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23-Ind-B1 Reliability and Maintainability · December 2016

Question 7 of 11: Power Outages — Poisson Goodness-of-Fit Test

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

Notes on this paper

National Exams — December 2016 — 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 4 (40 marks) — a 6-question, 100-mark paper as printed. All eleven questions across the three sections are solved below for completeness.

Reference texts: Montgomery & Runger, Applied Statistics and Probability for Engineers (7th ed., Wiley) — joint distributions and covariance (ch. 5), point/interval estimation (ch. 8), hypothesis testing incl. two-sample and goodness-of-fit tests (ch. 9–10), simple linear regression (ch. 11), design and analysis of single-factor and factorial experiments (ch. 13–14). Montgomery, Peck & Vining, Introduction to Linear Regression Analysis (6th ed., Wiley) — multiple regression by matrices, confidence/prediction intervals (ch. 2–3). Montgomery, Design and Analysis of Experiments (9th ed., Wiley) — two-way factorial ANOVA and $2^k$ designs (ch. 5, 6–7).

Question 7 (Section B.3): Power Outages — Poisson Goodness-of-Fit Test (15 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. $n=39$ billing periods (28 days each), hypothesized rate $\lambda=0.05$/day:

Outages/period0123456
Periods observed81295221

Find. Whether $\chi^2$ GOF supports $\text{Poisson}(\lambda\cdot28=1.4/\text{period})$ at $\alpha=0.05$.

Approach. Convert the daily rate to a per-billing-period rate, compute Poisson-expected counts for each category, combine tail categories until every expected count $\ge5$, then compute $\chi^2=\sum(O-E)^2/E$ against $\chi^2_{0.05,\,k-1}$ (no parameter is estimated from the sample, since $\lambda$ is externally hypothesized).

  1. Hypothesized per-period rate. $$\lambda_{period}=0.05\times28=\boxed{1.4}$$ (Consistency check: $1.4\times39=54.6$ vs. the sample's own implied rate $69/39=1.77$ — the sample runs somewhat higher than the claim, which the test below will assess formally.)
  2. Expected counts. $E_k=39\times\text{Poisson}(k;1.4)$: $E_0=9.617$, $E_1=13.464$, $E_2=9.425$, $E_3=4.398$, $E_4=1.539$, $E_5=0.431$, $E_{\ge6}=0.125$. Categories 3–6 all have $E\lt5$, so combine them into a single "$3^+$" tail: $O_{3^+}=5+2+2+1=10$, $E_{3^+}=4.398+1.539+0.431+0.125=6.494$. This leaves 4 usable categories $\{0,1,2,3^+\}$.
  3. Chi-square statistic. $$\chi^2=\sum\frac{(O-E)^2}{E}=\frac{(8-9.617)^2}{9.617}+\frac{(12-13.464)^2}{13.464}+\frac{(9-9.425)^2}{9.425}+\frac{(10-6.494)^2}{6.494}=\boxed{2.344}$$ With $\lambda$ given (not estimated), $df=k-1=3$, and $\chi^2_{0.05,3}=7.815$.
  4. Decision. Since $2.344\lt7.815$, fail to reject $H_0$: $$\boxed{\text{the data are consistent with Poisson}(\lambda=0.05/\text{day}); NSPI's claim is supported}$$
Summary
QuantityValue
$\lambda$/period1.4
$\chi^2$2.344
$df$, critical value3, 7.815
VerdictFail to reject — Poisson model fits