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Re: Simulation for the standard deviation
Posted:
Feb 23, 2013 7:58 AM
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On 22/02/2013 21:06, Ray Koopman wrote:
> On Feb 22, 6:05 am, Cristiano <cristi...@NSgmail.com> wrote:
>> On 22/02/2013 6:15, Ray Koopman wrote:
>>> On Feb 20, 3:54 am, Cristiano <cristi...@NSgmail.com> wrote:
>>>
>>>> Short question: does anybody know how to calculate the confidence
>>>> interval of the standard deviation for the uniform distribution?
>>>
>>> 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.
>>
>> Suppose I randomly pick 0.1, 0.4 and 0.2 (n = 3);
>> what should I write to calculate a 99% confidence interval?
>
> F(x) = 3 x^2 - 2 x^3 = p
>
> F(.941097) = .99
>
> SD >= (.4 - .1)/(.941097 * sqrt(12))
That lower bound doesn't work.
According to the practical definition of CI given here:
http://www.itl.nist.gov/div898/handbook/eda/section3/eda352.htm
I wrote a simulation which counts how many times the sample SD exceeds
the calculated lower bound.
Using 10^6 trials and a confidence level of .9, I see that the
calculated SD is greater than your lower bound 10^6 times, while the
calculated SD is greater than my lower bound 900221 times (I find the
confidence limits as explained in my original post).
Cristiano
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