
Re: Simulation for the standard deviation
Posted:
Feb 24, 2013 11:51 AM


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^(n1)  (n1)*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 pvalue, because a good SD will have a "pvalue" around 0.5, while it should be 1.
Is there any way to use your procedure to calculate a pvalue for SD?
Thank you Cristiano

