[R-sig-ME] A consultation about lmer -- normality assumptions

[John Maindonald]

It is also assumed that the random effects are normally distributed.  Depending
on the relative magnitude of the variance components, and on the relative
contributions to the variance of the statistic that is of interest.  Where there are
a small number of random effects (alias BLUPs), normality can be a much more
serious matter for them than for the residuals, because there is less opportunity
for the Central Limit Theorem to kick in.

Unbalance in the data can make it quite a bit harder to check the relevant set(s)
of random effects for normality.  One may need to resort  to simulation.

John Maindonald             email: john.maindonald using anu.edu.au<mailto:john.maindonald using anu.edu.au>


On 24/12/2019, at 23:24, Thierry Onkelinx via R-sig-mixed-models <r-sig-mixed-models using r-project.org<mailto:r-sig-mixed-models using r-project.org>> wrote:

Dear anonymous,

The assumption of normality is on the residuals, not on the observations.
The assumption holds reasonably for the sleepstudy, except for a few
outliers.

library(lme4)
fm1 <- lmer(Reaction ~ Days + (Days | Subject), sleepstudy)
qqnorm(residuals(fm1))

Best regards,

ir. Thierry Onkelinx
Statisticus / Statistician

Vlaamse Overheid / Government of Flanders
INSTITUUT VOOR NATUUR- EN BOSONDERZOEK / RESEARCH INSTITUTE FOR NATURE AND
FOREST
Team Biometrie & Kwaliteitszorg / Team Biometrics & Quality Assurance
thierry.onkelinx using inbo.be<mailto:thierry.onkelinx using inbo.be>
Havenlaan 88 bus 73, 1000 Brussel
www.inbo.be

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Op ma 23 dec. 2019 om 17:36 schreef 何如梦 <18754808835 using 163.com>:

I have some confusion about the use of lmer. As I learned the data
should  be normal distributed when use liner mixed model. I tested the
exmple  data (Reaction$sleepstudy) by using shapiro.test. The result shows
that  it was not normal distributed.  So I have some confusion about the
requirment of data distribution in function lmer and glmmer.
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