We can estimate the standard error (SE) of the sample mean as the sample standard deviation divided by the square root of the number of samples, cf. https://en.wikipedia.org/wiki/Standard_error#Standard_error_of_the_sample_mean which writes it as:

${\displaystyle {\sigma }_{\bar {\mathbf{x}}}\ \approx {\frac {\sigma _{\mathbf{x}}}{\sqrt {N}}}}$

Does there exist a similar measure (where the number of samples is involved) to estimate the standard error of the mean based on mean absolute deviation (MAD)? By MAD I mean the mean of the absolute value of the deviation from the mean:

$\displaystyle\mathrm{MAD}(\mathbf{x}) = \frac{1}{N}\sum_i^{N} \mathrm{Abs}(x_i-\bar{\mathbf{x}})$

The MAD is, in my mind, a similar quantity to the SD (in that it measures some sort of spread), but is not equal to the SD in the limit $N\to\infty$. Would it be terribly wrong to write the MAD-based SE as something like $\displaystyle\frac{\mathrm{MAD}(\mathbf{x})}{\sqrt{N}}$?



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