1. **Problem Statement:** Test whether the number of arrivals per unit time at the hospital follows a Poisson distribution using the given observed frequencies and a 5% level of significance. 2. **Formula and Important Rules:** The Poisson probability mass function is given by: $$P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$$ where $\lambda$ is the mean number of arrivals per unit time. The chi-square goodness of fit test statistic is: $$\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$$ where $O_i$ is the observed frequency and $E_i$ is the expected frequency. We reject the null hypothesis if $\chi^2$ calculated $>$ $\chi^2$ critical value at $\alpha=0.05$ with degrees of freedom $df = \text{number of categories} - 1 - \text{parameters estimated}$. 3. **Calculate the sample mean $\lambda$:** $$\lambda = \frac{\sum (x_i \times f_i)}{\sum f_i}$$ where $x_i$ is the number of arrivals and $f_i$ is the observed frequency. Calculate total frequency: $$N = 10 + 30 + 40 + 50 + 35 + 20 + 10 + 5 = 200$$ Calculate $\sum x_i f_i$: $$0\times10 + 1\times30 + 2\times40 + 3\times50 + 4\times35 + 5\times20 + 6\times10 + 7\times5 = 0 + 30 + 80 + 150 + 140 + 100 + 60 + 35 = 595$$ So, $$\lambda = \frac{595}{200} = 2.975$$ 4. **Calculate expected frequencies $E_i$ using Poisson probabilities:** Calculate $P(X=k)$ for $k=0$ to $7$: $$P(X=k) = \frac{e^{-2.975} (2.975)^k}{k!}$$ Calculate $e^{-2.975} \approx 0.0509$ Calculate each $P(X=k)$: - $P(0) = 0.0509 \times \frac{2.975^0}{0!} = 0.0509$ - $P(1) = 0.0509 \times \frac{2.975^1}{1} = 0.1514$ - $P(2) = 0.0509 \times \frac{2.975^2}{2} = 0.2253$ - $P(3) = 0.0509 \times \frac{2.975^3}{6} = 0.2233$ - $P(4) = 0.0509 \times \frac{2.975^4}{24} = 0.1660$ - $P(5) = 0.0509 \times \frac{2.975^5}{120} = 0.0987$ - $P(6) = 0.0509 \times \frac{2.975^6}{720} = 0.0489$ - $P(7) = 0.0509 \times \frac{2.975^7}{5040} = 0.0208$ Calculate expected frequencies $E_i = N \times P(X=k)$: - $E_0 = 200 \times 0.0509 = 10.18$ - $E_1 = 200 \times 0.1514 = 30.28$ - $E_2 = 200 \times 0.2253 = 45.06$ - $E_3 = 200 \times 0.2233 = 44.66$ - $E_4 = 200 \times 0.1660 = 33.20$ - $E_5 = 200 \times 0.0987 = 19.74$ - $E_6 = 200 \times 0.0489 = 9.78$ - $E_7 = 200 \times 0.0208 = 4.16$ 5. **Calculate the chi-square test statistic:** $$\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$$ Calculate each term: - $\frac{(10 - 10.18)^2}{10.18} = 0.0032$ - $\frac{(30 - 30.28)^2}{30.28} = 0.0026$ - $\frac{(40 - 45.06)^2}{45.06} = 0.5673$ - $\frac{(50 - 44.66)^2}{44.66} = 0.6423$ - $\frac{(35 - 33.20)^2}{33.20} = 0.0976$ - $\frac{(20 - 19.74)^2}{19.74} = 0.0033$ - $\frac{(10 - 9.78)^2}{9.78} = 0.0050$ - $\frac{(5 - 4.16)^2}{4.16} = 0.1712$ Sum: $$\chi^2 = 0.0032 + 0.0026 + 0.5673 + 0.6423 + 0.0976 + 0.0033 + 0.0050 + 0.1712 = 1.4925$$ 6. **Degrees of freedom:** Number of categories = 8 Parameters estimated = 1 (mean $\lambda$) $$df = 8 - 1 - 1 = 6$$ 7. **Critical value:** From chi-square table at $\alpha=0.05$ and $df=6$, critical value $\chi^2_{0.05,6} = 12.592$ 8. **Decision:** Since $\chi^2_{calculated} = 1.4925 < 12.592 = \chi^2_{critical}$, we fail to reject the null hypothesis. 9. **Conclusion:** There is not enough evidence at the 5% significance level to reject the claim that the arrivals follow a Poisson distribution.