the study design of the data I have to analyse is simple. There is 1 control group (CTRL) and
2 different treatment groups (TREAT_1 and TREAT_2). The data also includes 2 covariates COV1 and COV2. I have been asked to check if there is a linear or quadratic treatment effect in the data.
I created a dummy data set to explain my situation:
df1 <- data.frame(
Observation = c(rep("CTRL",15), rep("TREAT_1",13), rep("TREAT_2", 12)),
COV1 = c(rep("A1", 30), rep("A2", 10)),
COV2 = c(rep("B1", 5), rep("B2", 5), rep("B3", 10), rep("B1", 5), rep("B2", 5), rep("B3", 10)),
Variable = c(3944133, 3632461, 3351754, 3655975, 3487722, 3644783, 3491138, 3328894,
3654507, 3465627, 3511446, 3507249, 3373233, 3432867, 3640888,
3677593, 3585096, 3441775, 3608574, 3669114, 4000812, 3503511, 3423968,
3647391, 3584604, 3548256, 3505411, 3665138,
4049955, 3425512, 3834061, 3639699, 3522208, 3711928, 3576597, 3786781,
3591042, 3995802, 3493091, 3674475)
)
plot(Variable ~ Observation, data = df1)
As you can see from the plot there is a linear relationship between the control and the treatment groups. To check if this linear effect is statistical significant I change the contrasts using the contr.poly() function and fit a linear model like this:
contrasts(df1$Observation) <- contr.poly(levels(df1$Observation))
lm1 <- lm(log(Variable) ~ Observation, data = df1)
summary.lm(lm1)
From the summary we can see that the linear effect is statistically significant:
Observation.L 0.029141 0.012377 2.355 0.024 *
Observation.Q 0.002233 0.012482 0.179 0.859
However, this first model does not include any of the two covariates. Including them results in a non-significant p-value for the linear relationship:
lm2 <- lm(log(Variable) ~ Observation + COV1 + COV2, data = df1)
summary.lm(lm2)
Observation.L 0.04116 0.02624 1.568 0.126
Observation.Q 0.01003 0.01894 0.530 0.600
COV1A2 -0.01203 0.04202 -0.286 0.776
COV2B2 -0.02071 0.02202 -0.941 0.354
COV2B3 -0.02083 0.02066 -1.008 0.320
So far so good. However, I have been told to conduct a Type II Anova rather than Type I. To conduct a Type II Anova I used the Anova() function from the car package.
Anova(lm2, type="II")
Anova Table (Type II tests)
Response: log(Variable)
Sum Sq Df F value Pr(>F)
Observation 0.006253 2 1.4651 0.2453
COV1 0.000175 1 0.0820 0.7763
COV2 0.002768 2 0.6485 0.5292
Residuals 0.072555 34
The problem here with using Type II is that you do not get a p-value for the linear and quadratic effect. So I do not know if the effect is statistically linear and or quadratic.
I found out that the following code produces the same p-value for Observation as the Anova() function. But the result also does not include any p-values for the linear or quadratic effect:
lm2 <- lm(log(Variable) ~ Observation + COV1 + COV2, data = df1)
lm3 <- lm(log(Variable) ~ COV1 + COV2, data = df1)
anova(lm2, lm3)
Does anybody know how to conduct a Type II anova and the contrasts function to obtain the p-values for the linear and quadratic effects?
Help would be very much appreciated.
Best
Peter
I found one partial workaround for this, but it may require further correction. The documentation for the function drop1() from the stats package indicates that this function produces Type II sums of squares (although this page: http://www.statmethods.net/stats/anova.html ) declares that drop1() produces Type III sums of squares, and I didn't spend too much time poring over this (http://afni.nimh.nih.gov/sscc/gangc/SS.html) to cross-check sums of squares calculations. You could use it to calculate everything manually, but I suspect you're asking this question because it would be nice if someone had already worked through it.
Anyway, I added a second vector to the dummy data called Observation2, and set it up with just the linear contrasts (you can only specify one set of contrasts for a given vector at a given time):
df1[,"Observation2"]<-df1$Observation
contrasts(df1$Observation2, how.many=1)<-contr.poly
Then created a third linear model:
lm3<-lm(log(Variable)~Observation2+COV1+COV2, data=df1)
And conducted F tests with drop1 to compare F statistics from Type II ANOVAs between the two models:
lm2, which contains both the linear and quadratic terms:
drop1(lm2, test="F")
lm3, which contains just the linear contrasts:
drop1(lm3, test="F")
This doesn't include a direct comparison of the models against each other, although the F statistic is higher (and p value accordingly lower) for the linear model, which would lead one to rely upon it instead of the quadratic model.
