Wonnacott, Brown & Nation
Elizabeth Wonnacott
August, 2016
- Load packages and helper functions
- Load dataset
- Experiment 1 (children), input nouns
- Experiment 1 (children), minimal exposure nouns
- Experiment 1 (children), reported awareness of lexicality
- Experiment 2 (adults), trained nouns
- Experiment 2 (adults), minimal exposure nouns
- Experiment 2 (adults), reported awareness of lexicality
Load packages and helper functions
Packages.
rm(list=ls())
library(ggplot2)
library(plotrix)
#suppressPackageStartupMessages(library(dplyr))
library(plyr)
#library(bootstrap)
suppressPackageStartupMessages(library(lme4))
library(knitr)
theme_set(theme_bw())
opts_chunk$set(fig.width=8, fig.height=5,
echo=TRUE, warning=FALSE, message=FALSE, cache=TRUE)Helper functions.
SummarySE
This function can be found on the website “Cookbook for R”.
http://www.cookbook-r.com/Graphs/Plotting_means_and_error_bars_(ggplot2)/#Helper
It summarizes data, giving count, mean, standard deviation, standard error of the mean, and confidence interval (default 95%).
data: a data frame.
measurevar: the name of a column that contains the variable to be summarized. groupvars: a vector containing names of columns that contain grouping variables. na.rm: a boolean that indicates whether to ignore NA’s. conf.interval: the percent range of the confidence interval (default is 95%).
summarySE <- function(data=NULL, measurevar, groupvars=NULL, na.rm=FALSE,
conf.interval=.95, .drop=TRUE) {
require(plyr)
# New version of length which can handle NA's: if na.rm==T, don't count them
length2 <- function (x, na.rm=FALSE) {
if (na.rm) sum(!is.na(x))
else length(x)
}
# This does the summary. For each group's data frame, return a vector with
# N, mean, and sd
datac <- ddply(data, groupvars, .drop=.drop,
.fun = function(xx, col) {
c(N = length2(xx[[col]], na.rm=na.rm),
mean = mean (xx[[col]], na.rm=na.rm),
sd = sd (xx[[col]], na.rm=na.rm)
)
},
measurevar
)
# Rename the "mean" column
datac <- rename(datac, c("mean" = measurevar))
datac$se <- datac$sd / sqrt(datac$N) # Calculate standard error of the mean
# Confidence interval multiplier for standard error
# Calculate t-statistic for confidence interval:
# e.g., if conf.interval is .95, use .975 (above/below), and use df=N-1
ciMult <- qt(conf.interval/2 + .5, datac$N-1)
datac$ci <- datac$se * ciMult
return(datac)
}myCenter
This function ouputs the centered values of an variable, which can be a numeric variable, a factor, or a data frame. It was taken from Florian Jaegers blog https://hlplab.wordpress.com/2009/04/27/centering-several-variables/.
From his blog:
-If the input is a numeric variable, the output is the centered variable.
-If the input is a factor, the output is a numeric variable with centered factor level values. That is, the factor’s levels are converted into numerical values in their inherent order (if not specified otherwise, R defaults to alphanumerical order). More specifically, this centers any binary factor so that the value below 0 will be the 1st level of the original factor, and the value above 0 will be the 2nd level.
-If the input is a data frame or matrix, the output is a new matrix of the same dimension and with the centered values and column names that correspond to the colnames() of the input preceded by “c” (e.g. “Variable1” will be “cVariable1”).
myCenter= function(x) {
if (is.numeric(x)) { return(x - mean(x, na.rm=T)) }
if (is.factor(x)) {
x= as.numeric(x)
return(x - mean(x, na.rm=T))
}
if (is.data.frame(x) || is.matrix(x)) {
m= matrix(nrow=nrow(x), ncol=ncol(x))
colnames(m)= paste("c", colnames(x), sep="")
for (i in 1:ncol(x)) {
m[,i]= myCenter(x[,i])
}
return(as.data.frame(m))
}
}lizCenter
This function provides a wrapper around myCenter allowing you to center a specific list of variables from a dataframe.
x: data frame listfname: a list of the variables to be centered (e.g. list(variable1,variable2))
The output is a copy of the data frame with a column (always a numeric variable) added for each of the centered variables. These columns are labelled with the each column’s previous name, but with “.ct” appended (e.g., “variable1” will become “variable1.ct”).
lizCenter= function(x, listfname)
{
for (i in 1:length(listfname))
{
fname = as.character(listfname[i])
x[paste(fname,".ct", sep="")] = myCenter(x[fname])
}
return(x)
}lizContrasts
This function can be used used to create two centered dummy variables which stand in place of a three way factor (condition). This allows us to inspect each contrast separately, as well as their interactions with other factors. Other fixed effects in the model in the model can be evaluated as the average effects across all levels of the factor.
The function takes a dataframe (d), a factor from that database (condition), which must have three levels, and the name of the level of the factor which is to be used as the baseline for the contrasts (baselevel).
For example, if d is dataframe with a factor “condition” with three levels “lex_skew” lex_noskew" “mixed” then lizContrasts(d, d$condition, “lex_no_skew”) returns a dataframe with two (numeric) columns added labelled “lex_noskew_VERSUS_lex_mixed” and “lex_noskew_VERSUS_lex_skew”. Wherever you would normally use “condition” in a formula in an lme, it can be replaced by (lex_noskew_VERSUS_lex_mixed + “lex_noskew_VERSUS_lex_skew) e.g. ~ (a * condition) becomes ~ (a * (lex_noskew_VERSUS_lex_mixed + lex_noskew_VERSUS_lex_skew)).
lizContrasts= function(d, condition, baselevel)
{
condition = factor(condition)
condition = relevel(condition, baselevel)
a= (contrasts(condition)-apply(contrasts(condition),2,mean))
d$dummy1[condition== rownames(a)[1]] <- a[1]
d$dummy1[condition== rownames(a)[2]] <- a[2]
d$dummy1[condition== rownames(a)[3]] <- a[3]
d$dummy2[condition== rownames(a)[1]] <- a[4]
d$dummy2[condition== rownames(a)[2]] <- a[5]
d$dummy2[condition== rownames(a)[3]] <- a[6]
name1 = paste(baselevel, rownames(a)[2],sep="_VERSUS_")
name2 = paste(baselevel, rownames(a)[3],sep="_VERSUS_")
d[name1] = d$dummy1
d[name2] = d$dummy2
d$dummy1 <-NULL
d$dummy2 <-NULL
return(d)
}lizContrasts4
This function is a version of the previous lizContrasts function but to be used to create three centered dummy variables which stand in place of a four way factor (condition).
