Equation of standard error of a sample mean In the equation of the standard error of a sample mean, it's given by: standard deviation divided by root-n.
My major concern is: is it the standard deviation of population or the sample on which this is based?
In some books, it is mentioned to be standard deviation of population while some books mentioned it is standard deviation of sample. In Wikipedia, it is the standard Deviation of the population which is impossible to know, hence standard deviation of sample is used. 
 A: In the usual situation, the actual standard error of the mean is $\sigma/\sqrt{n}$, but you almost never know $\sigma$ (the population standard deviation), so you have to estimate it. Consequently, the estimated standard error of the mean is usually taken to be $s/\sqrt{n}$. 
Some sources call that "the standard error of the mean", but strictly it's an estimate of that.
A: When you divide by n, it's the "population". But that's theoretical hogwash, you never know the SD of a population or if you did why the heck wouldn't you know the mean?
Using the sample SD is fine, but the SD/root-n estimator is biased due to using the same data to estimate two quantities. The degrees-of-freedom correction: SD / root(n-1) gives the minimum variance unbiased estimator.
A: It was already answered in two previous answers, but as a comment, notice that what Wikipedia says is:

Estimate
Since the population standard deviation is seldom known, the standard
  error of the mean is usually estimated as the sample standard
  deviation divided by the square root of the sample size (assuming
  statistical independence of the values in the sample).
  $$ {\sigma}_\bar{x}\ \approx \frac{s}{\sqrt{n}} $$

So the definition is given in terms of the unknown population parameter, but the estimator (function to estimate it from the data you have) is given in terms of sample estimate.
