I should like to calculate - by bootstrapping Krippendorff's Alpha outcomes - a 95% confidence interval for Krippendorff's Alpha coefficient of Raters Reliability using R package irr.
Let's use "C data from Krippendorff" in the package irr and the R script for calculating Krippendorff's Alpha once:
# the "C" data from Krippendorff
#rater per row; rated subject per column; NAs allowed
library(irr)
nmm<-matrix(c(1,1,NA,1,2,2,3,2,3,3,3,3,3,3,3,3,2,2,2,2,1,2,3,4,4,4,4,4,
1,1,2,1,2,2,2,2,NA,5,5,5,NA,NA,1,1,NA,NA,3,NA),nrow=4)
kripp.alpha(nmm,"ordinal")
You can use the boot function from the boot package to bootstrap values. Here I'll bootstrap the set of subjects but keep the raters fixed:
library(boot)
library(irr)
ka <- function(data, indices) kripp.alpha(nmm[,indices], "ordinal")$value
b <- boot(seq(ncol(nmm)), ka, 1000)
Now you can use the boot.ci function to compute a 95% confidence interval for the bootstrapped value; I'll use the percentile confidence interval, but others are available (check out ?boot.ci):
boot.ci(b, type="perc")
# BOOTSTRAP CONFIDENCE INTERVAL CALCULATIONS
# Based on 1000 bootstrap replicates
#
# CALL :
# boot.ci(boot.out = b, type = "perc")
#
# Intervals :
# Level Percentile
# 95% ( 0.4297, 1.0000 )
# Calculations and Intervals on Original Scale
Unfortunately, the bootstrap solution given by josliber doesn't do what you think it does. The problem is that boot() expects data in an nXm matrix while kripp.alpha() expects data in an mXn matrix. The solution given will run, as shown, but the resampling being done is not by subjects, but by raters, so with 4 raters in the example data set we have a small number of possible samples, with the possibility that the resampled set will come from a single rater (hence the conf interval includes 1.0).
One solution is to keep your data in the nXm form that boot uses, and add a matrix transpose before giving it to kripp.alpha().
alpha.boot <- function(data,x) {
d <- t(data[x,])
kripp.alpha(d,method="nominal")$value
}
Probably too late to be helpful for Paul, but for future reference note that none of the proposed methods is consistent with the bootstrap algorithm described by Klaus Krippendorff (http://web.asc.upenn.edu/usr/krippendorff/boot.c-Alpha.pdf).
The repeated samples do neither draw raters nor units, but "hypothetical reliability data from the matrix of observed coincidences among pairs of values assigned to units in independent replications" (Krippendorff 2016). Thus, conventional bootstrap implementations will not give the answer intended by Krippendorff.
Best
Daniel
Related
I'm wondering how to use replication weights and the confint implementation of the survey package to construct bootstrap confidence intervals/standard errors.
Looking at the survey package's implementation of confint, it seems as though it's simply taking the standard error of the theta list generated after replication, and multiplying it by the statistic corresponding to a given alpha range.
But that doesn't really correspond with any bootstrap implementation I'm aware of. Typically, you'd instead be using a percentile of the distribution of theta to get the sample mean's distribution's confidence interval. The T and BCa intervals are another matter.
Here is my R Code. I have not used the provided weights, instead letting repweights weights be generated as "sample with replacement" weights with equal probability.
data(api)
d <- apiclus1 %>% select(fpc, dnum,api99)
dclus1<-svydesign(id=~dnum, data=d, fpc=~fpc)
rclus1<-as.svrepdesign(dclus1,type="bootstrap", replicates=100)
To test the confidence intervals, we can use:
test_mean <- svymean(~api99, rclus1)
confint(test_mean, df=degf(rclus1))
confint(test_mean, df=degf(rclus1)) - mean(d$api99)
Which results:
2.5 % 97.5 %
api99 554.2971 659.6592
2.5 % 97.5 %
api99 -52.68107 52.68107
So clearly the interval is symmetric, which defeats some of the purpose of using the bootstrap.
So let's try this:
test_bs <- withReplicates(rclus1, function(w, data) weighted.mean(data$api99, w), return.replicates=T)
This will bootstrap the replicates, where the weights are the repweights (which I assume are with replacement weights). Here are the intervals using BCa intervals on the replications:
bca(test$replicates) - mean(d$api99)
-43.2878 49.1148
Clearly not symmetric.
