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 (n−1)S^2/σ^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: b.rowlingson@lancaster.ac.uk
> To: csardi@rmki.kfki.hu
> CC: r-help@r-project.org
> Subject: Re: [R] Simulation
> 
> On Wed, May 13, 2009 at 4:26 PM, Gábor Csárdi <csardi@rmki.kfki.hu> wrote:
> > On Wed, May 13, 2009 at 5:13 PM, Debbie Zhang <debbie0621@hotmail.com> 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
> 
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> 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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