I have been asked to see if there is a linear trend in 3 groups of data (5 points each) by using ANOVA and linear contrasts. The 3 groups represent data collected in 2010, 2011 and 2012. I want to use R for this procedure and I have tried both of the following:
contrasts(data$groups, how.many=1) <- contr.poly(3)
contrasts(data$groups) <- contr.poly(3)
Both ways seem to work fine but give slightly different answers in terms of their p-values. I have no idea which is correct and it is really tricky to find help for this on the web. I would like help figuring out what is the reasoning behind the different answers. I'm not sure if it has something to do with partitioning sums of squares or whatnot.
Both approaches differ with respect to whether a quadratic polynomial is used.
For illustration purposes, have a look at this example, both x and y are a factor with three levels.
x <- y <- gl(3, 2)
# [1] 1 1 2 2 3 3
# Levels: 1 2 3
The first approach creates a contrast matrix for a quadratic polynomial, i.e., with a linear (.L) and a quadratic trend (.Q). The 3 means: Create the 3 - 1th polynomial.
contrasts(x) <- contr.poly(3)
# [1] 1 1 2 2 3 3
# attr(,"contrasts")
# .L .Q
# 1 -7.071068e-01 0.4082483
# 2 -7.850462e-17 -0.8164966
# 3 7.071068e-01 0.4082483
# Levels: 1 2 3
In contrast, the second approach results in a polynomial of first order (i.e., a linear trend only). This is due to the argument how.many = 1. Hence, only 1 contrast is created.
contrasts(y, how.many = 1) <- contr.poly(3)
# [1] 1 1 2 2 3 3
# attr(,"contrasts")
# .L
# 1 -7.071068e-01
# 2 -7.850462e-17
# 3 7.071068e-01
# Levels: 1 2 3
If you're interested in the linear trend only, the second option seems more appropriate for you.
Changing the contrasts you ask for changes the degrees of freedom of the model. If one model requests linear and quadratic contrasts, and a second specifies only, say, the linear contrast, then the second model has an extra degree of freedom: this will increase the power to test the linear hypothesis, (at the cost of preventing the model fitting the quadratic trend).
Using the full ("nlevels - 1") set of contrasts creates an orthogonal set of contrasts which explore the full set of (independent) response configurations. Cutting back to just one prevents the model from fitting one configuration (in this case the quadratic component which our data in fact possess.
To see how this works, use the built-in dataset mtcars, and test the (confounded) relationship of gears to gallons. We'll hypothesize that the more gears the better (at least up to some point).
df = mtcars # copy the dataset
df$gear = as.ordered(df$gear) # make an ordered factor
Ordered factors default to polynomial contrasts, but we'll set them here to be explicit:
contrasts(df$gear) <- contr.poly(nlevels(df$gear))
Then we can model the relationship.
m1 = lm(mpg ~ gear, data = df);
summary.lm(m1)
# Estimate Std. Error t value Pr(>|t|)
# (Intercept) 20.6733 0.9284 22.267 < 2e-16 ***
# gear.L 3.7288 1.7191 2.169 0.03842 *
# gear.Q -4.7275 1.4888 -3.175 0.00353 **
#
# Multiple R-squared: 0.4292, Adjusted R-squared: 0.3898
# F-statistic: 10.9 on 2 and 29 DF, p-value: 0.0002948
Note we have F(2,29) = 10.9 for the overall model and p=.038 for our linear effect with an estimated extra 3.7 mpg/gear.
Now let's only request the linear contrast, and run the "same" analysis.
contrasts(df$gear, how.many = 1) <- contr.poly(nlevels(df$gear))
m1 = lm(mpg ~ gear, data = df)
summary.lm(m1)
# Coefficients:
# Estimate Std. Error t value Pr(>|t|)
# (Intercept) 21.317 1.034 20.612 <2e-16 ***
# gear.L 5.548 1.850 2.999 0.0054 **
# Multiple R-squared: 0.2307, Adjusted R-squared: 0.205
# F-statistic: 8.995 on 1 and 30 DF, p-value: 0.005401
The linear effect of gear is now bigger (5.5 mpg) and p << .05 - A win? Except the overall model fit is now significantly worse: variance accounted for is now just 23% (was 43%)! Why is clear if we plot the relationship:
plot(mpg ~ gear, data = df) # view the relationship
So, if you're interested in the linear trend, but also expect (or are unclear about) additional levels of complexity, you should also test these higher polynomials. The quadratic (or, in general, trends up to levels-1).
