Two samples are taken with the following sample means, sizes, and standard deviations given below. Assume the populations are normal and the samples are independent.
\[
\begin{array}{ll}
\bar{x}_{1}=31 & \bar{x}_{2}=38 \\
n_{1}=58 & n_{2}=68 \\
s_{1}=2 & s_{2}=5
\end{array}
\]

Find a $86 \%$ confidence interval, round answers to the nearest hundredth.
\[
\square< \mu_{1}-\mu_{2}< \square
\]

Step 1 ：Given the sample means \(\bar{x}_{1}=31\) and \(\bar{x}_{2}=38\), sample sizes \(n_{1}=58\) and \(n_{2}=68\), and standard deviations \(s_{1}=2\) and \(s_{2}=5\). We are asked to find a \(86\%\) confidence interval for the difference between the two means.

Step 2 ：First, we calculate the standard error, which is the square root of the sum of the variances divided by their respective sample sizes. The formula for the standard error is \(\sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}\).

Step 3 ：Substituting the given values into the formula, we get \(\sqrt{\frac{2^{2}}{58} + \frac{5^{2}}{68}}\), which simplifies to approximately \(0.66\).

Step 4 ：Next, we calculate the margin of error by multiplying the standard error by the z-score for an \(86\%\) confidence level, which is \(1.47\). So, the margin of error is \(1.47 \times 0.66\), which is approximately \(0.97\).

Step 5 ：Finally, we calculate the confidence interval by subtracting and adding the margin of error from the difference of the sample means. The lower limit of the confidence interval is \((31 - 38) - 0.97\), which is approximately \(-7.97\). The upper limit of the confidence interval is \((31 - 38) + 0.97\), which is approximately \(-6.03\).

Step 6 ：So, the \(86\%\) confidence interval for the difference between the two means is \(\boxed{-7.97 < \mu_{1} - \mu_{2} < -6.03}\).