Linear mixed effects model standardization with Z score not giving consistent results
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HI, I am running a mixed effects model with interaction terms, and when I run the model with the raw data, I essentialy get what I would expect
Linear mixed-effects model fit by ML
Model information:
Number of observations 44
Fixed effects coefficients 5
Random effects coefficients 22
Covariance parameters 2
Formula:
outcome ~ 1 + intervals + exposure*sessions + (1 | subject)
Model fit statistics:
AIC BIC LogLikelihood Deviance
264 276.49 -125 250
Fixed effects coefficients (95% CIs):
Name Estimate SE tStat DF pValue Lower Upper
'(Intercept)' 26.899 2.6919 9.9926 39 2.6128e-12 21.454 32.344
'exposure' -1.6766 1.0159 -1.6504 39 0.10689 -3.7313 0.3782
'intervals' -0.19934 0.033912 -5.8783 39 7.6441e-07 -0.26794 -0.13075
'sessions' -0.73017 0.44687 -1.634 39 0.11031 -1.6341 0.17371
'exposure:sessions' 0.4013 0.14279 2.8105 39 0.0076969 0.11249 0.69012
Random effects covariance parameters (95% CIs):
Group: subject (22 Levels)
Name1 Name2 Type Estimate Lower Upper
'(Intercept)' '(Intercept)' 'std' 3.1621 1.9751 5.0624
Group: Error
Name Estimate Lower Upper
'Res Std' 3.1433 2.3339 4.2334
But when I standardize the predictor using the matlab zscore function I get something totally different, whether I also standardize the dependent variable or not
Linear mixed-effects model fit by ML
Model information:
Number of observations 44
Fixed effects coefficients 5
Random effects coefficients 22
Covariance parameters 2
Formula:
drinking ~ 1 + intervals + exposure*sessions + (1 | subject)
Model fit statistics:
AIC BIC LogLikelihood Deviance
127.45 139.94 -56.727 113.45
Fixed effects coefficients (95% CIs):
Name Estimate SE tStat DF pValue Lower Upper
'(Intercept)' 0.0078404 0.17882 0.043845 39 0.96525 -0.35386 0.36954
'exposure' -0.020592 0.14613 -0.14091 39 0.88867 -0.31618 0.27499
'intervals' -0.1539 0.21666 -0.71034 39 0.48172 -0.59213 0.28433
'sessions' 0.3547 0.21217 1.6717 39 0.10258 -0.07446 0.78385
'exposure:sessions' -0.099309 0.24164 -0.41098 39 0.68334 -0.58807 0.38945
Random effects covariance parameters (95% CIs):
Group: subject (22 Levels)
Name1 Name2 Type Estimate Lower Upper
'(Intercept)' '(Intercept)' 'std' 0.6939 0.43148 1.1159
Group: Error
Name Estimate Lower Upper
'Res Std' 0.65418 0.4809 0.8899
When I look at plots of the individual predictors before and after z transformation, everything looks similar, just the values are (expectedly) different. I would not expect such a drastic difference in LME models. Grateful for any assistance! Thank you!
3 Comments
the cyclist
on 28 Sep 2022
Can you upload the data and the code you used to fit the models?
bsriv
on 28 Sep 2022
the cyclist
on 28 Sep 2022
Edited: the cyclist
on 28 Sep 2022
If anyone else happens to investigate, be aware that the attached files are MAT, not CSV.
Accepted Answer
the cyclist
on 28 Sep 2022
Edited: the cyclist
on 28 Sep 2022
I'm not sure why you expect the model to be unchanged (or what specific aspects of the model you expect to be unchanged).
Even with a simple linear model, it is obvious to me that the fitted parameters will be different under a zscore transform of the predictors. For example,
rng default
x1 = [-2; -1; 0; 1; 2] + 10;
x2 = zscore(x1); % new predictor is zscore of original predictor
y = 2*x1 + randn(5,1);
tbl = table(x1,x2,y);
mdl1 = fitlm(tbl,"y ~ x1") % fit original predictor
mdl2 = fitlm(tbl,"y ~ x2") % fit new predictor
In this simpler case, it is pretty easy to understand why certain things (e.g. beta for the x1 vs. x2 ) is different. It may or may not be obvious why the R-squared is unchanged.
In your more complicated model, especially because of the interactions, I think it could be tricky what could be expected to remain constant across the two models.
What did you expect to remain constant after this transform? Presumably not the beta coefficients.
4 Comments
bsriv
on 28 Sep 2022
the cyclist
on 28 Sep 2022
First, I have reason to believe that the zscore was not done correctly. When I zscored the ocds variable, I did not replicate the value in the other table. (I did this both with the built-in MATLAB function, and also just directly.)
When I used the correctly z-scored values on an OLS model (not LME), I get exactly what I expected (as I did in the small example I posted here). When I then used your LME model, some of the p-values were identical between the zscored and non-zscored models, but some were not. (For example, the interaction term was identical, as was the intervals term.) Note that I did zscore the response variable.
I can't say that I have figured out the entire theory behind this, but I speculate that your expectation could be wrong. In particular, I can imagine that allowing for different intercepts for each subject could result in non-significant slope betas, where they had been significant previously, given the scaling.
load("tbl_roc")
tbl_roc_z = tbl_roc;
tbl_roc_z.alc_cue = zscore(tbl_roc.alc_cue);
tbl_roc_z.intervals = zscore(tbl_roc.intervals);
tbl_roc_z.cbt = zscore(tbl_roc.cbt);
tbl_roc_z.drinking = zscore(tbl_roc.drinking);
lme = fitlme(tbl_roc,'drinking ~ alc_cue+cbt+intervals+alc_cue*cbt+(1|subject)')
lme_z = fitlme(tbl_roc_z,'drinking ~ alc_cue+cbt+intervals+alc_cue*cbt+(1|subject)')
bsriv
on 28 Sep 2022
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