Score: $0 / 3$ Penalty: none Question Watch Video Show Examples A survey was given to a rancom sample of 300 voters in the United States to ask about their preference for a presidential candidate. Of those surveyed, $49 \%$ of the people said they preferred Candidate A. Determine a $95 \%$ confidence interval for the percentage of people who prefer Candidate A, rounding values to the nearest tenth. Answer Attempt 1 out of 2 Submit Answer

Step 1 ：The problem is asking for a 95% confidence interval for the percentage of people who prefer Candidate A. To calculate a confidence interval, we need to know the sample proportion and the sample size. The sample proportion is the percentage of people who prefer Candidate A, which is 49%. The sample size is the number of people surveyed, which is 300.

Step 2 ：The formula for a confidence interval is: \[\bar{x} \pm z \sqrt{\frac{\bar{x}(1-\bar{x})}{n}}\] where: - \(\bar{x}\) is the sample proportion, - \(z\) is the z-score corresponding to the desired confidence level (for a 95% confidence level, \(z\) is approximately 1.96), - \(n\) is the sample size.

Step 3 ：We can plug in the given values into this formula to calculate the confidence interval. The sample proportion is 0.49, the sample size is 300, and the z-score is 1.96.

Step 4 ：The standard error is calculated as 0.028861739379323625 and the margin of error is 0.0565690091834743.

Step 5 ：The lower bound of the confidence interval is calculated as 43.3 and the upper bound is 54.7.

Step 6 ：The final answer is: The 95% confidence interval for the percentage of people who prefer Candidate A is \(\boxed{[43.3, 54.7]}\).