PROC TRANSREG AND ANOVA

[ANKH1]

Thank you!

[Rick_SAS · Accepted solution]

The documentation for the TRANSREG section on the Box-Cox transformation says,

"This family of transformations of the positive [emphasis added] dependent variable y is controlled by the parameter ...."

It also states that you can "specify the PARAMETER=c transformation option when you want to shift the values of y, usually to avoid negatives." So when DV=0, you can to get rid of the error and apply the BC transformation by adding any positve value, such as the following (which adds 1):

proc transreg data=Have;
   model BoxCox(DV / lambda=1 parameter=1) = class(cat);
run;

[ANKH1]

Hi! 

Thanks, by adding the parameter=1 it ran. 

proc transreg details data=sample1 ;
model boxcox(NFSDDFI / lambda = -3 to 3 by 0.25 parameter =1) = CLASS(ANIMALCAT);
run;

I looked in "Details" and this was the output:

It recommends lambda=0. I log transformed but it didn't work since I have 11 zeros. 

Below is the code:

DATA sample2;
SET sample1;
TRANSFI=log(NFSDDFI);
RUN;

proc glm data=sample2;
class ANIMALCAT;
model TRANSFI= ANIMALCAT;
output out = notrans r= resid;
run;
symbol1 i=sm70;

proc univariate data=notrans noprint;
var resid;
histogram resid/normal kernel;
qqplot resid/normal (mu = est sigma=est);
run;

I tried using another DV that had no zeros but when I transformed the DV with the recommended lambda the DV was not normalized. 

Am I reading the output from proc transreg correctly?

[Rick_SAS]

Yes, you are reading the output correctly, but you need to incorporate the PARAMETER=1 information:

DATA sample2;
SET sample1;
TRANSFI=log(NFSDDFI + 1);
RUN;

[ANKH1]

Perfect! But according to K-S the resids are still not normal, even though the histogram looks pretty normal to me. What should I do? 

[Rick_SAS]

If you have 19,900 observations, the test will always fail. Very large samples invariably produce statistically significant GOF tests.

I don't know what you are trying to accomplish, but you probably don't need to "do" anything. Your parameter estimates and contrasts are probably spot-on. From the graph, I would expect the p-values and CIs for the estimates to be conservative, since it looks like the distribution of the residuals has a smaller tail than normality. So your 95% CI might actually be a 96% or 97% CI.. 

[ANKH1]

Ok, I will carry with our analyses with the transformed variable. Thank you very much!