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Re: Simulation for the standard deviation
Posted:
Feb 24, 2013 8:08 PM
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On 24/02/2013 21:49, Ray Koopman wrote:
> 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?
Sorry, but I mistakenly wrote that the lower bound is 0.595019, while I
find p which gives that lower bound, so n= 3, p= 0.595019 and lower
bound = sd = 0.224508.
> 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.
I randomly pick 3 numbers in U(0,1) and I get, for example, sd= .1234,
but we know that the expected sd is .288675. How good .1234 is? Is there
any way to calculate a p-value which says how good is .1234?
Cristiano
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