lizContrasts4= function(d, condition, baselevel)
{
condition = factor(condition)
condition = relevel(condition, baselevel)
a= (contrasts(condition)-apply(contrasts(condition),2,mean))
d$dummy1[condition== rownames(a)[1]] <- a[1]
d$dummy1[condition== rownames(a)[2]] <- a[2]
d$dummy1[condition== rownames(a)[3]] <- a[3]
d$dummy1[condition== rownames(a)[4]] <- a[4]
d$dummy2[condition== rownames(a)[1]] <- a[5]
d$dummy2[condition== rownames(a)[2]] <- a[6]
d$dummy2[condition== rownames(a)[3]] <- a[7]
d$dummy2[condition== rownames(a)[4]] <- a[8]
d$dummy3[condition== rownames(a)[1]] <- a[9]
d$dummy3[condition== rownames(a)[2]] <- a[10]
d$dummy3[condition== rownames(a)[3]] <- a[11]
d$dummy3[condition== rownames(a)[4]] <- a[12]
name1 = paste(baselevel, rownames(a)[2],sep="_VERSUS_")
name2 = paste(baselevel, rownames(a)[3],sep="_VERSUS_")
name3 = paste(baselevel, rownames(a)[4],sep="_VERSUS_")
d[name1] = d$dummy1
d[name2] = d$dummy2
d[name3] = d$dummy3
d$dummy1 <-NULL
d$dummy2 <-NULL
d$dummy3 <-NULL
return(d)
}Bf
This function is a version of the Dienes (2008) calculator. It can be found at this webstie: http://www.lifesci.sussex.ac.uk/home/Zoltan_Dienes/inference/Bf%20with%20t%20likelihood.html
From that website:
The following R code is based on the Baguley and Kaye (2010) R code for the Dienes (2008) calculator. The likelihood function for the data has been changed from a Normal distribution to a t-distribution. This is a way of accounting for the variance of observations being unknown in advance of the data - thus no correction factor need be applied to the SE when this calculator is used.
Bf<-function(sd, obtained, dfdata, uniform, lower=0, upper=1, meanoftheory=0, sdtheory=1, tail=2)
{
area <- 0
if(identical(uniform, 1)){
theta <- lower
range <- upper - lower
incr <- range / 2000
for (A in -1000:1000){
theta <- theta + incr
dist_theta <- 1 / range
height <- dist_theta * dnorm(obtained, theta, sd)
area <- area + height * incr
}
}else{
theta <- meanoftheory - 5 * sdtheory
incr <- sdtheory / 200
for (A in -1000:1000){
theta <- theta + incr
dist_theta <- dnorm(theta, meanoftheory, sdtheory)
if(identical(tail, 1)){
if (theta <= 0){
dist_theta <- 0
} else {
dist_theta <- dist_theta * 2
}
}
height <- dist_theta * dt((obtained-theta)/sd, df=dfdata)
area <- area + height * incr
}
}
LikelihoodTheory <- area
Likelihoodnull <- dt(obtained/sd, df = dfdata)
BayesFactor <- LikelihoodTheory / Likelihoodnull
ret <- list("LikelihoodTheory" = LikelihoodTheory, "Likelihoodnull" = Likelihoodnull, BayesFactor)
ret
}Bf_powercalc
This works with the Bf funciton above. It requires the same values as that function (i.e. the obtained mean and SE for the current sample, a value for the predicted mean, which is set to be sdtheory (with meanoftheory=0), and the current number of participants N). However rather than return BF for current sample, it works out what the BF would be for a range of different subject numbers (assuming that the SE scales with sqrt(N)),
Bf_powercalc<-function(sd, obtained, dfdata, uniform, lower=0, upper=1, meanoftheory=0, sdtheory=1, tail=2, N, min, max)
{
x = c(0)
y = c(0)
dfadjust = N - dfdata
# note: working out what the difference between N and df is (for the contrast between two groups, this is 2; for constraints where there is 4 groups this will be 3, etc.)
for(newN in min : max)
{
B = as.numeric(Bf(sd = sd*sqrt(N/newN), obtained, dfdata = newN-dfadjust, uniform, lower, upper, meanoftheory, sdtheory, tail)[3])
x= append(x,newN)
y= append(y,B)
output = cbind(x,y)
}
output = output[-1,]
return(output)
}Load dataset
The data file contains data for both input nouns and minimal exposure nouns. Analyses were conducted with non-alteranting nouns only, and therefore productions with alternating nouns have been removed from this data set. The data has also been filtered to remove trials not meeting the baseline criteria described in the paper (baseline criteria: trials were excluded if children had initially used an incorrect noun and been corrected by the experimenter, if they inserted an alternative word for a particle, or if they failed to include a particle. Children were not penalized for omitting to repeat the initial “moop”.)
alldata = read.csv("skewdata_Aug2016.csv")Experiment 1 (children), input nouns
select appropriate data
exp1.child.trained = subset(alldata, agegroup == "child" & old_me == "old")check proportion of missing data
For the lexical-skew and lexical-noskew conditions, the number of possible trials contributing to the analysis is 15 children * 4 nouns * 4 productions per noun making 240 trials; for the mixed condition it’s 15 children * 2 nouns * 4 productions per noun making 120 trials. There are no trials for the generalist condition. Tthe proportion of missing trials in each condition is therefore:
1-(round(table(exp1.child.trained$condition)/c(0,15*4*4,15*4*4,15*2*4),2)) ##
## gen_noskew lex_noskew lex_skew mixed
## 0.05 0.03 0.02get the means
means for table 3
kable(summarySE(aggregate(correct ~ participantid + condition , FUN = mean, data = exp1.child.trained), measurevar="correct", groupvars = "condition"), digits = 2)| condition | N | correct | sd | se | ci |
|---|---|---|---|---|---|
| lex_noskew | 15 | 0.74 | 0.16 | 0.04 | 0.09 |
| lex_skew | 15 | 0.96 | 0.06 | 0.02 | 0.03 |
| mixed | 15 | 0.70 | 0.26 | 0.07 | 0.14 |
Separating out the nouns with the majority/ minority particle in the lexicalist-skewed condition (note: majority = 1, minority = 0).
kable(summarySE(aggregate(correct ~ participantid + majority_det, FUN = mean, data = subset(exp1.child.trained, condition == "lex_skew")), measurevar="correct", groupvars = "majority_det"), digits = 2)| majority_det | N | correct | sd | se | ci |
|---|---|---|---|---|---|
| 0 | 15 | 0.90 | 0.23 | 0.06 | 0.13 |
| 1 | 15 | 0.99 | 0.03 | 0.01 | 0.02 |
statistical analyses
contrast between conditions 1
lme analysis all data
set up the data frame so that we can look at the contrasts bewteen lex_noskew condition and each of the other conditions using the lizContrasts helper function, and center the control variable using the lizCenter function.
