# What role does the sampling fraction play in the simple random sample variance formula?

Consider a survey where the population being estimated is finite. Using a simple random sample, you draw $n<N$ data. The sampling fraction is $f=\frac{n}{N}$. The variance will be $(1-f)\frac{s^2}{n}$. Can someone explain how the sampling fraction effects this formula?

• Can you clarify your question? Is it about the finite population correction? – gung - Reinstate Monica Apr 4 '15 at 1:16
• @gung Yes! How does it affect the variance of a simple random sample? – Jnyeboah93 Apr 4 '15 at 1:34
• @gung Do you understand my question? – Jnyeboah93 Apr 4 '15 at 1:47
• Did I get it wrong? If so, please edit to correct it. Just make sure it is clear what you are asking. – gung - Reinstate Monica Apr 4 '15 at 1:51

In many cases, $f$ is neglected when evaluating the sampling variance. Think about it : if you sample $n = 10 000~$ units among $~N = 300 000 000~$, then $~ f \approx 3 \cdot 10^{-6} \ll 1$.

The most important terms of this formula are : $n$, the sample size, and $s^2$, the variance of the variable you're trying to measure.

Nevertheless, it is somehow logical to see $f$ appear in the formula. Indeed, if $f = 1$ (census), variance of the SRS has to be $0$.