The variance of Xbar decreases as 1/n; the sample variance
of X does not.
- tom blackwell - u michigan medical school - ann arbor -
On Tue, 19 Aug 2003, Padmanabhan, Sudharsha wrote:
I am running a few simulations for clinical trial anlysis. I want some help
regarding the following.
On 08/19/03 17:42, Padmanabhan, Sudharsha wrote:
Hello,
I am running a few simulations for clinical trial anlysis. I want some help
regarding the following.
We know trhat as the sample size increases, the variance should decrease, but
I am getting some unexpected results. SO I ran a code
First of all, your subscripting is wrong. The first index is for row, and
the second for column. Thus large[i,] refers to the i-th row of large,
rather than the i-th column. Also, the code as you provided contain syntax
error.
Try:
set.seed(311) ## Always a good idea to set seed for
I think you are confused. As sample size increases, the variance of an
estimate based on that sample will decrease asymtotically to zero (e.g.,
the standard error of the mean will go to zero). However the variance of
the sample itself will not change. Any difference you see in your data
is simply
Perhaps you were trying for as sample size increases, variance *of the
mean* decreases (a least when variance is finite). If you swap mean and
var in your code, I think you will get what you are looking for.
-- Tony Plate
At Tuesday 05:42 PM 8/19/2003 +, Padmanabhan, Sudharsha wrote:
Hi.
There is no reason the variance of a normal should decrease as you take
larger samples. Indeed, in your call itself, you say that you want a sample
from a normal with a standard deviation of 3, and so a variance of 9. As
expected, both of your estimates of variance are close to 9.
What
Padmanabhan, Sudharsha [EMAIL PROTECTED]
We know trhat as the sample size increases, the variance should
decrease,
Should it?
I can paraphrase his test case thus:
v100 - sapply(1:100, function(i) var(rnorm(100, 0, 3)))
# We expect the elements of v100 to cluster