exp1.child.trained= lizContrasts(exp1.child.trained, exp1.child.trained$condition, "lex_noskew")
exp1.child.trained = lizCenter(exp1.child.trained, list("correct_det"))run the lmer model:
child.trained.lmer = glmer(correct ~ (lex_noskew_VERSUS_mixed+ lex_noskew_VERSUS_lex_skew ) * correct_det.ct + (correct_det.ct|participantid),control=glmerControl(optimizer = "bobyqa"), family = binomial, data = exp1.child.trained)
kable(summary(child.trained.lmer)$coefficients,
digits = 3)| Estimate | Std. Error | z value | Pr(>|z|) | |
|---|---|---|---|---|
| (Intercept) | 2.413 | 0.328 | 7.350 | 0.000 |
| lex_noskew_VERSUS_mixed | -0.050 | 0.579 | -0.087 | 0.931 |
| lex_noskew_VERSUS_lex_skew | 3.114 | 0.741 | 4.204 | 0.000 |
| correct_det.ct | -0.348 | 0.706 | -0.492 | 0.622 |
| lex_noskew_VERSUS_mixed:correct_det.ct | 0.124 | 1.327 | 0.093 | 0.926 |
| lex_noskew_VERSUS_lex_skew:correct_det.ct | 0.705 | 1.600 | 0.440 | 0.660 |
BF analyses corresponding to lme analysis all data
For each contrast, we require an estimate of the mean difference between the conditions and of the corresponding standard error. These are taken from the model described above:
lex_noskew_VERSUS_mixed_child_trained_MEANDIFF = abs(summary(child.trained.lmer)$coefficients[2])
lex_noskew_VERSUS_mixed_child_trained_MEANDIFF## [1] 0.05044683lex_noskew_VERSUS_mixed_child_trained_SE = abs(summary(child.trained.lmer)$coefficients[8])
lex_noskew_VERSUS_mixed_child_trained_SE ## [1] 0.5792097lex_noskew_VERSUS_lex_skew_child_trained_MEANDIFF = abs(summary(child.trained.lmer)$coefficients[3])
lex_noskew_VERSUS_lex_skew_child_trained_MEANDIFF## [1] 3.11429lex_noskew_VERSUS_lex_skew_child_trained_SE = abs(summary(child.trained.lmer)$coefficients[9])
lex_noskew_VERSUS_lex_skew_child_trained_SE## [1] 0.7407162We now use these values in the Bf calculator along with the predicted difference of 2.757657 (obtained from equivalent analsyes over data from Wonnacott (2011))
Bf(sd=lex_noskew_VERSUS_mixed_child_trained_SE , obtained=lex_noskew_VERSUS_mixed_child_trained_MEANDIFF , dfdata = 28, uniform = 0, sdtheory = 2.757657, meanoftheory = 0, tail=1)## $LikelihoodTheory
## [1] 0.08815955
##
## $Likelihoodnull
## [1] 0.3938469
##
## [[3]]
## [1] 0.2238422Bf(sd=lex_noskew_VERSUS_lex_skew_child_trained_SE, obtained=lex_noskew_VERSUS_lex_skew_child_trained_MEANDIFF , dfdata = 28, uniform = 0, sdtheory = 2.757657, meanoftheory = 0, tail=1)## $LikelihoodTheory
## [1] 0.1142007
##
## $Likelihoodnull
## [1] 0.0003274736
##
## [[3]]
## [1] 348.7324comparing accuracy with majority and minority particle nouns within skewed language
lme analyis skewed data
select the subset of data from skewed langauge and center (i) the variable of interest (majority_det – which codes for whether this is a noun where the more frequent particle is the correct one) and (ii) the control variable
exp1.child.trained.skew = subset(exp1.child.trained, condition == "lex_skew")
exp1.child.trained.skew= lizCenter(exp1.child.trained.skew, list("correct_det", "majority_det"))Run an lme to see if participants are more accurate with the majority particle. Note that majority_det is within participants. The model with the interaction with the control variable and full random slopes structure didn’t converge and was systematically simplified.
Non-converging models:
child.trained.skew.lmer = glmer(correct ~ (majority_det.ct * correct_det.ct) + (majority_det.ct * correct_det.ct|participantid), control=glmerControl(optimizer = "bobyqa"), family = binomial, data = exp1.child.trained.skew)
child.trained.skew.lmer = glmer(correct ~ (majority_det.ct * correct_det.ct) + (majority_det.ct + correct_det.ct|participantid), control=glmerControl(optimizer = "bobyqa"), family = binomial, data = exp1.child.trained.skew)
child.trained.skew.lmer = glmer(correct ~ (majority_det.ct * correct_det.ct) + (majority_det.ct |participantid), control=glmerControl(optimizer = "bobyqa"), family = binomial, data = exp1.child.trained.skew)Converging model:
child.trained.skew.lmer = glmer(correct ~ (majority_det.ct + correct_det.ct) + (majority_det.ct |participantid), control=glmerControl(optimizer = "bobyqa"), family = binomial, data = exp1.child.trained.skew)
kable(summary(child.trained.skew.lmer)$coefficients,
digits = 3)| Estimate | Std. Error | z value | Pr(>|z|) | |
|---|---|---|---|---|
| (Intercept) | 5.610 | 1.475 | 3.803 | 0.000 |
| majority_det.ct | -1.855 | 3.666 | -0.506 | 0.613 |
| correct_det.ct | -2.525 | 2.299 | -1.098 | 0.272 |
BF analysis corresponding to lme analysis for skewed data (comparing accuracy with majority and minority particle nouns within skewed language)
We require an estimate of the mean difference between the conditions and corresponding standard error. These are taken from the model above:
majority_VERSUS_minority_child_trained_MEANDIFF = abs(summary(child.trained.skew.lmer)$coefficients[2])
majority_VERSUS_minority_child_trained_MEANDIFF ## [1] 1.85548majority_VERSUS_minority_child_trained_SE = abs(summary(child.trained.skew.lmer)$coefficients[5])
majority_VERSUS_minority_child_trained_SE ## [1] 3.666009We now use these values in the Bf calculator along with the predicted difference of 2.161343 (obtained from equivalent analsyes over data from Wonnacott (2011))
Bf(sd=majority_VERSUS_minority_child_trained_SE , obtained=majority_VERSUS_minority_child_trained_MEANDIFF , dfdata = 28, uniform = 0, sdtheory = 2.161343, meanoftheory = 0, tail=1)## $LikelihoodTheory
## [1] 0.371252
##
## $Likelihoodnull
## [1] 0.3464833
##
## [[3]]
## [1] 1.071486contrast between conditions 2
lme analysis - balanced data set
Remove two of the three majority particle nouns (specifically, those coded as 3 and 4) from the lex_skew condition. This will leave just two nouns - coded 1 and 2 - which go with the minority particle and majority particle respectively, so that majority and minority particle nouns contribute equally to the input.
