The latest code I have put together is this. Could you point out what is 
missing here ? 

#Reference values plotted on x-axis. These are constant.
 #These values could be time of day. So every day at the same
 #time we could collect other measurements
 referenceset <- data.frame(c(5,10,15,20,25,30,35,40,50,60))
 colnames( referenceset) <- c("reference")

 #These are the sets of measurements. So every day at the same
 #time we could collect several samples. This is simulated now.
 sampleset <- data.frame( matrix(sample(1:2, c(20000), replace = TRUE), 
ncol = 2000) )
 
 sampleset <- cbind( sampleset, referenceset )
 
 #Calculate mean
 sampleset$mean <- apply(sampleset[,1:10],2,mean)
 
 #Calculate Standard Deviation
 sampleset$sd <- apply(sampleset[,c(1:10)],2,sd)
 
 #Calculate Standard Error
 sampleset$se <- sampleset$sd / sqrt(2000)
 
 #print(sampleset)
 
 plot( sampleset$reference,
       sampleset$mean,
           las=1,
           ylab="Mean of 'y' values",
           xlab="x",
      );
 
arrows(sampleset$reference,
       sampleset$mean-sampleset$se,
           sampleset$reference,
           sampleset$mean+sampleset$se,
           code = 3,
           angle=90,
           length=0.2)






Thanks.



From:   Rolf Turner <r.turner@auckland.ac.nz>
To:     Jim Lemon <jim@bitwrit.com.au>
Cc:     mohan.radhakrishnan@polarisft.com, r-help@r-project.org
Date:   12/06/2013 02:53 PM
Subject:        Re: [R] Simple Error Bar




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.radhakrishnan@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 the
>> number of samples are very small. Is that correct ? I am plotting the 
>> mean
>> of '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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