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Background

I am running a mixed linear model with four fixed predictors and (most of) their interactions. I am quite new to these models and I am trying to understand how I should interpret my results and whether I am using the correct approach to generate them.

My Model

My model contains two factor variables: (1) Group_Variable with three levels: 'Group_1_H', 'Group_2_L','Group_3_T'; (2) Variable_C with two levels: 'level1_O','Level2_S'.

And it contains two continuous variables: Variable_A and Variable_B. The dependent variable is continuous.

My prediction is that there will be an interaction between Group_Variable and Variable_C, whereby Group_1_H will show have a significantly different score on the outcome variable based on whether they are in condition 'level1_O' compared to 'level2_S' relative to the other two groups.

Here is the model specification:

model <- my_outcome ~  lmerTest::lmer(Group_Variable * 
    Variable_A * Variable_C + Group_Variable * Variable_B * 
    Variable_C + (1 | ID_Variable), data = mydata)

Here is my output when I use the summary() function:

> summary(model)
Linear mixed model fit by REML. t-tests use Satterthwaite's method ['lmerModLmerTest']
Formula: Outcome_Variable ~ Group_Variable * Variable_A * Variable_C +      Group_Variable * Variable_B * Variable_C + (1 | Variable_ID)
   Data: crossvalidate

REML criterion at convergence: -2102.4

Scaled residuals: 
    Min      1Q  Median      3Q     Max 
-7.2535 -0.4783  0.0113  0.5460  4.3982 

Random effects:
 Groups      Name        Variance Std.Dev.
 Variable_ID (Intercept) 0.01559  0.1249  
 Residual                0.03115  0.1765  
Number of obs: 3950, groups:  Variable_ID, 88

Fixed effects:
                                                        Estimate Std. Error         df t value Pr(>|t|)    
(Intercept)                                            1.939e-01  5.085e-02  1.181e+03   3.813 0.000144 ***
Group_VariableGroup_2_L                               -9.593e-02  6.816e-02  1.140e+03  -1.407 0.159566    
Group_VariableGroup_3_T                               -5.413e-02  6.571e-02  1.017e+03  -0.824 0.410242    
Variable_A                                            -1.011e-02  1.155e-02  3.851e+03  -0.875 0.381359    
Variable_CLevel2_S                                    -4.811e-02  5.423e-02  3.853e+03  -0.887 0.375013    
Variable_B                                             8.054e-02  8.273e-03  3.862e+03   9.736  < 2e-16 ***
Group_VariableGroup_2_L:Variable_A                     2.542e-02  1.542e-02  3.851e+03   1.648 0.099371 .  
Group_VariableGroup_3_T:Variable_A                     3.246e-02  1.493e-02  3.851e+03   2.175 0.029705 *  
Group_VariableGroup_2_L:Variable_CLevel2_S             4.883e-02  7.282e-02  3.852e+03   0.671 0.502523    
Group_VariableGroup_3_T:Variable_CLevel2_S             2.971e-02  7.060e-02  3.851e+03   0.421 0.673929    
Variable_A:Variable_CLevel2_S                          3.740e-02  1.397e-02  3.850e+03   2.678 0.007441 ** 
Group_VariableGroup_2_L:Variable_B                     1.457e-02  1.119e-02  3.860e+03   1.303 0.192717    
Group_VariableGroup_3_T:Variable_B                     2.173e-02  1.092e-02  3.860e+03   1.990 0.046710 *  
Variable_CLevel2_S:Variable_B                          1.579e-02  1.005e-02  3.856e+03   1.570 0.116423    
Group_VariableGroup_2_L:Variable_A:Variable_CLevel2_S -2.434e-02  1.875e-02  3.850e+03  -1.298 0.194355    
Group_VariableGroup_3_T:Variable_A:Variable_CLevel2_S -4.374e-02  1.836e-02  3.850e+03  -2.382 0.017255 *  
Group_VariableGroup_2_L:Variable_CLevel2_S:Variable_B -1.081e-02  1.363e-02  3.855e+03  -0.793 0.427704    
Group_VariableGroup_3_T:Variable_CLevel2_S:Variable_B  7.997e-03  1.344e-02  3.854e+03   0.595 0.551886    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

And here is the output when I use the Anova function from the car package:

> car::Anova(model)
Analysis of Deviance Table (Type II Wald chisquare tests)

Response: Outcome_Variable
                                         Chisq Df Pr(>Chisq)    
Group_Variable                          6.3807  2   0.041157 *  
Variable_A                             31.3062  1  2.204e-08 ***
Variable_C                            110.3747  1  < 2.2e-16 ***
Variable_B                           1553.9462  1  < 2.2e-16 ***
Group_Variable:Variable_A               1.0300  2   0.597500    
Group_Variable:Variable_C              26.7630  2  1.543e-06 ***
Variable_A:Variable_C                   2.8659  1   0.090477 .  
Group_Variable:Variable_B              18.5980  2  9.152e-05 ***
Variable_C:Variable_B                   7.7037  1   0.005511 ** 
Group_Variable:Variable_A:Variable_C    5.6792  2   0.058450 .  
Group_Variable:Variable_C:Variable_B    2.1613  2   0.339370    
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

And the post hoc pair-wise contrasts:

Group_Variable_pairwise Variable_C_pairwise estimate SE df t.ratio p.value
Group_3_T - Group_1_H level1_O - Level2_S 0.0767201 0.0156317 3889.770 4.907988 0.0000029
Group_3_T - Group_2_L level1_O - Level2_S 0.0156633 0.0148981 3889.107 1.051362 0.5444832
Group_1_H - Group_2_L level1_O - Level2_S -0.0610568 0.0163169 3898.048 -3.741947 0.0005430

Questions

I have some questions based on this.

  1. Is the Anova() function appropriate here? From what I have read online, people use the Anova function to compare models to find out whether the addition of a variable of interest significantly improves the model fit (e.g. Model interpretation in R (anova vs summary output)). Can I also use it just to summarise the model I have specified and interpret the results?

  2. Why are the p values for the interaction term (highlighted in both screenshots) different between the summary and Anova outputs? I have been trying to read about this online, and from what I understand the Anova is an omnibus test (i.e. compares several parameters at once) whereas the summary is step-wise. However, I would still expect there to be a significant result for 'Group_VariableGroup_1_H:Variable_CLevel2_S' in the summary as this is (I think?) essentially the same comparison as group 3 vs group 1 in the post hoc comparison table.

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    $\begingroup$ Please type your question as text, do not just post a photograph or screenshot (see here). When you retype the question, add the self-study tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. Please make these changes as just posting your homework & hoping someone will do it for you is grounds for closing. $\endgroup$ Commented Oct 30, 2021 at 13:45
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    $\begingroup$ Thanks for your feedback to improve my question. I have now replaced the screenshots with code. The self study tag is not appropriate here as this is not a homework question, it is related to my research. $\endgroup$
    – Zcjth84
    Commented Nov 1, 2021 at 9:37

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