For the provided sample mean, sample size, and population standard deviation, complete parts (a) through (c) Assume that $\mathrm{x}$ is normally distributed.
$$\overline{x}=29,n=9,\sigma =6$$
a. Find a $95\mathrm{\%}$ confidence interval for the population mean

The $95\mathrm{\%}$ 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 $◻$.
(Round to two decimal places as needed.)

Step 1 ：Given the sample mean $\overline{x}=29$, 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 $\overline{x}-z\times SE$ and $\overline{x}+z\times SE$, 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 $\boxed{{\displaystyle 25.08}}$ to $\boxed{{\displaystyle 32.92}}$. The margin of error is $\boxed{{\displaystyle 3.92}}$.