Questions: Refer to the accompanying data display that results from a sample of airport data speeds in Mbps. The results in Tinterval the screen display are based on a 95% confidence level. Write a statement that correctly interprets the confidence interval. (13.046,22.15)/x=17.598 S x=16.01712719 n=50 Choose the correct answer below. A. We have 95% confidence that the limits of 13.05 Mbps and 22.15 Mbps contain the sample mean of the data speeds at the airports. B. 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. C. 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. D. The limits of 13.05 Mbps and 22.15 Mbps contain 95% of all of the data speeds at the airports.

Solution Steps

We have the following data from the sample of airport data speeds in Mbps:

• Sample mean (\(\bar{x}\)) = 17.598
• Sample size (\(n\)) = 50
• Sample standard deviation (\(s\)) = 16.0171
• Confidence interval limits = (13.046, 22.15)

Using the formula for the confidence interval for the mean of a single population with unknown variance and large sample size at a 95% confidence level, we have:

\[ \bar{x} \pm z \cdot \frac{s}{\sqrt{n}} \]

Where:

• \(z\) for a 95% confidence level is approximately 1.96.
• The calculation yields:

\[ 17.6 \pm 1.96 \cdot \frac{16.02}{\sqrt{50}} = (13.16, 22.04) \]

The calculated confidence interval is (13.16, 22.04). This means we have 95% confidence that the true mean of the population of all data speeds at the airports lies within this interval.

Among the provided options, the correct interpretation of the confidence interval is:

C. 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.

Final Answer