exp1.child.trained$remove = 0
exp1.child.trained$remove[exp1.child.trained$condition == "lex_skew" & exp1.child.trained$noun == "trainednoun3" ]=1
exp1.child.trained$remove[exp1.child.trained$condition == "lex_skew" & exp1.child.trained$noun == "trainednoun4" ]=1
exp1.child.trained.balanced = subset(exp1.child.trained, remove == 0)table(exp1.child.trained.balanced\(noun,exp1.child.trained.balanced\)condition )
set up the data frame so that we can look at the contrasts bewteen lex_noskew condition and each of the other conditions using the lizContrasts helper function, and center the control variable using the lizCenter function
exp1.child.trained.balanced= lizContrasts(exp1.child.trained.balanced, exp1.child.trained.balanced$condition, "lex_noskew")
exp1.child.trained.balanced = lizCenter(exp1.child.trained.balanced, list("correct_det"))run the lmer model:
child.trained.lmer.balanced = glmer(correct ~ (lex_noskew_VERSUS_mixed+ lex_noskew_VERSUS_lex_skew ) * correct_det.ct + (correct_det.ct|participantid),control=glmerControl(optimizer = "bobyqa"), family = binomial, data = exp1.child.trained.balanced)
kable(summary(child.trained.lmer.balanced)$coefficients,
digits = 3)| Estimate | Std. Error | z value | Pr(>|z|) | |
|---|---|---|---|---|
| (Intercept) | 2.394 | 0.368 | 6.512 | 0.000 |
| lex_noskew_VERSUS_mixed | -0.033 | 0.626 | -0.053 | 0.957 |
| lex_noskew_VERSUS_lex_skew | 3.004 | 0.838 | 3.585 | 0.000 |
| correct_det.ct | 0.092 | 0.773 | 0.119 | 0.906 |
| lex_noskew_VERSUS_mixed:correct_det.ct | 0.181 | 1.389 | 0.130 | 0.896 |
| lex_noskew_VERSUS_lex_skew:correct_det.ct | 1.714 | 1.759 | 0.975 | 0.330 |
BF analyses corresponding to lme analysis with balanced data set
For the contrast between lex_skew and lex_noskew, we require an estimate of the mean difference between the conditions and of the corresponding standard error and this is taken from the model above
lex_noskew_VERSUS_lex_skew_child_trained_balanced_MEANDIFF = abs(summary(child.trained.lmer.balanced)$coefficients[3])
lex_noskew_VERSUS_lex_skew_child_trained_balanced_MEANDIFF## [1] 3.004239lex_noskew_VERSUS_lex_skew_child_trained_balanced_SE = abs(summary(child.trained.lmer.balanced)$coefficients[9])
lex_noskew_VERSUS_lex_skew_child_trained_balanced_SE## [1] 0.8378964We now use these values in the Bf calculator along with the predicted difference of 2.757657 (obtained from equivalent analsyes over data from Wonnacott (2011))
Bf(sd=lex_noskew_VERSUS_lex_skew_child_trained_balanced_SE, obtained=lex_noskew_VERSUS_lex_skew_child_trained_balanced_MEANDIFF , dfdata = 28, uniform = 0, sdtheory = 2.757657, meanoftheory = 0, tail=1)## $LikelihoodTheory
## [1] 0.1345978
##
## $Likelihoodnull
## [1] 0.0016508
##
## [[3]]
## [1] 81.53485Experiment 1 (children), minimal exposure nouns
select appropriate data
exp1.child.me = subset(alldata, agegroup == "child" & old_me == "m.e.")check proportion of missing data
In each condition, the number of possible trials contributing to the analysis is 15 children * 2 nouns * 4 productions per noun making 120 trials. The proportion of missing trials in each condition is therefore:
1-(round(table(exp1.child.me$condition)/c(15*2*4,15*2*4,15*2*4,15*2*4),2)) ##
## gen_noskew lex_noskew lex_skew mixed
## 0.05 0.05 0.06 0.05get the means
means for table 1
kable(summarySE(aggregate(correct ~ participantid + condition , FUN = mean, data = exp1.child.me), measurevar="correct", groupvars = "condition"), digits = 2)| condition | N | correct | sd | se | ci |
|---|---|---|---|---|---|
| gen_noskew | 15 | 0.58 | 0.27 | 0.07 | 0.15 |
| lex_noskew | 15 | 0.64 | 0.30 | 0.08 | 0.17 |
| lex_skew | 15 | 0.86 | 0.22 | 0.06 | 0.12 |
| mixed | 15 | 0.58 | 0.17 | 0.04 | 0.10 |
Separating out the nouns with the majority/ minority particle in the lexicalist-skewed condition.
kable(summarySE(aggregate(correct ~ participantid + majority_det, FUN = mean, data = subset(exp1.child.me, condition == "lex_skew")), measurevar="correct", groupvars = "majority_det"), digits = 2)| majority_det | N | correct | sd | se | ci |
|---|---|---|---|---|---|
| 0 | 15 | 0.77 | 0.42 | 0.11 | 0.23 |
| 1 | 15 | 0.92 | 0.22 | 0.06 | 0.12 |
statistical analyses
contrast between conditions
lme analysis
set up the data frame so that we can look at the contrasts bewteen lex_noskew condition and each of the other conditions using the lizContrasts helper function, and center the control variable using the lizCenter function.
exp1.child.me= lizContrasts4(exp1.child.me, exp1.child.me$condition, "lex_noskew")
exp1.child.me = lizCenter(exp1.child.me, list("correct_det"))run the lmer model:
```r child.me.lmer = glmer(correct ~ (lex_noskew_VERSUS_mixed+ lex_noskew_VERSUS_lex_skew + lex_noskew_VERSUS_gen_noskew ) * correct_det.ct + (correct_det.ct|participantid),control=glmerControl(optimizer = “bobyqa”), family = binomial, data = exp1.child.me)
kable(summary(child.me.lmer)$coefficients, digits = 3) ```
| Estimate | Std. Error | z value | Pr(>|z|) | |
|---|---|---|---|---|
| (Intercept) | 1.637 | 0.355 | 4.607 | 0.000 |
| lex_noskew_VERSUS_mixed | -0.488 | 0.821 | -0.595 | 0.552 |
| lex_noskew_VERSUS_lex_skew | 2.664 | 0.926 | 2.876 | 0.004 |
| lex_noskew_VERSUS_gen_noskew | -0.655 | 0.821 | -0.798 | 0.425 |
| correct_det.ct | -0.364 | 0.792 | -0.460 | 0.646 |
| lex_noskew_VERSUS_mixed:correct_det.ct | 0.379 | 1.927 | 0.197 | 0.844 |
| lex_noskew_VERSUS_lex_skew:correct_det.ct | -0.537 | 2.102 | -0.256 | 0.798 |
| lex_noskew_VERSUS_gen_noskew:correct_det.ct | -0.790 | 1.933 | -0.409 | 0.683 |
BF analyses corresponding to lme analysis all data
For each contrast, we require an estimate of the mean difference between the conditions and of the corresponding standard error. These are taken from the model described above:
r lex_noskew_VERSUS_mixed_child_me_MEANDIFF = abs(summary(child.me.lmer)$coefficients[2]) lex_noskew_VERSUS_mixed_child_me_MEANDIFF
## [1] 0.4882844
r lex_noskew_VERSUS_mixed_child_me_SE = abs(summary(child.me.lmer)$coefficients[10]) lex_noskew_VERSUS_mixed_child_me_SE
## [1] 0.8213186
r lex_noskew_VERSUS_lex_skew_child_me_MEANDIFF = abs(summary(child.me.lmer)$coefficients[3]) lex_noskew_VERSUS_lex_skew_child_me_MEANDIFF
## [1] 2.663743
r lex_noskew_VERSUS_lex_skew_child_me_SE = abs(summary(child.me.lmer)$coefficients[11]) lex_noskew_VERSUS_lex_skew_child_me_SE
## [1] 0.9260443
r lex_noskew_VERSUS_gen_noskew_child_me_MEANDIFF = abs(summary(child.me.lmer)$coefficients[4]) lex_noskew_VERSUS_gen_noskew_child_me_MEANDIFF
## [1] 0.6548326
r lex_noskew_VERSUS_gen_noskew_child_me_SE = abs(summary(child.me.lmer)$coefficients[12]) lex_noskew_VERSUS_gen_noskew_child_me_SE
## [1] 0.8208123
We now use these values in the Bf calculator along with the predicted difference of 2.757657 (obtained from equivalent analsyes over data from Wonnacott (2011))
Bf(sd=lex_noskew_VERSUS_mixed_child_me_SE , obtained=lex_noskew_VERSUS_mixed_child_me_MEANDIFF , dfdata = 28, uniform = 0, sdtheory = 2.757657, meanoftheory = 0, tail=1)## $LikelihoodTheory
## [1] 0.1603362
##
## $Likelihoodnull
## [1] 0.3296401
##
## [[3]]
## [1] 0.4863978Bf(sd=lex_noskew_VERSUS_lex_skew_child_me_SE, obtained=lex_noskew_VERSUS_lex_skew_child_me_MEANDIFF , dfdata = 28, uniform = 0, sdtheory = 2.757657, meanoftheory = 0, tail=1)## $LikelihoodTheory
## [1] 0.165898
##
## $Likelihoodnull
## [1] 0.009261317
##
## [[3]]
## [1] 17.91301Bf(sd=lex_noskew_VERSUS_gen_noskew_child_me_SE, obtained=lex_noskew_VERSUS_gen_noskew_child_me_MEANDIFF , dfdata = 28, uniform = 0, sdtheory = 2.757657, meanoftheory = 0, tail=1)## $LikelihoodTheory
## [1] 0.1718672
##
## $Likelihoodnull
## [1] 0.2854262
##
## [[3]]
## [1] 0.6021424BF power cacluation for minimal exposure lexical_unskewed versus mixed contrast
We found no substantial preference for either H1 over H0, or H0 over H1. We now consider how many more participants we might need in order to have evidence in each direction.
