Here is the exact text as provided: --- \[_a : \mu_1 -mu_2 \Test Statistic Formula\[ t \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\{s_1^2}{n1} + \frac{s_2^2}{n_}}} \Substitute: [ t = \frac{23.6 - 20.1}{\sqrt{\frac{4.8^2}{50} + \frac{5.^2}{55}}} \] \[ = \frac{3.5}{\sqrt{\frac{23.04}{50} + \frac{25}{55}}} \] \[ = \frac{3.5}{\sqrt{.4608 + .4545}} = \frac{3.5}{\sqrt{.9153}} \] \[ t = \frac{3.5}{.9567} \approx 3.66 \] --- \[ t = \frac{3.5}{.9567} \approx 3.66 \] --- **Step 3 — Degrees of Freedom (Welch–Satterthwaite)** \[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} \] \[ df = \frac{\left( \frac{.4608 + .4545}{1} \right)^2}{\frac{\left(\frac{.4608}{50}\right)^2}{49} + \frac{\left(\frac{.4545}{55}\right)^2}{54}} \] \[ df = \frac{(.9153)^2}{\frac{(.009216)^2}{49} + \frac{(.008264)^2}{54}} \approx \frac{.8378}{.000188 + .000126} \approx 102.68 \] \[ d_f \approx 103 \] --- **Step 4 — P-value** Now, \[ t = 3.66, \quad d_f = 103 \] For a one-tailed test: \[ p = P(T_{df} > 3.66) \] Using Excel: (not shown here, but implied) ---