I am trying to replicate Stata's marginal effects from multinomial logit models in R but with no success. For the multinomial logit model, I used the multinom() function from the nnet package and for the marginal effects I used the margins package but the marginal_effects function seems to only display effects of a single variable. What if I want to have the marginal effects of the variable conditioned on another variable? Here is the output from Stata:
. margins, dydx(male) at(site=(1 2 3)) #male conditioned on site
Average marginal effects Number of obs = 615
Model VCE : OIM
dy/dx w.r.t. : 1.male
1._predict : Pr(insure==Indemnity), predict(pr outcome(1))
2._predict : Pr(insure==Prepaid), predict(pr outcome(2))
3._predict : Pr(insure==Uninsure), predict(pr outcome(3))
1._at : site = 1
2._at : site = 2
3._at : site = 3
------------------------------------------------------------------------------
| Delta-method
| dy/dx Std. Err. z P>|z| [95% Conf. Interval]
-------------+----------------------------------------------------------------
1.male |
_predict#_at |
1 1 | -.1492951 .0728108 -2.05 0.040 -.2920016 -.0065885
1 2 | -.159346 .0723512 -2.20 0.028 -.3011517 -.0175403
1 3 | -.055138 .0875712 -0.63 0.529 -.2267745 .1164984
2 1 | .0763095 .0765406 1.00 0.319 -.0737074 .2263264
2 2 | .1747759 .0730055 2.39 0.017 .0316877 .3178641
2 3 | .0861997 .0843816 1.02 0.307 -.0791852 .2515846
3 1 | .0729855 .0516839 1.41 0.158 -.0283131 .1742842
3 2 | -.0154299 .0104982 -1.47 0.142 -.036006 .0051462
3 3 | -.0310617 .0495625 -0.63 0.531 -.1282025 .0660791
------------------------------------------------------------------------------
Note: dy/dx for factor levels is the discrete change from the base level.
My attempt to calculate the marginal effects of male using the marginal_effects function:
library(nnet)
sysdsn1$insure <- as.factor(sysdsn1$insure)
sysdsn1$male <- as.factor(sysdsn1$male)
sysdsn1$site <- as.factor(sysdsn1$site)
sysdsn1$nonwhite <- as.factor(sysdsn1$nonwhite)
sysdsn1$insure <- relevel(sysdsn1$insure, ref = "3") #set the reference level
mn0 <- multinom(insure ~ age + male*site + nonwhite, data = sysdsn1) #multinomial logit model
head(marginal_effects(mn0, variables = "male")) #this only calculate marginal effects of male, how to condition on site?
dydx_male1
1 -0.01310874
2 -0.01744213
3 0.07911846
4 -0.03386199
5 -0.01728126
6 -0.01638176
Data
Data can be downloaded from http://www.stata-press.com/data/r13/sysdsn1.dta and imported into R
Related
I have a data frame with post and follow-up measurements for approximately 200 people. In the study, we try to find out if there is a correlation between sports participation and distress symptoms. We have two measurement periods (post and follow-up) that are conducted after a workshop about health and sports. Post was conducted 6 months after the Workshop and followup one year after the workshop. We formed the following hypothesis: „Participation in sport for obese people within one year after a workshop correlates significantly positively with psychological distress symptoms at follow up.“ I assume, the dependent variable is psychological distress and the independent is the participation in sports activities. The data structure looks like:
Df
$ measurement_period : Factor w/ 2 levels "0","1": 1 1 1 1
$ psychological_distress ; int 12 45 32 85
$ participation : Factor w/ 2 levels "0","1": 1 1 1 1
$ id : num 1 2 3 4
After reading some posts here, we believe that there are 2 levels in the model: 1 ) measurement period (post and follow up) 2) id
At first we conductet a unconditiional Model (intercept only Model for confirming if a multilevel Model fits, hope that this is right) with following code:
test <-lmer(psychological_distress ~1+(1|id),data=Df
But we are not sure if the model is appropriate given the data structure and, whether the level 1 and level 2 classification is correct.
Thank you very much in advance!
