.122)\ \text{person-months}}$$
(c) Is $-3.0$ an anomaly? Standardize against the fitted $N(1.001,\,0.950)$ model: $$z=\frac{-3.0-1.001}{0.9747}=-4.10.$$ A project $4.10$ standard deviations below the mean has a two-tailed probability of only about $0.004\%$ under the normal model — by any conventional significance threshold this is a $$\boxed{\text{statistical anomaly (outlier)}}.$$ One caution is worth flagging: the sample's own recorded minimum, $-5.081$, is itself $z=(-5.081-1.001)/0.9747\approx-6.24$ under the same normal model — a value so extreme that a true Normal population of size $250$ would essentially never produce it (expected count $\ll1$). That the sample nonetheless contains it suggests the real distribution of estimate errors may have slightly heavier tails than a pure Normal (a few genuinely exceptional projects), so while $-3.0$ is confidently flagged as unusual relative to the bulk of the data, the exact numeric tail probability from the Normal model should be read as directionally right rather than exact.

Item	Result
(a) suggested distribution	Normal, $\mu\approx1.00,\ \sigma^2\approx0.95$ (symmetric: mean≈median, skewness≈0; support unrestricted in sign)
(b) 95% CI on mean difference	$(0.880,\ 1.122)$ person-months
(c) $-3.0$-month case	$z\approx-4.10$ — a statistical anomaly

Concept & Theory

Topic. Distribution diagnosis from summary statistics, the $t$-interval for a mean, and $z$-score anomaly screening.

Key relations. Symmetry diagnostics: mean$\approx$median, skewness$\approx0$; $CI=\bar x\pm t_{\alpha/2,n-1}s/\sqrt n$; $z=(x-\mu)/\sigma$.

Why this works. Summary statistics alone (mean, median, skewness) are enough to rule out badly-skewed or one-sided families and to justify a symmetric model; once a distribution is fitted, any single new observation can be screened for plausibility purely from its distance (in standard deviations) from the fitted mean, without needing the raw underlying data.

Common pitfall. Reading “3.0 person months ahead of schedule” as a difference of $+3.0$ instead of $-3.0$ (actual finishing early means Actual $<$ Estimated, so Actual−Estimated is negative); or quoting the CI half-width without converting the sample variance to a standard deviation first.

Source. Banks, Carson, Nelson & Nicol, Discrete-Event System Simulation, ch. 4 (probability review, distribution shape) & ch. 12 (input/output statistical analysis, confidence intervals).

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