First, assume ACTUAL mean difference is ZERO: with this amount of variance, approx how many participants would we need to have substantial evidnece for the NULL (given that standard error scales with number of participants)?
x=Bf_powercalc(sd=lex_noskew_VERSUS_mixed_child_me_SE , obtained=0 , dfdata = 28, uniform = 0, sdtheory = 2.757657, meanoftheory = 0, tail=1,N=30, min=10, max=600)
plot(x)
abline(h=(1/3))
x[12:14,] ## x y
## [1,] 21 0.3398171
## [2,] 22 0.3328486
## [3,] 23 0.3262900suggests we need an N of 22 (note that this is LESS than the current N of 30)
Second, assume the ACTUAL mean difference is the current estimate: with this amount of variance, approx how many participants would we need to have substantial evidnece for H1 (given that standard error scales with number of participants)?
x=Bf_powercalc(sd=lex_noskew_VERSUS_lex_skew_child_me_SE, obtained=lex_noskew_VERSUS_lex_skew_child_me_MEANDIFF , dfdata = 28, uniform = 0, sdtheory = 2.757657, meanoftheory = 0, tail=1,N=30, min=10, max=600)
plot(x)
abline(h=(3))
x[514:516,] ## x y
## [1,] 523 5.099459e+26
## [2,] 524 5.754409e+26
## [3,] 525 6.493489e+26suggests we need an N of 524 (i.e. 262 per condition)
comparing accuracy with majority and minority particle nouns within skewed language
lme analyis skewed data
select the subset of data from skewed langauge and center (i) the variable of interest (majority_det – which codes for whether this is a noun where the more frequent particle is the correct one) and (ii) the control variable
exp1.child.me.skew = subset(exp1.child.me, condition == "lex_skew")
exp1.child.me.skew= lizCenter(exp1.child.me.skew, list("correct_det", "majority_det"))Run an lme to see if participants are more accurate with the majority particle. Note that majority_det is within participants. The model with the interaction with the control variable and full random slopes structure didn’t converge and was systematically simplified.
Non-converging models:
child.me.skew.lmer = glmer(correct ~ (majority_det.ct * correct_det.ct ) + (majority_det.ct* correct_det.ct|participantid), control=glmerControl(optimizer = "bobyqa"), family = binomial, data = exp1.child.me.skew)
child.me.skew.lmer = glmer(correct ~ (majority_det.ct * correct_det.ct ) + (majority_det.ct+ correct_det.ct|participantid), control=glmerControl(optimizer = "bobyqa"), family = binomial, data = exp1.child.me.skew)Converging model:
child.me.skew.lmer = glmer(correct ~ (majority_det.ct * correct_det.ct ) + (majority_det.ct|participantid), control=glmerControl(optimizer = "bobyqa"), family = binomial, data = exp1.child.me.skew)
kable(summary(child.me.skew.lmer)$coefficients,
digits = 3)| Estimate | Std. Error | z value | Pr(>|z|) | |
|---|---|---|---|---|
| (Intercept) | 9.636 | 2.448 | 3.936 | 0.000 |
| majority_det.ct | -2.186 | 4.751 | -0.460 | 0.646 |
| correct_det.ct | -0.712 | 3.292 | -0.216 | 0.829 |
| majority_det.ct:correct_det.ct | 1.650 | 6.361 | 0.259 | 0.795 |
BF analysis corresponding to lme analysis for skewed data (comparing accuracy with majority and minority particle nouns within skewed language)
We require an estimate of the mean difference between the conditions and corresponding standard error. These are taken from the model above:
majority_VERSUS_minority_child_me_MEANDIFF = abs(summary(child.me.skew.lmer)$coefficients[2])
majority_VERSUS_minority_child_me_MEANDIFF ## [1] 2.185635majority_VERSUS_minority_child_me_SE = abs(summary(child.me.skew.lmer)$coefficients[6])
majority_VERSUS_minority_child_me_SE ## [1] 4.751256We now use these values in the Bf calculator along with the predicted difference of 1.261336 (obtained from equivalent analsyes over data from Wonnacott (2011))
Bf(sd=majority_VERSUS_minority_child_me_SE , obtained=majority_VERSUS_minority_child_me_MEANDIFF , dfdata = 28, uniform = 0, sdtheory = 1.261336, meanoftheory = 0, tail=1)## $LikelihoodTheory
## [1] 0.3774785
##
## $Likelihoodnull
## [1] 0.3545033
##
## [[3]]
## [1] 1.064809Experiment 1 (children), reported awareness of lexicality
means for table 4 (awareness of lexicality)
trained nouns (aware = 1, unaware = 0)
kable(summarySE(aggregate(correct ~ participantid + condition + lexicallyaware, FUN = mean, data = exp1.child.trained), measurevar="correct", groupvars = c("condition", "lexicallyaware")), digits = 2)| condition | lexicallyaware | N | correct | sd | se | ci |
|---|---|---|---|---|---|---|
| lex_noskew | 0 | 12 | 0.73 | 0.16 | 0.04 | 0.10 |
| lex_noskew | 1 | 3 | 0.75 | 0.22 | 0.12 | 0.54 |
| lex_skew | 0 | 7 | 0.99 | 0.03 | 0.01 | 0.02 |
| lex_skew | 1 | 8 | 0.94 | 0.07 | 0.03 | 0.06 |
| mixed | 0 | 11 | 0.73 | 0.28 | 0.08 | 0.19 |
| mixed | 1 | 4 | 0.63 | 0.20 | 0.10 | 0.32 |
minimal exposure nouns
kable(summarySE(aggregate(correct ~ participantid + condition + lexicallyaware, FUN = mean, data = exp1.child.me), measurevar="correct", groupvars = c("condition", "lexicallyaware")), digits = 2)| condition | lexicallyaware | N | correct | sd | se | ci |
|---|---|---|---|---|---|---|
| gen_noskew | 0 | 11 | 0.62 | 0.25 | 0.07 | 0.17 |
| gen_noskew | 1 | 4 | 0.46 | 0.34 | 0.17 | 0.55 |
| lex_noskew | 0 | 12 | 0.63 | 0.29 | 0.09 | 0.19 |
| lex_noskew | 1 | 3 | 0.67 | 0.38 | 0.22 | 0.95 |
| lex_skew | 0 | 7 | 0.82 | 0.24 | 0.09 | 0.22 |
| lex_skew | 1 | 8 | 0.89 | 0.21 | 0.07 | 0.17 |
| mixed | 0 | 11 | 0.61 | 0.19 | 0.06 | 0.13 |
| mixed | 1 | 4 | 0.48 | 0.04 | 0.02 | 0.06 |
Experiment 2 (adults), trained nouns
select appropriate data
exp1.adult.trained = subset(alldata, agegroup == "adult" & old_me == "old")check proportion of missing data
For the lexical-skew and lexical-noskew conditions, the number of possible trials contributing to the analysis is 15 adults * 4 nouns * 4 productions per noun making 240 trials; for the mixed condition it’s 15 adults * 2 nouns * 4 productions per noun making 120 trials. There was no generalist condition for adults. The proportion of missing trials in each condition is therefore:
r 1-(round(table(exp1.adult.trained$condition)/c(0,15*4*4,15*4*4,15*2*4),2))
## ## gen_noskew lex_noskew lex_skew mixed ## 0 0 0
get the means
means for table 3
kable(summarySE(aggregate(correct ~ participantid + condition , FUN = mean, data = exp1.adult.trained), measurevar="correct", groupvars = "condition"), digits = 2)| condition | N | correct | sd | se | ci |
|---|---|---|---|---|---|
| lex_noskew | 15 | 0.78 | 0.24 | 0.06 | 0.13 |
| lex_skew | 15 | 0.93 | 0.14 | 0.04 | 0.08 |
| mixed | 15 | 0.61 | 0.17 | 0.04 | 0.09 |
Separating out the nouns with the majority/ minority particle in the lexicalist-skewed condition.
