--- title: "Chi-square goodness of fit: Poisson errors" subject: "Probability and Statistics" chapter: 7 key: "semester_5/probability_and_statistics:chi-square-goodness-fit-poisson-pages" source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/chi-square-goodness-fit-poisson-pages site: BCT Engineering --- # Chi-square goodness of fit: Poisson errors Probability and Statistics · Chapter 7: Inference concerning proportion - Asked in: 2074 Chaitra, 6 marks; 2071 Shrawan, 5 marks (asked 2 times) - Video: [Poisson Random Variables: Chi Square for Poisson Distribution](https://www.youtube.com/watch?v=ABcBPTef_vE) — StatsResource - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/chi-square-goodness-fit-poisson-pages - Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-7 ## Question Define chi-square distribution. A book containing 500 pages, was thoroughly checked. The distribution of number of error page was given below as | Number of errors | 0 | 1 | 2 | 3 | 4 | 5 | |---|---|---|---|---|---|---| | Number of pages | 275 | 138 | 75 | 7 | 4 | 1 | Using chi-square test of goodness of fit, verify whether the arrivals follow a poison distribution at 5% level of significance. ## Short answer Estimate Poisson mean λ=0.66, compute expected frequencies, pool classes with expected<5 to get χ²=14.007 with d.f.=2. Critical χ²(0.05,2)=5.991. Reject H0; the number of errors per page does not follow a Poisson distribution. ## Answer **Chi-square distribution:** if Z₁, Z₂, …, Z_k are independent standard normal variables, then χ² = Z₁² + Z₂² + … + Z_k² follows the chi-square distribution with k degrees of freedom. It is a right-skewed continuous distribution (mean = k, variance = 2k) used for goodness-of-fit tests, tests of independence and tests of a variance. #### Goodness of fit to a Poisson distribution 1. H0: the number of errors per page follows a Poisson distribution. H1: it does not follow a Poisson distribution. 2. Level of significance: α = 0.05 3. Estimate the Poisson mean from the data: ```text λ = Σfx/Σf = (0×275 + 1×138 + 2×75 + 3×7 + 4×4 + 5×1)/500 = 330/500 = 0.66 ``` 4. Expected frequencies E = 500 × e^(−0.66)·0.66^x/x!: ```text Errors Observed P(x) Expected 0 275 0.51685 258.43 1 138 0.34112 170.56 2 75 0.11257 56.29 3 7 0.02477 12.38 4 4 0.00409 2.04 5+ 1 0.00060 0.30 ``` The classes x = 4 and x = 5+ have expected frequencies below 5, and even together they stay below 5 (2.34). So they are pooled with x = 3: ```text Errors O E (O−E)²/E 0 275 258.43 1.063 1 138 170.56 6.216 2 75 56.29 6.223 3 or more 12 14.73 0.505 Total 500 500.00 χ² = 14.007 ``` 5. Degrees of freedom = (number of classes) − 1 − (parameters estimated) = 4 − 1 − 1 = 2. Critical value: χ²(0.05; 2) = **5.991**. 6. Decision: calculated χ² = 14.007 > 5.991 (p-value = 0.0009), so **reject H0**. **Conclusion:** the errors per page do not follow a Poisson distribution at the 5% level. The observed data have too many pages with no error and too many with 2 errors compared with the Poisson model. Note: without pooling the classes the calculated χ² is 19.33 (d.f. = 4, critical 9.488); the decision is the same. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is chi-square with k d.f. defined? A. Square root of a t variable B. Sum of k normal variables C. Ratio of two normals D. Sum of squares of k independent standard normal variables Answer: D. χ² = Z1² + ... + Zk², with mean k and variance 2k. (Key point: Chi-square definition) 2. How is the Poisson mean estimated from the data? A. Σfx/Σf = 330/500 = 0.66 B. Σf/Σfx C. The largest frequency D. Total pages/6 Answer: A. lambda = Σfx/Σf = 0.66. (Key point: Parameter estimate) 3. After pooling to 4 classes with one estimated parameter, what are the d.f.? A. 3 B. 1 C. 2 D. 4 Answer: C. d.f. = classes − 1 − parameters estimated = 4 − 1 − 1 = 2; critical value 5.991. (Key point: Degrees of freedom) 4. Calculated chi-square is 14.007 > 5.991. What follows? A. Errors follow Poisson B. No conclusion C. Accept H0 D. Reject H0; errors per page do not follow a Poisson distribution Answer: D. The calculated value is in the rejection region, so the Poisson fit is rejected. 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