Using percentile intervals:
c(quantile(test$replicates, 0.025),quantile(test$replicates, 0.975)) - mean(d$api99)
2.5% 97.5%
-45.50944 48.06085
Valliant implements percentile intervals this way, which should be equivalent to my percentile intervals:
smho.boot.a <- as.svrepdesign(design = smho.dsgn,
type = "subbootstrap",
replicates = 500)
# total & CI for EOYCNT based on RW bootstrap
a1 <- svytotal( ̃EOYCNT, design = smho.boot.a,
na.rm=TRUE,
return.replicates = TRUE)
# Compute CI based on bootstrap percentile method.
ta1 <- quantile(a1$replicates, c(0.025, 0.975))
I'm looking for clarifications on
a. How to construct bootstrap CIs with survey package for statistics of interest
b. If my withReplication implementation of the bootstrap is correct
The percentile stuff doesn't actually work with multistage survey data (or, at least, it isn't known to work (or, at least, it isn't known to me to work)). Survey bootstraps just estimate the variance. You don't get the higher order accuracy for smooth functions that way, but you do get asymptotically the right variance. To get asymmetric intervals you need to bootstrap some suitable function of the parameter, as svyciprop and svyquantile (and in a sense svyglm) do.
If you assume simple random sampling with replacement of clusters then you could make the percentile bootstrap and extensions work, but that's not a common structure for real-world surveys (and it doesn't really need the survey package)
I'm sure the maintainer of the package would be happy to implement a bootstrap that gave better asymmetric intervals and worked for multistage samples, if someone pointed out suitable references to him.
I am using the useful gratia package by Gavin Simpson to extract the difference in two smooths for two different levels of a factor variable. The smooths are generated by the wonderful mgcv package. For example
library(mgcv)
library(gratia)
m1 <- gam(outcome ~ s(dep_var, by = fact_var) + fact_var, data = my.data)
diff1 <- difference_smooths(m1, smooth = "s(dep_var)")
draw(diff1)
This give me a graph of the difference between the two smooths for each level of the "by" variable in the gam() call. The graph has a shaded 95% credible interval (CI) for the difference.
Statistical significance, or areas of statistical significance at the 0.05 level, is assessed by whether or where the y = 0 line crosses the CI, where the y axis represents the difference between the smooths.
Here is an example from Gavin's site where the "by" factor variable had 3 levels.
The differences are clearly statistically significant (at 0.05) over nearly all of the graphs.
Here is another example I have generated using a "by" variable with 2 levels.
The difference in my example is clearly not statistically significant anywhere.
In the mgcv package, an approximate p value is outputted for a smooth fit that tests the null hypothesis that the coefficients are all = 0, based on a chi square test.
My question is, can anyone suggest a way of calculating a p value that similarly assesses the difference between the two smooths instead of solely relying on graphical evidence?
The output from difference_smooths() is a data frame with differences between the smooth functions at 100 points in the range of the smoothed variable, the standard error for the difference and the upper and lower limits of the CI.
Here is a link to the release of gratia 0.4 that explains the difference_smooths() function
enter link description here
but gratia is now at version 0.6
enter link description here
Thanks in advance for taking the time to consider this.
Don
One way of getting a p value for the interaction between the by factor variables is to manipulate the difference_smooths() function by activating the ci_level option. Default is 0.95. The ci_level can be manipulated to find a level where the y = 0 is no longer within the CI bands. If for example this occurred when ci_level = my_level, the p value for testing the hypothesis that the difference is zero everywhere would be 1 - my_level.
This is not totally satisfactory. For example, it would take a little manual experimentation and it may be difficult to discern accurately when zero drops out of the CI. Although, a function could be written to search the accompanying data frame that is outputted with difference_smooths() as the ci_level is varied. This is not totally satisfactory either because the detection of a non-zero CI would be dependent on the 100 points chosen by difference_smooths() to assess the difference between the two curves. Then again, the standard errors are approximate for a GAM using mgcv, so that shouldn't be too much of a problem.
Here is a graph where the zero first drops out of the CI.
Zero dropped out at ci_level = 0.88 and was still in the interval at ci_level = 0.89. So an approxiamte p value would be 1 - 0.88 = 0.12.
Can anyone think of a better way?
Reply to Gavin Simpson's comments Feb 19
Thanks very much Gavin for taking the time to make your comments.
I am not sure if using the criterion, >= 0 (for negative diffs), is a good way to go. Because of the draws from the posterior, there is likely to be many diffs that meet this criterion. I am interpreting your criterion as sample the posterior distribution and count how many differences meet the criterion, calculate the percentage and that is the p value. Correct me if I have misunderstood. Using this approach, I consistently got p values at around 0.45 - 0.5 for different gam models, even when it was clear the difference in the smooths should be statistically significant, at least at p = 0.05, because the confidence band around the smooth did not contain zero at a number of points.
