standard error of known population values i tried to get a clearer understanding of the standard error by constructing a hypothetical population consisting of the following values: 1, 3, 5, 7. i calculated the sample means of all samples with a size = 2. as expected, the mean of all sample means was equal to the population mean (4). however, the standard deviation of the sample means (1.633)--also known as the standard error of the mean was not equal to the population standard deviation (2.58) divided by the square root of the sample size (square root of 2) which was 1.825. but, i discovered that the variance of the sample variances was 5.33 and the standard deviation of the samples variances was 2.31. dividing 2.31 by the square root of the sample size (square root of 2) yielded a value of 1.633 which is identical to the standard deviation of the sample means.
i am bit confused because the standard error according to most authors is the population standard deviation divided by the square root of the sample size. however, in this case the standard error is equal to the standard deviation of the sample variances divided by the square root of the sample size. 
i would appreciate if you could help me on this matter.
thank you.
 A: I think you are running into problems here with "sample variance" and "population variance".
(remembering that s.d. is simply the square root of these)
The actual variance of a population is
$\sigma^2 = \frac{1}{N}\sum_{i=1}^N (x_i-\mu)^2$
while the sample estimate of the variance of a population is
$\hat{\sigma^2} = \frac{1}{N-1} \sum_{i=1}^N (x_i-\bar{x})^2$
Now in every day statistical life it is very rare that we actually have the full population to deal with, so most statistical packages give you the second definition by default.  It would appear that you have given ${1,3,5,7}$ to a statistical package and asked for the sample standard deviation not the population standard deviation.
For example, in R:
> sd(c(1,3,5,7))
[1] 2.581989    #sample estimate

The actual figure you want is
> sqrt(sum((c(1,3,5,7)-4)^2)/4)
[1] 2.236068   # population standard deviation

Then for the standard deviation of your two sample means you need to do this:
> samples <- expand.grid(c(1,3,5,7),c(1,3,5,7))
> samples$m <- 0.5*(samples$Var1 + samples$Var2)
    > sd(samples$m)
[1] 1.632993
> sqrt(sum((samples$m-4)^2)/length(samples$m))
[1] 1.581139

And notice that
> 2.236068/sqrt(2)
[1] 1.581139

The key point to note here is that you have evaluated all possible sample means, and so again you have the population and must use the population variance calculation.
