23-Ind-B1 Reliability and Maintainability · December 2017
Question 2 of 10: Job Time in System — Sample-Size Design, CI, Anomaly Check, Two-Sample Comparison
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2017 — 98-Ind-B1 Applied Probability & Statistics. Three-hour, closed-book exam; one of two permitted calculators (Sharp or Casio), one 8.5″×11.0″ aid sheet (both sides), statistical tables supplied. Format: three sections — Section A: do 2 of 4 (30 marks); Section B: do 2 of 3 (30 marks); Section C: do 2 of 3 (40 marks) — a 6-question, 100-mark paper as printed. All ten questions across the three sections are solved below for completeness.
Reference texts: Montgomery & Runger, Applied Statistics and Probability for Engineers (7th ed., Wiley) — discrete/continuous distributions (ch. 3–4), joint distributions (ch. 5), point/interval estimation and sample size (ch. 8), hypothesis testing incl. two-sample tests (ch. 9–10), simple linear regression (ch. 11), design and analysis of single-factor experiments (ch. 13). Montgomery, Peck & Vining, Introduction to Linear Regression Analysis (6th ed., Wiley) — multiple regression by matrices (ch. 2–3). Montgomery, Design and Analysis of Experiments (9th ed., Wiley) — multi-factor and $2^k$ factorial designs (ch. 5–6).
Question 2 (Section A.2): Job Time in System — Sample-Size Design, CI, Anomaly Check, Two-Sample Comparison (15 marks)
Given. Pilot run, $n=10$ replications of job time in system (minutes):
Rep
1
2
3
4
5
6
7
8
9
10
TIS (min)
112.6
105.6
95.9
94.4
97.1
92.1
109.2
120.5
110.6
98.8
Later run: $n=57$, $\bar x=104.0$, $s^2=100$.
Find. (a) sample size for a $\pm3$-minute, 95% margin of error; (b) 95% CI for the later-run mean; (c) is a future 110-minute average an anomaly; (d) do the pilot and later-run results differ significantly.
Approach. (a) inverts the standard $z$-based margin-of-error formula using the pilot's own sample standard deviation as the planning estimate of $\sigma$; (b) is a routine $t$-based CI for a mean; (c) treats 110 as a single future draw and standardizes against the later run's own mean/sd; (d) is a two-sample (Welch) $t$-test comparing the pilot and later-run means.
Pilot statistics. From the 10 pilot values, $\bar x_0=103.68$ min, $s_0=9.383$ min.
(a) Sample size for $\pm3$ min at 19/20 (95%) confidence. Using the pilot $s_0$ as the planning estimate of $\sigma$ and $z_{0.025}=1.96$: $$n=\left(\frac{z_{0.025}\,s_0}{E}\right)^2=\left(\frac{1.96\times9.383}{3}\right)^2=37.58\ \Rightarrow\ \boxed{n=38\ \text{replications}}$$ (rounded up, since a fractional replication count is not achievable).
(b) 95% CI for the later-run mean. With $n=57$, $\bar x=104.0$, $s=\sqrt{100}=10$, $t_{0.025,56}=2.003$: $$\bar x\pm t_{0.025,56}\frac{s}{\sqrt n}=104.0\pm 2.003\left(\frac{10}{\sqrt{57}}\right)=\boxed{(101.35,\ 106.65)\ \text{min}}$$
(c) Is a future 110-minute average an anomaly? Treating 110 as a draw from the later run's own $N(104,100)$ description: $$z=\frac{110-104}{10}=\boxed{0.60}$$ Since $|0.60|\ll 2$, a 110-minute average is comfortably within one standard deviation of the mean and is not a statistical anomaly — it is a routine, unremarkable outcome for this process.
(d) Is the pilot statistically different from the later run? Two independent samples with unequal $n$ and different sample variances call for Welch's $t$-test: $$se=\sqrt{\frac{s_0^2}{n_0}+\frac{s^2}{n}}=\sqrt{\frac{9.383^2}{10}+\frac{100}{57}}=3.235,\qquad t=\frac{103.68-104.0}{3.235}=\boxed{-0.099}$$ Welch–Satterthwaite $df\approx12.9$, giving $t_{crit}\approx2.16$ ($p=0.923$). Since $|t|\ll t_{crit}$, there is no evidence the pilot and later-run means differ — the two samples are statistically consistent with the same underlying process.