Instead, I was thinking perhaps it would be better to compare the means of the posterior distribution of each of the diffs. For example
# get coefficients for the by smooths
coeff.level1 <- coef(gam.model1)[31:38]
coeff.level0 <- coef(gam.model1)[23:30]
# these indices are specific to my multi-variable gam.model1
# in my case 8 coefficients per smooth
# get posterior coefficients variances for the by smooths' coefficients
vp_level1 <- gam.model1$Vp[31:38, 31:38]
vp_level0 <- gam.model1$Vp[23:30, 23:30]
#run the simulation to get the distribution of each
#difference coefficient using the joint variance
library(MASS)
no.draws = 1000
sim <- mvrnorm(n = no.draws, (coeff.level1 - coeff.level0),
(vp_level1 + vp_level0))
# sim is a no.draws X no. of coefficients (8 in my case) matrix
# put the results into a data.frame.
y.group <- data.frame(y = as.vector(sim),
group = c(rep(1,no.draws), rep(2,no.draws),
rep(3,no.draws), rep(4,no.draws),
rep(5,no.draws), rep(6,no.draws),
rep(7,no.draws), rep(8,no.draws)) )
# y has the differences sampled from their posterior distributions.
# group is just a grouping name for the 8 sets of differences,
# (one set for each difference in coefficients)
# compare means with a linear regression
lm.test <- lm(y ~ as.factor(group), data = y.group)
summary(lm.test)
# The p value for the F statistic tells you how
# compatible the data are with the null hypothesis that
# all the group means are equal to each other.
# Same F statistic and p value from
anova(lm.test)
One could argue that if all coefficients are not equal to each other then they all can't be equal to zero but that isn't what we want here.
The basis of the smooth tests of fit given by summary(mgcv::gam.model1)
is a joint test of all coefficients == 0. This would be from a type of likelihood ratio test where model fit with and without a term are compared.
I would appreciate some ideas how to do this with the difference between two smooths.
Now that I got this far, I had a rethink of your original suggestion of using the criterion, >= 0 (for negative diffs). I reinterpreted this as meaning for each simulated coefficient difference distribution (in my case 8), count when this occurs and make a table where each row (my case, 8) is for one of these distributions with two columns holding this count and (number of simulation draws minus count), Then on this table run a chi square test. When I did this, I got a very low p value when I believe I shouldn't have as 0 was well within the smooth difference CI across almost all the levels of the exposure. Maybe I am still misunderstanding your suggestion.
Follow up thought Feb 24
In a follow up thought, we could create a variable that represents the interaction between the by factor and continuous variable
library(dplyr)
my.dat <- my.dat %>% mutate(interact.var =
ifelse(factor.2levels == "yes", 1, 0)*cont.var)
Here I am assuming that factor.2levels has the levels ("no", "yes"), and "no" is the reference level. The ifelse function creates a dummy variable which is multiplied by the continuous variable to generate the interactive variable.
Then we place this interactive variable in the GAM and get the usual statistical test for fit, that is, testing all the coefficients == 0.
#GavinSimpson actually posted a method of how to get the difference between two smooths and assess its statistical significance here in 2017. Thanks to Matteo Fasiolo for pointing me in that direction.
In that approach, the by variable is converted to an ordered categorical variable which causes mgcv::gam to produce difference smooths in comparison to the reference level. Statistical significance for the difference smooths is then tested in the usual way with the summary command for the gam model.
However, and correct me if I have misunderstood, the ordered factor approach causes the smooth for the main effect to now be the smooth for the reference level of the ordered factor.
The approach I suggested, see the main post under the heading, Follow up thought Feb 24, where the interaction variable is created, gives an almost identical result for the p value for the difference smooth but does not change the smooth for the main effect. It also does not change the intercept and the linear term for the by categorical variable which also both changed with the ordered variable approach.
Maybe a dumb question, but I just started, so thanks for any help.
I created 99% confidence interval for a proportion, but I'm not sure if it is correct, how can I make sure, (when we're calculating confidence interval for mean, we're using t-score, and we can test the results by using t.test function and degrees of freedom)
Is there any similar function to do the same thing for z, proportions? or I can do the same thing by using t.test?
