Re: st: lnskew0 and bcskew0

[tdavis7@emory.edu]

Hi -

Thank you for your response.  But I have one last question.  The latter 
transformation appears to be a simple z score?  Is that correct?  I ask 
because this changes my regression results a bit and I want to make 
sure that I haven't performed some obscure transformation that I am 
unable to explain.

I am concerned about normality because I am creating multiple imputed 
data sets using Amelia with the data that I currently have.  One of the 
assumptions of multiple imputation is normality (univariate at the 
least but mulitvariate ideally).  I plan on using STATA to estimated 
OLS regression coeffients with the imputed data, but I also plan to do 
some SEM and HLM with the imputed data.  Can I still use the "g 
z=invnorm(_n/(_N+1))" transformation or should I stick with lnskew0 
even though the histogram appears skewed despite the acutal skew 
statistic?

Thanks.


Quoting Austin Nichols <austinnichols@gmail.com>:


> tdavis7@emory.edu --
> The histogram may "be skewed" to your eye, but I'm betting the
> skewness is very very close to zero.
> sysuse auto, clear
> qui su mpg, d
> di r(skewness)
> lnskew0 z=mpg
> qui su z, d
> di r(skewness)
>
> I suppose the "right" method depends on the desired result--here is
> another way to transform your variable to make its skewness zero (and
> its mean zero, and its standard deviation one):
> sysuse auto, clear
> sort mpg
> g z=invnorm(_n/(_N+1))
>
> On 4/19/07, tdavis7@emory.edu <tdavis7@emory.edu> wrote:
> > How certain can I be that lnskew0 and bcskew0 have normalized the
> > distribution for the variable have transformed with them (a percent)?
> > After using these transformations, the histograms are still skewed.
> > What could be the problem? Is it possible that they don't work in all
> > instances?
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