why not simply
vars=list()
for (i in 1:1000) vars[[i]] = var(z[[i]])
On Mon, May 18, 2009 at 6:51 AM, Kon Knafelman konk2...@hotmail.com wrote:
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.rowling...@lancaster.ac.uk
To: csa...@rmki.kfki.hu
CC: r-help@r-project.org
Subject: Re: [R] Simulation
On Wed, May 13, 2009 at 4:26 PM, Gábor Csárdi csa...@rmki.kfki.hu wrote:
On Wed, May 13, 2009 at 5:13 PM, Debbie Zhang debbie0...@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
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