Answer
df=19, $X_{L}^2=6.844$, $ X_{R}^2=38.582$, $\mu$ is between 0.0288 and 0.0685.
Work Step by Step
$\alpha=1-0.99=0.01.$ By using the table we can find the critical chi-square values with with $df=sample \ size-1=20-1=19$.
$X_{L}^2= X_{0.995}^2=6.844$
$ X_{R}^2= X_{0.005}^2=38.582$
Hence the confidence interval:$\mu$ is between $\sqrt{\frac{(n-1)\cdot s^2}{ X_{R}^2}}=\sqrt{\frac{(19)\cdot 0.04111^2}{38.582}}=0.0288$ and $\sqrt{\frac{(n-1)\cdot s^2}{ X_{L}^2}}=\sqrt{\frac{(19)\cdot 0.04111^2}{6.844}}=0.0685.$
