# Thread: Finding sample standard deviation

1. ## Finding sample standard deviation

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?

2. you need n, and then just look up either of those chi-square percentiles
then solve for s.
If, you really don't have n, just go down the tables and see which of those ratio of percentiles
Which F-Distribution table should I use? In other words, how do I infer the alpha level from my work so far? Is it just the same as for $\displaystyle \sigma^2$, which is .10?