Related
Due to an interesting turn of events, I'm trying use the lme4 package in R to fit a model in which the random slopes are not allowed to correlate with each other or the random intercept. Effectively, I want to estimate the variance parameter for each random slope, but none of the correlations/covariances. From the reading I've done so far, I think what I want is effectively a diagonal variance/covariance structure for the random effects.
An answer to a similar question here provides a workaround to specify a model where slopes are correlated with intercepts, but not with each other. I also know the || syntax in lme4 makes slopes that are correlated with each other, but not with the intercepts. Neither of these seems to fully accomplish what I'm looking to do.
Borrowing the example from the earlier post, if my model is:
m1 <- lmer (Y ~ A + B + (1+A+B|Subject), data=mydata)
is there a way to specify the model such that I estimate variance parameters for A and B while constraining all three correlations to 0? I would like to achieve a result that looks something like this:
VarCorr(m1)
## Groups Name Std.Dev. Corr
## Subject (Intercept) 1.41450
## A 1.49374 0.000
## B 2.47895 0.000 0.000
## Residual 0.96617
I'd prefer a solution that could achieve this for an arbitrary number of random slopes. For example, if I were to add a random effect for a third variable C, there would be 6 correlation parameters to fix at 0 rather than 3. However, anything that could get me started in the right direction would be extremely helpful.
Edit:
On asking this question, I misunderstood what the || syntax does in lme4. Struck through the incorrect statement above to avoid misleading anyone in the future.
This is exactly what the double-bar notation does. However, note that the || in lme4 does not work as one might expect for factor variables. It does work 'properly' in glmmTMB, and the afex::mixed() function is a wrapper for [g]lmer which does implement a fully functional version of ||. (I have meant to import this into lme4 for years but just haven't gotten around to it yet ...)
simulated example
library(lme4)
set.seed(101)
dd <- data.frame(A = runif(500), B = runif(500),
Subject = factor(rep(1:25, 20)))
dd$Y <- simulate(~ A + B + (1 + A + B|Subject),
newdata = dd,
family = gaussian,
newparams = list(beta = rep(1,3), theta = rep(1,6), sigma = 1))[[1]]
solution
summary(m <- lmer (Y ~ A + B + (1+A+B||Subject), data=dd))
The correlations aren't listed because they are structurally absent (internally, the random effects term is expanded to (1|Subject) + (0 + A|Subject) + (0+B|Subject), which is also why the groups are listed as Subject, Subject.1, Subject.2).
Random effects:
Groups Name Variance Std.Dev.
Subject (Intercept) 0.8744 0.9351
Subject.1 A 2.0016 1.4148
Subject.2 B 2.8718 1.6946
Residual 0.9456 0.9724
Number of obs: 500, groups: Subject, 25
How can I use the rms package in R to execute a negative binomial regression? (I originally posted this question on Statistics SE, but it was closed apparently because it is a better fit here.)
With the MASS package, I use the glm.nb function, but I am trying to switch to the rms package because I sometimes get weird errors when bootstrapping with glm.nb and some other functions. But I cannot figure out how to do a negative binomial regression with the rms package.
Here is sample code of what I would like to do (copied from the rms::Glm function documentation):
library(rms)
## Dobson (1990) Page 93: Randomized Controlled Trial :
counts <- c(18,17,15,20,10,20,25,13,12)
outcome <- gl(3,1,9)
treatment <- gl(3,3)
f <- Glm(counts ~ outcome + treatment, family=poisson())
f
anova(f)
summary(f, outcome=c('1','2','3'), treatment=c('1','2','3'))
So, instead of using family=poisson(), I would like to use something like family=negative.binomial(), but I cannot figure out how to do this.
In the documentation for family {stats}, I found this note in the "See also" section:
For binomial coefficients, choose; the binomial and negative binomial distributions, Binomial, and NegBinomial.
But even after clicking the link for ?NegBinomial, I cannot make any sense of this.
I would appreciate any help on how to use the rms package in R to execute a negative binomial regression.
opinion up front You might be better off posting (as a separate question) a reproducible example of the "weird errors" from your bootstrap attempts and seeing whether people have ideas for resolving them. It's fairly common for NB fitting procedures to throw warnings or errors when data are equi- or underdispersed, as the estimates of the dispersion parameter become infinite in this case ...
