tdavis7 --
My suggestion was a bit tongue in cheek--it is not a z-score, as that
is an affine transformation (-ssc inst center- to use -center- to make
z-scores using the s option), whereas "sort mpg // g
z=invnorm(_n/(_N+1))" is nonlinear and makes a variable look very
normal indeed...
You should probably not transform your variable at all.
Why aren't you are using -ice-and -mim- from SSC? -ssc inst ice- and
-ssc inst mim- put both at your disposal.
On 4/24/07, [email protected] <[email protected]> wrote:
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 <[email protected]>:
> [email protected] --
> 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, [email protected] <[email protected]> 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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