1. **State the problem:** We want to test if the gender of a professor is independent of the department using the given data. 2. **Data:** | Dept. | Math | Physics | Chemistry | Linguistics | English | |------------|-------|---------|-----------|-------------|---------| | Men | 51 | 80 | 28 | 18 | 35 | | Women | 7 | 7 | 9 | 20 | 24 | 3. **Hypotheses:** - Null hypothesis $H_0$: Gender and department are independent. - Alternative hypothesis $H_a$: Gender and department are not independent. 4. **Test used:** Chi-square test for independence. 5. **Calculate totals:** - Total men $= 51 + 80 + 28 + 18 + 35 = 212$ - Total women $= 7 + 7 + 9 + 20 + 24 = 67$ - Total professors $= 212 + 67 = 279$ - Department totals: - Math $= 51 + 7 = 58$ - Physics $= 80 + 7 = 87$ - Chemistry $= 28 + 9 = 37$ - Linguistics $= 18 + 20 = 38$ - English $= 35 + 24 = 59$ 6. **Expected counts:** Expected count for each cell $= \frac{(\text{row total})(\text{column total})}{\text{grand total}}$ Calculate expected counts for men: - Math men $= \frac{212 \times 58}{279} \approx 44.06$ - Physics men $= \frac{212 \times 87}{279} \approx 66.10$ - Chemistry men $= \frac{212 \times 37}{279} \approx 28.11$ - Linguistics men $= \frac{212 \times 38}{279} \approx 28.88$ - English men $= \frac{212 \times 59}{279} \approx 44.85$ Expected counts for women: - Math women $= \frac{67 \times 58}{279} \approx 13.94$ - Physics women $= \frac{67 \times 87}{279} \approx 20.90$ - Chemistry women $= \frac{67 \times 37}{279} \approx 8.89$ - Linguistics women $= \frac{67 \times 38}{279} \approx 9.12$ - English women $= \frac{67 \times 59}{279} \approx 14.15$ 7. **Calculate the test statistic:** $$\chi^2 = \sum \frac{(O - E)^2}{E}$$ Where $O$ is observed count and $E$ is expected count. Calculate for men: - Math: $\frac{(51 - 44.06)^2}{44.06} = \frac{6.94^2}{44.06} \approx 1.09$ - Physics: $\frac{(80 - 66.10)^2}{66.10} = \frac{13.9^2}{66.10} \approx 2.92$ - Chemistry: $\frac{(28 - 28.11)^2}{28.11} = \frac{(-0.11)^2}{28.11} \approx 0.0004$ - Linguistics: $\frac{(18 - 28.88)^2}{28.88} = \frac{(-10.88)^2}{28.88} \approx 4.10$ - English: $\frac{(35 - 44.85)^2}{44.85} = \frac{(-9.85)^2}{44.85} \approx 2.16$ Calculate for women: - Math: $\frac{(7 - 13.94)^2}{13.94} = \frac{(-6.94)^2}{13.94} \approx 3.45$ - Physics: $\frac{(7 - 20.90)^2}{20.90} = \frac{(-13.9)^2}{20.90} \approx 9.24$ - Chemistry: $\frac{(9 - 8.89)^2}{8.89} = \frac{0.11^2}{8.89} \approx 0.0014$ - Linguistics: $\frac{(20 - 9.12)^2}{9.12} = \frac{10.88^2}{9.12} \approx 12.99$ - English: $\frac{(24 - 14.15)^2}{14.15} = \frac{9.85^2}{14.15} \approx 6.85$ Sum all values: $$\chi^2 \approx 1.09 + 2.92 + 0.0004 + 4.10 + 2.16 + 3.45 + 9.24 + 0.0014 + 12.99 + 6.85 = 42.80$$ **Answer (a):** The test statistic is approximately $42.80$. 8. **Degrees of freedom:** $$df = (\text{number of rows} - 1)(\text{number of columns} - 1) = (2 - 1)(5 - 1) = 4$$ 9. **Critical value:** At significance level $\alpha = 0.05$ and $df=4$, the critical value from chi-square table is approximately $9.488$. **Answer (b):** The critical value is $9.488$. 10. **Decision:** Since the test statistic $42.80$ is greater than the critical value $9.488$, we reject the null hypothesis. **Answer (c):** Yes, there is sufficient evidence to reject the claim that gender is independent of department. --- **Summary:** - (a) Test statistic $\chi^2 \approx 42.80$ - (b) Critical value $9.488$ - (c) Decision: **Yes**, reject independence claim.