1. **State the problem:** We want to determine if there is an association between packaging material and spoilage status, i.e., whether they are independent. 2. **Method:** Use the Chi-square test of independence. 3. **Formula:** $$\chi^2 = \sum \frac{(O - E)^2}{E}$$ where $O$ is the observed frequency and $E$ is the expected frequency under independence. 4. **Calculate expected frequencies:** Expected frequency for each cell = $\frac{(\text{row total}) \times (\text{column total})}{\text{grand total}}$ For example, for Retort pouch and Spoilage: Yes: $$E = \frac{120 \times 42}{300} = 16.8$$ Calculate all expected values: - Retort pouch: Yes: 16.8, No: 87.2, Minor: 16 - Glass jar: Yes: 15.4, No: 79.9, Minor: 14.7 - Plastic bottle: Yes: 9.8, No: 50.9, Minor: 9.3 5. **Calculate $\chi^2$ contributions for each cell:** $$\frac{(O - E)^2}{E}$$ - Retort pouch: Yes: $\frac{(10 - 16.8)^2}{16.8} = 2.77$ - Retort pouch: No: $\frac{(82 - 87.2)^2}{87.2} = 0.31$ - Retort pouch: Minor: $\frac{(20 - 16)^2}{16} = 0.75$ - Glass jar: Yes: $\frac{(9 - 15.4)^2}{15.4} = 2.66$ - Glass jar: No: $\frac{(91 - 79.9)^2}{79.9} = 1.54$ - Glass jar: Minor: $\frac{(10 - 14.7)^2}{14.7} = 1.52$ - Plastic bottle: Yes: $\frac{(15 - 9.8)^2}{9.8} = 2.81$ - Plastic bottle: No: $\frac{(45 - 50.9)^2}{50.9} = 0.68$ - Plastic bottle: Minor: $\frac{(10 - 9.3)^2}{9.3} = 0.05$ 6. **Sum all contributions:** $$\chi^2 = 2.77 + 0.31 + 0.75 + 2.66 + 1.54 + 1.52 + 2.81 + 0.68 + 0.05 = 12.09$$ 7. **Degrees of freedom:** $$(\text{rows} - 1) \times (\text{columns} - 1) = (3 - 1) \times (3 - 1) = 4$$ 8. **Interpretation:** Compare $\chi^2 = 12.09$ with critical value from Chi-square table at $\alpha=0.05$ and 4 degrees of freedom (9.488). Since $12.09 > 9.488$, reject the null hypothesis. **Conclusion:** There is evidence to suggest packaging material and spoilage status are not independent; they are associated.