Step 1 :Given that the sample mean (\(\bar{x}\)) is 22.6 miles, the sample standard deviation (s) is 7.4 miles, and the sample size (n) is 6 people.
Step 2 :We are asked to find a 99% confidence interval for the population mean (\(\mu\)). This can be calculated using the formula: \(\bar{x} \pm t \frac{s}{\sqrt{n}}\), where t is the t-score corresponding to the desired level of confidence.
Step 3 :Since the sample size is 6, the degrees of freedom is 5 (since \(n - 1 = 5\)). The t-score for a 99% confidence interval with 5 degrees of freedom can be found using a t-distribution table or a statistical calculator. The t-score is approximately 4.032.
Step 4 :Substituting the given values into the formula, we get the margin of error to be approximately 12.2 miles.
Step 5 :Subtracting and adding the margin of error from the sample mean, we get the 99% confidence interval for the population mean to be approximately (10.4, 34.8) miles.
Step 6 :Interpreting the results, we can say that with 99% confidence, the population mean driving distance to work is between 10.4 miles and 34.8 miles. This is because a confidence interval estimates the range of values within which the population parameter (in this case, the mean driving distance to work) is likely to fall.
Step 7 :Final Answer: The 99% confidence interval for the population mean is \(\boxed{(10.4, 34.8)}\). This means that we are 99% confident that the true population mean driving distance to work is between 10.4 miles and 34.8 miles.