fm1 <- lmer(Reaction ~ Days + (Days | Subject), sleepstudy)

The notation (Days | Subject) says to allow the intercept and Days to vary randomly for each level of Subject .

Can you please explain me the result of the following commands ?

(Intercept)        Days 
 24.740448    5.922133 

[1] 21.595943  5.455217

[1] 25.59182


[1] 0.9183965

What is the INTERPRETATION of the results found from various commands?

That is , if one asks me what is the meaning of the results that you have found from sd(ranef(fm1)$Subject[,1]) and attr(summary(fm1)$varcor$Subject,"stddev")[1] ? Both are standard deviation of Intercept but of course there is difference between these two results . But I don't know what is this ?

In ?getMe , it is said that from summary(fm1)$sigma , we found residual standard error . But why doesn't the result match with sd(residuals(summary(fm1))) ?

Also , In summary(fm1)$varcor there is value 0.066 under the column Corr . Does it mean correlation between two random effects (Intercept) and Days is 0.066 ?

Any help is appreciated . Thank you .

  • $\begingroup$ You probably want to calculate sd(residuals(fm1)) and not sd(residuals(summary(fm1)). $\endgroup$
    – Qaswed
    Apr 27 '16 at 10:08

attr(summary(fm1)$varcor$Subject,"stddev")[1] gives you the estimated standard deviation of your random intercepts. This is the square-roots of the first main diagonal element of your estimate for the random effects variance-matrix $G$ (in Wikipedia's notation). You get $\hat{G}$ directly with VarCorr(fm1)$Subject[,]. With sd(ranef(fm1)$Subject[,1]) you manually calculate the standard deviation from your estimated random intercepts. A fundamental difference is, that $\hat{G}$ estimates a population parameter, and sd(ranef(fm1)$Subject[,1]) a parameter in a subset of the population. The same for 25.59182 = summary(fm1)$sigma vs. sd(residuals(fm1)) = 23.50343 (and not sd(residuals(summary(fm1))! I would say, this in an error in your code).

Yes, 0.066 is the correlation between the random effects.


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.