I'm trying to figure out the model for a fully factorial experiment.
I have the following factors
Treatment Day Hour Subject ResponseVariable
10 days of measurements, 4 different time points within each day, 2 different treatments measured, 12 subjects )6 subjects within treatment 1, and 6 different subjects in treatment 2)
for each day I measured: 6 subjects in treatment 1, the other 6 in treatment 2, at 4 different time points.
For Subjects, I have 12 different subjects, but Subjects 1-6 are in Treatment-1 and Subjects 7-12 are in Treatment-2. The subjects did not change treatments, thus I measured the same set of subjects for each treatment each of the 10 days
So what's tripping me up is specifying the correct error term.
I thought I had the general model down but R is giving me "Error() model is singular"
aov(ResponseVariable ~ T + R + S + TR + TS + RS + Error(T/S)
any thoughts would help?
I've gotten the same error, and I think my problem was missing observations. Are you missing any observations? I believe they're less of a problem for linear mixed effects, and I've read that some people use lme instead of repeated-measures ANOVA for those cases.
Your error term can be interpreted as "the S effect within each T". It sounds from your description as though that's what you want, so I don't think that's what's causing your error message.
One note: I see you've got a variable named "T". R let you do that? T is normally reserved for meaning "TRUE". That might be part of your problem.
Related
I've been working on the following problem recently: We sent 18 people, 9 each, several times to two different clubs "N" and "O". These people arrived at the club either between 8 and 10 am (10) or between 10 and 12 pm (12). Each club consists of four sectors with ascending price classes. At the end of each test run, the subjects filled out a questionnaire reflecting a score for their satisfaction depending on the different parameters. The aim of the study is to find out how satisfaction can be modelled as a function of the club. You can download the data as csv for one week with this link (without spaces): https: // we.tl/t-I0UXKYclUk
After some try and error, I fitted the following model using the lme4 package in R (the other models were singular, had too strong internal correlations or higher AIC/BIC):
mod <- lmer(Score ~ Club + (1|Sector:Subject) + (1|Subject), data = dl)
Now I wanted to create some diagnostic plots as indicated here.
plot(resid(mod), dl$Score)
plot(mod, col=dl$Club)
library(lattice)
qqmath(mod, id=0.05)
Unfortunately, it turns out that there are still patterns in the residuals that can be attributed to the club but are not captured by the model. I have already tried to incorporate the club into the random effects, but this leads to singularities. Does anyone have a suggestion on how I can deal with these patterns in the residuals? Thank you!
I´m having problems when I try to fix heteroscedasticity in mixed models with lme.
My experimental design consists of 12 independent enclosures (Encl) with populations of lizards (Subject_ID). We applied 2 crossed treatments (Lag: 3 levels and Var: 2 levels). And we repeated the experiment two years (Year), so individuals that survived the first year, were measured again the next year. We analyse the snout vent length (SVL) in mm. Sex (males and females). This was my model:
ctrl <- lmeControl(maxIter=200, msMaxIter=200, msVervose=TRUE, opt="optim")
options (contrast=c(factor="contr.sum", ordered="contr.poly"))
model.SVL <- lme(SVL~Lag*Var*Sex*Year, random=list(~1|Subject_ID, ~1|Encl), control=ctrl, data=data)
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The model showed heteroscedasticity in several triple interactions using bartlett.test, so I corrected it with varIdent. However, heteroscedasticity was not fixed, and now, qqplot indicates leptokurtic distribution.
model.SVL2 <- lme(SVL~Lag*Var*Sex*Year, random=list(~1|Subject_ID, ~1|Encl), control=ctrl, weights=varIdent (form=~1|Lag*Var*Sex*Year), data=data)
What could be the problem?
I think the problem is using varIdent when I include Subject_ID as a random factor. If I remove it, this doesn't happen. Maybe it is because many individuals do not survive two years, and it is a random factor with many levels but few replications
I understand I can use lmer but I would like to undertake a repeated measures anova in order to carry out both a within group and a between group analysis.
So I am trying to compare the difference in metabolite levels between three groups ( control, disease 1 and disease 2) over time ( measurements collected at two timepoints), and to also make a within group comparison, comparing time point 1 with time point 2.
Important to note - these are subjects sending in samples not timed trial visits where samples would have been taken on the same day or thereabouts. For instance time point 1 for one subject could be 1995, time point 1 for another subject 1996, the difference between timepoint 1 and timepoint 2 is also not consistent. There is an average of around 5 years, however max is 15, min is .5 years.
I have 43, 45, and 42 subjects respectively in each group. My response variable would be say metabolite 1, the predictor would be Group. I also have covariates I would like to be accounted for such as age, BMI, and gender. I would also need to account for family ID (which I have as a random effect in my lmer model). My column with Time has a 0 to mark the time point 1 and 1 is timepoint 2). I understand I must segregate the within and between subjects command, however, I am unsure how to do this. From my understanding so far;
If I am using the anova_test, my formula that needs to be specified for between subjects would be;
Metabolite1 ~ Group*Time
Whilst for within subjects ( seeing whether there is any difference within each group at TP1 vs TP2), I am unsure how I would specify this ( the below is not correct).
Metabolite1 ~ Time + Error(ID/Time)
The question is, how do I combine this altogether to specify the between and within subject comparisons I would like and accounting for the covariates such as gender, age and BMI? I am assuming if I specify covariates it will become an ANCOVA not an ANOVA?
