1. **State the problem:** We want to test if there is a dependent relationship between buying beer and buying diapers using the given contingency table and a significance level of 0.01. 2. **Given data:** | | Bought Diapers | Did Not Buy Diapers | Totals | |---------------|----------------|---------------------|--------| | Beer | 7 | 50 | 57 | | No Beer | 9 | 61 | 70 | | Totals | 16 | 111 | 127 | 3. **Formula for expected frequency:** $$E_{ij} = \frac{(\text{row total}) \times (\text{column total})}{\text{grand total}}$$ 4. **Calculate expected frequencies:** - For Beer and Bought Diapers: $$E_{11} = \frac{57 \times 16}{127} = \frac{912}{127} \approx 7.18$$ - For Beer and Did Not Buy Diapers: $$E_{12} = \frac{57 \times 111}{127} = \frac{6327}{127} \approx 49.82$$ - For No Beer and Bought Diapers: $$E_{21} = \frac{70 \times 16}{127} = \frac{1120}{127} \approx 8.82$$ - For No Beer and Did Not Buy Diapers: $$E_{22} = \frac{70 \times 111}{127} = \frac{7770}{127} \approx 61.18$$ 5. **Calculate the test statistic (Chi-square):** $$\chi^2 = \sum \frac{(O - E)^2}{E}$$ Where $O$ is observed frequency and $E$ is expected frequency. Calculate each term: - $\frac{(7 - 7.18)^2}{7.18} = \frac{(-0.18)^2}{7.18} = \frac{0.0324}{7.18} \approx 0.0045$ - $\frac{(50 - 49.82)^2}{49.82} = \frac{0.18^2}{49.82} = \frac{0.0324}{49.82} \approx 0.00065$ - $\frac{(9 - 8.82)^2}{8.82} = \frac{0.18^2}{8.82} = \frac{0.0324}{8.82} \approx 0.0037$ - $\frac{(61 - 61.18)^2}{61.18} = \frac{(-0.18)^2}{61.18} = \frac{0.0324}{61.18} \approx 0.00053$ Sum these: $$\chi^2 \approx 0.0045 + 0.00065 + 0.0037 + 0.00053 = 0.00938$$ 6. **Find the critical value:** Degrees of freedom $df = (rows - 1)(columns - 1) = (2-1)(2-1) = 1$ At significance level $\alpha = 0.01$ and $df=1$, the critical value from Chi-square table is approximately 6.635. 7. **Decision:** Since $\chi^2 = 0.00938 < 6.635$, we fail to reject the null hypothesis. **Conclusion:** There is not sufficient evidence at the 0.01 significance level to support a dependent relationship between buying beer and buying diapers. **Final answers:** (a) Expected frequencies: | | Bought Diapers | Did Not Buy Diapers | |---------------|----------------|---------------------| | Beer | 7.18 | 49.82 | | No Beer | 8.82 | 61.18 | (b) Test statistic: $0.00938$ (c) Critical value: $6.635$ (d) Sufficient data to support claim? No