Answer
df=24, $X_{L}^2=9.886$, $ X_{R}^2=45.559$, $\mu$ is between 0.174 and 0.374.
Work Step by Step
$\alpha=1-0.99=0.01.$ By using the table we can find the critical chi-square values with with $df=sample \ size-1=25-1=24$.
$X_{L}^2= X_{0.995}^2=9.886$
$ X_{R}^2= X_{0.005}^2=45.559$
Hence the confidence interval:$\mu$ is between $\sqrt{\frac{(n-1)\cdot s^2}{ X_{R}^2}}=\sqrt{\frac{(24)\cdot 0.24^2}{45.559}}=0.174$ and $\sqrt{\frac{(n-1)\cdot s^2}{ X_{L}^2}}=\sqrt{\frac{(24)\cdot 0.24^2}{9.886}}=0.374$
