# Should one apply bias correction for the standard deviation, for small sample sizes, as a matter of course?

If one is dealing with small sample sizes, let's say $$8-16$$ observations per sample, and we are interested in estimates of the standard-deviation (let us also assume Gaussian statistics), is there a reason not to apply sample-size bias correction?

For example the arithmetic sample standard deviation, $$s = \sqrt{\frac{1}{N-1} \sum_{i}^{N} {(x_{i} - \bar{x})^2}} \text{,}$$ is a biased estimator -- especially for small sample sizes. This can be corrected with the $$c_4(N)$$ bias correction factor. This gives an unbiased estimate of the standard deviation as $$\sigma_{\rm{est}} = s /c_{4}(N) \text{.}$$

There are bias correction factors for small-sample size estimates attained from MLE methods, or the $$\rm{MAD}$$, $$Q_{n}$$, and $$S_{n}$$ dispersion estimators which can also be used to make estimates on the standard deviation.

I ask because if we use an estimate of the standard-deviation to produce a standard error for the mean, i.e. $$\rm{S.E} = \sigma_{\rm{est}}/\sqrt{N}$$ for small sample sizes the correction factor can make quite a difference in the result, so:

1. Is there a good reason not to account for sample sizes, especially when they are small?
2. Does including bias correction for estimates which are used to produce standard errors and/or weights, have any undesirable consequences?
• Aren't these questions thoroughly addressed at stats.stackexchange.com/questions/3931 ?
– whuber
Jan 8, 2021 at 14:11
• I don't think so. These refer to Bessel's correction, but if you have small samples sizes you seem to need more, i.e. $c_{4}$. If I shouldn't be using $c_{4}$, why do we have this correction factor at all? For diagnostics of the estimators performance?
– Q.P.
Jan 9, 2021 at 6:51

1. If you were going to make a confidence interval or hypothesis test, the sample size is accounted for in the degrees of freedom for the t-distribution. There is a good reason for using the usual estimate of the variance- it is an unbiased estimate of $$\sigma^2$$. You can't have an estimate that is unbiased for $$\sigma^2$$ and simultaneously has its square root unbiased for $$\sigma$$. They had to choose one or the other and they chose to use the estimator that was unbiased for $$\sigma^2$$.