# When to use Bessel correction and and does it alter the standard error?

is there any condition about when to use the Bessel correction as Python uses it to find the Standard Deviation by default.

The standard error of the mean is $$\sigma/\sqrt{n}$$

but if $$\sigma$$ is calculated by using Bessel's correction, won't it affect the standard error of the mean?

• Maybe start by looking at Wikipedia. Then consider: If population mean $\mu$ is known, then $\sigma^2$ is estimated by $V = \frac 1n \sum_i (X-\mu)^2,$ with $E(V) = \sigma^2.$ If $\mu$ unknown and estimated by $\bar X,$ then $S^2 = \frac{1}{n-1}\sum_i (X_i - \bar X)^2$ has $E(S^2) = \sigma^2.$ – BruceET Oct 12 '19 at 17:53

It's essential to distinguish between $$\sigma$$ the standard deviation of the population distribution and a sample standard deviation $$s$$.
Bessel's correction corrects for bias in using $$s^2$$ to estimate $$\sigma^2$$; in particular if you use the natural $$n$$-denominator sample variance, $$E(s_n^2)<\sigma^2$$, but by just the amount that $$\frac{n}{n-1}s_n^2$$ would have the correct expected value (i.e. if you took many, many samples of size n and averaged their estimates, it would have the right value as long as the population variance was finite).
Consequently, you don't calculate $$\sigma$$ using Bessel's correction; if you can calculate $$\sigma$$ because you have the population, you do it from the definition of standard deviation for a distribution.
You can calculate the sample standard deviation $$s$$ or you can estimate $$\sigma$$ (typically by $$s$$).
You may well use Bessel's correction in calculating $$s$$. Whether you use Bessel's correction or not (or indeed use some other correction, such as one to attempt to unbias the standard deviation rather than the variance) would affect your estimate of the standard error, certainly. Simply choose an estimator with the properties you want.