1. **Problem Statement:** Test the surveyor’s claim using a chi-square goodness-of-fit test with significance level $\alpha = 0.01$. 2. **Chi-square goodness-of-fit test formula:** $$\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$$ where $O_i$ is the observed frequency and $E_i$ is the expected frequency for category $i$. 3. **Given data:** Day | Mon | Tues | Wed | Thu | Fri | Sat No of accidents | 15 | 19 | 13 | 12 | 16 | 15 4. **Null hypothesis ($H_0$):** Accidents are uniformly distributed over the days. 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:** $$\chi^2 = \frac{(15-15)^2}{15} + \frac{(19-15)^2}{15} + \frac{(13-15)^2}{15} + \frac{(12-15)^2}{15} + \frac{(16-15)^2}{15} + \frac{(15-15)^2}{15}$$ $$= 0 + \frac{16}{15} + \frac{4}{15} + \frac{9}{15} + \frac{1}{15} + 0 = \frac{30}{15} = 2$$ 7. **Degrees of freedom:** $df = 6 - 1 = 5$ 8. **Critical value at $\alpha=0.01$ and $df=5$:** Approximately 15.086 9. **Decision:** Since $\chi^2 = 2 < 15.086$, we fail to reject the null hypothesis. 10. **Conclusion:** There is not enough evidence to reject the claim that accidents are uniformly distributed over the week at the 1% significance level. **Final answer:** $\chi^2 = 2.0$, Null Hypothesis not rejected.