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 gallons per mile, and standard deviation of 1.2 miles per gallon. What is the \( 95 \% \) confidence interval estimate of the mean mileage? \( (28.42,28.58) \) \( (28.1,28.9) \) \( (28.46,28.54) \) \( (27.36,29.64) \) \( (27.3,29.7) \)

To find the \( 95\% \) confidence interval for the mean mileage, we can use the formula: \[ \text{Confidence Interval} = \bar{x} \pm z \left( \frac{s}{\sqrt{n}} \right) \] where: - \( \bar{x} \) is the sample mean (28.5 miles per gallon). - \( z \) is the z-value for a 95% confidence level (approximately 1.96). - \( s \) is the sample standard deviation (1.2 miles per gallon). - \( n \) is the sample size (30 cars). Calculating the margin of error: \[ \frac{s}{\sqrt{n}} = \frac{1.2}{\sqrt{30}} \approx 0.219 \] \[ \text{Margin of Error} = 1.96 \times 0.219 \approx 0.429 \] Now we find the interval: \[ 28.5 - 0.429 \approx 28.071 \] \[ 28.5 + 0.429 \approx 28.929 \] Thus, the \( 95\% \) confidence interval estimate of the mean mileage is approximately \( (28.07, 28.93) \), which rounds to \( (28.1, 28.9) \). So, the correct answer is \( (28.1, 28.9) \).