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 2 of 3 : Find the standard error of the sampling distribution to be used in constructing the confidence interval. Round your answer 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 standard error of the sampling distribution to be used in constructing the confidence interval. The standard error of the difference in means can be calculated using the formula: \[SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\] where \(s_1\) and \(s_2\) are the standard deviations of the two samples, and \(n_1\) and \(n_2\) are the sizes of the two samples.

Step 3 ：Substituting the given values into this formula, we get: \[SE = \sqrt{\frac{15^2}{13} + \frac{16^2}{7}}\]

Step 4 ：Solving the above expression, we find that the standard error of the difference in means is approximately 7.34.

Step 5 ：However, the question asks for the answer to be rounded to the nearest whole number. So, rounding 7.34 to the nearest whole number, we get 7.

Step 6 ：Final Answer: The standard error of the sampling distribution used in constructing the confidence interval, rounded to the nearest whole number, is \(\boxed{7}\).