Thread: Confidence inteval

Let X be a random variable following a normal distribution.
A sample is taken from a population.
If n= 8
sample mean = 7.2

(i=1) E (n=8) [x(i) - x-bar ]^2 = 3.44

Calculate a 99% confidence interval for the population standard deviation?

Do you do this by first solving for the population variance confidence interval and then taking the square root of the upper and lower limits?

Is that E really a sum?
You need to use the chi-square distribution for a CI about sigma.

Originally Posted by matheagle

Is that E really a sum?
You need to use the chi-square distribution for a CI about sigma.

Yes, E = sum.

You mean the confidence interval is
(n-1)s^2/ X^2 (n-1);(a/2)

and

(n-1)s^2/ X^2 (n-1);(1 - a/2)?

Like (3.44/20.28 ; 3.44/0.989)?

But I thought the standard deviation CI was being asked for, not the variance CI, or could you maybe explain how this works? Thanks!

I didn't check your CI formula
BUT IF (a,b) is your CI for
Then is the CI for

Originally Posted by matheagle

I didn't check your CI formula
BUT IF (a,b) is your CI for
Then is the CI for

Ok, thanks!