1
$\begingroup$

I have a dataset in which the response variable is non normal, but on log transforming, it follows the normal distribution. I constructed a mixed effect model using lme4::lmer() as below (multiple measurements within each region), where A is a continuous variable and there are 7 regions

model<-lmer(log(response)~A + (1|Region), data, REML=FALSE)

My output is

Linear mixed model fit by maximum likelihood . t-tests use Satterthwaite's method ['lmerModLmerTest']
Formula: logresponse ~ A + (1 | Region)
Data: df[, -c(6)]

 AIC      BIC   logLik deviance df.resid 
151.8    159.8    -71.9    143.8       51 

Scaled residuals: 
 Min       1Q   Median       3Q      Max 
-2.24856 -0.63631  0.02206  0.59999  1.94895 

Random effects:
Groups   Name        Variance Std.Dev.
Region   (Intercept) 0.8611   0.9280  
Residual             0.5814   0.7625  
Number of obs: 55, groups:  Region, 7

Fixed effects:
              Estimate Std. Error        df t value Pr(>|t|)  
(Intercept) -1.112047   0.747373 19.651457  -1.488   0.1526  
 A           0.008585   0.004250 29.157266   2.020   0.0526 .

Correlation of Fixed Effects:
(Intr)
A -0.871

1) Would I infer that for A effects the response variable and the response increases by exp(0.008585) +/- exp(0.004250) (std error)? Would that be the right interpretation?

2) exp(Intercept) is not equal to the mean(log response). They are close but not equal. Why are they not equal and should I be concerned?

I know that this is a hard question to answer without knowing the dataset. So please suggest how I can improve the question.

$\endgroup$
  • $\begingroup$ Your resulting model is $y = \exp(-1.112 + 0.009 x)$. This is a non-linear function and thus there can't be a constant increase. Given the model above, I don't understand why you'd expect an equality in your second question. $\endgroup$ – Roland Aug 17 '18 at 8:44
2
$\begingroup$
  1. Even if the model is quite complicated, interpretation of the effect of A on the response variable is straightforward, as is any other linear model: according to your output, all other things being equal, a variation of magnitude $x$ in your independent variable A is associated with a variation of $ e^{ x * 0.008585}$ of your response variable. However, standard error is not at play when making prediction from your fitted model. It gives you informations about the accuracy of your coefficient estimate and allows you to compute a confidence interval around your estimate, but cannot be referred as a "more or less modifier" of the prediction.
  2. Regarding the interpretation of the intercept, it's a bit more complicated to translate it clearly, because it's related to the Region random effect. Here you have a random effect on the intercept, hence its value depends on the value of the categorical Region variable. The value -1.112047 showed in your output is the intercept value for the base value of the Region variable (i.e. Region category encoded as 0). In any case, there is no need for the intercept to be equal to the mean of the reponse variable, intercept is simply the expected value of your response variable when all other dependent variables are equal to zero.
$\endgroup$
2
$\begingroup$

The standard error can't be used after you transform, because exp(x)+exp(y)~=exp(x+y).

I find it easier to think in terms of confidence intervals instead: you can exponentiate your lower and upper bound confidence intervals to get the correct confidence intervals.

In addition, you should interpret your transformed beta coefficients (and associated confidence intervals) as multiplicative changes once you transform to the original units.

exp(.008585) ~= 1.0086

meaning you have a .86% increase for 1 unit of A. However, you cannot simply multiply this number for other values of A: you need to multiply before exponentiation, i.e., exp(.008585 * A). In practice, for a continuous predictor value and especially because you have just one fixed effect, it makes the most sense to think about predictions and prediction intervals of your response as a function of A, and then transform these to create an informative plot.

exp(Intercept) should be equal to mean(log response when A==0) only, they will not be equal if you average over all of your responses.

$\endgroup$
  • $\begingroup$ Thanks for pointing out the interpretation of the confidence limits. However, considering the answer below that the interpretation is in terms of % increase and % decrease, is that true for CIs too? With above example, upper CI will be exp(0.008585+ 0.004250)=1.012918 and hence upper CI is 1.2%? How do I interpret it in units of the response? $\endgroup$ – tg110 Aug 17 '18 at 13:51
  • $\begingroup$ Yes, the CIs also indicate % increase and decrease. $\endgroup$ – Bryan Krause Aug 17 '18 at 16:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.