1. **State the problem:** We want to find the 95% confidence interval (CI) for the difference between two population means, $\mu_1 - \mu_2$, based on two independent samples. 2. **Given data:** - Sample 1: $n_1 = 40$, $\bar{x}_1 = 105$, $s_1 = 12$ - Sample 2: $n_2 = 50$, $\bar{x}_2 = 98$, $s_2 = 15$ - Confidence level = 95% 3. **Formula for the CI of difference of means (assuming unequal variances):** $$\left(\bar{x}_1 - \bar{x}_2\right) \pm t^* \times \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$$ where $t^*$ is the critical value from the $t$-distribution with degrees of freedom calculated by the Welch-Satterthwaite equation. 4. **Calculate the difference of sample means:** $$\bar{x}_1 - \bar{x}_2 = 105 - 98 = 7$$ 5. **Calculate the standard error (SE):** $$SE = \sqrt{\frac{12^2}{40} + \frac{15^2}{50}} = \sqrt{\frac{144}{40} + \frac{225}{50}} = \sqrt{3.6 + 4.5} = \sqrt{8.1} \approx 2.846$$ 6. **Calculate degrees of freedom (df) using Welch-Satterthwaite formula:** $$df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} = \frac{(3.6 + 4.5)^2}{\frac{3.6^2}{39} + \frac{4.5^2}{49}} = \frac{8.1^2}{\frac{12.96}{39} + \frac{20.25}{49}} = \frac{65.61}{0.3323 + 0.4133} = \frac{65.61}{0.7456} \approx 88.0$$ 7. **Find the critical $t^*$ value for 95% CI and $df \approx 88$:** From $t$-tables or calculator, $t^* \approx 1.987$ 8. **Calculate margin of error (ME):** $$ME = t^* \times SE = 1.987 \times 2.846 \approx 5.655$$ 9. **Construct the confidence interval:** $$7 \pm 5.655 = (7 - 5.655, 7 + 5.655) = (1.345, 12.655)$$ **Final answer:** The 95% confidence interval for $\mu_1 - \mu_2$ is approximately $$\boxed{(1.35, 12.66)}$$.