Note too that in this example the physical mechanism is confounded: We've forgotten that number of gears is confounded with automatic vs manual transmission, and also with weight, and sedan vs sports car.
If someone wants to test the hypothesis that 4 gears is better than 3, they could answer this question :-)
Related
I'm trying to get the degrees of freedom from emmeans of a glmer model for reporting reasons, but they just show Inf.
Here's some sample data. In the real data, there is no nesting structure, this is just a consequence of how I built the data frame:
set.seed(1234)
dat <- data.frame(
dv=c(rnorm(mean=1, sd=0.2, n=12000)),
id=c(rep(c("1", "2", "3"), times=c(4000, 4000, 4000))),
region=c(rep(rep(c("1", "2"), times=c(2000, 2000)), 3)),
intervention=c(rep(c("1", "2", "1"), times=c(4000, 4000, 4000))),
timepoint=c(rep(rep(c("1", "2"), times=c(2000, 2000)), times=3)),
direction=c(rep(rep(c("1", "2"), times=c(2000, 2000)), 3))
)
glmm_1 <- glmer(dv ~ intervention*timepoint*region + direction + (1|id), data=dat, family=gaussian(link="log"))
glmm_1_emm <- emmeans::emmeans(glmm_1, pairwise ~ intervention*region*timepoint, type = "response")
glmm_1_emm$emmeans
NOTE: A nesting structure was detected in the fitted model:
timepoint %in% (direction*region), region %in% direction
region timepoint direction intervention response SE df asymp.LCL asymp.UCL
1 1 1 1 1 0.00313 Inf 0.994 1.01
2 2 2 1 1 0.00313 Inf 0.998 1.01
1 1 1 2 1 0.00442 Inf 0.992 1.01
2 2 2 2 1 0.00442 Inf 0.995 1.01
Confidence level used: 0.95
Intervals are back-transformed from the log scale
This is really more of a statistical (i.e. for CrossValidated) than a computational question. tl;dr finite-size corrections are rarely considered for GLMs or GLMMs, and for GLMMs in particular there is little theoretical work I'm aware of that would even specify how to compute them. That's why emmeans etc. report df as Inf.
df in emmeans output represents the "denominator degrees of freedom" (i.e. the nu2 value you would use if testing against an F distribution F_{nu1,nu2}), which is something like (number of observations - number of parameters estimated) for simple (non-mixed) models like a linear regression or simple ANOVA, but which is considerably harder to define for multilevel models (i.e. linear mixed models). For generalized linear (and linear mixed) models, it gets even worse. Quoting from the "degrees of freedom" section of the GLMM FAQ (see there for full references):
When the responses are not normally distributed (as in GLMs and GLMMs), and when the scale parameter is not estimated (as in standard Poisson- and binomial-response models), then the deviance differences are only asymptotically F- or chi-square-distributed (i.e. not for our real, finite-size samples). In standard GLM practice, we usually ignore this problem; there is some literature on finite-size corrections for GLMs under the rubrics of “Bartlett corrections” and “higher order asymptotics” (see McCullagh and Nelder (1989), Cordeiro, Paula, and Botter (1994), Cordeiro and Ferrari (1998) and the cond package (on CRAN) [which works with GLMs, not GLMMs]), but it’s rarely used. (The bias correction/Firth approach implemented in the brglm package attempts to address the problem of finite-size bias, not finite-size non-chi-squaredness of the deviance differences.)
When the scale parameter in a GLM is estimated rather than fixed (as in Gamma or quasi-likelihood models), it is sometimes recommended to use an F
test to account for the uncertainty of the scale parameter (e.g. Venables and Ripley (2002) recommend anova(...,test="F") for quasi-likelihood models)
Combining these issues, one has to look pretty hard for information on small-sample or finite-size corrections for GLMMs: Feng, Braun, and McCulloch (2004) and Bell and Grunwald (2010) look like good starting points, but it’s not at all trivial.
Apparently emmeans::emmeans calculates Inf degrees of freedom, not sure why. But I've spotted this:
str(glmm_1_emm$emmeans)
# 'emmGrid' object with variables:
# region = 1, 2
# timepoint = 1, 2
# direction = 1, 2
# intervention = 1, 2
# Nesting structure: timepoint %in% (direction*region), region %in% direction
# Transformation: “log”
# Some things are non-estimable (null space dim = 5) ## <--------------------- !!!
There's a summary method involved, emmeans:::summary.emmGrid, since the summary is not calculated until you print it, i.e. evaluate glmm_1_emm$emmeans.
