# Re: st: lnskew0 and bcskew0

 From [email protected] To [email protected] Subject Re: st: lnskew0 and bcskew0 Date Tue, 24 Apr 2007 13:15:08 -0400

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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```
```

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