#coffeinjunky is correct that using family = negative.binomial(theta=VALUE) will work (where VALUE is a numeric constant, e.g. theta=1 for the geometric distribution [a special case of the NB]). However: you won't be able (without significantly more work) be able to fit the general NB model, i.e. the model where the dispersion parameter (theta) is estimated as part of the fitting procedure. That's what MASS::glm.nb does, and AFAICS there is no analogue in the rms package.
There are a few other packages/functions in addition to MASS::glm.nb that fit the negative binomial model, including (at least) bbmle and glmmTMB — there may be others such as gamlss.
## Dobson (1990) Page 93: Randomized Controlled Trial :
dd < data.frame(
counts = c(18,17,15,20,10,20,25,13,12)
outcome = gl(3,1,9),
treatment = gl(3,3))
MASS::glm.nb
library(MASS)
m1 <- glm.nb(counts ~ outcome + treatment, data = dd)
## "iteration limit reached" warning
glmmTMB
library(glmmTMB)
m2 <- glmmTMB(counts ~ outcome + treatment, family = nbinom2, data = dd)
## "false convergence" warning
bbmle
library(bbmle)
m3 <- mle2(counts ~ dnbinom(mu = exp(logmu), size = exp(logtheta)),
parameters = list(logmu ~outcome + treatment),
data = dd,
start = list(logmu = 0, logtheta = 0)
)
signif(cbind(MASS=coef(m1), glmmTMB=fixef(m2)$cond, bbmle=coef(m3)[1:5]), 5)
MASS glmmTMB bbmle
(Intercept) 3.0445e+00 3.04540000 3.0445e+00
outcome2 -4.5426e-01 -0.45397000 -4.5417e-01
outcome3 -2.9299e-01 -0.29253000 -2.9293e-01
treatment2 -1.1114e-06 0.00032174 8.1631e-06
treatment3 -1.9209e-06 0.00032823 6.5817e-06
These all agree fairly well (at least for the intercept/outcome parameters). This example is fairly difficult for a NB model (5 parameters + dispersion for 9 observations, data are Poisson rather than NB).
Based on this, the following seems to work:
library(rms)
library(MASS)
counts <- c(18,17,15,20,10,20,25,13,12)
outcome <- gl(3,1,9)
treatment <- gl(3,3)
Glm(counts ~ outcome + treatment, family = negative.binomial(theta = 1))
General Linear Model
rms::Glm(formula = counts ~ outcome + treatment, family = negative.binomial(theta = 1))
Model Likelihood
Ratio Test
Obs 9 LR chi2 0.31
Residual d.f.4 d.f. 4
g 0.2383063 Pr(> chi2) 0.9892
Coef S.E. Wald Z Pr(>|Z|)
Intercept 3.0756 0.2121 14.50 <0.0001
outcome=2 -0.4598 0.2333 -1.97 0.0487
outcome=3 -0.2962 0.2327 -1.27 0.2030
treatment=2 -0.0347 0.2333 -0.15 0.8819
treatment=3 -0.0503 0.2333 -0.22 0.8293
I am analysing routinely collected substance use data during the first 12 months' of treatment in a large sample of outpatients attending drug and alcohol treatment services. I am interested in whether differing levels of methamphetamine use (no use, low use, and high use) at the outset of treatment predicts different levels after a year in treatment, but the data is very irregular, with different clients measured at different times and different numbers of times during their year of treatment.
The data for the high and low use group seem to suggest that drug use at outset reduces during the first 3 months of treatment and then asymptotes. Hence I thought I would try a non-linear exponential decay model.
