1. **Problem Statement:** Test whether the number of arrivals per unit time at the hospital follows a Poisson distribution using the chi-square goodness of fit test at 5% significance level. 2. **Given Data:** Number of arrivals (x): 0, 1, 2, 3, 4, 5, 6, 7 Observed frequency (O): 10, 30, 40, 50, 35, 20, 10, 5 3. **Step 1: Calculate the sample mean \( \bar{x} \) (estimate of Poisson parameter \( \lambda \))** $$\bar{x} = \frac{\sum x_i O_i}{\sum O_i} = \frac{0\times10 + 1\times30 + 2\times40 + 3\times50 + 4\times35 + 5\times20 + 6\times10 + 7\times5}{10 + 30 + 40 + 50 + 35 + 20 + 10 + 5}$$ Calculate numerator: $$0 + 30 + 80 + 150 + 140 + 100 + 60 + 35 = 595$$ Calculate denominator: $$10 + 30 + 40 + 50 + 35 + 20 + 10 + 5 = 200$$ So, $$\bar{x} = \frac{595}{200} = 2.975$$ 4. **Step 2: Calculate expected frequencies \( E_i \) using Poisson probabilities:** Poisson probability mass function: $$P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!}$$ Calculate \( e^{-\lambda} = e^{-2.975} \approx 0.0509 \) Calculate probabilities for each \( x \): - \( P(0) = 0.0509 \times \frac{2.975^0}{0!} = 0.0509 \) - \( P(1) = 0.0509 \times \frac{2.975^1}{1!} = 0.1514 \) - \( P(2) = 0.0509 \times \frac{2.975^2}{2!} = 0.2253 \) - \( P(3) = 0.0509 \times \frac{2.975^3}{6} = 0.2233 \) - \( P(4) = 0.0509 \times \frac{2.975^4}{24} = 0.1660 \) - \( P(5) = 0.0509 \times \frac{2.975^5}{120} = 0.0987 \) - \( P(6) = 0.0509 \times \frac{2.975^6}{720} = 0.0490 \) - \( P(7) = 0.0509 \times \frac{2.975^7}{5040} = 0.0208 \) Calculate expected frequencies: $$E_i = P(x_i) \times 200$$ - \( E_0 = 0.0509 \times 200 = 10.18 \) - \( E_1 = 0.1514 \times 200 = 30.28 \) - \( E_2 = 0.2253 \times 200 = 45.06 \) - \( E_3 = 0.2233 \times 200 = 44.66 \) - \( E_4 = 0.1660 \times 200 = 33.20 \) - \( E_5 = 0.0987 \times 200 = 19.74 \) - \( E_6 = 0.0490 \times 200 = 9.80 \) - \( E_7 = 0.0208 \times 200 = 4.16 \) 5. **Step 3: Combine categories if expected frequencies are less than 5 (none here are less than 5, so no combination needed).** 6. **Step 4: Calculate chi-square test statistic:** $$\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$$ Calculate each term: - \( \frac{(10 - 10.18)^2}{10.18} = 0.0032 \) - \( \frac{(30 - 30.28)^2}{30.28} = 0.0026 \) - \( \frac{(40 - 45.06)^2}{45.06} = 0.566 \) - \( \frac{(50 - 44.66)^2}{44.66} = 0.642 \) - \( \frac{(35 - 33.20)^2}{33.20} = 0.097 \) - \( \frac{(20 - 19.74)^2}{19.74} = 0.0033 \) - \( \frac{(10 - 9.80)^2}{9.80} = 0.0041 \) - \( \frac{(5 - 4.16)^2}{4.16} = 0.171 \) Sum: $$\chi^2 = 0.0032 + 0.0026 + 0.566 + 0.642 + 0.097 + 0.0033 + 0.0041 + 0.171 = 1.4892$$ 7. **Step 5: Determine degrees of freedom:** $$df = k - 1 - m = 8 - 1 - 1 = 6$$ where \(k=8\) categories and \(m=1\) estimated parameter (\(\lambda\)). 8. **Step 6: Find critical value from chi-square table at \(\alpha=0.05\) and \(df=6\):** $$\chi^2_{critical} = 12.592$$ 9. **Step 7: Conclusion:** Since \(\chi^2 = 1.4892 < 12.592\), we fail to reject the null hypothesis. **Interpretation:** There is no significant evidence at 5% level to reject the claim that the arrivals follow a Poisson distribution.