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I can calculate the standard error of the mean as:

$SEM=\frac{\sigma }{\sqrt{n}}$

where $\sigma$ is the standard deviation and $n$ the sample size.

I would imagine that the equivalent for the median is simply:

$SEMedian=\frac{MAD}{\sqrt{n}}$

where $MAD$ is the median absolute deviation.

But I cannot find any mention of this, so I imagine not. What am I getting wrong? Is there an equivalent to the SEM for the median?

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  • $\begingroup$ Why would you assume $\text{SE}_{\tilde{x}} = \frac{\tilde{x}}{\sqrt{n}}$, when $\text{SE}_{\bar{x}} \ne \frac{\bar{x}}{\sqrt{n}}$? $\endgroup$
    – Alexis
    Oct 20, 2022 at 18:13
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    $\begingroup$ My bad, dumb mistake. Now corrected. $\endgroup$ Oct 20, 2022 at 18:16
  • $\begingroup$ That other question does not explain what is wrong with my logic. That comment which just got deleted was much better. But never mind. $\endgroup$ Oct 20, 2022 at 18:32
  • $\begingroup$ (1) What logic? (2) Another take on your question would be to consider what, if anything in general, is estimated by the median absolute deviation of the sample over the square root of the sample size, & I've added another link. $\endgroup$ Oct 20, 2022 at 19:09
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    $\begingroup$ +1 @francoiskroll Thank you for the correction! Just a mistake, no 'dumb' about it. :) I hope the duplicate links are informative. $\endgroup$
    – Alexis
    Oct 20, 2022 at 20:46

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