One gallon of gas is put into each of 30 test cars. The resulting gas-mileage values of the sample are computed with mean of 28.5 miles per gallon, and standard deviation of 1.2 miles per gallon. What is the $95 \%$ confidence interval estimate of the mean mileage? $(28.42,28.58)$ $(27.3,29.7)$ $(28.1,28.9)$ $(27.36,29.64)$ $(28.46,28.54)$

Step 1 ：We are given that one gallon of gas is put into each of 30 test cars. The resulting gas-mileage values of the sample are computed with a mean of 28.5 miles per gallon, and a standard deviation of 1.2 miles per gallon. We are asked to find the $95 \%$ confidence interval estimate of the mean mileage.

Step 2 ：To solve this problem, we need to use the formula for the confidence interval of the mean. The formula is given by: \(\bar{x} \pm z \frac{s}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(z\) is the z-score corresponding to the desired confidence level, \(s\) is the standard deviation of the sample, and \(n\) is the sample size.

Step 3 ：In this case, we have \(\bar{x} = 28.5\) miles per gallon, \(s = 1.2\) miles per gallon, and \(n = 30\) cars. The z-score for a $95\%$ confidence level is approximately \(1.96\).

Step 4 ：We can plug these values into the formula to find the confidence interval. The lower bound of the confidence interval is calculated as \(28.5 - 1.96 \times \frac{1.2}{\sqrt{30}} = 28.07\) and the upper bound is calculated as \(28.5 + 1.96 \times \frac{1.2}{\sqrt{30}} = 28.93\).

Step 5 ：The $95 \%$ confidence interval estimate of the mean mileage is \(\boxed{(28.07, 28.93)}\). However, this exact interval is not one of the options given in the question. The closest option is \(\boxed{(28.1, 28.9)}\), so this would be the best choice given the options.