If a 90% confidence interval for $\displaystyle \sigma^2$ is reported to be (51.47,261.90), what is the value of the sample standard deviation.

Attempt:

$\displaystyle 51.47<\sigma^2<261.90$

So

$\displaystyle \frac{(n-1)s^2}{\chi_{.95,n-1}^2}=51.47$ and $\displaystyle \frac{(n-1)s^2}{\chi_{.05,n-1}^2}=261.90$

Then

$\displaystyle s=\sqrt{\frac{51.47*\chi_{.95,n-1}^2}{n-1}}=\sqrt{\frac{261.90*\chi_{.05,n-1}^2}{n-1}}$

$\displaystyle 5.0884=\frac{\chi_{.95,n-1}^2}{\chi_{.05,n-1}^2}$

So this looks like it should have an F-distribution with same numerator and denominator degrees of freedom, which should help me figure out what those degrees of freedom are.

I can't figure out how to do this. Or am I way off?