1. **Problem Statement:** We have data on the number of car accidents per week over 80 weeks with frequencies for 0, 1, 2, and 3 accidents. We want to test if this data follows a Poisson distribution at a significance level of 0.01. 2. **Poisson Distribution Formula:** The probability mass function (pmf) of a Poisson distribution is: $$P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$$ where $\lambda$ is the average rate (mean) of occurrences. 3. **Calculate the sample mean $\lambda$:** $$\lambda = \frac{\sum (x_i \times f_i)}{n} = \frac{0\times3 + 1\times22 + 2\times33 + 3\times16}{80}$$ Calculate numerator: $$0 + 22 + 66 + 48 = 136$$ So, $$\lambda = \frac{136}{80} = 1.7$$ 4. **Calculate expected frequencies using Poisson probabilities:** For each $k=0,1,2,3$: $$P(k) = \frac{e^{-1.7} 1.7^k}{k!}$$ Calculate $e^{-1.7} \approx 0.1827$. - For $k=0$: $$P(0) = 0.1827 \times \frac{1.7^0}{0!} = 0.1827$$ Expected frequency: $$E_0 = 80 \times 0.1827 = 14.62$$ - For $k=1$: $$P(1) = 0.1827 \times \frac{1.7^1}{1} = 0.1827 \times 1.7 = 0.3106$$ Expected frequency: $$E_1 = 80 \times 0.3106 = 24.85$$ - For $k=2$: $$P(2) = 0.1827 \times \frac{1.7^2}{2} = 0.1827 \times \frac{2.89}{2} = 0.1827 \times 1.445 = 0.2640$$ Expected frequency: $$E_2 = 80 \times 0.2640 = 21.12$$ - For $k=3$: $$P(3) = 0.1827 \times \frac{1.7^3}{6} = 0.1827 \times \frac{4.913}{6} = 0.1827 \times 0.8188 = 0.1495$$ Expected frequency: $$E_3 = 80 \times 0.1495 = 11.96$$ 5. **Perform Chi-square goodness-of-fit test:** $$\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$$ Where $O_i$ are observed frequencies and $E_i$ expected frequencies. Calculate each term: - For $k=0$: $$\frac{(3 - 14.62)^2}{14.62} = \frac{(-11.62)^2}{14.62} = \frac{135.05}{14.62} = 9.24$$ - For $k=1$: $$\frac{(22 - 24.85)^2}{24.85} = \frac{(-2.85)^2}{24.85} = \frac{8.12}{24.85} = 0.33$$ - For $k=2$: $$\frac{(33 - 21.12)^2}{21.12} = \frac{11.88^2}{21.12} = \frac{141.14}{21.12} = 6.68$$ - For $k=3$: $$\frac{(16 - 11.96)^2}{11.96} = \frac{4.04^2}{11.96} = \frac{16.32}{11.96} = 1.36$$ Sum: $$\chi^2 = 9.24 + 0.33 + 6.68 + 1.36 = 17.61$$ 6. **Degrees of freedom:** $$df = \text{number of categories} - 1 - \text{number of estimated parameters} = 4 - 1 - 1 = 2$$ 7. **Critical value at $\alpha=0.01$ and $df=2$:** From chi-square tables, critical value $\approx 9.21$. 8. **Decision:** Since $17.61 > 9.21$, we reject the null hypothesis. **Conclusion:** At the 0.01 significance level, the data does not follow a Poisson distribution.