1. **Problem statement:** We want to test if eye color and blood type are independent using the given data for 400 people. 2. **Data:** | Eye Color | A | B | AB | O | Total | |-----------|----|----|----|----|-------| | Blue | 95 | 40 | 80 | 25 | 240 | | Brown | 65 | 50 | 40 | 5 | 160 | | Total |160 | 90 |120 | 30 | 400 | 3. **Hypotheses:** - Null hypothesis $H_0$: Eye color and blood type are independent. - Alternative hypothesis $H_a$: Eye color and blood type are not independent. 4. **Test used:** Chi-square test for independence. 5. **Formula for expected frequency:** $$E_{ij} = \frac{(\text{row total}_i)(\text{column total}_j)}{\text{grand total}}$$ 6. **Calculate expected frequencies:** - For Blue & A: $E_{Blue,A} = \frac{240 \times 160}{400} = 96$ - For Blue & B: $E_{Blue,B} = \frac{240 \times 90}{400} = 54$ - For Blue & AB: $E_{Blue,AB} = \frac{240 \times 120}{400} = 72$ - For Blue & O: $E_{Blue,O} = \frac{240 \times 30}{400} = 18$ - For Brown & A: $E_{Brown,A} = \frac{160 \times 160}{400} = 64$ - For Brown & B: $E_{Brown,B} = \frac{160 \times 90}{400} = 36$ - For Brown & AB: $E_{Brown,AB} = \frac{160 \times 120}{400} = 48$ - For Brown & O: $E_{Brown,O} = \frac{160 \times 30}{400} = 12$ 7. **Calculate chi-square statistic:** $$\chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}}$$ Where $O_{ij}$ are observed frequencies. Calculate each term: - Blue & A: $\frac{(95-96)^2}{96} = \frac{1}{96} \approx 0.0104$ - Blue & B: $\frac{(40-54)^2}{54} = \frac{196}{54} \approx 3.6296$ - Blue & AB: $\frac{(80-72)^2}{72} = \frac{64}{72} \approx 0.8889$ - Blue & O: $\frac{(25-18)^2}{18} = \frac{49}{18} \approx 2.7222$ - Brown & A: $\frac{(65-64)^2}{64} = \frac{1}{64} \approx 0.0156$ - Brown & B: $\frac{(50-36)^2}{36} = \frac{196}{36} \approx 5.4444$ - Brown & AB: $\frac{(40-48)^2}{48} = \frac{64}{48} \approx 1.3333$ - Brown & O: $\frac{(5-12)^2}{12} = \frac{49}{12} \approx 4.0833$ Sum all terms: $$\chi^2 \approx 0.0104 + 3.6296 + 0.8889 + 2.7222 + 0.0156 + 5.4444 + 1.3333 + 4.0833 = 17.1277$$ 8. **Degrees of freedom:** $df = (\text{number of rows} - 1)(\text{number of columns} - 1) = (2-1)(4-1) = 3$ 9. **Critical value at 5% significance level:** From chi-square table, $\chi^2_{0.05,3} = 7.815$ 10. **Decision:** Since $17.1277 > 7.815$, we reject the null hypothesis. 11. **Conclusion:** There is sufficient evidence at the 5% significance level to conclude that eye color and blood type are not independent.