I have a data set with positive skewness when I log tranform it tends to be negatively skewed. Is there any other transformation that I can use or any statistical method works? Thanks!!!
$\begingroup$
$\endgroup$
14
-
$\begingroup$ Cube root is often quite good for distributions near the gamma (Wilson-Hilferty); but in many situations you simply won't find a suitable transformation. More importantly, why do you need to make data normal? $\endgroup$– Glen_bCommented Sep 10, 2014 at 23:48
-
$\begingroup$ Hello Glen_b, I am using linear mixed model for data analysis which assumes data to be normally distributed. I could use generalized linear mixed model but I don't know the distribution. $\endgroup$– AlphCommented Sep 11, 2014 at 1:49
-
$\begingroup$ (i) Are you looking at the marginal distribution of $y$? (ii) The discussion here - by invoking Gauss Markov - seems to suggest you don't need to assume normality of even the errors to estimate the model (though the usual inference - hypotheses and tests - presumably relies on it in small samples). $\endgroup$– Glen_bCommented Sep 11, 2014 at 2:09
-
$\begingroup$ Hello Glen_b, I guess I don't understand your point here about marginal distribution of y. Are you referring to population averaged model that is estimated by generalized estimating equations? I know linear mixed model is one of regression model. I appreciate your clarification. $\endgroup$– AlphCommented Sep 11, 2014 at 2:37
-
$\begingroup$ I mean "are you looking at the raw y's rather than conditional y's". So how did you decide "the data was skewed"? I'm reasonably sure there's no assumption - even for hypothesis tests - about the raw y's in linear mixed models. $\endgroup$– Glen_bCommented Sep 11, 2014 at 2:42
|
Show 9 more comments
2 Answers
$\begingroup$
$\endgroup$
2
The log is considered part of a whole continuum of power transformations...
Power result
-1 1/y
-.5 1/sqrt(y)
0 log y
.5 sqrt(y)
1 y
2 y^2
(The $0$ case is confusing because we all know that $y^0=1$. But it works out if you look at the limit of $(y^p-1)/p$ as $p$ approaches zero.)
Anyway, note that $\sqrt y$ corresponds to $p=\frac12$ which is between the identity and the log -- so that might give you pretty good results.
-
$\begingroup$ Hello rvl, how is the interpretation after analysis with transnformed variable? can we interpret in terms of original scale of variable? $\endgroup$– AlphCommented Sep 10, 2014 at 21:24
-
$\begingroup$ Well, you can square the mean to put it back on the original scale - it won't be the same as the mean, it's a different kind of mean. And you can back- transform confidence limits as well. But don't just back-transform the standard deviations or standard errors. Those will be wrong. (Look up the delta method if you want to know how to do this right.) finally, similar questions came up in another posting: stats.stackexchange.com/questions/114259/… $\endgroup$ Commented Sep 10, 2014 at 22:57
$\begingroup$
$\endgroup$
In other words, you need to get familiar with Box-Cox transformation that can be done via this R package, among other things.
