Given two independent random samples with the following results:
\[
\begin{array}{ll}
n_{1}=13 & n_{2}=7 \\
\bar{x}_{1}=94 & \bar{x}_{2}=128 \\
s_{1}=15 & s_{2}=16
\end{array}
\]
Use this data to find the $95 \%$ confidence interval for the true difference between the population means. Assume that the population variances are not equal and that the two populations are normally distributed.

Copy Data
Step 3 of 3: Construct the $95 \%$ corfidence interval. Round your answers to the nearest whole number.

Step 1 ：Given two independent random samples with the following results: \[\begin{array}{ll} n_{1}=13 & n_{2}=7 \\ \bar{x}_{1}=94 & \bar{x}_{2}=128 \\ s_{1}=15 & s_{2}=16 \end{array}\]

Step 2 ：We are asked to find the $95 \%$ confidence interval for the true difference between the population means. We assume that the population variances are not equal and that the two populations are normally distributed.

Step 3 ：The formula for the confidence interval in this case is: \[(\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\] where: \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means, \(s_1\) and \(s_2\) are the sample standard deviations, \(n_1\) and \(n_2\) are the sample sizes, and \(t_{\alpha/2}\) is the t-score for a 95% confidence interval.

Step 4 ：The degrees of freedom for this t-distribution is the smaller of \(n_1 - 1\) and \(n_2 - 1\). In this case, the degrees of freedom is 6.

Step 5 ：Plugging in the given values into the formula, we find that the lower bound of the confidence interval is -52 and the upper bound is -16.

Step 6 ：Thus, the $95 \%$ confidence interval for the true difference between the population means is \(\boxed{(-52, -16)}\).