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Re: Simulation for the standard deviation
Posted:
Feb 24, 2013 3:49 PM
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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.
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