Some example code that I found that had both a between and within subject comparison design (termed mixed anova).
aov1 <- aov(Recall~(Task*Valence*Gender*Dosage)+Error(Subject/(Task*Valence))+(Gender*Dosage),ex5)
Where he specifies that the within subject comparison is within the Error term. Also explained here https://rpkgs.datanovia.com/rstatix/reference/anova_test.html
However, mine, which I realise is very wrong currently ( is missing a correct within subject comparison).
repmes<-anova_test(data=mets, Metabolite1~ Group*Time + Error(ID/Time), covariate=c("Age", "BMI",
"Gender", "FamilyID")
I ultimately would like to determine from this with appropriate post hoc tests ( if p < 0.05) whether there are any significant differences in Metabolite 1 expression between groups between the two time points (i.e over time), and whether there are any significant differences between subjects comparing TP1 with TP2. Please can anybody help.
I have a set of data that came from a psychological experiment where subjects were randomly assigned to one of four treatment conditions and their wellbeing w measured on six different occasions. The exact day of measurement on each occasion differs slightly from subject to subject. The first measurement occasion for all subjects is day zero.
I analyse this with lmer :
model.a <- lmer(w ~ day * treatment + (day | subject),
REML=FALSE,
data=exper.data)
Following a simple visual inspection of the change-trajectories of subjects, I'd now like to include (and examine the effect of including) the possibility that the slope of the line for each subject changes at a point mid-way between measurement occasion 3 and 4.
I'm familiar with modeling the alteration in slope by including an additional time-variable in the lmer specification. The approach is described in chapter 6 ('Modeling non-linear change') of the book Applied Longitudinal Data Analysis by Singer and Willett (2005). Following their advice, for each measurement, for each subject, there is now an additional variable called latter.day. For measurements up to measurement 3, the value of latter.day is zero; for later measurements, latter.day encodes the number of days after day 40 (which is the point at which I'd like to include the possible slope-change).
What I cannot see is how to adjust the lmer coding of the examples in the Singer and Willett cases to suit my own problem ... which includes the same point-of-slope-change for all subjects as well as a between-subjects factor (treatment). I'd appreciate help on how to write the specification for lmer.
I have a question concerning multi level regression models in R, specifically how to add predictors for my level 2 "measure".
Please consider the following example (this is not a real dataset, so the values might not make much sense in reality):
date id count bmi poll
2012-08-05 1 3 20.5 1500
2012-08-06 1 2 20.5 1400
2012-08-05 2 0 23 1500
2012-08-06 2 3 23 1400
The data contains
different persons ("id"...so it's two persons)
the body mass index of each person ("bmi", so it doesn't vary within an id)
the number of heart problems each person has on a specific day ("count). So person 1 had three problems on August the 5th, whereas person 2 had no difficulties/problems on that day
the amount of pollutants (like Ozon or sulfit dioxide) which have been measured on that given day
My general research question is, if the amount of pollutants effects the numer of heart problems in the population.
In a first step, this could be a simple linear regression:
lm(count ~ poll)
However, my data for each day is so to say clustered within persons. I have two measures from person 1 and two measures from person 2.
So my basic idea was to set up a multilevel model with persons (id) as my level 2 variable.
I used the nlme package for this analysis:
lme(fixed=count ~ poll, random = ~poll|id, ...)
No problems so far.
However, the true influence on level 2 might not only come from the fact that I have different persons. Rather it would be much more likely that the effect WITHIN a person might come from his or her bmi (and many other person related variables, like age, amount of smoking and so on).
To make a longstory short:
How can I specify such level two predictors in the lme function?
Or in other words: How can I setup a model, where the relationship between heart problems and pollution is different/clustered/moderated by the body mass index of a person (and as I said maybe additionally by this person's amount of smoking or age)
Unfortunately, I don't have a clue, how to tell R, what I want. I know oif other software (one of them called HLM), which is capable of doing waht I want, but I'm quite sure that R can this as well...
So, many thanks for any help!
deschen
Short answer: you do not have to, as long as you correctly specify random effects. The lme function automatically detects which variables are level 1 or 2. Consider this example using Oxboys where each subject was measured 9 times. For the time being, let me use lmer in the lme4 package.
library(nlme)
library(dplyr)
library(lme4)
library(lmerTest)
Oxboys %>% #1
filter(as.numeric(Subject)<25) %>% #2
mutate(Group=rep(LETTERS[1:3], each=72)) %>% #3
lmer(height ~ Occasion*Group + (1|Subject), data=.) %>% #4
anova() #5
Here I am picking 24 subjects (#2) and arranging them into 3 groups (#3) to make this data balanced. Now the design of this study is a split-plot design with a repeated-measures factor (Occasion) with q=9 levels and a between-subject factor (Group) with p=3 levels. Each group has n=8 subjects. Occasion is a level-1 variable while Group is level 2.
In #4, I did not specify which variable is level 1 or 2, but lmer gives you correct output. How do I know it is correct? Let us check the multi-level model's degrees of freedom for the fixed effects. If your data is balanced, the Kenward–Roger approximation used in the lmerTest will give you exact dfs and F/t-ratios according to this article. That is, in this example dfs for the test of Group, Occasion, and their interaction should be p-1=2, q-1=8, and (p-1)*(q-1)=16, respectively. The df for the Subject error term is (n-1)p = 21 and the df for the Subject:Occasion error term is p(n-1)(q-1)=168. In fact, these are the "exact" values we get from the anova output (#5).
I do not know what algorithm lme uses for approximating dfs, but lme does give you the same dfs. So I am assuming that it is accurate.