How to combine standard deviations from independent samples? I have a population of 7954 subjects and multiple random samples from it:

*

*1st sample size 2,665 with a standard deviation 1.386 and mean -7.45

*2nd sample size 2,399 with std dev 1.53 and mean -7.33

*3rd sample size 802 with std dev 3.02 and mean -7.45.

I want to calculate a standard deviation for the population.  Also, if I take another sample of 2,133 what can be my standard deviation be for it?
σ = s * √(n / (n - 1)) and σₓ = σ/√n I couldn't yield any results or validate them.
 A: Taking this question (and comments) at face value, it concerns a distribution $F$ (the empirical distribution of a population) and statistics from multiple independent random samples of $F.$ This formulation is possible because the sampling was done with replacement (as stated in a comment to the question).

To establish a notation, let the samples be indexed by $i = 1, 2, 3$ (the results will obviously generalize to other than three samples); designate their sizes as $n_i;$ let their means and standard deviations be $m_i$ and $s_i,$ respectively; and suppose the observations in sample $i$ are $x_{ij},$ $j = 1, 2, \ldots, n_i.$
Combining the $x_{ij}$ into a single sample yields a random sample with replacement from $F.$  The question asks how to estimate the standard deviation of $F$ from the combined sample.
The answer is an algebraic identity that expressed the formulas for the counts, sums, and sum of squares  of the combined sample in terms of the individual sample statistics.
Most likely you used a formula equivalent to
$$m_i  n_i = \sum_{j=1}^{n_i} x_{ij}$$
for the means, a relation used to simplify the standard deviation below.  This implies the residuals sum to zero,
$$\sum_{j=1}^{n_i} (x_{ij} - m_i) =\sum_{j=1}^{n_i} x_{ij} - \sum_{j=1}^{n_i}m_i = \sum_{j=1}^{n_i} x_{ij} - m_i n_i = 0,$$
which will also be exploited next.
Your formula for the standard deviations probably was something like
$$\begin{aligned}
(n_i-1)s_i^2 &= \sum_{j=1}^{n_i} \left(x_{ij} - m_i\right)^2 \\
&=\sum_j x_{ij}(x_{ij} -m_i) - \sum_j m_i(x_{ij} - m_i)\\
&= \sum_j (x_{ij}^2 -x_{ij}m_i) - m_i\sum_j (x_{ij} - m_i)\\
&=\sum_j x_{ij}^2 - m_i \sum_j x_{ij} - m_i(0)\\
&= \sum_{j=1}^{n_i} x_{ij}^2 - n_i m_i^2.
\end{aligned}$$
We thereby obtain an expression for the sum of squares of all the data in terms of the means and standard deviations of the individual samples:

$$\sum_{j=1}^{n_i} x_{ij}^2  = (n_i-1)s_i^2 + n_i m_i^2.$$


*

*The combined count is $$n = \sum_{i=1}^3 n_i.$$


*The combined sum is $$S = \sum_{i=1}^3 n_i m_i.$$


*The combined sum of squares is$$SS = \sum_{i=1}^3 (n_i-1)s_i^2 + n_i m_i^2.$$
Thus, the combined statistics are
$$m = \frac{1}{n}\sum_{i=1}^3 S$$
and
$$s^2 = \frac{1}{n-1}\sum_{i=1}^3 (SS - nm^2).$$

Remarks
$s^2$ is an unbiased estimator of the population variance.  Thus, on average, any random sample (of any size greater than $1$) will have a variance equal to $s^2.$  This implies its standard deviation will, on average, not equal $s,$ but it ought to be close.  Without more detailed information about the data, this is the best we can do.
If instead you used different formulas for the $m_i$ and $s_i,$ you can work through the algebra in the same way.  You won't be successful with some formulas (such as those based on order statistics), but with the commonest ones the algebra will be equally simple.
Statistics based on higher moments can be combined with the same technique.  For instance, various formulas for skewness involve the counts, sums, sums of squares, and sums of cubes of the data.  Solve for these sums in terms of the counts, means, standard deviations, and skewnesses of the individual samples; combine them; and apply the desired formula.
