# Comparing GLM/Lmer Models

I am struggling with choosing the correct model for my study, and I hoped that maybe someone would be able to help me, or shine some light please :)

I have lots of data about vegetation and individual preferences that I am trying to analyse. I tried using a mixed model (lmer) to begin with, with the 3 different fields where I repeated the study as a random effect. I began by making my model including every single interaction, but it was too much, and R gave me this error message:

"fixed-effect model matrix is rank deficient so dropping 151 columns / coefficients. Error: Dropping columns failed to produce full column rank design matrix"

So I dropped the interactions and just did the other factors. I did one version as a glm without the random effect, and one as an lmer with the random effect. I then tried to compare them using anova, but I don't understand the results, I'll put the code and results below.

Please can you guys have a look and tell me what you think, that would be great, Thank you! (Sorry this is so long)

mod2 <- glm(Buffer ~ Age + Sex + Captures
+ PC1+ PC2+ Lvl1_Av + Lvl1_Med
+ Lvl1_SD+ Lvl1_Sum+ Lvl2_Av+ Lvl2_Med
+ Lvl2_SD+ Lvl2_Sum+ Lvl3_Av
+ Lvl3_Med
+ Lvl3_SD
+ Lvl3_Sum
+ Lvl4_Av
+ Lvl4_Med
+ Lvl4_SD
+ Lvl4_Sum )

mod3 <- lmer(Buffer ~ Age + Sex + Captures
+ PC1+ PC2+ Lvl1_Av + Lvl1_Med
+ Lvl1_SD+ Lvl1_Sum+ Lvl2_Av+ Lvl2_Med
+ Lvl2_SD+ Lvl2_Sum+ Lvl3_Av
+ Lvl3_Med
+ Lvl3_SD
+ Lvl3_Sum
+ Lvl4_Av
+ Lvl4_Med
+ Lvl4_SD
+ Lvl4_Sum + (1|Fence))

anova(mod2, mod3, test="Chisq")

#And this is what I got

> anova(mod2, mod3, test="Chisq")
Analysis of Deviance Table

Response: Buffer

Terms added sequentially (first to last)

Df Deviance Resid. Df Resid. Dev  Pr(>Chi)
NULL                        80    7541772
Age       1   335703        79    7206069 < 2.2e-16 ***
Sex       1  1225073        78    5980996 < 2.2e-16 ***
Captures  1  1365027        77    4615968 < 2.2e-16 ***
PC1       1     9632        76    4606337 0.0001964 ***
PC2       1   194968        75    4411369 < 2.2e-16 ***
Lvl1_Av   1      883        74    4410486 0.2596526
Lvl1_Med  1    24511        73    4385975 2.848e-09 ***
Lvl1_SD   1    69605        72    4316370 < 2.2e-16 ***
Lvl1_Sum  1  4229768        71      86602 < 2.2e-16 ***
Lvl2_Av   1      250        70      86352 0.5485363
Lvl2_Med  1      360        69      85992 0.4713995
Lvl2_SD   1      237        68      85755 0.5589011
Lvl2_Sum  1    24078        67      61676 3.922e-09 ***
Lvl3_Av   1     1330        66      60346 0.1664550
Lvl3_Med  1      345        65      60001 0.4810493
Lvl3_SD   1        2        64      59999 0.9524658
Lvl3_Sum  1     1395        63      58604 0.1564064
Lvl4_Av   1      304        62      58300 0.5085230
Lvl4_Med  1     3928        61      54372 0.0174052 *
Lvl4_SD   1       85        60      54286 0.7260341
Lvl4_Sum  1    13301        59      40985 1.210e-05 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

• Hi there and welcome. Where can we find the data? – Reviewer – Jim Jul 17 '18 at 23:04
• Hi, thanks :) I don't really know how to include data on here, so I didn't put any! I see now that that might not be very helpful. I tried copying some into here but that doesn't work at all – Maddy Jul 18 '18 at 20:02

You are specifying the model comparison wrong.

1. There is no reason to use glm with a Gaussian family. Use lm as it is fully equivalent but computationally superior.

2. You need to ensure that R uses the correct method for the anova generic. Since this is an S3 generic and method dispatch works only on the first argument, the lmer model must be first. Your code actually calls anova.glm, which does not do the intended model comparison.

So, in summary, with an easy example based on the iris dataset:

mod1 <- lm(Sepal.Length ~ Sepal.Width, data = iris)
mod2 <- lmer(Sepal.Length ~ Sepal.Width + (1 | Species), data = iris)

anova(mod2, mod1, test="Chisq")
#refitting model(s) with ML (instead of REML)
#Data: iris
#Models:
#fit1: Sepal.Length ~ Sepal.Width
#fit2: Sepal.Length ~ Sepal.Width + (1 | Species)
#     Df    AIC    BIC   logLik deviance  Chisq Chi Df Pr(>Chisq)
#fit1  3 371.99 381.02 -182.996   365.99
#fit2  4 200.53 212.57  -96.265   192.53 173.46      1  < 2.2e-16 ***
#---
#Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

• Thank you very much, that's very helpful, I'll give that a go! – Maddy Jul 18 '18 at 19:58
• And just one more for clarification sorry; the model didn't work when I included all the interactions, is it okay to just not write them into the model, or is that inaccurate and I should find a way to include them in the model? Thanks! – Maddy Jul 18 '18 at 20:00
• @Maddy I can't really comment on that without much more information. However, you seem to have many predictors and their names are suspicious. You should check for collinearity and look into methods like LASSO. I believe you are already over-fitting without interactions. – Roland Jul 19 '18 at 6:03