Questions: At a certain plant, cars are being produced with a gas mileage that has a standard deviation of 1.8 miles per gallon. Suppose the average gas mileage in a simple random sample of 19 tested cars is 30.01 miles per gallon. Construct and interpret a 92% confidence interval for the average gas mileage for all cars produced at the plant. You may also assume that the sample is from a normal population.

Solution Steps

To construct a confidence interval, we first need to determine the Z-score corresponding to the desired confidence level of \(92\%\). The Z-score for a \(92\%\) confidence level is approximately \(Z = 1.7507\).

The margin of error (ME) can be calculated using the formula:

\[ \text{Margin of Error} = Z \times \frac{\sigma}{\sqrt{n}} \]

Substituting the known values:

\[ \text{Margin of Error} = 1.7507 \times \frac{1.8}{\sqrt{19}} \approx 0.7229 \]

The confidence interval for the mean can be expressed as:

\[ \bar{x} \pm \text{Margin of Error} \]

Where \(\bar{x} = 30.01\). Thus, the confidence interval is:

\[ 30.01 \pm 0.7229 \]

Calculating the lower and upper bounds:

\[ \text{Lower Bound} = 30.01 - 0.7229 \approx 29.29 \] \[ \text{Upper Bound} = 30.01 + 0.7229 \approx 30.73 \]

Final Answer