Hi,
> g=list()
> for(i in 1:1000){z[[i]]=rnorm(15,0,1)}
I've attempted a similar problem based on the above method. Now, if i want to
find the sample variance, do i go about it like this?
> for (i in 1:1000)vars[[i]] = sum(z[[i]])
> vars[[i]]
the overall sigma squared will just be 1, because the distribution is standard
normal. Is this correct?
if so, then to find (nâ1)S^2/Ï^2,
i will need s=999*sum(vars[[i]]))/1?
Is this correct, or am i getting lost along the way?
Thank you
> Date: Wed, 13 May 2009 16:45:22 +0100
> From: [email protected]
> To: [email protected]
> CC: [email protected]
> Subject: Re: [R] Simulation
>
> On Wed, May 13, 2009 at 4:26 PM, Gábor Csárdi <[email protected]> wrote:
> > On Wed, May 13, 2009 at 5:13 PM, Debbie Zhang <[email protected]>
> > wrote:
> >>
> >>
> >> Dear R users,
> >>
> >> Can anyone please tell me how to generate a large number of samples in R,
> >> given certain distribution and size.
> >>
> >> For example, if I want to generate 1000 samples of size n=100, with a
> >> N(0,1) distribution, how should I proceed?
> >>
> >> (Since I dont want to do "rnorm(100,0,1)" in R for 1000 times)
> >
> > Why not? It took 0.05 seconds on my 5 years old laptop.
>
> Second-guessing the user, I think she maybe doesn't want to type in
> 'rnorm(100,0,1)' 1000 times...
>
> Soln - "for" loop:
>
> > z=list()
> > for(i in 1:1000){z[[i]]=rnorm(100,0,1)}
>
> now inspect the individual bits:
>
> > hist(z[[1]])
> > hist(z[[545]])
>
> If that's the problem, then I suggest she reads an introduction to R...
>
> Barry
>
> ______________________________________________
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
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