sample variance from simulation

Hey,

when i type in either of those formulas into R, i dont really get the answer
im looking for. For such large samples, isnt the sample variance meant to
approach the actual variance, which is 1 for a standard normal?

also, when i use sapply, i 1000 results for variance, where i think i just
need one number.

I've worked on this problem for so long. The initial problem is as follows

"Use the simulation capacity of R to generate m = 1 000
samples of size n = 15 from a N(0,1) distribution. Compute the statistic
(n-1)S^2/?^2 for the normally generated values, labelling as NC14. Produce
probability histogram for NC14 and superimpose the theoretical distribution
for a ?2 (14 degrees of freedom)"

g=list()
for(i in 1:1000){z[[i]]=rnorm(15,0,1)}

for (i in 1:1000)vars[[i]] = sum(z[[i]])

vars[[i]]

sum(var(z[[i]]))

[1] 0.9983413

Does this make sense?  my logic is that i use the loop again to add up all
the individual variances. im not really sure if i did it correctly, but if
someone could make the necessary corrections, i'd be very very greatful.

Thanks heaps guys for taking the time to look at this

Date: Mon, 18 May 2009 15:06:47 +0200
From: Waclaw.Marcin.Kusnierczyk at idi.ntnu.no
To: konk2001 at hotmail.com
CC: Mike.Lawrence at dal.ca; r-help at r-project.org
Subject: Re: [R] sample variance from simulation

Mike Lawrence wrote:

why not simply

vars=list()
for (i in 1:1000) vars[[i]] = var(z[[i]])

... or, much simpler,

vars = sapply(z, var)

vQ