The university data center has two main computers. The center wants to examine whether computer 1 is receiving tasks that require processing times comparable to those of computer 2 . A random sample of 10 processing times from computer 1 showed a mean of 68 seconds with a standard deviation of 18 seconds, while a random sample of 9 processing times from computer 2 (chosen independently of those for computer 1 ) showed a mean of 63 seconds with a standard deviation of 20 seconds. Assume that the populations of processing times are normally distributed for each of the two computers and that the variances are equal. Construct a $95 \%$ confidence interval for the difference $\mu_{1}-\mu_{2}$ between the mean processing time of computer 1 , $\mu_{1}$, and the mean processing time of computer $2, \mu_{2}$. Then find the lower limit and upper limit of the $95 \%$ confidence interval. Carry your intermedlate computations to at least three decimal places. Round your responses to at least two decimal places. (If necessary, consult a list of formulas.) Explanation Check

Step 1 ：We are given that the sample mean processing time for computer 1, denoted as \(\bar{x}_1\), is 68 seconds, and the sample standard deviation, denoted as \(s_1\), is 18 seconds. The sample size, denoted as \(n_1\), is 10.

Step 2 ：For computer 2, the sample mean processing time, denoted as \(\bar{x}_2\), is 63 seconds, and the sample standard deviation, denoted as \(s_2\), is 20 seconds. The sample size, denoted as \(n_2\), is 9.

Step 3 ：We are asked to construct a 95% confidence interval for the difference between the mean processing times of the two computers, denoted as \(\mu_1 - \mu_2\). The formula for the confidence interval for the difference between two means is given by: \(CI = (\bar{x}_1 - \bar{x}_2) \pm z \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\), where \(z\) is the z-score corresponding to the desired level of confidence. For a 95% confidence interval, \(z\) is approximately 1.96.

Step 4 ：Substituting the given values into the formula, we get: \(CI = (68 - 63) \pm 1.96 \sqrt{\frac{18^2}{10} + \frac{20^2}{9}}\).

Step 5 ：Solving the above expression, we find the lower and upper limits of the 95% confidence interval for the difference between the mean processing times of the two computers.

Step 6 ：The lower limit of the 95% confidence interval is \(\boxed{-12.18}\) and the upper limit is \(\boxed{22.18}\). This means that we are 95% confident that the true difference between the mean processing times of the two computers lies within this interval.