You want to take a trip to Paris. You randomly select 225 flights to Europe and find a mean and sample standard deviation of \$1500 and \$900, respectively. Construct and interpret a 95% confidence interval for the true mean price for a trip to Paris.

• A. (1381.74, 1618.26)
• B. (702.9, 1097.1)
• C. (897.5, 902.5)
• D. (1498.5, 1501.5)

Answer: (A)

Verified step by step guidance

1. Identify the sample mean (\(\bar{x}\)) and sample standard deviation (s) from the problem. Here, \(\bar{x} = 1500\) and \(s = 900\).
2. Determine the sample size (n), which is given as 225 flights.
3. Since the sample size is large (n > 30), use the Z-distribution to construct the confidence interval. For a 95% confidence level, the Z-score is approximately 1.96.
4. Calculate the standard error of the mean (SE) using the formula: \(SE = \frac{s}{\sqrt{n}}\). Substitute the values to find SE.
5. Construct the confidence interval using the formula: \(\bar{x} \pm Z \times SE\). Substitute the values to find the lower and upper bounds of the confidence interval.