sample variance from simulation

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/?^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 at lancaster.ac.uk
To: csardi at rmki.kfki.hu
CC: r-help at r-project.org
Subject: Re: [R] Simulation

On Wed, May 13, 2009 at 4:26 PM, G?bor Cs?rdi <csardi at rmki.kfki.hu> wrote:

On Wed, May 13, 2009 at 5:13 PM, Debbie Zhang <debbie0621 at 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