kable(summarySE(aggregate(correct ~ participantid + majority_det, FUN = mean, data = subset(exp1.adult.trained, condition == "lex_skew")), measurevar="correct", groupvars = "majority_det"), digits = 2)| majority_det | N | correct | sd | se | ci |
|---|---|---|---|---|---|
| 0 | 15 | 0.88 | 0.27 | 0.07 | 0.15 |
| 1 | 15 | 0.94 | 0.12 | 0.03 | 0.06 |
statistical analyses
contrast between conditions 1
lme analysis all data
set up the data frame so that we can look at the contrasts between the lex_noskew condition and each of the other conditions using the lizContrasts helper function, and center the control variable using the lizCenter function.
exp1.adult.trained= lizContrasts(exp1.adult.trained, exp1.adult.trained$condition, "lex_noskew")
exp1.adult.trained = lizCenter(exp1.adult.trained, list("correct_det"))run the lmer model:
```r adult.trained.lmer = glmer(correct ~ (lex_noskew_VERSUS_mixed+ lex_noskew_VERSUS_lex_skew ) * correct_det.ct + (correct_det.ct|participantid),control=glmerControl(optimizer = “bobyqa”), family = binomial, data = exp1.adult.trained)
kable(summary(adult.trained.lmer)$coefficients, digits = 3) ```
| Estimate | Std. Error | z value | Pr(>|z|) | |
|---|---|---|---|---|
| (Intercept) | 1.996 | 0.304 | 6.557 | 0.000 |
| lex_noskew_VERSUS_mixed | -1.296 | 0.617 | -2.099 | 0.036 |
| lex_noskew_VERSUS_lex_skew | 1.759 | 0.691 | 2.544 | 0.011 |
| correct_det.ct | -0.220 | 0.406 | -0.541 | 0.589 |
| lex_noskew_VERSUS_mixed:correct_det.ct | -0.447 | 0.625 | -0.715 | 0.475 |
| lex_noskew_VERSUS_lex_skew:correct_det.ct | -0.948 | 0.787 | -1.205 | 0.228 |
BF analyses corresponding to lme analysis all data
For each contrast, we require an estimate of the mean difference between the conditions and of the corresponding standard error. These are taken from the model described above:
r lex_noskew_VERSUS_mixed_adult_trained_MEANDIFF = abs(summary(adult.trained.lmer)$coefficients[2]) lex_noskew_VERSUS_mixed_adult_trained_MEANDIFF
## [1] 1.295501
r lex_noskew_VERSUS_mixed_adult_trained_SE = abs(summary(adult.trained.lmer)$coefficients[8]) lex_noskew_VERSUS_mixed_adult_trained_SE
## [1] 0.6171106
r lex_noskew_VERSUS_lex_skew_adult_trained_MEANDIFF = abs(summary(adult.trained.lmer)$coefficients[3]) lex_noskew_VERSUS_lex_skew_adult_trained_MEANDIFF
## [1] 1.759348
r lex_noskew_VERSUS_lex_skew_adult_trained_SE = abs(summary(adult.trained.lmer)$coefficients[9]) lex_noskew_VERSUS_lex_skew_adult_trained_SE
## [1] 0.6914691
We now use these values in the Bf calculator along with the predicted difference of 2.757657 (obtained from equivalent analsyes over data from Wonnacott (2011))
Bf(sd=lex_noskew_VERSUS_mixed_adult_trained_SE , obtained=lex_noskew_VERSUS_mixed_adult_trained_MEANDIFF , dfdata = 28, uniform = 0, sdtheory = 2.757657, meanoftheory = 0, tail=1)## $LikelihoodTheory
## [1] 0.15276
##
## $Likelihoodnull
## [1] 0.04748425
##
## [[3]]
## [1] 3.217067Bf(sd=lex_noskew_VERSUS_lex_skew_adult_trained_SE, obtained=lex_noskew_VERSUS_lex_skew_adult_trained_MEANDIFF , dfdata = 28, uniform = 0, sdtheory = 2.757657, meanoftheory = 0, tail=1)## $LikelihoodTheory
## [1] 0.1584024
##
## $Likelihoodnull
## [1] 0.01937472
##
## [[3]]
## [1] 8.175725comparing accuracy with majority and minority particle nouns within skewed language
lme analyis skewed data
select the subset of data from skewed langauge and center (i) the variable of interest (majority_det – which codes for whether this is a noun where the more frequent particle is the correct one) and (ii) the control variable
exp1.adult.trained.skew = subset(exp1.adult.trained, condition == "lex_skew")
exp1.adult.trained.skew= lizCenter(exp1.adult.trained.skew, list("correct_det", "majority_det"))Run an lme to see if participants are more accurate with the majority particle. Note that majority_det is within participants.
adult.trained.skew.lmer = glmer(correct ~ (majority_det.ct * correct_det.ct) + (majority_det.ct * correct_det.ct|participantid), control=glmerControl(optimizer = "bobyqa"), family = binomial, data = exp1.adult.trained.skew)
kable(summary(adult.trained.skew.lmer)$coefficients,
digits = 3)| Estimate | Std. Error | z value | Pr(>|z|) | |
|---|---|---|---|---|
| (Intercept) | 6.231 | 2.209 | 2.821 | 0.005 |
| majority_det.ct | 2.426 | 3.392 | 0.715 | 0.474 |
| correct_det.ct | 4.205 | 4.345 | 0.968 | 0.333 |
| majority_det.ct:correct_det.ct | 3.940 | 7.532 | 0.523 | 0.601 |
BF analysis corresponding to lme analysis for skewed data (comparing accuracy with majority and minority particle nouns within skewed language)
We require an estimate of the mean difference between the conditions and corresponding standard error. These are taken from the model above:
r majority_VERSUS_minority_adult_trained_MEANDIFF = abs(summary(adult.trained.skew.lmer)$coefficients[2]) majority_VERSUS_minority_adult_trained_MEANDIFF
## [1] 2.426323
r majority_VERSUS_minority_adult_trained_SE = abs(summary(adult.trained.skew.lmer)$coefficients[6]) majority_VERSUS_minority_adult_trained_SE
## [1] 3.391853
We now use these values in the Bf calculator along with the predicted difference of 2.161343 (obtained from equivalent analsyes over data from Wonnacott (2011))
Bf(sd=majority_VERSUS_minority_adult_trained_SE , obtained=majority_VERSUS_minority_adult_trained_MEANDIFF , dfdata = 28, uniform = 0, sdtheory = 2.161343, meanoftheory = 0, tail=1)## $LikelihoodTheory
## [1] 0.3596668
##
## $Likelihoodnull
## [1] 0.3040798
##
## [[3]]
## [1] 1.182804contrast between conditions 2
lme analysis - balanced data set
Remove two of the three majority particle nouns (specifically, those coded as 3 and 4) from the lex_skew condition. This will leave just two nouns - coded 1 and 2 - which go with the minority particle and majority particle respectively, so that majority and minority particle nouns contribute equally to the input.
