Is that E really a sum?
You need to use the chi-square distribution for a CI about sigma.
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?
You mean the confidence interval is
(n-1)s^2/ X^2 (n-1);(a/2)
(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!