I've got a se of the mean that is less than the se of the median, what does this tell me about my sampling distr.
Each standard error calculation was used with same sample size.
1
-
$\begingroup$ Almost a duplicate of For what (symmetric) distributions is sample mean a more efficient estimator than sample median?. It specifies symmetry, which is useful if we want the sample mean and median to be converging to the same quantity in large samples, given that otherwise it's relatively unlikely to be the case that the two population quantities coincide; the analysis done there would seem to apply here. [The answer is mine, so I'll leave the judgement to others] $\endgroup$– Glen_bCommented Nov 6, 2016 at 3:31
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
3
Both the sample mean and the sample median (for a distribution with a smooth pdf) have a normal distribution for large sample size.
The variance of the sample mean is $\frac{\sigma^2}{n}$
The variance of the sample median is $\frac{1}{4nf(m)^2}$ where $f(m)$ is the probability density function evaluated at the median (source).
So for large sample sizes a comparison between the standard error of the mean and standard error of the median only gives you a comparison between the variance and the pdf evaluated at the median.
-
$\begingroup$ Good reasoning--but didn't you intend to conclude that it gives you a comparison between the variance and the PDF evaluated at the median? $\endgroup$– whuber ♦Commented Nov 5, 2016 at 20:22
-
-
$\begingroup$ That variance of the sample median is an asymptotic result -- it could probably be made clearer in the linked answer. $\endgroup$– Glen_bCommented Nov 6, 2016 at 3:28
$\begingroup$
$\endgroup$
2
As per @Hugh, I do not think it is measure of a distribution's normality.
I think you may want to compare coefficient of variation $\frac{\sigma}{\mu}$ with interquartile ratio $\frac{IQR}{median}$, which only for a normal distribution would be $1.349\frac{\sigma}{\mu}=\frac{IQR}{median}$ or some variation thereof. For example, comparison of standard deviation and IQR forms the basis of a simple test for normality.
-
1$\begingroup$ In your first paragraph you seem to confuse a result that holds only for Normal distributions for a definition. See the earlier reply by Hugh or consult stats.stackexchange.com/questions/45124. $\endgroup$– whuber ♦Commented Nov 5, 2016 at 20:24
-