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 due to chance. If you repeat, the larger set may or may not have a larger variance.
> var(rnorm(10000, 0, 3)) [1] 8.958727 > var(rnorm(10000, 0, 3)) [1] 9.155332 > var(rnorm(10000, 0, 3)) [1] 9.050894 > var(rnorm(10000, 0, 3)) [1] 9.282509 > var(rnorm(100000, 0, 3)) [1] 8.990778 > var(rnorm(100000, 0, 3)) [1] 9.024343 > var(rnorm(100000, 0, 3)) [1] 8.999064 > > var(rnorm(100000, 0, 3)) [1] 9.088034 HTH Jim James W. MacDonald Affymetrix and cDNA Microarray Core University of Michigan Cancer Center 1500 E. Medical Center Drive 7410 CCGC Ann Arbor MI 48109 734-647-5623 >>> "Padmanabhan, Sudharsha" <[EMAIL PROTECTED]> 08/19/03 01:42PM >>> 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 (shown below) to check the validity of this. large<-array(1,c(1000,1000)) small<-array(1,c(100,1000)) for(i in 1:1000){large[i,]<-rnorm(1000,0,3)} for(i in 1:1000){small[i,]<-rnorm(100,0,3)}} yy<-array(1,100) for(i in 1:100){yy[i]<-var(small[i,])} y1y<-array(1,1000) for(i in 1:1000){y1y[i]<-var(large[i,])} mean(yy);mean(y1y); [1] 8.944 [1] 9.098 This shows that on an average,for 1000 such samples of 1000 Normal numbers, the variance is higher than that of a 100 samples of 1000 random numbers. Why is this so? Can someone please help me out???? Thanks. Regards ~S. ______________________________________________ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help ______________________________________________ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help