Provided that the degrees of freedom are correct later on, then you could extract them using a rather artless capture.output approach:
tmp <- capture.output(glmm_1_emm$emmeans)
res <- read.table(text=tmp[1:(which(tmp == '') - 1)], header=TRUE)
res
# region timepoint direction intervention response SE df asymp.LCL asymp.UCL
# 1 1 1 1 1 1 0.00313 Inf 0.994 1.01
# 2 2 2 2 1 1 0.00313 Inf 0.998 1.01
# 3 1 1 1 2 1 0.00442 Inf 0.992 1.01
# 4 2 2 2 2 1 0.00442 Inf 0.995 1.01
res[, 7] ## degrees of freedom
# [1] Inf Inf Inf Inf
I have a multinomial logit model created with the nnet R package, using the multinom command. The dependent variable has three categories/choice options. I am modelling the probability of selecting a certain irrigation type (no irrigation, surface irrigation, drip irrigation) based on farmer characteristics.
I would like to estimate marginal effects, i.e. by how much does the probability of selecting irrigation type Y change when I increase independent variable X by one unit? I have tried doing this with the margins package (marginal_effects), but this gives only 1 value per observation in the dataset. I was expecting three values, since I want the marginal effect for each of the three irrigation types.
Does someone know if there is a better R package to use for this? Or whether I am doing something wrong with the margins packages? Thank you.
You can use the marginaleffects
package to do
that (disclaimer: I am the maintainer). Please note the warning.
library(nnet)
library(marginaleffects)
mod <- multinom(factor(cyl) ~ hp + mpg, data = mtcars, quiet = true)
mfx <- marginaleffects(mod, type = "probs")
## Warning in sanity_model_specific.multinom(model, ...): The standard errors
## estimated by `marginaleffects` do not match those produced by Stata for
## `nnet::multinom` models. Please be very careful when interpreting the results.
summary(mfx)
## Average marginal effects
## type Group Term Effect Std. Error z value Pr(>|z|) 2.5 %
## 1 probs 6 hp 2.792e-04 0.000e+00 Inf < 2.22e-16 2.792e-04
## 2 probs 6 mpg -1.334e-03 0.000e+00 -Inf < 2.22e-16 -1.334e-03
## 3 probs 8 hp 2.396e-05 1.042e-126 2.298e+121 < 2.22e-16 2.396e-05
## 4 probs 8 mpg -2.180e-04 1.481e-125 -1.472e+121 < 2.22e-16 -2.180e-04
## 97.5 %
## 1 2.792e-04
## 2 -1.334e-03
## 3 2.396e-05
## 4 -2.180e-04
##
## Model type: multinom
## Prediction type: probs
The marginaleffects package should work in theory, but my example doesn't compile because of file size restrictions (meaning I don't have enough RAM for the 1.5 GB vector it tries to use). It's not even that large of a dataset, which is odd.
If you use marginal_effects() (margins package) for multinomial models, it only displays the output for a default category. You have to manually set each category you want to see. You can clean up the output with broom and then combine some other way. It's clunky, but it can work.
marginal_effects(model, category = 'cat1')
In the process of studying logistic regression using carret's mdrr data, questions arise.
I created a full model using a total of 19 variables, and I have questions about the notation of the categorical variable.
In my regression model, the categorical variables are:
nDB : 0 or 1 or 2
nR05 : 0 or 1
nR10 : 1 or 2
I created a full model using glm, but I do not know why the names of categorical variables have one of the numbers in the category.
-------------------------------------------------------------------------------
glm(formula = mdrrClass ~ ., family = binomial, data = train)
#Coefficients:
#(Intercept) nDB1 nDB2 nX nR051 nR101 nBnz2
#5.792e+00 5.287e-01 -3.103e-01 -2.532e-01 -9.291e-02 9.259e-01 -2.108e+00
#SPI BLI PW4 PJI2 Lop BIC2 VRA1
#3.222e-05 -1.201e+01 -3.754e+01 -5.467e-01 1.010e+00 -5.712e+00 -2.424e-04
# PCR H3D FDI PJI3 DISPm DISPe G.N..N.
# -6.397e-02 -4.360e-04 3.458e+01 -6.579e+00 -5.690e-02 2.056e-01 -7.610e-03
#Degrees of Freedom: 263 Total (i.e. Null); 243 Residual
#Null Deviance: 359.3
#Residual Deviance: 232.6 AIC: 274.6
-------------------------------------------------------------------------------
The above results show that nDB is numbered, and nR05 and nR10 are related to categories.
