test for significance of interaction in linear mixed models in nlme in R

Question

I use lme function in the nlme R package to test if levels of factor items has significant interaction with levels of factor conditi

Answer 1

You'll need to address your second question about the interaction first. You can certainly set up the likelihood ratio test as in Jan van der Laan's answer. You can also use anova directly on a fitted lme object. See the help page for anova.lme for more information.

In terms of interpreting your coefficients, I often find that it sometimes takes making a summary table of the group means in order to appropriately figure out which linear combination of coefficients in the model represents each group. I will show an example with the intercept removal like in your question, although I find this rarely helps me figure out my coefficients once I have two factors in a model. Here is an example of what I mean with the Orthodont dataset (which I decided to make balanced):

require(nlme)

# Make dataset balanced
Orthodont2 = Orthodont[-c(45:64),]
# Factor age
Orthodont2$fage = factor(Orthodont2$age)

# Create a model with an interaction using lme; remove the intercept
fit1 = lme(distance ~ Sex*fage - 1, random = ~1|Subject, data = Orthodont2)
summary(fit1)

Here are the estimated fixed effects. But what do each of these coefficients represent?

Fixed effects: distance ~ Sex * fage - 1 
                     Value Std.Error DF  t-value p-value
SexMale          23.636364 0.7108225 20 33.25213  0.0000
SexFemale        21.181818 0.7108225 20 29.79903  0.0000
fage10            0.136364 0.5283622 61  0.25809  0.7972
fage12            2.409091 0.5283622 61  4.55954  0.0000
fage14            3.727273 0.5283622 61  7.05439  0.0000
SexFemale:fage10  0.909091 0.7472171 61  1.21664  0.2284
SexFemale:fage12 -0.500000 0.7472171 61 -0.66915  0.5059
SexFemale:fage14 -0.818182 0.7472171 61 -1.09497  0.2778

Making a summary of group means helps figure this out.

require(plyr)
ddply(Orthodont2, .(Sex, age), summarise, dist = mean(distance) )

     Sex fage     dist
1   Male    8 23.63636
2   Male   10 23.77273
3   Male   12 26.04545
4   Male   14 27.36364
5 Female    8 21.18182
6 Female   10 22.22727
7 Female   12 23.09091
8 Female   14 24.09091

Notice that the first fixed effect coefficient, called SexMale, is the mean distance for the age 8 males. The fixed effect SexFemale is the age 8 female mean distance. Those are the easiest to see (I always start with the easy ones), but the rest aren't too bad to figure out. The mean distance for age 10 males is the first coefficient plus the third coefficient (fage10). The mean distance for age 10 females is the sum of the coefficients SexFemale, fage10, and SexFemale:fage10. The rest follow along the same lines.

Once you know how to create linear combinations of coefficients for the group means, you can use estimable to calculate any comparisons of interest. Of course there are a bunch of caveats here about main effects, statistical evidence of an interactions, theoretical reasons to leave interactions in, etc. That is for you to decide. But if I was leaving the interaction in the model (note there is no statistical evidence of an interaction, see anova(fit1)) and yet wanted to compare the overall mean of Male to Female, I would write out the following linear combinations of coefficients:

# male/age group means
male8 = c(1, 0, 0, 0, 0, 0, 0, 0)
male10 = c(1, 0, 1, 0, 0, 0, 0, 0)
male12 = c(1, 0, 0, 1, 0, 0, 0, 0)
male14 = c(1, 0, 0, 0, 1, 0, 0, 0)

# female/age group means
female8 = c(0, 1, 0, 0, 0, 0, 0, 0)
female10 = c(0, 1, 1, 0, 0, 1, 0, 0)
female12 = c(0, 1, 0, 1, 0, 0, 1, 0)
female14 = c(0, 1, 0, 0, 1, 0, 0, 1)

# overall male group mean
male = (male8 + male10 + male12 +male14)/4
# overall female group mean
female = (female8 + female10 + female12 + female14)/4

require(gmodels)
estimable(fit1, rbind(male - female))

You can check your overall group means to make sure you made your linear combinations of coefficients correctly.

ddply(Orthodont2, .(Sex), summarise, dist = mean(distance) )

     Sex     dist
1   Male 25.20455
2 Female 22.64773

Answer 2

I respectfully disagree with @sven-hohenstein

In R, the default coding for categorial variables is treatment contrast coding. In treatment contrasts, the first level is the reference level. All remaining factor levels are compared with the reference level.

First, the fixed effects are specified here with a zero intercept, ... ~ 0 + .... This means that the condition coding is no longer contr.treatment. If I'm not mistaken, the main effects of Control and Treatment are now interpretable as their respective deviations from the group mean...

In your model, the factor items has three levels: E1, E2, and E3. The two contrasts test the difference between (a) E2 and E1, and (b) E3 and E1. The main effects of these contrasts are estimated for the level Control of the factor condition, since this is the reference category of this factor.

...when the value of items is at its reference level of E1! Therefore:

• Main effect Control = how much Control:E1 observations deviate from the mean of item E1.
• Main effect Treatment = how much Treatment:E1 observations deviate from the mean of item E1.
• Main effect E2 = how much Control:E2 observations deviate from the mean of item E2.
• Main effect E3 = how much Control observations deviate from the mean of item E3.
• Interaction Treatment:E2 = how much Treatment:E2 observations deviate from the mean of item E2
• Interaction Treatment:E3 = how much Treatment:E3 observations deviate from the mean of item E3.

Thanks for the pointer to estimable, I haven't tried it before. For custom contrasts, I've been (ab)using glht from the multcomp package.

Answer 3

The usual way to test if the interaction is significant is to do a likelihood ratio test (e.g. see discussion on R-Sig-ME).

To do that you have to also estimate a model without interaction and you'll also have to use method="ML":

f0 = lme(response ~ 0 + factor(condition) * factor(items),
    random = ~1|subject, method="ML")
f1 = lme(response ~ 0 + factor(condition) + factor(items),
    random = ~1|subject, method="ML")

You can then compare using anova:

anova(f0, f1)

Also see this blog post