For the provided sample mean, sample size, and population standard deviation, complete parts (a) through (c) Assume that $\mathrm{x}$ is normally distributed. \[ \bar{x}=29, n=9, \sigma=6 \] a. Find a $95 \%$ confidence interval for the population mean The $95 \%$ confidence interval is from 25.08 to 32.92 . (Round to two decimal places as needed.) b. Identify and interpret the margin of error. The margin of error is $\square$. (Round to two decimal places as needed.)
Solution
Step 1 ：Given the sample mean \(\bar{x} = 29\), sample size \(n = 9\), and population standard deviation \(\sigma = 6\).
Step 2 ：Calculate the standard error (SE) using the formula \(\sigma / \sqrt{n}\), which gives SE = 2.0.
Step 3 ：Find the z-score for a 95% confidence level, which is \(z = 1.96\).
Step 4 ：Calculate the lower and upper bounds of the confidence interval using the formulas \(\bar{x} - z \times SE\) and \(\bar{x} + z \times SE\), respectively. This gives a confidence interval from 25.08 to 32.92.
Step 5 ：Calculate the margin of error (ME) using the formula \(z \times SE\), which gives ME = 3.92.
Step 6 ：The final answer is: The 95% confidence interval for the population mean is from \(\boxed{25.08}\) to \(\boxed{32.92}\). The margin of error is \(\boxed{3.92}\).