I started with the following nonlinear generalised least squares model using the gnls() function in the nlme package:
fitExp <- gnls(outcome ~ C*exp(-k*yearsFromStart),
params = list(C ~ atsBase_fac, k ~ atsBase_fac),
data = dfNL,
start = list(C = c(nsC[1], lsC[1], hsC[1]),
k = c(nsC[2], lsC[2], hsC[2])),
weights = varExp(-0.8, form = ~ yearsFromStart),
control = gnlsControl(nlsTol = 0.1))
where outcome is number of days of drug use in the 28 days previous to measurement, atsBase_fac is a three-level categorical predictor indicating level of amphetamine use at baseline (noUse, lowUse, and highUse), yearsFromStart is a continuous predictor indicating time from start of treatment in years (baseline = 0, max - 1), C is a parameter indicating initial level of drug use, and k is the rate of decay in drug use. The starting values of C and k are taken from nls models estimating these parameters for each group. These are the results of that model
Generalized nonlinear least squares fit
Model: outcome ~ C * exp(-k * yearsFromStart)
Data: dfNL
AIC BIC logLik
27672.17 27725.29 -13828.08
Variance function:
Structure: Exponential of variance covariate
Formula: ~yearsFromStart
Parameter estimates:
expon
0.7927517
Coefficients:
Value Std.Error t-value p-value
C.(Intercept) 0.130410 0.0411728 3.16738 0.0015
C.atsBase_faclow 3.409828 0.1249553 27.28839 0.0000
C.atsBase_fachigh 20.574833 0.3122500 65.89218 0.0000
k.(Intercept) -1.667870 0.5841222 -2.85534 0.0043
k.atsBase_faclow 2.481850 0.6110666 4.06150 0.0000
k.atsBase_fachigh 9.485155 0.7175471 13.21886 0.0000
So it looks as if there are differences between groups in initial rate of drug use and in rate of reduction in drug use. I would like to go a step further and fit a nonlinear mixed effects model.I tried consulting Pinhiero and Bates' book accompanying the nlme package but the only models I could find that used irregular, sparse data like mine used a self-starting function, and my model does not do that.
I tried to adapt the gnls() model to nlme like so:
fitNLME <- nlme(model = outcome ~ C*exp(-k*yearsFromStart),
data = dfNL,
fixed = list(C ~ atsBase_fac, k ~ atsBase_fac),
random = pdDiag(yearsFromStart ~ id),
groups = ~ id,
start = list(fixed = c(nsC[1], lsC[1], hsC[1], nsC[2], lsC[2], hsC[2])),
weights = varExp(-0.8, form = ~ yearsFromStart),
control = nlmeControl(optim = "optimizer"))
bit I keep getting error message, I presume through errors in the syntax specifying the random effects.
Can anyone give me some tips on how the syntax for the random effects works in nlme?
The only dataset in Pinhiero and Bates that resembled mine used a diagonal variance-covariance matrix. Can anyone filled me in on the syntax of this nlme function, or suggest a better one?
p.s. I wish I could provide a reproducible example but coming up with synthetic data that re-creates the same errors is way beyond my skills.
I try to analyze some simulated longitudinal data in R using a mixed-effects model (lme4 package).
Simulated data: 25 subjects have to perform 2 tasks at 5 consecutive time points.
#Simulate longitudinal data
N <- 25
t <- 5
x <- rep(1:t,N)
#task1
beta1 <- 4
e1 <- rnorm(N*t, mean = 0, sd = 1.5)
y1 <- 1 + x * beta1 + e1
#task2
beta2 <- 1.5
e2 <- rnorm(N*t, mean = 0, sd = 1)
y2 <- 1 + x * beta2 + e2
data1 <- data.frame(id=factor(rep(1:N, each=t)), day = x, y = y1, task=rep(c("task1"),length(y1)))
data2 <- data.frame(id=factor(rep(1:N, each=t)), day = x, y = y2, task=rep(c("task2"),length(y2)))
data <- rbind(data1, data2)
Question1: How to analyze how a subject learns each task?
library(lme4)
m1 <- lmer(y ~ day + (1 | id), data=data1)
summary(m1)
...
Fixed effects:
Estimate Std. Error df t value Pr(>|t|)
(Intercept) 1.2757 0.3561 123.0000 3.582 0.000489 ***
day 3.9299 0.1074 123.0000 36.603 < 2e-16 ***
With ranef(m1) I get the random intercept for each subject, which I think reflects the baseline value for each subject at day = 1. But I don't understand how I can tell how an individual learns a task, or whether subjects differ in the way how they learn the task.
Question2: How can I analyze whether the way subjects learn differ between task1 and task2.
I expanded on your example to answer your questions briefly, but I can recommend reading chapter 15 of Snijders & Bosker (2012) or the book by Singer & Willet (2003) for a better explanation. Day is treated as a continuous variable in your model, seeing as you have panel data (i.e. everyone is measured at the same day) and day has no meaning apart from indicating the different measurement occasions, it may be better to treat day as a factor (i.e. use dummy variables).
