ANCOVA by SPSS with outliers and non-homogeneity

Question

I’m working on ANCOVA to compare post- vs pre-treatment percentage change of a blood test value, L, between 10 groups (10 dosages from 3 drugs, all of which can lower L), with the pre-treatment (baseline) value of L as a covariate. I aim to get the baseline-adjusted least-square means for % change of L, i.g. the efficacy of those drugs and dosages. This method is backed by some papers.

The n varies greatly between groups, from 100+ to 10k+, as some dosages are prescribed more often. Usually, L is right-skew distributed, so I Ln-transformed it both at pre- and post-treatment, and the dependent became LnPost-LnPre = Ln(Post/Pre) with the covariate LnPre.

By transformation, the dependents of each group looked normally distributed, except for several outliers. I checked, and those are not typos. So their “normal curves” on the histogram looked like a bell shape but with prolonged lines at the bottom on either side or both. Moreover, the dependents didn’t pass the homogeneity test in ANOVA, suggesting variance inequality. Plus, the group-by-covariate interaction was positive.

So, I got outliers besides bell shapes, non-homogeneity, and an interaction between my dependent and covariate, which broke the assumptions of ANCOVA.

Q1: I’m unsure if I could still use ANCOVA; if not, can I use other models to achieve my goal?

Q2: Can the transformed dependent be a bell shape with outliers on the histogram?

Answer 1

Percent change is not a good response variable for reasons detailed here. And the baseline adjustment often needs to be nonlinear. Use of log ratios as you did is much better than using percent change but still makes too many assumptions in some cases.

Instead of an ad hoc approach to examining normality and equal variance assumptions consider semiparametric regression models that treat Y as ordinal, making the result much more robust, and invariant to transformations of Y. The proportional odds model is one such model. See the Nonparametrics chapter of BBR.