# Thread: Finding sample standard deviation

1. ## Finding sample standard deviation

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

Attempt:
$51.47<\sigma^2<261.90$

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

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

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

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
gives you your number 5.0884

3. I really don't have n. The question typed is precisely as it is posed in the text.

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 $\sigma^2$, which is .10?

4. no, use the chi-square table.
Let n=2,3,4,5... find the .05 and .95 percentiles and take that ratio
see when the ratio is 5.0884
(But I didn't check the math to see if 5.0884 was correct.)