exp1.adult.trained$remove = 0
exp1.adult.trained$remove[exp1.adult.trained$condition == "lex_skew" & exp1.adult.trained$noun == "trainednoun3" ]=1
exp1.adult.trained$remove[exp1.adult.trained$condition == "lex_skew" & exp1.adult.trained$noun == "trainednoun4" ]=1
exp1.adult.trained.balanced = subset(exp1.adult.trained, remove == 0)set up the data frame so that we can look at the contrasts bewteen lex_noskew condition and each of the other conditions using the lizContrasts helper function, and center the control variable using the lizCenter function
exp1.adult.trained.balanced= lizContrasts(exp1.adult.trained.balanced, exp1.adult.trained.balanced$condition, "lex_noskew")
exp1.adult.trained.balanced = lizCenter(exp1.adult.trained.balanced, list("correct_det"))run the lmer model:
```r adult.trained.lmer.balanced = glmer(correct ~ (lex_noskew_VERSUS_mixed+ lex_noskew_VERSUS_lex_skew ) * correct_det.ct + (correct_det.ct|participantid),control=glmerControl(optimizer = “bobyqa”), family = binomial, data = exp1.adult.trained.balanced)
kable(summary(adult.trained.lmer.balanced)$coefficients, digits = 3) ```
| Estimate | Std. Error | z value | Pr(>|z|) | |
|---|---|---|---|---|
| (Intercept) | 2.086 | 0.331 | 6.294 | 0.000 |
| lex_noskew_VERSUS_mixed | -1.241 | 0.565 | -2.196 | 0.028 |
| lex_noskew_VERSUS_lex_skew | 2.149 | 0.808 | 2.660 | 0.008 |
| correct_det.ct | -0.665 | 0.529 | -1.258 | 0.209 |
| lex_noskew_VERSUS_mixed:correct_det.ct | -0.322 | 0.660 | -0.487 | 0.626 |
| lex_noskew_VERSUS_lex_skew:correct_det.ct | -1.955 | 1.262 | -1.549 | 0.121 |
BF analyses corresponding to lme analysis with balanced data set
For the contrast between lex_skew and lex_noskew, we require an estimate of the mean difference between the conditions and of the corresponding standard error and this is taken from the model above
lex_noskew_VERSUS_lex_skew_adult_trained_balanced_MEANDIFF = abs(summary(adult.trained.lmer.balanced)$coefficients[3])
lex_noskew_VERSUS_lex_skew_adult_trained_balanced_MEANDIFF## [1] 2.148548lex_noskew_VERSUS_lex_skew_adult_trained_balanced_SE = abs(summary(adult.trained.lmer.balanced)$coefficients[9])
lex_noskew_VERSUS_lex_skew_adult_trained_balanced_SE## [1] 0.8077We now use these values in the Bf calculator along with the predicted difference of 2.757657 (obtained from equivalent analsyes over data from Wonnacott (2011))
Bf(sd=lex_noskew_VERSUS_lex_skew_adult_trained_balanced_SE, obtained=lex_noskew_VERSUS_lex_skew_adult_trained_balanced_MEANDIFF , dfdata = 28, uniform = 0, sdtheory = 2.757657, meanoftheory = 0, tail=1)## $LikelihoodTheory
## [1] 0.1679098
##
## $Likelihoodnull
## [1] 0.01507205
##
## [[3]]
## [1] 11.14048Experiment 2 (adults), minimal exposure nouns
select appropriate data
exp1.adult.me = subset(alldata, agegroup == "adult" & old_me == "m.e.")check proportion of missing data
In each condition, the number of possible trials contributing to the analysis is 15 children * 2 nouns * 4 productions per noun making 120 trials. There is no generalist condition for adults. The proportion of missing trials in each condition is therefore:
r 1-(round(table(exp1.adult.me$condition)/c(0,15*2*4,15*2*4,15*2*4),2))
## ## gen_noskew lex_noskew lex_skew mixed ## 0.00 0.00 0.01
get the means
means for table 1
kable(summarySE(aggregate(correct ~ participantid + condition , FUN = mean, data = exp1.adult.me), measurevar="correct", groupvars = "condition"), digits = 2)| condition | N | correct | sd | se | ci |
|---|---|---|---|---|---|
| lex_noskew | 15 | 0.89 | 0.27 | 0.07 | 0.15 |
| lex_skew | 15 | 0.97 | 0.13 | 0.03 | 0.07 |
| mixed | 15 | 0.84 | 0.26 | 0.07 | 0.14 |
Separating out the nouns with the majority/ minority particle in the lexicalist-skewed condition.
kable(summarySE(aggregate(correct ~ participantid + majority_det, FUN = mean, data = subset(exp1.adult.me, condition == "lex_skew")), measurevar="correct", groupvars = "majority_det"), digits = 2)| majority_det | N | correct | sd | se | ci |
|---|---|---|---|---|---|
| 0 | 15 | 1.00 | 0.00 | 0.00 | 0.00 |
| 1 | 15 | 0.93 | 0.26 | 0.07 | 0.14 |
statistical analyses
contrast between conditions
lme analysis
set up the data frame so that we can look at the contrasts bewteen lex_noskew condition and each of the other conditions using the lizContrasts helper function, and center the control variable using the lizCenter function.
