I recently came across what I think may be a problem in how the `anova()` function from the `lmerTest` packages computes its F-statistics and corresponding P-values for fixed effects from mixed-effects models. Let me start by saying that I know of the controversy surrounding calculating P-values from mixed effects models (for reason discussed [here](https://stat.ethz.ch/pipermail/r-help/2006-May/094765.html)). Nonetheless, many folks still want P-values and thus a number of ways have been developed to accommodate this (see [here](http://stats.stackexchange.com/questions/118416/getting-p-value-with-mixed-effect-with-lme4-package)). Here I want to show the results of a commonly used approach — namely, the `anova` function from the `lmerTest` package — and hope that someone has an idea of why the results are not quite making sense.
First [here](https://www.dropbox.com/sh/iaio65nq9mu0owx/AAD1C4-l58ldfMS0-9XAf-_ba?dl=0) is my data. I had to link to it because of its size. Note that the biomass column has been standardized (mean = 0, sd = 1), hence the negative values. This does not alter the output. Once downloaded and the working directory has been specified, the file can be read in as follows:
dat <- read.csv("StackOverflow_Data.csv", header = T)
Below is my model using the `lmer` function from `lme4`. In this model I have plant biomass as a response variable and three factors — A, B, and C — each with two levels, as predictors. Plant Genotype and spatial block are included as random effects.
model <- lmer(Biomass ~ A + B + C +
A:B + A:C +
B:C + A:B:C +
(1 | Genotype) + (1 | Block) ,
data = dat, REML = T)
Summarizing the above model using `summary(model)` we get:
Linear mixed model fit by maximum likelihood t-tests use Satterthwaite approximations
to degrees of freedom [lmerMod]
Formula: Biomass ~ A + B + C + A:B + A:C + B:C + A:B:C + (1 | Genotype) +
(1 | Block)
Data: dat
AIC BIC logLik deviance df.resid
1059.7 1111.0 -518.8 1037.7 776
Scaled residuals:
Min 1Q Median 3Q Max
-3.04330 -0.63914 0.00315 0.69108 2.82368
Random effects:
Groups Name Variance Std.Dev.
Genotype (Intercept) 0.07509 0.2740
Block (Intercept) 0.01037 0.1018
Residual 0.19038 0.4363
Number of obs: 787, groups: Genotype, 50; Block, 6
Fixed effects:
Estimate Std. Error df t value Pr(>|t|)
(Intercept) 2.27699 0.08162 47.50000 27.897 < 2e-16 ***
AYes -0.02308 0.09958 99.30000 -0.232 0.81719
BReduced -0.11036 0.06232 733.00000 -1.771 0.07700 .
CSupp -0.02152 0.06243 733.70000 -0.345 0.73039
AYes:BReduced 0.25113 0.08838 733.70000 2.841 0.00462 **
AYes:CSupp 0.02179 0.08854 734.50000 0.246 0.80567
BReduced:CSupp 0.19436 0.08838 733.10000 2.199 0.02817 *
AYes:BReduced:CSupp -0.21746 0.12507 734.20000 -1.739 0.08251 .
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Correlation of Fixed Effects:
(Intr) AYes BRedcd CSupp AYs:BR AYs:CS BRd:CS
AYes -0.607
BReduced -0.379 0.311
CSupp -0.379 0.311 0.498
AYes:BRedcd 0.269 -0.444 -0.706 -0.354
AYes:CSupp 0.268 -0.444 -0.352 -0.708 0.503
BRedcd:CSpp 0.267 -0.219 -0.706 -0.705 0.498 0.500
AYs:BRdc:CS -0.190 0.315 0.500 0.502 -0.709 -0.709 -0.708
The summary above uses the `lmerTest` package to compute P-values from the t-statistic using Satterthwaites's approximation to the denominator degrees of freedom. From this we see that both the `A:B`and `B:C` interaction are significant at the p = 0.05 level. In theory, these results should be consistent, at the very least qualitatively, with those produced from the `anova()` function in the `lmerTest` package, which computes P-values in the same way. However this isn't the case; Here is the output from `anova(model, type = 3)`. **Notice the type 3 test for SS**
Analysis of Variance Table of type III with Satterthwaite
approximation for degrees of freedom
Sum Sq Mean Sq NumDF DenDF F.value Pr(>F)
A 0.09492 0.09492 1 49.87 0.4986 0.48342
B 0.66040 0.66040 1 732.66 3.4688 0.06294 .
C 0.20207 0.20207 1 733.90 1.0614 0.30324
A:B 0.99470 0.99470 1 732.56 5.2247 0.02255 *
A:C 0.36903 0.36903 1 733.66 1.9383 0.16427
B:C 0.35867 0.35867 1 733.20 1.8839 0.17031
A:B:C 0.57552 0.57552 1 734.23 3.0230 0.08251 .
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
These results clearly differ. The `B:C` interaction is no longer significant and the P-value for the `A:B` interaction is quite a bit higher. Both models should be computing the P-values in similar ways and so it's hard to imagine them being so different. In fact, it seems that the `anova(model, type = 3)` function is actually using type 2 SS, which we can verify by running `anova(model, type = 2)`.
Analysis of Variance Table of type II with Satterthwaite
approximation for degrees of freedom
Sum Sq Mean Sq NumDF DenDF F.value Pr(>F)
A 0.09526 0.09526 1 49.87 0.5004 0.48263
B 0.65996 0.65996 1 732.66 3.4665 0.06302 .
C 0.19639 0.19639 1 733.91 1.0315 0.31013
A:B 0.99282 0.99282 1 732.56 5.2148 0.02268 *
A:C 0.37018 0.37018 1 733.65 1.9444 0.16362
B:C 0.35523 0.35523 1 733.20 1.8659 0.17237
A:B:C 0.57552 0.57552 1 734.23 3.0230 0.08251 .
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
The results are very similar. Also, we can use the `Anova()` function from the `car` package to verify this and we find that `Anova(model, type = 2, test.statistic = 'F')` produces:
Analysis of Deviance Table (Type II Wald F tests with Kenward-Roger df)
Response: Biomass
F Df Df.res Pr(>F)
A 0.4857 1 48.28 0.48917
B 3.4537 1 726.63 0.06351 .
C 1.0337 1 727.77 0.30962
A:B 5.1456 1 726.54 0.02360 *
A:C 1.9302 1 727.55 0.16517
B:C 1.8776 1 727.12 0.17103
A:B:C 2.9915 1 728.06 0.08413 .
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Note that the use of Kenward-Roger ddf does not change the results by much for my data. What's clear is that the type 2 SS results from the `Car` packaged produced results analogous to the type 3 SS results from the `lmerTest` package. This suggests that the `lmerTest` package is in fact computing type 2 SS. I struggle trying to figure out why this would be the case unless there is a problem in the computation of P-values from the `lmerTest` package. Am I missing something?
Any suggestions or ideas are welcome. Thanks a bunch!