Your model:
lmer(psychological_distress ~ 1 + (1|id) , data = Df)
is a variance components model. It will tell you how much of the variation in psychological_distress is attributable to the id level, and how much is attributable to the unit/residual level. That isn't going to answer your research question:
we try to find out if there is a correlation between sports participation and distress symptoms
To look into this, you need to include the participation variable as a fixed effect, and also the time variable, and their interaction. So in the first instance I would consider this:
lmer(psychological_distress ~ measurement_period*participation + (1|id) , data = Df)
A good website on how to fit longitudinal and growth models using lme4 is https://rpsychologist.com/r-guide-longitudinal-lme-lmer
As Robert pointed out, and as demonstrated on the website, it is often useful to fit an interaction between "time" and "group" (e.g., treatment vs. control), to see how the outcome changes for each group over time. You can see this change by looking at the coefficients, but it's usually easier to plot (adjusted) predictions.
Here's a toy example:
library(parameters)
library(datawizard)
library(lme4)
library(ggeffects)
data("qol_cancer")
# filter two time points
qol_cancer <- data_filter(qol_cancer, time %in% c(1, 2))
# create fake treatment/control variable
set.seed(123)
treatment <- sample(unique(qol_cancer$ID), size = length(unique(qol_cancer$ID)) / 2, replace = FALSE)
qol_cancer$treatment <- 0
qol_cancer$treatment[qol_cancer$ID %in% treatment] <- 1
qol_cancer$time <- as.factor(qol_cancer$time)
qol_cancer$treatment <- factor(qol_cancer$treatment, labels = c("control", "treatment"))
m <- lmer(QoL ~ time * treatment + (1 + time | ID),
data = qol_cancer,
control = lmerControl(check.nobs.vs.nRE = "ignore"))
model_parameters(m)
#> # Fixed Effects
#>
#> Parameter | Coefficient | SE | 95% CI | t(368) | p
#> ----------------------------------------------------------------------------------------
#> (Intercept) | 70.74 | 2.15 | [66.52, 74.97] | 32.90 | < .001
#> time [2] | 0.27 | 2.22 | [-4.10, 4.64] | 0.12 | 0.905
#> treatment [treatment] | 4.88 | 3.04 | [-1.10, 10.86] | 1.60 | 0.110
#> time [2] * treatment [treatment] | 1.95 | 3.14 | [-4.23, 8.13] | 0.62 | 0.535
#>
#> # Random Effects
#>
#> Parameter | Coefficient
#> ---------------------------------------
#> SD (Intercept: ID) | 15.14
#> SD (time2: ID) | 7.33
#> Cor (Intercept~time2: ID) | -0.62
#> SD (Residual) | 14.33
#>
#> Uncertainty intervals (equal-tailed) and p-values (two-tailed) computed
#> using a Wald t-distribution approximation.
ggpredict(m, c("time", "treatment")) |> plot()
Regarding the statistical significance of the interaction term: the p-values from the summary might be misleading. If you're really interested in statistically significant differences either between time points, or between groups (treatment vs. control), it is recommended to calculate pairwise contrasts including p-values. You can do this, e.g., with the emmeans-package.
library(emmeans)
emmeans(m, c("time", "treatment")) |> contrast(method = "pairwise", adjust = "none")
#> contrast estimate SE df t.ratio p.value
#> time1 control - time2 control -0.266 2.22 186 -0.120 0.9049
#> time1 control - time1 treatment -4.876 3.04 186 -1.604 0.1105
#> time1 control - time2 treatment -7.092 2.89 316 -2.453 0.0147
#> time2 control - time1 treatment -4.610 2.89 316 -1.594 0.1118
#> time2 control - time2 treatment -6.826 2.73 186 -2.497 0.0134
#> time1 treatment - time2 treatment -2.216 2.22 186 -0.997 0.3199
#>
#> Degrees-of-freedom method: kenward-roger
Created on 2022-05-22 by the reprex package (v2.0.1)
Here you can see, e.g., that treatment and control do not differ regarding their QoL at time point 1, but they do at time point 2.
I ran an ordinal logistic regression (using the function clmm from the R package ordinal) with a two-factor interaction and a random effect.
The response is a factor w/ 5 levels (Liker scale: 1 2 3 4 5), the independent variables are a factor w/ 2 levels (time) and a factor w/ 3 levels (group)
The code looks like this:
library(ordinal)
# dataset
ID time group response
person1 1 a 3
person2 1 a 5
person3 1 c 5
person4 1 b 2
person5 1 c 2
person6 1 a 4
person1 2 a 2
person2 2 a 2
person3 2 c 1
person4 2 b 4
person5 2 c 3
person6 2 a 4
... ... ... ...
