why sample variance has has n-1 in the denominator?

Question

Sample variance is calculated according to:

$s^2=\frac{\sum{(x-\bar{x})^2}}{n-1}$

Population variance is calculated according to:

$\sigma^2=\frac{\sum{(x-\mu)^2}}{n}$

Why denominator for sample variance is $n-1$ and not $n$?

Any help is appreciated.

Answer 1

To put it simply $(n-1)$ is a smaller number than $(n)$. When you divide by a smaller number you get a larger number. Therefore when you divide by $(n-1)$ the sample variance will work out to be a larger number.

Let's think about what a larger vs. smaller sample variance means. If the sample variance is larger than there is a greater chance that it captures the true population variance. That is why when you divide by $(n-1)$ we call that an unbiased sample estimate. Whereas dividing by $(n)$ is called a biased sample estimate.

Because we are trying to reveal information about a population by calculating the variance from a sample set we probably do not want to underestimate the variance. Basically by just dividing by $(n)$ we are underestimating the true population variance, that is why it is called a biased estimate.

Basically comes down to calculating a biased vs. unbiased sample variance estimate.

Also because you asked what as estimator is.

There was a good post here on CV that will give you some good insight. Hope this helps!