This question: 1 point(s) possible Submit quiz K Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. The results in the Tinterval screen display are based on a $95 \%$ confidence level. Write a statement that correctly interprets the confidence interval. \[ \begin{array}{l} (13.046,22.15) \\ x=17.598 \\ S x=16.01712719 \\ n=50 \end{array} \] Choose the correct answer below. A. We have $95 \%$ confidence that the limits of $13.05 \mathrm{Mbps}$ and $22.15 \mathrm{Mbps}$ contain the true value of the mean of the population of all data speeds at the airports. B. The limits of $13.05 \mathrm{Mbps}$ and $22.15 \mathrm{Mbps}$ contain the true value of the mean of the population of all data speeds at the airports. C. We have $95 \%$ confidence that the limits of $13.05 \mathrm{Mbps}$ and $22.15 \mathrm{Mbps}$ contain the sample mean of the data speeds at the airports. D. The limits of $13.05 \mathrm{Mbps}$ and $22.15 \mathrm{Mbps}$ contain $95 \%$ of all of the data speeds at the airports. Next
Solution
Step 1 ：The question is asking for the interpretation of a confidence interval. A confidence interval is a range of values, derived from a data set, that is likely to contain the value of an unknown population parameter. In this case, the confidence interval is (13.046, 22.15) with a 95% confidence level. This means that we are 95% confident that the true population mean lies within this interval.
Step 2 ：Therefore, the correct answer is A. We have 95% confidence that the limits of 13.05 Mbps and 22.15 Mbps contain the true value of the mean of the population of all data speeds at the airports.
Step 3 ：Final Answer: \boxed{\text{A. We have 95\% confidence that the limits of 13.05 Mbps and 22.15 Mbps contain the true value of the mean of the population of all data speeds at the airports.}}