I am wondering why numbers are attached as above.
When you have categorical predictors in any regression model you need to create dummy variables. R does this for you and the output you see are the contrasts
Your variable nDB has 3 levels: 0, 1, 2
One of those needs to be chosen as the reference level (R was chosen 0 for you in this case, but this can also be specified manually). Then dummy variables are created to compare every other level against your reference level: 0 vs 1 and 0 vs 2
R names these dummy variables nDB1 and nDB2. nDB1 is for the 0 vs 1 contrast, and nDB2 is for the 0 vs 2 contrast. The numbers after the variable names are just to indicate which contrast you're looking at
The coefficient values are interpreted as the difference in your y (outcome) value between groups 0 and 1 (nDB1), and separately between groups 0 and 2 (nDB2). In other words, what change in the outcome would you expect when moving from one group to the other?
Your other categorical variables have 2 levels and are just a simpler case of the above
For example, nR05 only has 0 and 1 as values. 0 was chosen as your reference, and because theres only 1 possible contrast here, a single dummy variable is created comparing 0 vs 1. In the output that dummy variable is called nR051
It's always the case for categorical variables, espacially when they are not binary (like your nDB). It's so that you know for which value you have the coefficient. For the nDB variable the model has created two new variables: nDB_1 which equals 1 if nDB=1 and equals 0 if nDB= 0 or nDB=2.
To analyze a binary variable (whose values would be TRUE / FALSE, 0/1, or YES / NO) according to a quantitative explanatory variable, a logistic regression can be used.
Consider for example the following data, where x is the age of 40 people, and y the variable indicating if they bought a death metal album in the last 5 years (1 if "yes", 0 if "no" )
Graphically, we can see that, more likely, the older people are, the less they buy death metal.
Logistic regression is a special case of the Generalized Linear Model (GLM).
With a classical linear regression model, we consider the following model:
Y = αX + β
The expectation of Y is therefore predicted as follows:
E (Y) = αX + β
Here, because of the binary distribution of Y, the above relations can not apply. To "generalize" the linear model, we therefore consider that
g (E (Y)) = αX + β
where g is a link function.
In this case, for a logistic regression, the link function corresponds to the logit function:
logit (p) = log (p/(1-p))
Note that this logit function transforms a value (p) between 0 and 1 (such as a probability for example) into a value between - ∞ and + ∞.
Here's how to do the logistic regression under R:
myreg=glm(y~x, family=binomial(link=logit))
summary(myreg)
glm(formula = y ~ x, family = binomial(link = logit))
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.8686 -0.7764 0.3801 0.8814 2.0253
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 5.9462 1.9599 3.034 0.00241 **
## x -0.1156 0.0397 -2.912 0.00360 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 52.925 on 39 degrees of freedom
## Residual deviance: 39.617 on 38 degrees of freedom
## AIC: 43.617
##
## Number of Fisher Scoring iterations: 5
We obtain the following model:
logit (E (Y)) = - 0.12X + 5.95
and we note that the (negative) influence of age on the purchase of death metal albums is significant at the 5% level (p(>[Z| ----> < 5%).
Thus, logistic regression is often used to bring out risk factor (like Age but also BMI, Sex and so on …)
I am use to conducting Tukey post-hoc tests in minitab. When I do, I usually get family grouping of the dependent/predictor variables.
In R, using TukeyHSD() the family grouping is not displayed (or calculated?). It only displays the relationship between each of the dependent/predictor variables. Is it possible to display the family groupings like in minitab?
Using the diamonds data set:
av <- aov(price ~ cut, data = diamonds)
tk <- TukeyHSD(av, ordered = T, which = "cut")
plot(tk)
Output:
Fit: aov(formula = price ~ cut, data = diamonds)
$cut
diff lwr upr p adj
Good-Ideal 471.32248 300.28228 642.3627 0.0000000
Very Good-Ideal 524.21792 401.33117 647.1047 0.0000000
Fair-Ideal 901.21579 621.86019 1180.5714 0.0000000
Premium-Ideal 1126.71573 1008.80880 1244.6227 0.0000000
Very Good-Good 52.89544 -130.15186 235.9427 0.9341158
Fair-Good 429.89331 119.33783 740.4488 0.0014980
Premium-Good 655.39325 475.65120 835.1353 0.0000000
Fair-Very Good 376.99787 90.13360 663.8622 0.0031094
Premium-Very Good 602.49781 467.76249 737.2331 0.0000000
Premium-Fair 225.49994 -59.26664 510.2665 0.1950425
Picture added to help clarify my response to Maruits's comment:
Here is a step-by-step example on how to reproduce minitab's table for the ggplot2::diamonds dataset. I've included details/explanation as much as possible.
