# Degrees of freedom for standard deviation of sample

would someone please explain why the degrees of freedom for a random sample is n-1 instead of n ?

I'm looking for an explanation that is intuitive and easily understood by a high school student. ## 1 Answer

The short answer is that dividing by n returns a biased approximation of the population standard deviation (which is usually what we are trying to estimate from our sample.) Such a calculation for sample standard deviation will be biased low (i.e. an underestimate) relative to the population standard deviation.

Dividing by n-1 makes the sample variance an unbiased estimator, and the sample standard deviation a less biased estimator (this bias is still an issue while n is small.)

Wikipedia provides some details here.

• See also stats.stackexchange.com/q/11707/16974 for more technical explanations (which is not what the OP wanted, but for those interested in theoretical underpinnings) – James Stanley May 6 '13 at 1:14
• I'd suggest avoiding talking about bias in the standard deviation as a motivation; the $n-1$ denominator is related to unbiasing the variance. Consequently - by the Jensen inequality - that guarantees bias in the standard deviation. – Glen_b May 6 '13 at 3:31
• Thank you both for the help! I'll definitely take a look at the information linked above later this week. – Quaxton Hale May 6 '13 at 4:15
• @Glen_b Fair point -- most of the (applied biostats) textbooks I've seen simply say that using n-1 gives a better estimate of variance, and leave it at that. – James Stanley May 6 '13 at 4:22
• Interesting though, dividing by $n$ gives a smaller mean square error $E [(s^2 - \sigma^2) ^2]$, and dividing by $n+1$ smaller still! Also, you don't get an unbiased estimate for $\sigma$ by dividing by $n-1$ (which is what $s^2$ is used for more likely) – probabilityislogic May 24 '16 at 10:30