# Variance of slope after fitting a mixed linear model

I have a question regarding variance components I obtained after fitting a mixed linear model with random slope and random intercept using R software. My model is the following:

$y_{ij} = β_0 + β_1 time_{ij}+ u_{0j} + u_{1j} time_i + ε_{ij}$

Results from fitting this linear mixed model is:

rs <- lmer(y ~ time + (time| id), data=dataset, REML=FALSE)

rs
Linear mixed model fit by maximum likelihood  ['lmerMod']
Formula: y ~ time + (time | id)
Data: dataset
AIC       BIC    logLik  deviance  df.resid
66.2413   70.0756   -27.1206  54.2413         8
Random effects:
Groups   Name       Variance  Std.Dev.   Corr
id      (Intercept) 8.846024. 2.97423
time       0.006426  0.08016   -1.00
Residual            1.473614. 1.21393
Number of obs: 14, groups:  id, 4

Fixed Effects:
Estimate Std.Error t value
(Intercept)   99.04580  1.57192   63.01
time          -0.16086  0.04457  -3.61


My questions are:

1. What is the slope variance $σ^2(β_1)$? It is 0.006426 or the result from vcov(rs)[2,2]?
2. Is it correct to say that the variance of residuals $σ^2(ε)$ is 1.473614?

You have answered your own questions. Indeed, the estimated variance of the random slope $u_{1j}$ (and not $\beta_1$ as you write) is 0.006426 and the estimated residual variance, i.e. the estimated variance of $\varepsilon_{ij}$ is 1.473614. Obviously, the estimated variance of the random intercept $u_{0j}$ is 8.846024. Notice that the variance of the estimate of the fixed effect $\beta_1$ is $0.04457^2$, which you can also get by the command vcov(rs)[2,2].