Calculate the margin of error of a confidence interval for the difference between two population means using the given information. Round your answer to six decimal places. \[ \sigma_{1}=14.14, n_{1}=98, \sigma_{2}=10.88, n_{2}=86, c=0.85 \] Answer How to enter your answer (opens in new window)

Step 1 ：We are given the standard deviations of the two populations as \(\sigma_1 = 14.14\) and \(\sigma_2 = 10.88\), the sizes of the two samples as \(n_1 = 98\) and \(n_2 = 86\), and the confidence level as \(c = 0.85\).

Step 2 ：We need to find the z-score corresponding to a confidence level of 0.85. This can be done using a standard normal distribution table or a z-score calculator. The z-score corresponding to a confidence level of 0.85 is approximately \(z_c = 1.4395314709384563\).

Step 3 ：We can now calculate the margin of error using the formula: \[E = z_c \sqrt{\frac{{\sigma_1}^2}{n_1} + \frac{{\sigma_2}^2}{n_2}}\]

Step 4 ：Substituting the given values into the formula, we get \[E = 1.4395314709384563 \sqrt{\frac{{14.14}^2}{98} + \frac{{10.88}^2}{86}}\]

Step 5 ：Solving the above expression, we get \(E = 2.660855\)

Step 6 ：Thus, the margin of error of the confidence interval for the difference between the two population means, rounded to six decimal places, is \(\boxed{2.660855}\)