I have a question about using linear mixed model effects in R using lmer.
I have a repeated measure experiment with 117 participants. They all perform a task with 5 categories (Prime_Names). The dependent variable is reaction times (Score). I want to compare those 5 categories with each other. There is a lot of missing data so I think a RM anova is not an option.
I have two questions:
Am I using the correct analysis if I do a linear mixed model effect analysis in R with lmer?
I am not sure if my model is completely correct, especially for the random effects. When do you use only "+ (1|Resp_ID)" and when do you use "+ (Prime_Name|Resp_ID)"
Two options:
Option 1:
model <- lmer(Score ~ Prime_Name + (1|Resp_ID), data=df)
Option 2:
model <- lmer(Score ~ Prime_Name + (Prime_Name|Resp_ID), data=df)
Any help will be appreciated.
Thank you
Related
I'm using R to run a logistic multilevel model with random intercepts. I'm using the frequentist approach (glmer). I'm not able to use Bayesian methods due to the research centre's policy.
When I run my code it says that my model is singular. I'm not sure why or how to fix the issue. Any advice would be appreciated!
More information about the multilevel model I used:
I'm using a multilevel modelling method used in intersectionality research called multilevel analysis of individual heterogeneity and discriminatory accuracy (MAIHDA). The method uses individual level data as level 2 (the intersection group) and nests individuals within their intersections.
My outcome is binary and I have three categorical variables as fixed effects (gender, martial status, and disability). The random effect (level 2) is called intersect1 which includes each unique combination of the categorical variables (gender x marital x disability).
This is the code:
MAIHDA_full <- glmer(IPV_pos ~ factor(sexgender) + factor(marital) + factor(disability) + (1|intersect1), data=Data, family=binomial, control=glmerControl(optimizer=”bobyqa”,optCtrl=list(maxfun=2e5)))
The usual reason for a singular fit with mixed effects models is that either the random structure is overfitted - typically because of the inclusion of random slopes, or in the case such as this where we only have random intercepts, then the variation in the intercepts is so small that the model cannot detect it.
Looking at your model formula I suspect the issue is:
The random effect (level 2) is called intersect1 which includes each unique combination of the categorical variables (gender x marital x disability).
If I have understood this correctly, the model is equivalent to:
IPV_pos ~ sexgender + marital + disability + (1 | sexgender:marital:disability)
It is likely that any variation in sexgender:marital:disability is captured by the fixed effects, leading to near-zero variation in the random intercepts.
I suspect you will find almost identical results if you don't use any random effect.
I am trying to predict depression by using two quantitative variables and their interaction. However, before I want to see how much variance they explain, I want to control for a few variables.
My plan was to build a logistic regression model:
Depression = Covariates + IV1 + IV2 + IV1:IV2
Unfortunately, R doesn't seem to care about the order in which you add the variables to the model (Type III sum of squares?). Is there a way to build a logistic regression model in which the order does matter?
Thanks in advance!
-Lukas
I am using the lmer() function (lme4 package) in R for analysing a longitudinal study in which I measured 120 subjects, 6 times. In first instance, I specified a model like this:
library(lme4)
model1 = lmer(DV ~ 1 + X1*X2 + (1+X1|SubjectID), REML="false")
X1 is a time-varying variable (level-1) and X2 is a subject-level variable (level-2).
Because these subjects are nested within several teams, I was advised to include a random intercept at the team-level (level-3). However, I only find how to include both random intercept and slope:
model2 = lmer(DV ~ 1 + X1*X2 + (1+X1|TeamID/SubjectID), REML="false")
Does anyone know how to add only a level-3 random intercept to model 1?
By using the term (1|SubjectID) you're telling the model to expect differing baselines only for different instances of SubjectID. To tell the model to expect different responses of this to the fixed effect X1, we use (1+X1|SubjectID). Therefore, you just need the terms
(1|TeamID) + (1+X1|SubjectID)
in your model.
By the way there's plenty of good information about this on Cross Validated.
I am currently working through Andy Field's book, Discovering Statistics Using R. Chapter 14 is on Mixed Modelling and he uses the lme function from the nlme package.
The model he creates, using speed dating data, is such:
speedDateModel <- lme(dateRating ~ looks + personality +
gender + looks:gender + personality:gender +
looks:personality,
random = ~1|participant/looks/personality)
I tried to recreate a similar model using the lmer function from the lme4 package; however, my results are different. I thought I had the proper syntax, but maybe not?
speedDateModel.2 <- lmer(dateRating ~ looks + personality + gender +
looks:gender + personality:gender +
(1|participant) + (1|looks) + (1|personality),
data = speedData, REML = FALSE)
Also, when I run the coefficients of these models I notice that it only produces random intercepts for each participant. I was trying to then create a model that produces both random intercepts and slopes. I can't seem to get the syntax correct for either function to do this. Any help would be greatly appreciated.
