Message-ID: <52A1978F.9080606@auckland.ac.nz>
Date: 2013-12-06T09:23:27Z
From: Rolf Turner
Subject: Simple Error Bar
In-Reply-To: <52A17C35.5030007@bitwrit.com.au>
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 at 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).