# model
model <- clmm(response ~ time*group + (1|ID))
# model results
formula: response ~ time * group + (1 | ID)
data: dataset
link threshold nobs logLik AIC niter max.grad cond.H
logit flexible 168 -226.76 473.52 508(4150) 9.42e-05 1.8e+02
Random effects:
Groups Name Variance Std.Dev.
ID (Intercept) 5.18 2.276
Number of groups: ID 84
Coefficients:
Estimate Std. Error z value Pr(>|z|)
time2 0.2837 0.5289 0.536 0.59170
group_b 1.8746 0.6946 2.699 0.00695 **
group_c 4.0023 0.9383 4.265 2e-05 ***
time2:group_b -0.5100 0.7294 -0.699 0.48447
time2:group_c -0.8830 0.9749 -0.906 0.36508
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Threshold coefficients:
Estimate Std. Error z value
1|2 -2.6223 0.6440 -4.072
2|3 0.2474 0.5427 0.456
3|4 2.5384 0.5824 4.359
4|5 4.6786 0.7143 6.550
As you can see, the model results show only whether there are differences compared to the intercept (i.e. time1:group_a). However, what I am also interested in is to check if the difference between time1:group_b and time2:group_b is statistically significant, same for group_c.
Since I have to account for the random effect, I cannot use a simple chi-square test to check for statistically significant differences between groups. I therefore tried to run the function contrast from the R package emmeans, which uses the output of the function emmeans, see the code below:
library(emmeans)
em <- emmeans(model, ~ time | group) #calculates the estimated marginal means
contrast(em, "consec", simple = "each")
# contrast results
$`simple contrasts for time`
group = a:
contrast estimate SE df z.ratio p.value
2 - 1 0.284 0.529 Inf 0.536 0.5917
group = b:
contrast estimate SE df z.ratio p.value
2 - 1 -0.226 0.482 Inf -0.470 0.6386
group = c:
contrast estimate SE df z.ratio p.value
2 - 1 -0.599 0.816 Inf -0.734 0.4629
Note: contrasts are still on the as.factor scale
$`simple contrasts for group`
time = 1:
contrast estimate SE df z.ratio p.value
b - a 1.87 0.695 Inf 2.699 0.0137
c - b 2.13 0.871 Inf 2.443 0.0284
time = 2:
contrast estimate SE df z.ratio p.value
b - a 1.36 0.687 Inf 1.986 0.0897
c - b 1.75 0.838 Inf 2.095 0.0695
Note: contrasts are still on the as.factor scale
P value adjustment: mvt method for 2 tests
My questions are:
a) Is this a correct and valid method to check whether the differences are significant?
b) If not, what is the correct way to do this?
Of course any other suggestion is extremely welcome! Thanks a lot.
I normally work with lme4 package, but the glmmTMB package is increasingly becoming better suited to work with highly complicated data (think overdispersion and/or zero-inflation).
Is there a way to extract posterior modes and credible intervals from glmmTMB models, similar to how it is done for lme4 models (example here).
Details:
I am working with count data (available here) that are zero-inflated and overdispersed and have random effects. The package best suited to work with this sort of data is the glmmTMB (details here). (Note two outliers: euc0==78 and np_other_grass==20).
