A student researcher compares the heights of American students and nonAmerican students from the student body of a certain college in order to estimate the difference in their mean heights. A random sample of 12 American students had a mean height of 70.4 inches with a standard deviation of 2.31 inches. A random sample of 17 non-American students had a mean height of 63.2 inches with a standard deviation of 2.68 inches. Determine the $95 \%$ confidence interval for the true mean difference between the mean height of the American students and the mean height of the non-American students. Assume that the population variances are equal and that the two populations are normally distributed. Step 1 of 3: Find the point estimate that should be used in constructing the confidence interval.

Step 1 ：The point estimate for the difference in means is simply the difference in sample means. So, we subtract the mean height of non-American students from the mean height of American students: \(70.4 - 63.2 = 7.2\) inches.

Step 2 ：Step 2 of 3: Find the standard error that should be used in constructing the confidence interval. The standard error for the difference in means, assuming equal variances, is given by the formula: \(\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\), where \(s_1\) and \(s_2\) are the sample standard deviations and \(n_1\) and \(n_2\) are the sample sizes. Substituting the given values, we get: \(\sqrt{\frac{2.31^2}{12} + \frac{2.68^2}{17}} = 0.96\) inches.

Step 3 ：Step 3 of 3: Construct the confidence interval. For a 95% confidence interval, the critical value from the t-distribution (with degrees of freedom min(\(n_1 - 1, n_2 - 1\)) = min(11, 16) = 11) is approximately 2.201. So, the confidence interval is given by: \(7.2 \pm 2.201 \times 0.96 = (5.07, 9.33)\) inches. Therefore, we are 95% confident that the true mean difference in height between American and non-American students is between 5.07 and 9.33 inches.