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 construct a 95% confidence interval for the mean mileage, we'll follow these steps: 1. **Given Data:** - Sample size (\( n \)) = 30 - Sample mean (\( \bar{x} \)) = 28.5 miles per gallon (mpg) - Sample standard deviation (\( s \)) = 1.2 mpg 2. **Determine the Appropriate Distribution:** - Since the sample size is moderate (\( n = 30 \)) and the population standard deviation is unknown, we'll use the **t-distribution**. 3. **Find the Critical t-Value:** - Degrees of freedom (\( df \)) = \( n - 1 = 29 \) - For a 95% confidence interval, the critical t-value (\( t_{\alpha/2, df} \)) ≈ **2.045** 4. **Calculate the Margin of Error (ME):** \[ \text{ME} = t_{\alpha/2, df} \times \left(\frac{s}{\sqrt{n}}\right) = 2.045 \times \left(\frac{1.2}{\sqrt{30}}\right) \approx 2.045 \times 0.2195 \approx 0.448 \] 5. **Construct the Confidence Interval:** \[ \text{CI} = \bar{x} \pm \text{ME} = 28.5 \pm 0.448 \approx (28.052, 28.948) \] Rounding to one decimal place gives approximately \( (28.1, 28.9) \). **Final Answer:** \( (28.1,\ 28.9) \)