The data looks like this:
euc0 ea_grass ep_grass np_grass np_other_grass month year precip season prop_id quad
3 5.7 0.0 16.7 4.0 7 2006 526 Winter Barlow 1
0 6.7 0.0 28.3 0.0 7 2006 525 Winter Barlow 2
0 2.3 0.0 3.3 0.0 7 2006 524 Winter Barlow 3
0 1.7 0.0 13.3 0.0 7 2006 845 Winter Blaber 4
0 5.7 0.0 45.0 0.0 7 2006 817 Winter Blaber 5
0 11.7 1.7 46.7 0.0 7 2006 607 Winter DClark 3
The glmmTMB model:
model<-glmmTMB(euc0 ~ ea_grass + ep_grass + np_grass + np_other_grass + (1|prop_id), data = euc, family= nbinom2) #nbimom2 lets var increases quadratically
summary(model)
confint(model) #this gives the confidence intervals
How I would normally extract the posterior mode and credible intervals for a lmer/glmer model:
#extracting model estimates and credible intervals
sm.model <-arm::sim(model, n.sim=1000)
smfixef.model = sm.model#fixef
smfixef.model =coda::as.mcmc(smfixef.model)
MCMCglmm::posterior.mode(smfixef.model) #mode of the distribution
coda::HPDinterval(smfixef.model) #credible intervals
#among-brood variance
bid<-sm.model#ranef$prop_id[,,1]
bvar<-as.vector(apply(bid, 1, var)) #between brood variance posterior distribution
bvar<-coda::as.mcmc(bvar)
MCMCglmm::posterior.mode(bvar) #mode of the distribution
coda::HPDinterval(bvar) #credible intervals
Most of an answer:
Getting a multivariate Normal sample of the parameters of the conditional model is pretty easy (I think this is what arm::sim() is doing.
library(MASS)
pp <- fixef(model)$cond
vv <- vcov(model)$cond
samp <- MASS::mvrnorm(1000, mu=pp, Sigma=vv)
(then use the rest of your method above).
I'm a little skeptical that your second example is doing what you want it to do. The variance of the conditional modes is not necessarily a good estimate of the between-group variance (e.g. see here). Furthermore, I'm nervous about the half-assed-Bayesian approach (e.g., why no priors? Why look at the posterior mode, which is rarely a meaningful value in a Bayesian context?) although I do sometimes use similar approaches myself!) However, it's not too hard to use glmmTMB results to do a proper Markov chain Monte Carlo analysis:
library(tmbstan)
library(rstan)
library(coda)
library(emdbook) ## for lump.mcmc.list(), or use runjags::combine.mcmc()
t2 <- system.time(m2 <- tmbstan(model$obj))
m3 <- rstan::As.mcmc.list(m2)
lattice::xyplot(m3,layout=c(5,6))
m4 <- emdbook::lump.mcmc.list(m3)
coda::HPDinterval(m4)
It may be helpful to know that the theta column of m4 is the log of the among-group standard standard deviation ...
(See vignette("mcmc", package="glmmTMB") for a little bit more information ...)
I think Ben has already answered your question, so my answer does not add much to the discussion... Maybe just one thing, as you wrote in your comments that you're interested in the within- and between-group variances. You can get these information via parameters::random_parameters() (if I did not misunderstand what you were looking for). See example below that first generates simulated samples from a multivariate normal (just like in Ben's example), and later gives you a summary of the random effect variances...
library(readr)
library(glmmTMB)
library(parameters)
library(bayestestR)
library(insight)
euc_data <- read_csv("D:/Downloads/euc_data.csv")
model <-
glmmTMB(
euc0 ~ ea_grass + ep_grass + np_grass + np_other_grass + (1 | prop_id),
data = euc_data,
family = nbinom2
) #nbimom2 lets var increases quadratically
# generate samples
samples <- parameters::simulate_model(model)
#> Model has no zero-inflation component. Simulating from conditional parameters.
# describe samples
bayestestR::describe_posterior(samples)
#> # Description of Posterior Distributions
#>
#> Parameter | Median | 89% CI | pd | 89% ROPE | % in ROPE
#> --------------------------------------------------------------------------------
#> (Intercept) | -1.072 | [-2.183, -0.057] | 0.944 | [-0.100, 0.100] | 1.122
#> ea_grass | -0.001 | [-0.033, 0.029] | 0.525 | [-0.100, 0.100] | 100.000
#> ep_grass | -0.050 | [-0.130, 0.038] | 0.839 | [-0.100, 0.100] | 85.297
#> np_grass | -0.020 | [-0.054, 0.012] | 0.836 | [-0.100, 0.100] | 100.000
#> np_other_grass | -0.002 | [-0.362, 0.320] | 0.501 | [-0.100, 0.100] | 38.945
# or directly get summary of sample description
sp <- parameters::simulate_parameters(model, ci = .95, ci_method = "hdi", test = c("pd", "p_map"))
sp
#> Model has no zero-inflation component. Simulating from conditional parameters.
