On Feb 24, 8:51 am, Cristiano <cristi...@NSgmail.com> wrote: > On 22/02/2013 6:15, Ray Koopman wrote: > >> For n iid samples from a continuous uniform distribution, >> Pr(r/R <= x) = F(x) = n*x^(n-1) - (n-1)*x^n, where >> r is the sample range, R is the true range, and 0 <= x <= 1. >> A 100p% confidence interval for R is R >= r/x, where F(x) = p. >> Divide that by sqrt(12) to get a lower bound for the SD. > > Your limit works, but now I'm trying to find a way which says > how good is the sample SD (w.r.t. 1/sqrt(12)). > I thought to calculate the above p for which I get the sample SD; > for example: > sd = 0.224508, xmin = 0.4087, xmax = 0.847092 > your lower bound = 0.595019. > But (as you know) I can't use that lower bound as p-value, because > a good SD will have a "p-value" around 0.5, while it should be 1. > > Is there any way to use your procedure to calculate a p-value for SD? > > Thank you > Cristiano
How did you get that lower bound? What were n and p?
I'm not sure what you're really trying to do. Are you trying to use the sample SD to make inferences about the true SD for samples from a uniform distribution? Why? If you know you're sampling from a uniform distribution then you should be using the range, not the SD.