1. **Problem Statement:** We want to test if the mean starting salary for UG graduates is higher than that for KNUST graduates at the 5% significance level. 2. **Hypotheses:** - Null hypothesis $H_0$: $\mu_{UG} \leq \mu_{KNUST}$ (mean salary of UG graduates is less than or equal to that of KNUST) - Alternative hypothesis $H_a$: $\mu_{UG} > \mu_{KNUST}$ (mean salary of UG graduates is greater) 3. **Test Selection:** Since we compare means from two independent samples and assume normality, we use a two-sample t-test for the difference of means. 4. **Test Statistic Formula:** $$ t = \frac{\bar{x}_{UG} - \bar{x}_{KNUST}}{\sqrt{\frac{s_{UG}^2}{n_{UG}} + \frac{s_{KNUST}^2}{n_{KNUST}}}} $$ where $\bar{x}$ is sample mean, $s^2$ is sample variance, and $n$ is sample size. 5. **Degrees of Freedom (Welch's approximation):** $$ df = \frac{\left(\frac{s_{UG}^2}{n_{UG}} + \frac{s_{KNUST}^2}{n_{KNUST}}\right)^2}{\frac{\left(\frac{s_{UG}^2}{n_{UG}}\right)^2}{n_{UG}-1} + \frac{\left(\frac{s_{KNUST}^2}{n_{KNUST}}\right)^2}{n_{KNUST}-1}} $$ 6. **Calculate sample means, variances, and sizes from the data provided (not shown here as data is missing).** 7. **Compute the test statistic $t$ using the formula above.** 8. **Find the critical t-value $t_{crit}$ from t-distribution tables at $\alpha=0.05$ and calculated degrees of freedom for a one-tailed test.** 9. **Decision Rule:** - If $t > t_{crit}$, reject $H_0$ and conclude mean salary for UG is significantly higher. - Otherwise, do not reject $H_0$. 10. **Interpretation:** Based on the comparison, conclude whether there is enough evidence at 5% significance level. *Note: Since the actual data values are not provided, numerical calculation cannot be completed here.*