Waterfall Heights The data show the heights in feet of waterfalls in Europe and in Asia. Find the $90 \%$ confidence interval for the difference of the means. Round the answers to one decimal place. \begin{tabular}{ccc|cc} \multicolumn{3}{c|}{ Europe } & \multicolumn{2}{c}{ Asia } \\ \hline 487 & 345 & 820 & 614 & 320 \\ 900 & 1385 & & 1137 & 722 \\ 1246 & 984 & & 722 & 830 \end{tabular} Send data to Excel Use $\mu_{1}$ for the mean height of waterfalls in Europe. Assume the variables are normally distributed and the variances are unequal. Do not round intermediate steps. \[ \square<\mu_{1}-\mu_{2}<\square \] Save For Later Submit Assign

Step 1 ：Calculate the mean for Europe: \(\mu_1 = \frac{487+345+820+900+1385+1246+984}{7} = 881.6\)

Step 2 ：Calculate the standard deviation for Europe: \(s_1 = \sqrt{\frac{(487-881.6)^2+(345-881.6)^2+(820-881.6)^2+(900-881.6)^2+(1385-881.6)^2+(1246-881.6)^2+(984-881.6)^2}{7-1}} = 367.3\)

Step 3 ：Calculate the mean for Asia: \(\mu_2 = \frac{614+320+1137+722+830}{5} = 724.6\)

Step 4 ：Calculate the standard deviation for Asia: \(s_2 = \sqrt{\frac{(614-724.6)^2+(320-724.6)^2+(1137-724.6)^2+(722-724.6)^2+(830-724.6)^2}{5-1}} = 293.3\)

Step 5 ：Calculate the standard error of the difference between the two means: \(SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{367.3^2}{7} + \frac{293.3^2}{5}} = 157.6\)

Step 6 ：The 90% confidence interval for the difference of the means is given by \((\mu_1 - \mu_2) \pm (t*SE)\), where t is the t-value from the t-distribution table with degree of freedom df = min(n1, n2) - 1 = min(7, 5) - 1 = 4. For a 90% confidence level, t = 2.132.

Step 7 ：Calculate the 90% confidence interval for the difference of the means: \((881.6 - 724.6) \pm (2.132*157.6) = 157.0 \pm 335.9\)

Step 8 ：Therefore, the 90% confidence interval for the difference of the means is \(\boxed{(-178.9, 492.9)}\), rounded to one decimal place.