Question $12-\quad$ of 19 Step 2 of 2 An automotive manufacturer wants to know the proportion of new car buyers who prefer foreign cars over domestic. Step 2 of 2: Suppose a sample of 675 new car buyers is drawn. Of those sampled, 270 preferred foreign over domestic cars. Using the data, construct the $90 \%$ confidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars. Round your answers to three decimal places. Answer How to enter your answer (opens in new window) 2 Points

Step 1 ：The problem is asking for a confidence interval for a population proportion. The formula for a confidence interval for a population proportion is given by: \(\hat{p} \pm Z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\) where \(\hat{p}\) is the sample proportion, \(Z\) is the Z-score corresponding to the desired level of confidence, and \(n\) is the sample size.

Step 2 ：In this case, \(\hat{p}\) is the proportion of new car buyers who prefer foreign cars over domestic cars in the sample, which is \(\frac{270}{675}\), \(Z\) is the Z-score corresponding to a $90\%$ confidence level, and \(n\) is the sample size, which is $675$.

Step 3 ：The Z-score for a $90\%$ confidence level is approximately $1.645$ (you can find this value in a standard normal distribution table).

Step 4 ：We can plug these values into the formula to find the confidence interval. After running the calculations, we find the lower and upper bounds of the confidence interval to be approximately $0.369$ and $0.431$ respectively.

Step 5 ：Final Answer: The $90\%$ confidence interval for the population proportion of new car buyers who prefer foreign cars over domestic cars is \(\boxed{[0.369, 0.431]}\).