Question 3 of 14: Question 3 (Part B, Q1 — Outlier Test on Firm A Data)
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 3 (Part B, Q1 — Outlier Test on Firm A Data) (5 marks)
Given. Ten reported shipping times (days) for each firm on the Nanticoke–Dartmouth oil route, used as a proxy for the Nanticoke–Halifax steel route:
Delivery #
Firm A (days)
Firm B (days)
1
10
8
2
8
9
3
10
7
4
2
18
5
11
8
6
8
12
7
6
4
8
11
5
9
11
6
10
9
11
Find. Whether Firm A's 4th observation (2 days) is a statistical outlier at the 5% level, assuming Normality.
Approach. Apply Grubbs' test for a single outlier: compute the sample mean/SD of all 10 Firm A values, form the standardized deviation of the suspect point, and compare it to the Grubbs critical value for $n=10$.
Sample mean and SD of Firm A (all 10 points).
$$\bar x_A = \frac{10+8+10+2+11+8+6+11+11+9}{10} = 8.6\ \text{days}, \qquad s_A = 2.836\ \text{days}.$$
Grubbs' test statistic. The suspect value is $x=2$:
$$G = \frac{|\bar x_A - x|}{s_A} = \frac{|8.6-2|}{2.836} = \boxed{2.327}.$$
Grubbs' critical value ($n=10$, two-sided, $\alpha=0.05$). Using $t_{\alpha/(2n),\,n-2}$ from the $t$-table (here $t_{0.0025,8}=3.833$):
$$G_{crit} = \frac{n-1}{\sqrt n}\sqrt{\frac{t^2}{n-2+t^2}} = \frac{9}{\sqrt{10}}\sqrt{\frac{3.833^2}{8+3.833^2}} = \boxed{2.290}.$$
Decision. Since $G=2.327 > G_{crit}=2.290$, the 2-day entry is flagged as a statistical outlier at the 5% level — though only just: it would NOT be flagged at $\alpha=0.025$ (which needs $G\gtrsim2.41$), so this is a borderline call, not an overwhelming one.
Quantity
Result
Firm A mean / SD (n=10)
8.6 / 2.836 days
Grubbs statistic $G$
2.327
Grubbs critical value ($n=10$, $\alpha=0.05$)
2.290
Conclusion
2-day entry IS flagged as an anomaly (borderline, $\alpha=0.05$)
Check: "anomaly" is answered here via a formal Grubbs test; a 3-sigma or Chauvenet's-criterion rule of thumb would reach the same conclusion (both flag any point more than roughly 2.2–2.3 SD from the mean at $n=10$), so the finding is not sensitive to which standard outlier test is chosen.
