# Explanation of finite correction factor

I understand that when sampling from a finite population and our sample size is more than 5% of the population, we need to make a correction on the sample's mean and standard error using this formula:

$$\hspace{10mm} FPC=\sqrt{\frac{N-n}{N-1}}$$

Where $$N$$ is the population size and $$n$$ is the sample size.

1. Why is the threshold set at 5%?
2. How was the formula derived?
3. Are there other online resources that comprehensively explain this formula besides this paper?
• You don't correct the mean! – whuber Dec 5 '10 at 21:42
• You only correct the variance. – HelloWorld Apr 10 '16 at 15:48

The threshold is chosen such that it ensures convergence of the hypergeometric distribution ($\sqrt{\frac{N-n}{N-1}}$ is its SD), instead of a binomial distribution (for sampling with replacement), to a normal distribution (this is the Central Limit Theorem, see e.g., The Normal Curve, the Central Limit Theorem, and Markov's and Chebychev's Inequalities for Random Variables). In other words, when $n/N\leq 0.05$ (i.e., $n$ is not 'too large' compared to $N$), the FPC can safely be ignored; it is easy to see how the correction factor evolves with varying $n$ for a fixed $N$: with $N=10,000$, we have $\text{FPC}=.9995$ when $n=10$ while $\text{FPC}=.3162$ when $n=9,000$. When $N\to\infty$, the FPC approaches 1 and we are close to the situation of sampling with replacement (i.e., like with an infinite population).
You will notice that some authors use $N$ instead of $N-1$ in the denominator of the FPC; in fact, it depends on whether you work with the sample or population statistic: for the variance, it will be $N$ instead of $N-1$ if you are interested in $S^2$ rather than $\sigma^2$.