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.
$\begingroup$
$\endgroup$
4
-
2$\begingroup$ stats.stackexchange.com/questions/16008/… $\endgroup$– ocramFeb 25, 2016 at 6:50
-
$\begingroup$ Answered many times on site. e.g here $\endgroup$– Glen_bFeb 25, 2016 at 6:52
-
$\begingroup$ I should say that I'm very new to stats and don't know what estimators are. $\endgroup$– flashburnFeb 25, 2016 at 7:09
-
$\begingroup$ See, for example, the definition and discussion here: What is the difference between an estimator and a statistic? $\endgroup$– Glen_bFeb 25, 2016 at 11:33
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
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!
-
15$\begingroup$ This doesn't explain for example why dividing by $(n - 2), (n - 3), \dots, 1, $ would not be even better. Nor does it explain why dividing by $n$ gives a biased estimator in the first place; that is just asserted. So, while you rightly focus on bias versus unbiased, the thread marked as duplicate is more precise. $\endgroup$– Nick CoxFeb 25, 2016 at 9:23
-
$\begingroup$ You are absolutely correct, the linked threads are more precise. They give great, well articulated answers. Like you said, I do not go into detail about why $(n-2)$ etc would lead to overestimating population variance. We can agree Fashburn should go through and read the other threads for a more complete answer. However I would argue there is quite a bit of math/higher level explanations found in those threads. I simply wanted to provide Flashburn with a really high level idea to chew on. As he said he is very new to stats. Thanks for the comment. $\endgroup$ Feb 25, 2016 at 10:59