However, for now I will continue with your example
Your first model (I think you want data instread of data1) gives a fixed linear slope (i.e. average slope, no difference in the tasks, no difference between individuals). The fixed intercept is the performance when day is 0, which has no meaning so you may want to consider centering the effect of day for a better interpretation (or indeed use dummies). The random effect gives the individual deviance from this intercept which has an estimated variance of 0.00 in your example so individuals hardly differ from each other in their starting position.
m1 <- lmer(y ~ day + (1 | id), data=data)
summary(m1)
Random effects:
Groups Name Variance Std.Dev.
id (Intercept) 0.00 0.000
Residual 18.54 4.306
Number of obs: 250, groups: id, 25
We can extend this model by adding an interaction with task. Meaning that the fixed slope is different for task1 and task2 which answers question 2 I believe (you can also use update() to update your model)
m2 <- lmer(y ~ day*task + (1|id), data = data)
summary(m2)
The effect of day in this model is the fixed slope of your reference category (task1) and the interaction is the difference between the slope of task1 and task2. The fixed effect of task is the difference in intercept.
model fit can be assessed with a deviance test, read Snijders & Boskers (2012) for an explanation of ML and REML estimates.
anova(m1,m2)
To add a random effect for the growth of individuals we can update the model again, which answers question 1
m3 <- lmer(y ~ day*task + (day|id), data = data)
summary(m3)
ranef(m3)
The random effects indicate the individual deviations in slope and intercept. A summary of the distribution of you random effects is included in the model summary (same as for m1).
Finally I think you could add a random effect on the day-task interaction to assess whether individuals differ in their performance growth on task1 and task2. But this depends very much on your data and the performance of the previous models.
m4 <- lmer(y ~ day*task + (day*task|id), data = data)
summary(m4)
ranef(m4)
Hope this helps. The books I recommended certainly should. Both provide excellent examples and explanation of theory (no R examples unfortunately). If you decide on a fixed occasion model (effect of day expressed by dummies) the nlme package provides excellent options to control the covariance structure of random effects. Good documentation of the package is provided by Pinheiro & Bates (2000).
I'm running a piecewise linear random coefficient model testing the influence of a covariate on the second piece. Thereby, I want to test whether the coefficient of the second piece under the influence of the covariate (piece2 + piece2:covariate) differs from the coefficient of the first piece (piece1), hence whether the growth rate differs.
I set up some exemplary data:
set.seed(100)
# set up dependent variable
temp <- rep(seq(0,23),50)
y <- c(rep(seq(0,23),50)+rnorm(24*50), ifelse(temp <= 11, temp + runif(1200), temp + rnorm(1200) + (temp/sqrt(temp))))
# set up ID variable, variables indicating pieces and the covariate
id <- sort(rep(seq(1,100),24))
piece1 <- rep(c(seq(0,11), rep(11,12)),100)
piece2 <- rep(c(rep(0,12), seq(1,12)),100)
covariate <- c(rep(0,24*50), rep(c(rep(0,12), rep(1,12)), 50))
# data frame
example.data <- data.frame(id, y, piece1, piece2, covariate)
# run piecewise linear random effects model and show results
library(lme4)
lmer.results <- lmer(y ~ piece1 + piece2*covariate + (1|id) , example.data)
summary(lmer.results)
I came across the linearHypothesis() command from the car package to test differences in coefficients. However, I could not find an example on how to use it when including interactions.
Can I even use linearHypothesis() to test this or am I aiming for the wrong test?
I appreciate your help.
Many thanks in advance!
Mac
Assuming your output looks like this
Estimate Std. Error t value
(Intercept) 0.26293 0.04997 5.3
piece1 0.99582 0.00677 147.2
piece2 0.98083 0.00716 137.0
covariate 2.98265 0.09042 33.0
piece2:covariate 0.15287 0.01286 11.9
if I understand correctly what you want, you are looking for the contrast:
piece1-(piece2+piece2:covariate)
or
c(0,1,-1,0,-1)
My preferred tool for this is function estimable in gmodels; you could also do it by hand or with one of the functions in Frank Harrel's packages.
library(gmodels)
estimable(lmer.results,c(0,1,-1,0,-1),conf.int=TRUE)
giving
Estimate Std. Error p value Lower.CI Upper.CI
(0 1 -1 0 -1) -0.138 0.0127 0 -0.182 -0.0928