1. **State the problem:** We are given the number of accidents in a city during a week and want to test if the accidents are uniformly distributed over the days of the week. 2. **Data given:** | Day | Mon | Tues | Wed | Thu | Fri | Sat | | No of accidents | 15 | 19 | 13 | 12 | 16 | 15 | 3. **Hypotheses:** - Null hypothesis $H_0$: Accidents are uniformly distributed over the days. - Alternative hypothesis $H_a$: Accidents are not uniformly distributed. 4. **Formula for chi-square goodness-of-fit test:** $$\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$$ where $O_i$ is observed frequency and $E_i$ is expected frequency. 5. **Calculate expected frequency:** Total accidents = $15 + 19 + 13 + 12 + 16 + 15 = 90$ Number of days = 6 Expected frequency per day $E = \frac{90}{6} = 15$ 6. **Calculate chi-square statistic:** $$\chi^2 = \frac{(15-15)^2}{15} + \frac{(19-15)^2}{15} + \frac{(13-15)^2}{15} + \frac{(12-15)^2}{15} + \frac{(16-15)^2}{15} + \frac{(15-15)^2}{15}$$ $$= 0 + \frac{16}{15} + \frac{4}{15} + \frac{9}{15} + \frac{1}{15} + 0 = \frac{30}{15} = 2.0$$ 7. **Degrees of freedom:** $df = n - 1 = 6 - 1 = 5$ 8. **Decision rule:** At significance level $\alpha = 0.05$, the critical value from chi-square table for $df=5$ is approximately 11.07. 9. **Conclusion:** Since calculated $\chi^2 = 2.0$ is less than critical value 11.07, we do not reject the null hypothesis. **Final answer:** $\chi^2 = 1.998$ (rounded), Null Hypothesis not rejected.