1. **State the problem:** We want to test if male students who participate in college athletics have a higher mean mass than those who do not. 2. **Given data:** - Group 1 (participants): $n_1=50$, mean $\bar{x}_1=68.2$, standard deviation $s_1=2.5$ - Group 2 (non-participants): $n_2=50$, mean $\bar{x}_2=67.5$, standard deviation $s_2=2.8$ - Significance level $\alpha=0.05$ 3. **Hypotheses:** - Null hypothesis $H_0$: $\mu_1 \leq \mu_2$ (participants are not more massive) - Alternative hypothesis $H_a$: $\mu_1 > \mu_2$ (participants are more massive) 4. **Test statistic:** Use two-sample t-test for means with unequal variances: $$ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} $$ 5. **Calculate:** $$ \text{Standard error} = \sqrt{\frac{2.5^2}{50} + \frac{2.8^2}{50}} = \sqrt{\frac{6.25}{50} + \frac{7.84}{50}} = \sqrt{0.125 + 0.1568} = \sqrt{0.2818} \approx 0.531 $$ $$ t = \frac{68.2 - 67.5}{0.531} = \frac{0.7}{0.531} \approx 1.317 $$ 6. **Degrees of freedom (Welch-Satterthwaite equation):** $$ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{(s_1^2/n_1)^2}{n_1-1} + \frac{(s_2^2/n_2)^2}{n_2-1}} = \frac{0.2818^2}{\frac{(0.125)^2}{49} + \frac{(0.1568)^2}{49}} = \frac{0.0794}{\frac{0.0156}{49} + \frac{0.0246}{49}} = \frac{0.0794}{0.000318 + 0.000502} = \frac{0.0794}{0.00082} \approx 96.8 $$ 7. **Critical value:** For $\alpha=0.05$ and $df \approx 97$, the one-tailed critical t-value is approximately $1.66$. 8. **Decision:** Since $t=1.317 < 1.66$, we fail to reject the null hypothesis. 9. **Conclusion:** There is not enough evidence at the 0.05 significance level to conclude that male students who participate in college athletics are more massive than those who do not.