1. **Problem Statement:** We want to test if the posters showing different milk types had an impact on students' drink choices at the 1% significance level. 2. **Data Given:** The observed frequencies of students choosing Regular, Strawberry, and Chocolate milk in three lines are: | | Regular | Strawberry | Chocolate | |----------|---------|------------|-----------| | Regular | 40 | 20 | 35 | | Strawberry| 50 | 12 | 33 | | Chocolate| 60 | 33 | 40 | 3. **Hypotheses:** - Null hypothesis $H_0$: The posters had no impact on students' drink choice (the variables are independent). - Alternative hypothesis $H_a$: The posters had some impact (the variables are dependent). 4. **Test Used:** Chi-square test for independence. 5. **Calculate row totals, column totals, and grand total:** - Row totals: Regular $= 40+20+35=95$, Strawberry $= 50+12+33=95$, Chocolate $= 60+33+40=133$ - Column totals: Regular $= 40+50+60=150$, Strawberry $= 20+12+33=65$, Chocolate $= 35+33+40=108$ - Grand total $= 95+95+133=323$ 6. **Calculate expected frequencies:** Expected frequency for each cell $= \frac{(\text{row total})(\text{column total})}{\text{grand total}}$ For example, expected frequency for Regular row and Regular column: $$E_{11} = \frac{95 \times 150}{323} = \frac{14250}{323} \approx 44.12$$ Calculate all expected frequencies: $$E = \begin{bmatrix} 44.12 & 19.14 & 31.74 \\ 44.12 & 19.14 & 31.74 \\ 61.76 & 26.72 & 44.52 \end{bmatrix}$$ 7. **Calculate the chi-square statistic:** $$\chi^2 = \sum \frac{(O - E)^2}{E}$$ Where $O$ is observed frequency and $E$ is expected frequency. Calculate each term: - $(40-44.12)^2/44.12 = 0.38$ - $(20-19.14)^2/19.14 = 0.04$ - $(35-31.74)^2/31.74 = 0.34$ - $(50-44.12)^2/44.12 = 0.79$ - $(12-19.14)^2/19.14 = 2.66$ - $(33-31.74)^2/31.74 = 0.05$ - $(60-61.76)^2/61.76 = 0.05$ - $(33-26.72)^2/26.72 = 1.48$ - $(40-44.52)^2/44.52 = 0.46$ Sum all terms: $$\chi^2 = 0.38 + 0.04 + 0.34 + 0.79 + 2.66 + 0.05 + 0.05 + 1.48 + 0.46 = 6.25$$ 8. **Degrees of freedom:** $$df = (\text{number of rows} - 1)(\text{number of columns} - 1) = (3-1)(3-1) = 4$$ 9. **Critical value at 1% significance level for $df=4$:** From chi-square tables, $\chi^2_{0.01,4} = 13.277$ 10. **Decision:** Since calculated $\chi^2 = 6.25 < 13.277$, we fail to reject the null hypothesis. 11. **Conclusion:** There is not enough evidence at the 1% significance level to conclude that the posters had an impact on students' drink choices. **Final answer:** A. The posters had no impact on students' drink choice