Please note that as far as I can tell, results shown in minitab's table are not dependent/related to results from Tukey's post-hoc test; they are based on results from the analysis of variance. Tukey's honest significant difference (HSD) test is a post-hoc test that establishes which comparisons (of all the possible pairwise comparisons of group means) are (honestly) statistically significant, given the ANOVA results.
In order to reproduce minitabs "mean-grouping" summary table (see the first table of "Interpret the results: Step 3" of the minitab Express Support), I recommend (re-)running a linear model to extract means and confidence intervals. Note that this is exactly how aov fits the analysis of variance model for each group.
Fit a linear model
We specify a 0 offset to get absolute estimates for every group (rather than estimates for the changes relative to an offset).
fit <- lm(price ~ 0 + cut, data = diamonds)
coef <- summary(fit)$coef;
coef;
# Estimate Std. Error t value Pr(>|t|)
#cutFair 4358.758 98.78795 44.12236 0
#cutGood 3928.864 56.59175 69.42468 0
#cutVery Good 3981.760 36.06181 110.41487 0
#cutPremium 4584.258 33.75352 135.81570 0
#cutIdeal 3457.542 27.00121 128.05137 0
Determine family groupings
In order to obtain something similar to minitab's "family groupings", we adopt the following approach:
Calculate confidence intervals for all parameters
Perform a hierarchical clustering analysis on the confidence interval data for all parameters
Cut the resulting tree at a height corresponding to the standard deviation of the CIs. This will gives us a grouping of parameter estimates based on their confidence intervals. This is a somewhat empirical approach but justifiable as the tree measures pairwise distances between the confidence intervals, and the standard deviation can be interpreted as a Euclidean distance.
We start by calculating the confidence interval and cluster the resulting distance matrix using hierarchical clustering using complete linkage.
CI <- confint(fit);
hc <- hclust(dist(CI));
We inspect the cluster dendrogram
plot(hc);
We now cut the tree at a height corresponding to the standard deviation of all CIs across all parameter estimates to get the "family groupings"
grps <- cutree(hc, h = sd(CI))
Summarise results
Finally, we collate all quantities and store results in a table similar to minitab's "mean-grouping" table.
library(tidyverse)
bind_cols(
cut = rownames(coef),
N = as.integer(table(fit$model$cut)),
Mean = coef[, 1],
Groupings = grps) %>%
as.data.frame()
# cut N Mean Groupings
#1 cutFair 1610 4358.758 1
#2 cutGood 4906 3928.864 2
#3 cutVery Good 12082 3981.760 2
#4 cutPremium 13791 4584.258 1
#5 cutIdeal 21551 3457.542 3
Note the near-perfect agreement of our results with those from the minitab "mean-grouping" table: cut = Ideal is by itself in group 3 (group C in minitab's table), while Fair+Premium share group 1 (minitab: group A ), and Good+Very Good share group 2 (minitab: group B).
See the cld function in the multcomp package, as explained here (copy-pasted below).
Example data set:
> data(ToothGrowth)
> ToothGrowth$treat <- with(ToothGrowth, interaction(supp,dose))
> str(ToothGrowth)
'data.frame': 60 obs. of 3 variables:
$ len : num 4.2 11.5 7.3 5.8 6.4 10 11.2 11.2 5.2 7 ...
$ supp: Factor w/ 2 levels "OJ","VC": 2 2 2 2 2 2 2 2 2 2 ...
$ dose: num 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 ...
$ treat: Factor w/ 6 levels "OJ.0.5","VC.0.5",..: 2 2 2 2 2 2 2 2 2 2 ...
