Uh, no. You are forgetting to take the square root of 10, and to divide by the square root of 12.
The variance of Y is (exactly) (56^2 - 1)/12, so the variance of Y-bar is this quantity over 10, so the standard deviation of Y-bar is sqrt((56^2 - 1)/12)/sqrt(10). Which is approximately
(ignoring the -1) 56/sqrt(12) * 1/sqrt(10). cheers, Rolf On 12/06/13 20:26, Jim Lemon wrote:
On 12/06/2013 04:16 PM, mohan.radhakrish...@polarisft.com wrote:Hi,Basic question with basic code. I am simulating a set of'y' values for a standard 'x' value measurement. So here the error bars are very long because thenumber of samples are very small. Is that correct ? I am plotting the meanof 'y' on the 'y' axis. Thanks, Mohan x<- data.frame(c(5,10,15,20,25,30,35,40,50,60)) colnames(x)<- c("x") y<- sample(5:60,10,replace=T) y1<- sample(5:60,10,replace=T) y2<- sample(5:60,10,replace=T) y3<- sample(5:60,10,replace=T) y4<- sample(5:60,10,replace=T) z<- data.frame(cbind(x,y,y1,y2,y3,y4)) z$mean<- apply(z[,c(2,3,4,5,6)],2,mean) z$sd<- apply(z[,c(2,3,4,5,6)],2,sd) z$se<- z$sd / sqrt(5)Hi Mohan,As your samples seem to follow a discrete uniform distribution, the standard deviation is approximately the number of integers in the range (56) divided by the number of observations (10).
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