Output interpretation of mixed anova with lme4 package in R

Question

I'm quite confused about interpreting the results from the lme4 package. When looking through this webpage and google, it seems like more people are generally confused. I cannot find a clear overview on what to do with all the output, so I hope this thread can function as a general overview for more people.

I'll use my data as an example: I have a 4 (between:groups)x 2 (within:time) design. I am looking for an interaction effect on reaction times of an attentional control task (e.g. stroop task). Since I have an unbalanced design (16, 19, 14 and 17 obs per group) after removing outliers, I wanted to use the lmer function to run a mixed anova (or mixed model now) in R.

I run the following command:

options(contrasts = c("contr.sum", "contr.poly"))

lmer_mixed_ANOVA <- lmer(control_out ~time*Groups + (1|ID), data=data_RT_control)

print(Anova(lmer_mixed_ANOVA,type=3))

print(anova(lmer_mixed_ANOVA,type=3))

print(summary(lmer_mixed_ANOVA))

and I get this as output:

   [1] "data_RT_control"
Analysis of Deviance Table (Type III Wald chisquare tests)

Response: control_out
               Chisq Df Pr(>Chisq)    
(Intercept) 291.1728  1  < 2.2e-16 ***
time          9.3639  1   0.002213 ** 
Groups        2.1204  3   0.547791    
time:Groups   6.5060  3   0.089427 .  
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Analysis of Variance Table of type III  with  Satterthwaite 
approximation for degrees of freedom
            Sum Sq Mean Sq NumDF  DenDF F.value   Pr(>F)   
time        8164.2  8164.2     1 58.039  9.3639 0.003348 **
Groups      1848.8   616.3     3 58.980  0.7068 0.551774   
time:Groups 5672.4  1890.8     3 57.995  2.1687 0.101448   
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Linear mixed model fit by REML 
t-tests use  Satterthwaite approximations to degrees of freedom ['lmerMod']
Formula: control_out ~ time * Groups + (1 | ID)
   Data: all_dataframes[[i]]

REML criterion at convergence: 1301

Scaled residuals: 
    Min      1Q  Median      3Q     Max 
-2.4262 -0.5228 -0.1374  0.5249  2.3927 

Random effects:
 Groups   Name        Variance Std.Dev.
 ID       (Intercept) 262.2    16.19   
 Residual             871.9    29.53   
Number of obs: 136, groups:  ID, 70

Fixed effects:
              Estimate Std. Error     df t value Pr(>|t|)    
(Intercept)     54.778      3.210 59.020  17.064  < 2e-16 ***
time1           -7.819      2.555 58.040  -3.060  0.00335 ** 
Groups1         -3.448      4.399 58.990  -0.784  0.43626    
Groups2          2.560      2.732 59.150   0.937  0.35263    
Groups3         -1.519      1.830 58.900  -0.830  0.40978    
time1:Groups1   -5.170      3.501 58.020  -1.477  0.14512    
time1:Groups2   -3.987      2.176 58.160  -1.832  0.07200 .  
time1:Groups3   -1.476      1.456 57.920  -1.014  0.31478    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Correlation of Fixed Effects:
            (Intr) time1  Grops1 Grops2 Grops3 tm1:G1 tm1:G2
time1       -0.031                                          
Groups1     -0.035  0.021                                   
Groups2      0.087 -0.004  0.027                            
Groups3     -0.022  0.002  0.020 -0.051                     
time1:Grps1  0.021 -0.044 -0.030 -0.016 -0.012              
time1:Grps2 -0.004  0.087 -0.016 -0.033  0.002  0.034       
time1:Grps3  0.002 -0.023 -0.012  0.002 -0.029  0.026 -0.051

This chunk of output is really overwhelming! Different tables give different p-values. Is there an overview, or would someone be willing to create an overview of how to interpret these outcomes step by step.

Answer 1

I assume that the order of your output matches the order of the code. The output object is lmer_mixed_ANOVA

The Anova function is a function from the car package. It appears to be doing calculations on deviances, not fixed effects t-tests or z-tests in an ANOVA table. The anova function is a method for output object from the lmer function. This table is comparable to SAS Type III sum of squares because of the type option.

The summary function applied to the output object gives the scaled residuals, the random effects, and the estimates of each fixed effect level, and the correlations of these effects. It sets the intercept to Group0, time0. 'time1' is the difference between the two time points, at Group 0. The three estimates called Group'n' are the difference between each Group and Group 0 at time 0. The following interaction terms are estimates at each Group level and time 1. The t-tests after the estimates test if these differences are equal to 0.

Yes, mixed models in R is confusing. I prefer commercial software for ease of use.