A simple random sample of size $n=13$ is drawn from a population that is normally distributed. The sample mean is found to be $\bar{x}=50$ and the sample standard deviation is found to be $s=10$. Construct a $90 \%$ confidence interval about the population mean.
The lower bound is The upper bound is (Round to two decimal plak as needed.)
Answer
Rounding to two decimal places, the final answer is: The \(90\%\) confidence interval about the population mean is approximately \(\boxed{[45.44, 54.56]}\).
Step 1 ：We are given a simple random sample of size \(n=13\) drawn from a population that is normally distributed. The sample mean is \(\bar{x}=50\) and the sample standard deviation is \(s=10\). We are asked to construct a \(90\%\) confidence interval about the population mean.
Step 2 ：The formula for a confidence interval is \(\bar{x} \pm z \frac{s}{\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(z\) is the z-score, \(s\) is the sample standard deviation, and \(n\) is the sample size.
Step 3 ：In this case, we have \(\bar{x} = 50\), \(s = 10\), and \(n = 13\). The z-score for a \(90\%\) confidence interval is approximately \(1.645\).
Step 4 ：Substituting these values into the formula, we get the margin of error as \(4.56240911395251\).
Step 5 ：Subtracting the margin of error from the sample mean, we get the lower bound of the confidence interval as \(45.43759088604749\).
Step 6 ：Adding the margin of error to the sample mean, we get the upper bound of the confidence interval as \(54.56240911395251\).
Step 7 ：Rounding to two decimal places, the final answer is: The \(90\%\) confidence interval about the population mean is approximately \(\boxed{[45.44, 54.56]}\).
