# finding standard deviation ?

• Jul 19th 2008, 07:30 PM
kscheels
finding standard deviation ?
GIven confidence level,11.24<u<12.01 and that n=90 + c= .90, find the sample standard devaiton o
• Jul 19th 2008, 07:31 PM
kscheels
[quote=kscheels;168572]GIven confidence level,11.24<u<12.01 and that n=90 + c= .90, find the sample standard devaiton o.
• Jul 19th 2008, 09:35 PM
mr fantastic
Quote:

Originally Posted by kscheels
GIven confidence level,11.24<u<12.01 and that n=90 + c= .90, find the sample standard devaiton o

I'm assuming that since you ask for sample standard deviation rather than population standard deviation, the confidence interval has been constructed using the t-distribution:

$L_1 = \bar{x} - \frac{s}{\sqrt{n}} \, t_{n-1, \, \alpha/2} < \mu < \bar{x} + \frac{s}{\sqrt{n}} \, t_{n-1, \, \alpha/2} = L_2$

Therefore $L_2 - L_1 = \frac{2 s}{\sqrt{n}} \, t_{n-1, ~ \alpha/2}$.

In your case $\alpha = 0.1 \Rightarrow \frac{\alpha}{2} = 0.05$ (I think ..... it's not really clear from what you posted), n = 90, $L_1 = 11.24$ and $L_2 = 12.01$. Use a table of critical values to get $t_{89, ~ 0.05}$.

Substitute all these values and solve for s.