There are a number of functions in R for computing confidence intervals for proportions. Personally, I like to use the binomCI function from the MKinfer package:
install.packages("MKinfer")
library(MKinfer)
x <- 50 # The number of "successes" in your data
n <- 100 # The number of observations in your data
binomCI(x, n, conf.level = 0.99, method = "wald")
Note however that the so-called Wald interval (which is presented in most introductory texts on statistics) that you probably computed usually is a poor choice. See this link for some other alternatives available through binomCI.
I'm new to R and trying to learn stats..
Here is one practice question that I'm trying to figure out
How should I use R code to create a function based on this math equation?
I have a dataframe like this
the "exposed" column from the df contains two groups, one is called"Test Group (Exposed)" the other one is called "Control Group". So the math function is referring to these two groups.
In another practice I have these codes here to calculate the confidence interval
# sample size
# OK for non normal data if n > 30
n <- 150
# calculate the mean & standard deviation
will_mean <- mean(will_sample)
will_s <- sd(will_sample)
# normal quantile function, assuming mean has a normal distribution:
qnorm(p=0.975, mean=0, sd=1) # 97.5th percentile for a N(0,1) distribution
# a.k.a. Z = 1.96 from the standard normal distribution
# calculate standard error of the mean
# standard error of the mean = mean +/- critical value x (s/sqrt(n))
# "q" functions in r give the value of the statistic at a given quantile
critical_value <- qt(p=0.975, df=n-1)
error <- critical_value * will_s/sqrt(n)
# confidence inverval
will_mean - error
will_mean + error
but I'm not sure how to do the exposed 2 groups
Don't worry it's quite easy if you have experience in at least one programming language, R is quite trivial.
The only remarkable difference between R and most of other programming languanges is that R was developed for statistical purposes.
You can compute what is the quantile for a certain significance level α (reminds to divide it by 2 for your formula) by using the function qnorm(). By default it is set up for standardized normal distribution, like in your case, but you can get more details using the documentation, reachable by the command ?qnorm().
Actually in the exercise you are not required to compute it, since you have to pass it as argument, but in reality you need to.
The code should be something like:
conf <- function(p1,p2,n1,n2,z){
part = z*(p1*(1-p1)/n1+p2*(1-p2)/n2)**(1/2)
return(c(p1-p2-part,
p1-p2+part))
}
Is it possible to/how can I generate a beta-binomial distribution from an existing vector?
My ultimate goal is to generate a beta-binomial distribution from the below data and then obtain the 95% confidence interval for this distribution.
My data are body condition scores recorded by a veterinarian. The values of body condition range from 0-5 in increments of 0.5. It has been suggested to me here that my data follow a beta-binomial distribution, discrete values with a restricted range.
set1 <- as.data.frame(c(3,3,2.5,2.5,4.5,3,2,4,3,3.5,3.5,2.5,3,3,3.5,3,3,4,3.5,3.5,4,3.5,3.5,4,3.5))
colnames(set1) <- "numbers"
I see that there are multiple functions which appear to be able to do this, betabinomial() in VGAM and rbetabinom() in emdbook, but my stats and coding knowledge is not yet sufficient to be able to understand and implement the instructions provided on the function help pages, at least not in a way that has been helpful for my intended purpose yet.
We can look at the distribution of your variables, y-axis is the probability:
x1 = set1$numbers*2
h = hist(x1,breaks=seq(0,10))
bp = barplot(h$counts/length(x1),names.arg=(h$mids+0.5)/2,ylim=c(0,0.35))
You can try to fit it, but you have too little data points to estimate the 3 parameters need for a beta binomial. Hence I fix the probability so that the mean is the mean of your scores, and looking at the distribution above it seems ok:
library(bbmle)
library(emdbook)
library(MASS)
mtmp <- function(prob,size,theta) {
-sum(dbetabinom(x1,prob,size,theta,log=TRUE))
}
m0 <- mle2(mtmp,start=list(theta=100),
data=list(size=10,prob=mean(x1)/10),control=list(maxit=1000))
THETA=coef(m0)[1]
We can also use a normal distribution:
normal_fit = fitdistr(x1,"normal")
MEAN=normal_fit$estimate[1]
SD=normal_fit$estimate[2]
Plot both of them:
lines(bp[,1],dbetabinom(1:10,size=10,prob=mean(x1)/10,theta=THETA),
col="blue",lwd=2)
lines(bp[,1],dnorm(1:10,MEAN,SD),col="orange",lwd=2)
legend("topleft",c("normal","betabinomial"),fill=c("orange","blue"))
I think you are actually ok with using a normal estimation and in this case it will be:
normal_fit$estimate
mean sd
6.560000 1.134196