1. **State the problem:** We want to determine if there is a significant difference between the performance of males and females in Math based on the given paired data. 2. **Hypotheses:** - Null hypothesis $H_0$: There is no difference in mean performance between males and females, i.e., $\mu_d = 0$ where $d$ is the difference. - Alternative hypothesis $H_a$: There is a difference, i.e., $\mu_d \neq 0$. 3. **Data:** | No | Male | Female | Difference $d = Male - Female$ | |---|---|---|---| | 1 | 34 | 32 | 2 | | 2 | 45 | 48 | -3 | | 3 | 38 | 40 | -2 | | 4 | 34 | 30 | 4 | | 5 | 43 | 41 | 2 | 4. **Calculate the mean difference $\bar{d}$:** $$\bar{d} = \frac{2 + (-3) + (-2) + 4 + 2}{5} = \frac{3}{5} = 0.6$$ 5. **Calculate the standard deviation of differences $s_d$:** First find squared deviations: $$(2 - 0.6)^2 = 1.96, \quad (-3 - 0.6)^2 = 12.96, \quad (-2 - 0.6)^2 = 6.76, \quad (4 - 0.6)^2 = 11.56, \quad (2 - 0.6)^2 = 1.96$$ Sum of squared deviations: $$1.96 + 12.96 + 6.76 + 11.56 + 1.96 = 35.2$$ Variance: $$s_d^2 = \frac{35.2}{5 - 1} = \frac{35.2}{4} = 8.8$$ Standard deviation: $$s_d = \sqrt{8.8} \approx 2.966$$ 6. **Calculate the test statistic $t$:** $$t = \frac{\bar{d} - 0}{s_d / \sqrt{n}} = \frac{0.6}{2.966 / \sqrt{5}} = \frac{0.6}{2.966 / 2.236} = \frac{0.6}{1.326} \approx 0.452$$ 7. **Degrees of freedom:** $df = n - 1 = 4$ 8. **Decision rule:** For a two-tailed test at significance level $\alpha = 0.05$, critical $t$ value from $t$-table with 4 df is approximately $\pm 2.776$. 9. **Conclusion:** Since $|t| = 0.452 < 2.776$, we fail to reject the null hypothesis. **Final answer:** There is no significant difference between the performance of males and females in Math based on this data.