An environmental group at a local college is conducting independent tests to determine the distance a particular make of automobile will tr while consuming only 1 gallon of gas. They test a sample of five cars obtain a mean of 28.2 miles. Assuming that the population standard deviation is 2.7 miles, find the 95 percent confidence interval for the mean distance traveled by all such cars using 1 gallon of gas. eBook [26.70, 35.70]

We are given the following information: - Sample mean: \( \overline{x} = 28.2 \) miles - Population standard deviation: \( \sigma = 2.7 \) miles - Sample size: \( n = 5 \) - Confidence level: 95% Because the population standard deviation is known, we use the standard normal (z) distribution to construct the confidence interval. **Step 1. Choose the Critical Value** For a 95% confidence interval, the critical value from the standard normal distribution is \[ z_{\alpha/2} = 1.96. \] **Step 2. Calculate the Standard Error** The standard error (SE) is given by \[ \text{SE} = \frac{\sigma}{\sqrt{n}}. \] Substitute the values: \[ \text{SE} = \frac{2.7}{\sqrt{5}} \approx \frac{2.7}{2.23607} \approx 1.207. \] **Step 3. Compute the Margin of Error** The margin of error (ME) is \[ \text{ME} = z_{\alpha/2} \times \text{SE} \approx 1.96 \times 1.207 \approx 2.364. \] **Step 4. Construct the Confidence Interval** The confidence interval is given by: \[ \overline{x} \pm \text{ME}. \] Substitute the known values: \[ \text{Lower bound} = 28.2 - 2.364 \approx 25.836, \] \[ \text{Upper bound} = 28.2 + 2.364 \approx 30.564. \] Thus, the 95% confidence interval for the mean distance traveled is approximately: \[ [25.84,\, 30.56]. \] *Note:* The eBook answer given as \([26.70,\, 35.70]\) does not align with the provided sample mean of \(28.2\) miles. The steps above show the interval computed directly from the given values.