1. **State the problem:** We want to test if the violent crime rates differ significantly among the four cities: Toronto, St. Catharines, Hamilton, and Sudbury. 2. **Write the hypotheses:** - Null hypothesis $H_0$: The crime rates are the same across all cities. - Alternative hypothesis $H_a$: At least one city has a different crime rate. 3. **Set up observed and expected frequencies:** - Observed frequencies $O$: Toronto = 92, St. Catharines = 34, Hamilton = 85, Sudbury = 49. - Total crimes $T = 92 + 34 + 85 + 49 = 260$. - Since no expected proportions are given, assume equal expected frequencies for each city: $$E = \frac{T}{4} = \frac{260}{4} = 65$$ - Expected frequencies $E$: Toronto = 65, St. Catharines = 65, Hamilton = 65, Sudbury = 65. 4. **Calculate the chi-square statistic:** - Formula: $$\chi^2 = \sum \frac{(O - E)^2}{E}$$ - Calculate each term: - Toronto: $\frac{(92 - 65)^2}{65} = \frac{27^2}{65} = \frac{729}{65} \approx 11.22$ - St. Catharines: $\frac{(34 - 65)^2}{65} = \frac{(-31)^2}{65} = \frac{961}{65} \approx 14.78$ - Hamilton: $\frac{(85 - 65)^2}{65} = \frac{20^2}{65} = \frac{400}{65} \approx 6.15$ - Sudbury: $\frac{(49 - 65)^2}{65} = \frac{(-16)^2}{65} = \frac{256}{65} \approx 3.94$ - Sum these values: $$\chi^2 = 11.22 + 14.78 + 6.15 + 3.94 = 36.09$$ 5. **Compare to critical value:** - Degrees of freedom $df = k - 1 = 4 - 1 = 3$. - At $\alpha = 0.05$ and $df=3$, the critical chi-square value is approximately 7.815. - Since $36.09 > 7.815$, we reject the null hypothesis. 6. **Conclusion:** - There is strong evidence to conclude that violent crime rates differ significantly among the four cities. - The observed distribution of crimes is unlikely due to chance if crime rates were equal.