Why does standard error not involve population size?

$$SE = \frac{SD}{\sqrt{N}}$$

• If N is the entire population, I would expect SE to be zero.
• If N is equal to population size - 1, I would expect SE to be lower than if N is a small fraction of the population size, yet SE is the same.

So why isn't population size part of SE?

While a mathematical explanation is great, I'd really like to get a gut feel intuition of the reason.

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Presumably, you're talking about the standard error of the mean and you have a dataset obtained as a simple random sample with replacement from a finite population of known size. If any of these are not the case, then please so indicate in the question. – whuber Sep 29 '12 at 21:26
Right, except shouldn't you say "without replacement," as Peter does below? – rolando2 Sep 29 '12 at 21:35
@rolando2: I am (pretty) sure whuber meant with not without, which explains why there would still be uncertainty after taking a sample of size $N$. (Also, there are two Peter's below.) I admit the link to SRS does allow for a bit of doubt. :-) – cardinal Sep 29 '12 at 21:40
@Rolando2, Cardinal is correct: The formula for simple random sampling without replacement involves a "finite population correction factor" which is not present in the question, whence the presumption that it is about sampling with replacement. – whuber Sep 29 '12 at 21:47
@whuber Yes, I am asking about standard error of the mean. But I don't know what with/out replacement is. – walrii Sep 29 '12 at 22:39

This formula assumes that the sample is a very small proportion of the population.

If there is a finite population and the sample is a substantial part of it, you can use the finite population correction:

$\text{FPC} = \sqrt{\frac{N-n}{N}}$

where $n$ is the sample size and $N$ is the population size. If $N = n$ then this will become 0, as you suspected. Ordinarily, though, it makes very little difference.

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FPC is usually defined as (N-n)/N as it represents the factor that reduces the variance. The square root of it would be given as the factor multiplying the standard deviation. So Peter almost got it right. It should be N in the denominator rather than N-1. – Michael Chernick Sep 29 '12 at 21:57
SE=(SD/√n) (√[(N-n)/N] – Michael Chernick Sep 29 '12 at 22:00
Why? That is, why N rather than N-1? I thought it would be analagous to the sample SD, but perhaps not. (It will make almost no difference in practical terms, of course, unless you have a very small population, in which case a census might work). – Peter Flom Sep 29 '12 at 22:04
Read Cochran's book. – Michael Chernick Sep 29 '12 at 22:11
This doesn't do the derivation the way Cochran would but it shows all the correct formulas. HERE – Michael Chernick Sep 29 '12 at 22:21