1. **State the problem:** We want to test if there is a significant difference in course enrollment between male and female students using the Chi-Square test at a 10% significance level. 2. **Set up hypotheses:** - Null hypothesis $H_0$: Gender and course enrollment are independent (no relationship). - Alternative hypothesis $H_a$: Gender and course enrollment are dependent (there is a relationship). 3. **Observed frequencies (O):** \begin{array}{c|ccc|c} \text{Gender} & \text{M\&E} & \text{Procurement} & \text{HRM} & \text{Total} \\ \hline \text{Male} & 15 & 23 & 10 & 48 \\ \text{Female} & 25 & 19 & 8 & 52 \\ \hline \text{Total} & 40 & 42 & 18 & 100 \end{array} 4. **Calculate expected frequencies (E) using:** $$E = \frac{(\text{row total}) \times (\text{column total})}{\text{grand total}}$$ For Male, M&E: $$E = \frac{48 \times 40}{100} = 19.2$$ For Male, Procurement: $$E = \frac{48 \times 42}{100} = 20.16$$ For Male, HRM: $$E = \frac{48 \times 18}{100} = 8.64$$ For Female, M&E: $$E = \frac{52 \times 40}{100} = 20.8$$ For Female, Procurement: $$E = \frac{52 \times 42}{100} = 21.84$$ For Female, HRM: $$E = \frac{52 \times 18}{100} = 9.36$$ 5. **Calculate Chi-Square statistic:** $$\chi^2 = \sum \frac{(O - E)^2}{E}$$ Calculate each term: - Male, M&E: $\frac{(15 - 19.2)^2}{19.2} = \frac{(-4.2)^2}{19.2} = \frac{17.64}{19.2} = 0.919$ - Male, Procurement: $\frac{(23 - 20.16)^2}{20.16} = \frac{2.84^2}{20.16} = \frac{8.066}{20.16} = 0.400$ - Male, HRM: $\frac{(10 - 8.64)^2}{8.64} = \frac{1.36^2}{8.64} = \frac{1.850}{8.64} = 0.214$ - Female, M&E: $\frac{(25 - 20.8)^2}{20.8} = \frac{4.2^2}{20.8} = \frac{17.64}{20.8} = 0.848$ - Female, Procurement: $\frac{(19 - 21.84)^2}{21.84} = \frac{(-2.84)^2}{21.84} = \frac{8.066}{21.84} = 0.369$ - Female, HRM: $\frac{(8 - 9.36)^2}{9.36} = \frac{(-1.36)^2}{9.36} = \frac{1.850}{9.36} = 0.198$ Sum all terms: $$\chi^2 = 0.919 + 0.400 + 0.214 + 0.848 + 0.369 + 0.198 = 2.948$$ 6. **Degrees of freedom (df):** $$df = (\text{number of rows} - 1) \times (\text{number of columns} - 1) = (2 - 1) \times (3 - 1) = 1 \times 2 = 2$$ 7. **Critical value:** At significance level $\alpha = 0.10$ and $df=2$, the critical value from Chi-Square table is approximately $4.605$. 8. **Decision:** Since $\chi^2 = 2.948 < 4.605$, we fail to reject the null hypothesis. **Conclusion:** There is no significant difference in course enrollment between male and female students at the 10% significance level.