# Mixed Models: Models solutions

This article contains solutions to exercises for an article in the mixed models series. For a list of topics covered by this series, see the Introduction.

There is often more than one approach to the exercises. Do not be concerned if your approach is different than the solution provided.

#### Exercise solutions

Write the R formula for a model with C as a fixed effect and two unobserved random variables. The random variables are random intercepts. One of the random variables is sampled as A, the other is sampled as B.

y ~ C + (1|A) + (1|B) results in the following model parameters- (intercept) (mean intercept associated with the groups of A and B at C = 0), \(\beta_0\)
- slope effect associated with C, \(\beta_1\)
- variance of intercept associated with the groups of A, \(\theta_1\)
- variance of intercept associated with the groups of B, \(\theta_2\)
- variance of residuals, \(\sigma^2\)

What is the mean and variance for both the unconditional y and the conditional y from problem 1?

Y\(_i\) \(| (\boldsymbol{B} = [b_j \ b_k ] )\) will have a mean of \(\beta_0 + \beta_1 c_i + b_{1,j[i]} + b_{2,k[i]}\) with variance of \(\sigma^2\), where \(c_i\) is the value of C from the \(i\)th observation, \(j[i]\) is the group of model variable A that observation \(i\) is a member of, \(k[i]\) is the group of model variable B that observation \(i\) is a member of, \(b_{1,j[i]}\) is the intercept effect associated with the \(j\)th sample of the model variable A, \(b_{2,k[i]}\) is the intercept effect associated with the \(j\)th sample of the model variable B.

Y\(_i\) will have a mean of \(\beta_0 + \beta_1\)c_i with variance of \(\theta_1 + \theta_2 + \sigma^2\)

Write the R formula for a model with a random slope for C and a correlated random intercept. The random slope is sampled as A.

y ~ C +(1+C|A) results in the following model parameters- (intercept) (mean intercept associated with the groups of A at C = 0), \(\beta_0\)
- slope effect associated with C (mean slope associate with the groups of A), \(\beta_1\)
- variance of intercept associated with A, \(\theta_1\)
- variance of slope of C associated with A, \(\theta_2\)
- covariance of the random intercept and the random slope of C within A, \(\theta_3\)
- variance of residuals, \(\sigma^2\)

This could also have been specified by y ~ C + (C|A).

What is the mean and variance for both the unconditional y and the conditional y from problem 3?

Y\(_i\) \(| (\boldsymbol{B} = \boldsymbol{b}_{j[i]})\) will have a mean of \(\beta_0 + \beta_1 c_i + b_{1,j[i]} + b_{2,j[i]} c_i\) with variance of \(\sigma^2\), where \(c_i\) is the value of C from the \(i\)th observation, \(j[i]\) is the group of model variable A that observation \(i\) is a member of, \(b_{1,j[i]}\) is the intercept effect associated with the \(j\)th sample of the model variable A, \(b_{2,j[i]}\) is the slope effect associated with the \(j\)th sample of the model variable A. There are two unobserved variables associated with with the population sampled by A. \(\boldsymbol{B}\) is the random vector \([ \boldsymbol{b}_1 \ \boldsymbol{b}_2]\).

Y\(_i\) will have a mean of \(\beta_0 + \beta_1\)c\(_i\) with variance of \(\theta_1 + \theta_2 + 2 \theta_3 + \sigma^2\)

Use the sleepstudy data set for the following exercises.

This data is load using

`library(lme4)`

`Loading required package: Matrix`

`data(sleepstudy)`

Create a model for Reaction time regressing on Days and accounting for the random effect of Subject.

`ssc <- lmer(Reaction ~ Days + (1 | Subject), data=sleepstudy) summary(ssc)`

`Linear mixed model fit by REML ['lmerMod'] Formula: Reaction ~ Days + (1 | Subject) Data: sleepstudy REML criterion at convergence: 1786.5 Scaled residuals: Min 1Q Median 3Q Max -3.2257 -0.5529 0.0109 0.5188 4.2506 Random effects: Groups Name Variance Std.Dev. Subject (Intercept) 1378.2 37.12 Residual 960.5 30.99 Number of obs: 180, groups: Subject, 18 Fixed effects: Estimate Std. Error t value (Intercept) 251.4051 9.7467 25.79 Days 10.4673 0.8042 13.02 Correlation of Fixed Effects: (Intr) Days -0.371`

The median reaction time is 288.7. Create variable for aboveAve. You can use the code provided below. Then create a model regressing on this new dichotomous variable

`aboveAve <- ifelse(sleepstudy$Reaction > 288.7, 1, 0) ssd <- glmer(aboveAve ~ Days + (1 | Subject), data=sleepstudy, family=binomial) summary(ssd)`

`Generalized linear mixed model fit by maximum likelihood (Laplace Approximation) [glmerMod] Family: binomial ( logit ) Formula: aboveAve ~ Days + (1 | Subject) Data: sleepstudy AIC BIC logLik deviance df.resid 158.6 168.2 -76.3 152.6 177 Scaled residuals: Min 1Q Median 3Q Max -7.5491 -0.3845 0.0003 0.3080 5.2636 Random effects: Groups Name Variance Std.Dev. Subject (Intercept) 7.951 2.82 Number of obs: 180, groups: Subject, 18 Fixed effects: Estimate Std. Error z value Pr(>|z|) (Intercept) -3.4601 0.9370 -3.693 0.000222 *** Days 0.7426 0.1299 5.719 1.07e-08 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Correlation of Fixed Effects: (Intr) Days -0.641`

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Last Revised: 3/23/2016