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My working population is non normally distributed and from time to time I need to take a sample of it. Sometimes my sample exists of 100 units, at other times of 150 units. Of my sample I calculate the mean, $\mu$, and with bootstrapping I determine the SE, $\sigma_{\bar x}$. Since $\sigma_{\bar x} = \sigma /\sqrt{N}$, I calculate $\sigma$. So far so good, by using control charts ($\mu,\sigma$) I can now see if there is anything happening to worry about.

By use of deviation calculation rules you can prove quite easily that the formula $\sigma_{\bar x} = \sigma /\sqrt{N}$ also can be used for non normal distributions.

I play with the idea to make control charts using the trimmed mean and the 70%-value. Now I am not so sure anymore. (Frankly, I have my doubts whether you can do this for normal distributions as well.) Can I use the same simple trick to remove the $n$ from my equation?

Let me put it like this: Is what I am writing valid? $$\sigma_{70\%} = cte * \sigma / \sqrt{N}$$ with a $cte$ independent of the value of N? $$\sigma_{\bar x_{70\%}} = cte * \sigma /\sqrt{N}$$ with a $cte$ independent of the value of $N$?

To use my control charts I don’t need to know the value of my $cte$. For my s-chart I can just plot out $cte* \sigma$ in function of my sample number. But I really need to get rid of the $N$!

I hope you understand what I am getting at. I would already be happy with a yes or a no, you can not do that anymore. If you can direct me to a place where I can find more information about this problem I would be even more grateful.

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    $\begingroup$ If you're going to invent a new control charting procedure, then there are more fundamental questions to ask: you need to ascertain the quality control analogs of false positive and false negative error rates. These are measured by the expected lengths of in-control runs when the system is in control and out of control, respectively. So I don't think it's productive to ask whether you can "do that;" instead, think in terms of how to achieve good run length properties for your procedure so that it works as intended. $\endgroup$
    – whuber ♦
    Commented Sep 3, 2012 at 20:24

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No. The reason the square root of n works for standard deviations is because the variance of a sum is the sum of variances. So for the sample mean,

var ({X$_1$+X$_2$+...+X$_n$}/n)= [var(X$_1$)+var(X$_2$)+...+var(X$_n$)]/n$^2$= n σ$^2$/n$^2$ = σ$^2$/n. Hence the standard error is σ/√n. The same result does not hold for trimmed means or percentiles of estimates.

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