#> # Description of Posterior Distributions
#>
#> Parameter | Coefficient | p_MAP | pd | CI
#> --------------------------------------------------------------
#> (Intercept) | -1.037 | 0.281 | 0.933 | [-2.305, 0.282]
#> ea_grass | -0.001 | 0.973 | 0.511 | [-0.042, 0.037]
#> ep_grass | -0.054 | 0.553 | 0.842 | [-0.160, 0.047]
#> np_grass | -0.019 | 0.621 | 0.802 | [-0.057, 0.023]
#> np_other_grass | 0.019 | 0.999 | 0.540 | [-0.386, 0.450]
plot(sp) + see::theme_modern()
#> Model has no zero-inflation component. Simulating from conditional parameters.
# random effect variances
parameters::random_parameters(model)
#> # Random Effects
#>
#> Within-Group Variance 2.92 (1.71)
#> Between-Group Variance
#> Random Intercept (prop_id) 2.1 (1.45)
#> N (groups per factor)
#> prop_id 18
#> Observations 346
insight::get_variance(model)
#> Warning: mu of 0.2 is too close to zero, estimate of random effect variances may be unreliable.
#> $var.fixed
#> [1] 0.3056285
#>
#> $var.random
#> [1] 2.104233
#>
#> $var.residual
#> [1] 2.91602
#>
#> $var.distribution
#> [1] 2.91602
#>
#> $var.dispersion
#> [1] 0
#>
#> $var.intercept
#> prop_id
#> 2.104233
Created on 2020-05-26 by the reprex package (v0.3.0)
I am a beginner with R. I am using glm to conduct logistic regression and then using the 'margins' package to calculate marginal effects but I don't seem to be able to exclude the missing values in my categorical independent variable.
I have tried to ask R to exclude NAs from the regression. The categorical variable is weight status at age 9 (wgt9), and it has three levels (1, 2, 3) and some NAs.
What am I doing wrong? Why do I get a wgt9NA result in my outputs and how can I correct it?
Thanks in advance for any help/advice.
Conduct logistic regression
summary(logit.phbehav <- glm(obese13 ~ gender + as.factor(wgt9) + aded08b,
data = gui, weights = bdwg01, family = binomial(link = "logit")))
Regression output
term estimate std.error statistic p.value
<chr> <dbl> <dbl> <dbl> <dbl>
1 (Intercept) -3.99 0.293 -13.6 2.86e- 42
2 gender 0.387 0.121 3.19 1.42e- 3
3 as.factor(wgt9)2 2.49 0.177 14.1 3.28e- 45
4 as.factor(wgt9)3 4.65 0.182 25.6 4.81e-144
5 as.factor(wgt9)NA 2.60 0.234 11.1 9.94e- 29
6 aded08b -0.0755 0.0224 -3.37 7.47e- 4
Calculate the marginal effects
effects_logit_phtotal = margins(logit.phtot)
print(effects_logit_phtotal)
summary(effects_logit_phtotal)
Marginal effects output
> summary(effects_logit_phtotal)
factor AME SE z p lower upper
aded08a -0.0012 0.0002 -4.8785 0.0000 -0.0017 -0.0007
gender 0.0115 0.0048 2.3899 0.0169 0.0021 0.0210
wgt92 0.0941 0.0086 10.9618 0.0000 0.0773 0.1109
wgt93 0.4708 0.0255 18.4569 0.0000 0.4208 0.5207
wgt9NA 0.1027 0.0179 5.7531 0.0000 0.0677 0.1377
First of all welcome to stack overflow. Please check the answer here to see how to make a great R question. Not providing a sample of your data, some times makes it impossible to answer the question. However taking a guess, I think that you have not set your NA values correctly but as strings. This behavior can be seen in the dummy data below.
