# Correction factor for calculating standard error when drawing sample without replacement - derivation

When drawing a sample of size n from population of size N

This relationship holds

$$SE\space when\space drawing\space sample\space without\space replacement\space =\space correction\space factor\space *\space SE\space when\space drawing\space with\space replacement$$

where

$$correction\space factor = \sqrt{\frac{N-n}{N-1}}$$

N = size of population

n = size of sample

I would like to know how this relationship was derived.

Suppose you have an urn with $$M$$ red marbles and $$N-M$$ blue marbles. You can say that $$N$$ is the population. Then you draw $$x$$ red marbles and $$n-x$$ blue marbles, where $$n$$ is the sample size. Then the random variable $$X$$ is hypergeometric distributed. Therefore the sample has a variance of $$Var(X)=n\cdot \frac{M}{N}\cdot \left(1-\frac{M}{N} \right)\cdot \frac{N-n}{N-1}$$. We can replace $$\frac{M}{N}$$ by $$p$$ and get $$Var(X)=n\cdot p\cdot \left(1-p \right)\cdot \frac{N-n}{N-1}$$. And the variance of the sample mean is $$Var\left(\frac{\sum\limits_{i=1}^n X_i}{n}\right)=Var(\overline X)=\frac{p\cdot \left(1-p \right)}{n}\cdot \frac{N-n}{N-1}$$
It is obvious that $$\frac{p\cdot \left(1-p \right)}{n}$$ is the sample mean variance of binomial distributed variables ($$\overline Y$$) divided by $$n$$. For a large $$n$$ the distribution of $$\overline Y$$ can approximated by the normal distribution (central limit theorem): $$Var\left(\overline Y \right)\approx \frac{\sigma^2}{ n}$$ The variance of $$\overline X$$ then is approximately $$Var\left(\overline Y \right)\cdot \frac{N-n}{N-1}$$. To obtain the approximated standard error of the mean we take the sqare root:
$$\sigma_{\overline x}\approx \underbrace{\frac{\sigma}{\sqrt n}}_{\text{SE with replacement} }\cdot \sqrt{\frac{N-n}{N-1}}$$
• For large N and M the fraction M/N can be regarded as constant and we replace M/N by p Is this assumption required? I think we can get to the next step without this assumption. – q126y Oct 8 '18 at 4:10
• @q126y Good catch. If M and N are large the variance become the variance of a binomial sample, since $\frac{N-n}{N-1}\approx 1$. But this is not needed here. I´ll correct it. – callculus Oct 8 '18 at 7:03