The only difference between the lme and the corresponding lmer formula should be that the random and fixed components are aggregated into a single formula:
dateRating ~ looks + personality +
gender + looks:gender + personality:gender +
looks:personality+ (1|participant/looks/personality)
using (1|participant) + (1|looks) + (1|personality) is only equivalent if looks and personality have unique values at each nested level.
It's not clear what continuous variable you want to define your slopes: if you have a continuous variable x and groups g, then (x|g) or equivalently (1+x|g) will give you a random-slopes model (x should also be included in the fixed-effects part of the model, i.e. the full formula should be y~x+(x|g) ...)
update: I got the data, or rather a script file that allows one to reconstruct the data, from here. Field makes a common mistake in his book, which I have made several times in the past: since there is only a single observation in the data set for each participant/looks/personality combination, the three-way interaction has one level per observation. In a linear mixed model, this means the variance at the lowest level of nesting will be confounded with the residual variance.
You can see this in two ways:
lme appears to fit the model just fine, but if you try to calculate confidence intervals via intervals(), you get
intervals(speedDateModel)
## Error in intervals.lme(speedDateModel) :
## cannot get confidence intervals on var-cov components:
## Non-positive definite approximate variance-covariance
If you try this with lmer you get:
## Error: number of levels of each grouping factor
## must be < number of observations
In both cases, this is a clue that something's wrong. (You can overcome this in lmer if you really want to: see ?lmerControl.)
If we leave out the lowest grouping level, everything works fine:
sd2 <- lmer(dateRating ~ looks + personality +
gender + looks:gender + personality:gender +
looks:personality+
(1|participant/looks),
data=speedData)
Compare lmer and lme fixed effects:
all.equal(fixef(sd2),fixef(speedDateModel)) ## TRUE
The starling example here gives another example and further explanation of this issue.
I would like to run repeated measure anova in R using regression models instead an 'Analysis of Variance' (AOV) function.
Here is an example of my AOV code for 3 within-subject factors:
m.aov<-aov(measure~(task*region*actiontype) + Error(subject/(task*region*actiontype)),data)
Can someone give me the exact syntax to run the same analysis using regression models? I want to make sure to respect the independence of residuals, i.e. use specific error terms as with AOV.
In a previous post I read an answer of the type:
lmer(DV ~ 1 + IV1*IV2*IV3 + (IV1*IV2*IV3|Subject), dataset))
I am really not sure about this solution since it still treats variables as between subjects, and I don't understand how adding random factors would change this.
Does someone know how to run repeated measure anova with lm/lmer taking into account residual independence?
Many thanks,
Solene
I have some worked examples with more detail here: https://keithlohse.github.io/mixed_effects_models/lohse_MER_chapter_02.html
But if you want to get a mixed model that is homologous to your ANOVA, you can include random intercepts for your each subject:factor with your within-subject factors. E.g.,
aov(DV~W1*W2*W3 + Error(SUBJECT/(W1*W2*W3)),data)
has a mixed-model equivalent of:
lmer(speed ~
# Fixed Effects
W1*W2*W3 +
# Random Effects
(1|SUBJECT) + (1|W1:SUBJECT) + (1|W2:SUBJECT) + (1|W3:SUBJECT),
data = DATA,
REML = TRUE)
With REML set to TRUE and a balanced design, you should get degrees of freedom and f-values that are identical to your ANOVA. ML tends to underestimate variance components, so if you are comparing nested models and need to use ML your results will not match precisely. If you are not comparing nested models and can use REML, then the ANOVA and mixed-model should match (again, in a balanced design).
To #skan's earlier answer and other ideas people might have, I am not saying this is THE random-effects structure (as it might be more appropriate to include random slopes for W1 compared to random-intercepts), but if you have one observation per subject:condition, then these random-effects produce an equivalent result.
If your aov example is right (maybe you don't want to nest things) you want this:
lmer(measure~(task*region*actiontype) + 1(1|subject/(task:region:actiontype))
If residual independence means intercept and slope independently calculated you need to specify them separately:
+(1|yourfactors)+(0+variable|yourfactors)
or use the symbol:
+(1||yourfactors)
Anyway if you read the help files you can find that lme4 can't deal with the most general problems.