First let's create the dummy data:
v1 <- c(2,3,3,3,2,2,2,2,NA,NA,NA)
v2 <- c(2,3,3,3,2,2,2,2,"NA","NA","NA")
v3 <- c(11,5,6,7,10,8,7,6,2,5,3)
obese <- c(0,1,1,0,0,1,1,1,0,0,0)
df <- data.frame(obese,v1,v2)
Using the variable named v1, does not include NA as a category:
glm(formula = obese ~ as.factor(v1) + v3, family = binomial(link = "logit"),
data = df)
Deviance Residuals:
1 2 3 4 5 6 7 8
-2.110e-08 2.110e-08 1.168e-05 -1.105e-05 -2.110e-08 3.094e-06 2.110e-08 2.110e-08
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) 401.48 898581.15 0 1
as.factor(v1)3 -96.51 326132.30 0 1
v3 -46.93 106842.02 0 1
While making the string "NA" to factor gives an output similar to the one in question:
glm(formula = obese ~ as.factor(v2) + v3, family = binomial(link = "logit"),
data = df)
Deviance Residuals:
Min 1Q Median 3Q Max
-1.402e-05 -2.110e-08 -2.110e-08 2.110e-08 1.472e-05
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) 394.21 744490.08 0.001 1
as.factor(v2)3 -95.33 340427.26 0.000 1
as.factor(v2)NA -327.07 613934.84 -0.001 1
v3 -45.99 84477.60 -0.001 1
Try the following to replace NAs that are strings:
gui$wgt9[ gui$wgt9 == "NA" ] <- NA
Don't forget to accept any answer that solved your problem.
I am trying to reproduce output from the PROC MIXED procedure using the Satterwaithe approximation in SAS using the lmerTest package in R.
This is my data:
Participant Condition Data
1 0 -1,032941629
1 0 0,869267841
1 0 -1,636722191
1 0 -1,15451393
1 0 0,340454836
1 0 -0,399315906
1 1 0,668983169
1 1 1,937817592
1 1 3,110013393
1 1 3,23409718
2 0 0,806881925
2 1 2,71020911
2 1 3,406864275
2 1 1,494288182
2 1 0,741827047
2 1 2,532062685
2 1 3,702118917
2 1 1,825046681
2 1 4,37167021
2 1 1,85125279
3 0 0,288743786
3 0 1,024396121
3 1 2,051281876
3 1 0,24543851
3 1 3,349677964
3 1 1,565395822
3 1 3,077031712
3 1 1,087494708
3 1 1,546150033
3 1 0,440249347
Using the following statement in SAS:
proc mixed data=mbd;
class participant;
model data = condition / solution ddfm=sat;
random intercept condition / sub=participant;
run;
I get this output:
My problem is that I can't seem to reproduce these results using lmerTest in R.
I thought that lmer(Data ~ Condition + (1 | Participant) + (Condition | Participant), REML=TRUE) was the equivalent statement of what I did in SAS but this gives different results. Note that the degrees of freedom are way off from the SAS output so I think I'm estimating parameters in R that I'm not estimating in SAS. I tried several other statements in R but I didn't manage to get the exact same output. However this should be possible as the lmer() function from the lmerTest package also uses the Satterwaithe approximation and should be exactly the same as the SAS PROC MIXED procedure.
Does anybody know what I'm doing wrong in R?
Thanks a lot!
Bart
You don't specify the same random effects as in your SAS example. (Condition | Participant) is expanded internally to (1 + Condition | Participant), which fits a random intercept, a random slope and the covariance between them [1]. So, you have two additional parameters (an intercept variance and the covariance) in your model. Uncorrelated random effects can be specified using || in lme4 syntax. Note how the formula gets expanded in the summary output.
library(lmerTest)
fit <- lmer(Data ~ Condition + (Condition || Participant), REML=TRUE, data = DF)
summary(fit)
#Linear mixed model fit by REML
#t-tests use Satterthwaite approximations to degrees of freedom ['lmerMod']
#Formula: Data ~ Condition + ((1 | Participant) + (0 + Condition | Participant))
# Data: DF
#
#REML criterion at convergence: 90.6
#
#Scaled residuals:
# Min 1Q Median 3Q Max
#-1.58383 -0.78970 -0.06993 0.87801 1.91237
#
#Random effects:
# Groups Name Variance Std.Dev.
# Participant (Intercept) 0.00000 0.000
# Participant.1 Condition 0.07292 0.270
# Residual 1.20701 1.099
#Number of obs: 30, groups: Participant, 3
#
#Fixed effects:
# Estimate Std. Error df t value Pr(>|t|)
#(Intercept) -0.09931 0.36621 26.50400 -0.271 0.788363
#Condition 2.23711 0.46655 12.05700 4.795 0.000432 ***
#---
#Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#
#Correlation of Fixed Effects:
# (Intr)
#Condition -0.785