## Essentials of Statistics (5th Edition)

$\mu$ is between 5.21 and 7.79. The interval doesn't contain 4 years, but there is an outlier so the result might not be correct.
The mean can be counted by summing all the data and dividing it by the number of data: $\frac{4+4+...+15}{20}=6.5.$ Standard deviation=$\sqrt{\frac{\sum (x-\mu)^2}{n-1}}=\sqrt{\frac{(4-6.5)^2+...+(15-6.5)^2}{19}}=3.51.$ $\alpha=1-0.9=0.1.$ $\sigma$ is 3.51 hence we use the z-distribution with $df=sample \ size-1=20-1=19$ in the table. $z_{\alpha/2}=z_{0.05}=1.645.$ Margin of error:$z_{\alpha/2}\cdot\frac{\sigma}{\sqrt {n}}=1.645\cdot\frac{3.51}{\sqrt{20}}=1.29.$ Hence the confidence interval:$\mu$ is between 6.5-1.29=5.21 and 6.5+1.29=7.79. The interval doesn't contain 4 years, but there is an outlier so the result might not be correct.