23-Ind-A6 Systems Simulation · December 2018

Question 2 of 8: Chi-Squared Goodness-of-Fit Test for the Hypothesized Poisson

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

Question 2 (Part A.2): Chi-Squared Goodness-of-Fit Test for the Hypothesized Poisson (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. The same 100-patrol survey of Question 1; hypothesized $H_0:$ data $\sim$ Poisson($\lambda=6$); $\alpha=0.05$.

Find. The $\chi^2$ test statistic, its critical value, and a reject/fail-to-reject decision.

Approach. Compute expected counts $E_x = n\cdot P(X=x)$ from the standard Poisson pmf (the exam's own printed formula, $f(x)=\lambda t e^{-\lambda t x}/x!$, is internally non-standard — treated as a transcription slip and replaced with $P(X=k)=\lambda^k e^{-\lambda}/k!$), combine adjacent cells so every expected count is $\ge 5$ (the standard chi-square validity rule), then compute $\chi^2=\sum (O-E)^2/E$ against $\chi^2_{0.95,df}$ with $df=(\text{bins})-1$ — not bins$-2$, because $\lambda=6$ is specified, not estimated from this sample.

1. Expected counts and binning. $E_x = 100\cdot\dfrac{6^x e^{-6}}{x!}$ for $x=0,\ldots,9$, with the open tail $E_{\ge 10}=100\,P(X\ge 10)$. The raw cells for $x=0,1,2$ (0.25, 1.49, 4.46) and $x=9,10^+$ (6.88, 8.39) fall below 5 and must be pooled with neighbours:

2. Test statistic. Eight bins, $df=8-1=7$: $$\chi^2 = \sum\frac{(O-E)^2}{E} = \frac{(4-6.20)^2}{6.20}+\cdots+\frac{(4-15.28)^2}{15.28} = \boxed{18.44}.$$
3. Critical value and decision. $$\chi^2_{0.95,7} = \boxed{14.07}.$$ Since $18.44 > 14.07$, reject $H_0$ — Robert's hypothesis is not correct: the data do not fit Poisson($\lambda=6$) at the 5% level. The dominant contributors to $\chi^2$ are the $\ge 9$ bin (observed 4 vs. expected 15.3, a large shortfall in the tail) and the $\le 2$ bin (observed 4 vs. expected 6.2) — the real distribution has noticeably less spread than a Poisson with the same rate, consistent with the mean/variance gap flagged as a check note in Question 1 (mean 5.74 well above sample variance 3.02, where a true Poisson requires equality).