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