exp1.adult.me= lizContrasts(exp1.adult.me, exp1.adult.me$condition, "lex_noskew")
exp1.adult.me = lizCenter(exp1.adult.me, list("correct_det"))run the lmer model:
```r adult.me.lmer = glmer(correct ~ (lex_noskew_VERSUS_mixed+ lex_noskew_VERSUS_lex_skew ) * correct_det.ct + (correct_det.ct|participantid),control=glmerControl(optimizer = “bobyqa”), family = binomial, data = exp1.adult.me)
kable(summary(adult.me.lmer)$coefficients, digits = 3) ```
| Estimate | Std. Error | z value | Pr(>|z|) | |
|---|---|---|---|---|
| (Intercept) | 12.843 | 14.428 | 0.890 | 0.373 |
| lex_noskew_VERSUS_mixed | -0.704 | 1.991 | -0.354 | 0.724 |
| lex_noskew_VERSUS_lex_skew | 11.826 | 43.104 | 0.274 | 0.784 |
| correct_det.ct | 7.554 | 28.962 | 0.261 | 0.794 |
| lex_noskew_VERSUS_mixed:correct_det.ct | -0.150 | 3.524 | -0.043 | 0.966 |
| lex_noskew_VERSUS_lex_skew:correct_det.ct | 22.292 | 86.282 | 0.258 | 0.796 |
BF analyses corresponding to lme analysis all data
For each contrast, we require an estimate of the mean difference between the conditions and of the corresponding standard error. These are taken from the model described above:
r lex_noskew_VERSUS_mixed_adult_me_MEANDIFF = abs(summary(adult.me.lmer)$coefficients[2]) lex_noskew_VERSUS_mixed_adult_me_MEANDIFF
## [1] 0.7039988
r lex_noskew_VERSUS_mixed_adult_me_SE = abs(summary(adult.me.lmer)$coefficients[8]) lex_noskew_VERSUS_mixed_adult_me_SE
## [1] 1.991358
r lex_noskew_VERSUS_lex_skew_adult_me_MEANDIFF = abs(summary(adult.me.lmer)$coefficients[3]) lex_noskew_VERSUS_lex_skew_adult_me_MEANDIFF
## [1] 11.82603
r lex_noskew_VERSUS_lex_skew_adult_me_SE = abs(summary(adult.me.lmer)$coefficients[9]) lex_noskew_VERSUS_lex_skew_adult_me_SE
## [1] 43.10405
We now use these values in the Bf calculator along with the predicted difference of 2.757657 (obtained from equivalent analsyes over data from Wonnacott (2011))
Bf(sd=lex_noskew_VERSUS_mixed_adult_me_SE , obtained=lex_noskew_VERSUS_mixed_adult_me_MEANDIFF , dfdata = 28, uniform = 0, sdtheory = 2.757657, meanoftheory = 0, tail=1)## $LikelihoodTheory
## [1] 0.2781582
##
## $Likelihoodnull
## [1] 0.3706698
##
## [[3]]
## [1] 0.7504205Bf(sd=lex_noskew_VERSUS_lex_skew_adult_me_SE, obtained=lex_noskew_VERSUS_lex_skew_adult_me_MEANDIFF , dfdata = 28, uniform = 0, sdtheory = 2.757657, meanoftheory = 0, tail=1)## $LikelihoodTheory
## [1] 0.3858014
##
## $Likelihoodnull
## [1] 0.3803004
##
## [[3]]
## [1] 1.014465comparing accuracy with majority and minority particle nouns within skewed language
lme analyis skewed data
select the subset of data from skewed langauge and center (i) the variable of interest (majority_det – which codes for whether this is a noun where the more frequent particle is the correct one) and (ii) the control variable
exp1.adult.me.skew = subset(exp1.adult.me, condition == "lex_skew")
exp1.adult.me.skew= lizCenter(exp1.adult.me.skew, list("correct_det", "majority_det"))Run an lme to see if participants are more accurate with the majority particle. Note that majority_det is within participants.
adult.me.skew.lmer = glmer(correct ~ (majority_det.ct * correct_det.ct) + (majority_det.ct * correct_det.ct|participantid), control=glmerControl(optimizer = "bobyqa"), family = binomial, data = exp1.adult.me.skew)
kable(summary(adult.me.skew.lmer)$coefficients,
digits = 3)| Estimate | Std. Error | z value | Pr(>|z|) | |
|---|---|---|---|---|
| (Intercept) | 26.934 | 16.232 | 1.659 | 0.097 |
| majority_det.ct | -12.148 | 28.535 | -0.426 | 0.670 |
| correct_det.ct | 0.139 | 29.639 | 0.005 | 0.996 |
| majority_det.ct:correct_det.ct | 23.984 | 25.044 | 0.958 | 0.338 |
BF analysis corresponding to lme analysis for skewed data (comparing accuracy with majority and minority particle nouns within skewed language)
We require an estimate of the mean difference between the conditions and corresponding standard error. These are taken from the model above:
r majority_VERSUS_minority_adult_me_MEANDIFF = abs(summary(adult.me.skew.lmer)$coefficients[2]) majority_VERSUS_minority_adult_me_MEANDIFF
## [1] 12.14758
r majority_VERSUS_minority_adult_me_SE = abs(summary(adult.me.skew.lmer)$coefficients[6]) majority_VERSUS_minority_adult_me_SE
## [1] 28.53479
We now use these values in the Bf calculator along with the predicted difference of 1.261336 (obtained from equivalent analsyes over data from Wonnacott (2011))
Bf(sd=majority_VERSUS_minority_adult_me_SE , obtained=majority_VERSUS_minority_adult_me_MEANDIFF , dfdata = 28, uniform = 0, sdtheory = 1.261336, meanoftheory = 0, tail=1)## $LikelihoodTheory
## [1] 0.3646299
##
## $Likelihoodnull
## [1] 0.3600854
##
## [[3]]
## [1] 1.012621Experiment 2 (adults), reported awareness of lexicality
means for table 5 (awareness of lexicality)
trained nouns (aware = 1, unaware = 0)
kable(summarySE(aggregate(correct ~ participantid + condition + lexicallyaware, FUN = mean, data = exp1.adult.trained), measurevar="correct", groupvars = c("condition", "lexicallyaware")), digits = 2)| condition | lexicallyaware | N | correct | sd | se | ci |
|---|---|---|---|---|---|---|
| lex_noskew | 1 | 15 | 0.78 | 0.24 | 0.06 | 0.13 |
| lex_skew | 0 | 1 | 1.00 | NA | NA | NA |
| lex_skew | 1 | 14 | 0.92 | 0.14 | 0.04 | 0.08 |
| mixed | 0 | 6 | 0.60 | 0.18 | 0.08 | 0.19 |
| mixed | 1 | 9 | 0.61 | 0.17 | 0.06 | 0.13 |
minimal exposure nouns
means for table 5 (awareness of lexicality)
kable(summarySE(aggregate(correct ~ participantid + condition + lexicallyaware, FUN = mean, data = exp1.adult.me), measurevar="correct", groupvars = c("condition", "lexicallyaware")), digits = 2)| condition | lexicallyaware | N | correct | sd | se | ci |
|---|---|---|---|---|---|---|
| lex_noskew | 1 | 15 | 0.89 | 0.27 | 0.07 | 0.15 |
| lex_skew | 0 | 1 | 1.00 | NA | NA | NA |
| lex_skew | 1 | 14 | 0.96 | 0.13 | 0.04 | 0.08 |
| mixed | 0 | 6 | 0.81 | 0.35 | 0.14 | 0.37 |
| mixed | 1 | 9 | 0.86 | 0.19 | 0.06 | 0.15 |
statistical comparison
Comparing the extent of reported awareness of lexical patterns in the different conditions using fishers exact test
awaretable = with(aggregate(correct ~ participantid + condition + lexicallyaware, FUN = mean, data = exp1.adult.me),
table(condition, lexicallyaware))
kable(awaretable)| 0 | 1 | |
|---|---|---|
| gen_noskew | 0 | 0 |
| lex_noskew | 0 | 15 |
| lex_skew | 1 | 14 |
| mixed | 6 | 9 |
# mixed versus lexicalist-unskewed
fisher.test(rbind(awaretable[2,], awaretable[4,]))##
## Fisher's Exact Test for Count Data
##
## data: rbind(awaretable[2, ], awaretable[4, ])
## p-value = 0.01686
## alternative hypothesis: true odds ratio is not equal to 1
## 95 percent confidence interval:
## 0.0000000 0.6740343
## sample estimates:
## odds ratio
## 0# mixed versus lexicalist-skewed
fisher.test(rbind(awaretable[3,], awaretable[4,]))##
## Fisher's Exact Test for Count Data
##
## data: rbind(awaretable[3, ], awaretable[4, ])
## p-value = 0.08008
## alternative hypothesis: true odds ratio is not equal to 1
## 95 percent confidence interval:
## 0.002183889 1.190995284
## sample estimates:
## odds ratio
## 0.1151766