Model fit:
> fit <- lm(len ~ treat, data=ToothGrowth)
All pairwise comparisons with Tukey test:
> apctt <- multcomp::glht(fit, linfct = multcomp::mcp(treat = "Tukey"))
Letter-based representation of all-pairwise comparisons (algorithm from Piepho 2004):
> lbrapc <- multcomp::cld(apctt)
> lbrapc
OJ.0.5 VC.0.5 OJ.1 VC.1 OJ.2 VC.2
"b" "a" "c" "b" "c" "c"
I'm analysing my binomial dataset with R using a generalized linear mixed model (glmer, lme4-package). I wanted to make the pairwise comparisons of a certain fixed effect ("Sound") using a Tukey's post-hoc test (glht, multcomp-package).
Most of it is working fine, but one of my fixed effect variables ("SoundC") has no variance at all (96 times a "1" and zero times a "0") and it seems that the Tukey's test cannot handle that. All pairwise comparisons with this "SoundC" give a p-value of 1.000 whereas some are clearly significant.
As a validation I changed one of the 96 "1"'s to a "0" and after that I got normal p-values again and significant differences where I expected them, whereas the difference had actually become smaller after my manual change.
Does anybody have a solution? If not, is it fine to use the results of my modified dataset and report my manual change?
Reproducible example:
Response <- c(1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,1,1,0,
0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,1,1,0,
1,1,0,1,1,0,1,1,1,1,0,0,1,1,0,1,1,0,1,1,0,1,1,0,1)
Data <- data.frame(Sound=rep(paste0('Sound',c('A','B','C')),22),
Response,
Individual=rep(rep(c('A','B'),2),rep(c(18,15),2)))
# Visual
boxplot(Response ~ Sound,Data)
# Mixed model
library (lme4)
model10 <- glmer(Response~Sound + (1|Individual), Data, family=binomial)
# Post-hoc analysis
library (multcomp)
summary(glht(model10, mcp(Sound="Tukey")))
This is verging on a CrossValidated question; you are definitely seeing complete separation, where there is a perfect division of your response into 0 vs 1 results. This leads to (1) infinite values of the parameters (they're only listed as non-infinite due to computational imperfections) and (2) crazy/useless values of the Wald standard errors and corresponding $p$ values (which is what you're seeing here). Discussion and solutions are given here, here, and here, but I'll illustrate a little more below.
To be a statistical grouch for a moment: you really shouldn't be trying to fit a random effect with only 3 levels anyway (see e.g. http://glmm.wikidot.com/faq) ...
Firth-corrected logistic regression:
library(logistf)
L1 <- logistf(Response~Sound*Individual,data=Data,
contrasts.arg=list(Sound="contr.treatment",
Individual="contr.sum"))
coef se(coef) p
(Intercept) 3.218876e+00 1.501111 2.051613e-04
SoundSoundB -4.653960e+00 1.670282 1.736123e-05
SoundSoundC -1.753527e-15 2.122891 1.000000e+00
IndividualB -1.995100e+00 1.680103 1.516838e-01
SoundSoundB:IndividualB 3.856625e-01 2.379919 8.657348e-01
SoundSoundC:IndividualB 1.820747e+00 2.716770 4.824847e-01
Standard errors and p-values are now reasonable (p-value for the A vs C comparison is 1 because there is literally no difference ...)
Mixed Bayesian model with weak priors:
library(blme)
model20 <- bglmer(Response~Sound + (1|Individual), Data, family=binomial,
fixef.prior = normal(cov = diag(9,3)))
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) 1.711485 2.233667 0.7662221 4.435441e-01
## SoundSoundB -5.088002 1.248969 -4.0737620 4.625976e-05
## SoundSoundC 2.453988 1.701674 1.4421024 1.492735e-01
The specification diag(9,3) of the fixed-effect variance-covariance matrix produces
$$
\left(
\begin{array}{ccc}
9 & 0 & 0 \
0 & 9 & 0 \
0 & 0 & 9
\end{array}
\right)
$$
In other words, the 3 specifies the dimension of the matrix (equal to the number of fixed-effect parameters), and the 9 specifies the variance -- this corresponds to a standard devation of 3 or a 95% range of about $\pm 6$, which is quite large/weak/uninformative for logit-scaled responses.
These are roughly consistent (the model is very different)
library(multcomp)
summary(glht(model20, mcp(Sound="Tukey")))
## Estimate Std. Error z value Pr(>|z|)
## SoundB - SoundA == 0 -5.088 1.249 -4.074 0.000124 ***
## SoundC - SoundA == 0 2.454 1.702 1.442 0.309216
## SoundC - SoundB == 0 7.542 1.997 3.776 0.000397 ***
As I said above, I would not recommend a mixed model in this case anyway ...