I am trying to summarise data from a household survey and as such most of my data is categorical (factor) data. I was looking to summarise it with plots of frequencies of responses to certain questions (e.g., a bar plot of percentages of households answering certain questions, with error bars showing confidence intervals). I found this excellent tutorial which I had thought was the answer to my prayers (http://www.cookbook-r.com/Manipulating_data/Summarizing_data/) but turns out this is only going to help with continuous data.
What I need is something similar that will allow me to calculate proportions of counts and standard errors / confidence intervals of these proportions.
Essentially I want to be able to produce summary tables that look like this for each of the questions asked in my survey data:
# X5employf X5employff N(count) proportion SE of prop. ci of prop
# 1 1 20 0.64516129 ? ?
# 1 2 1 0.03225806 ? ?
# 1 3 9 0.29032258 ? ?
# 1 NA 1 0.290322581 ? ?
# 2 4 1 0.1 ? ?
structure(list(X5employf = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L), .Label = c("1", "2", "3"), class = "factor"), X5employff = structure(c(1L, 2L, 3L, NA, 4L, 5L, 6L, 7L, 8L, 4L, 5L, 6L, 7L), .Label = c("1", "2", "3", "4", "5", "6", "7", "8"), class = "factor"), count = c(20L, 1L, 9L, 1L, 1L, 5L, 2L, 1L, 1L, 4L, 5L, 4L, 1L)), .Names = c("X5employf", "X5employff", "count"), row.names = c(NA, -13L), class = "data.frame")
I would then want to plot barplots in ggplot (or similar) using these summary data with error bars showing the confidence intervals.
I had thought to amend the code provided in the tutorial above to calculate the columns above, though as a relative newcomer to R, am struggling a little! I have been experimenting with the ggply package but not so great on the syntax so I have managed to get as far as this with the following code:
> X5employ_props <- ddply(X5employ_counts, .(X5employf), transform, prop=count/sum(count))
But I end up with this:
X5employf X5employff count prop
1 1 1 20 1.0000000
2 1 2 1 1.0000000
3 1 3 9 1.0000000
4 2 4 1 0.2000000
5 3 4 4 0.8000000
6 2 5 5 0.5000000
7 3 5 5 0.5000000
8 2 6 2 0.3333333
9 3 6 4 0.6666667
10 2 7 1 0.5000000
11 3 7 1 0.5000000
12 2 8 1 1.0000000
13 1 <NA> 1 1.0000000
With all my proportions being 1, presumably because they are being calculated across rows and not columns
I wondered if anyone could help or knows of packages / code that would do the job for me!
There are numerous methods to calculate binomial confidence intervals and I doubt there is much consensus on which method is best. That said, here is one approach to calculate binomial confidence intervals using several different methods. I am not sure whether this helps.
library(binom)
x <- c(3, 4, 5, 6, 7)
n <- rep(10, length(x))
binom.confint(x, n, conf.level = 0.95, methods = "all")
method x n mean lower upper
1 agresti-coull 3 10 0.3000000 0.10333842 0.6076747
2 agresti-coull 4 10 0.4000000 0.16711063 0.6883959
3 agresti-coull 5 10 0.5000000 0.23659309 0.7634069
4 agresti-coull 6 10 0.6000000 0.31160407 0.8328894
5 agresti-coull 7 10 0.7000000 0.39232530 0.8966616
6 asymptotic 3 10 0.3000000 0.01597423 0.5840258
7 asymptotic 4 10 0.4000000 0.09636369 0.7036363
8 asymptotic 5 10 0.5000000 0.19010248 0.8098975
9 asymptotic 6 10 0.6000000 0.29636369 0.9036363
10 asymptotic 7 10 0.7000000 0.41597423 0.9840258
11 bayes 3 10 0.3181818 0.09269460 0.6058183
12 bayes 4 10 0.4090909 0.15306710 0.6963205
13 bayes 5 10 0.5000000 0.22352867 0.7764713
14 bayes 6 10 0.5909091 0.30367949 0.8469329
15 bayes 7 10 0.6818182 0.39418168 0.9073054
16 cloglog 3 10 0.3000000 0.07113449 0.5778673
17 cloglog 4 10 0.4000000 0.12269317 0.6702046
18 cloglog 5 10 0.5000000 0.18360559 0.7531741
19 cloglog 6 10 0.6000000 0.25266890 0.8272210
20 cloglog 7 10 0.7000000 0.32871659 0.8919490
21 exact 3 10 0.3000000 0.06673951 0.6524529
22 exact 4 10 0.4000000 0.12155226 0.7376219
23 exact 5 10 0.5000000 0.18708603 0.8129140
24 exact 6 10 0.6000000 0.26237808 0.8784477
25 exact 7 10 0.7000000 0.34754715 0.9332605
26 logit 3 10 0.3000000 0.09976832 0.6236819
27 logit 4 10 0.4000000 0.15834201 0.7025951
28 logit 5 10 0.5000000 0.22450735 0.7754927
29 logit 6 10 0.6000000 0.29740491 0.8416580
30 logit 7 10 0.7000000 0.37631807 0.9002317
31 probit 3 10 0.3000000 0.08991347 0.6150429
32 probit 4 10 0.4000000 0.14933907 0.7028372
33 probit 5 10 0.5000000 0.21863901 0.7813610
34 probit 6 10 0.6000000 0.29716285 0.8506609
35 probit 7 10 0.7000000 0.38495714 0.9100865
36 profile 3 10 0.3000000 0.08470272 0.6065091
37 profile 4 10 0.4000000 0.14570633 0.6999845
38 profile 5 10 0.5000000 0.21765974 0.7823403
39 profile 6 10 0.6000000 0.30001552 0.8542937
40 profile 7 10 0.7000000 0.39349089 0.9152973
41 lrt 3 10 0.3000000 0.08458545 0.6065389
42 lrt 4 10 0.4000000 0.14564246 0.7000216
43 lrt 5 10 0.5000000 0.21762124 0.7823788
44 lrt 6 10 0.6000000 0.29997837 0.8543575
45 lrt 7 10 0.7000000 0.39346107 0.9154146
46 prop.test 3 10 0.3000000 0.08094782 0.6463293
47 prop.test 4 10 0.4000000 0.13693056 0.7263303
48 prop.test 5 10 0.5000000 0.20142297 0.7985770
49 prop.test 6 10 0.6000000 0.27366969 0.8630694
50 prop.test 7 10 0.7000000 0.35367072 0.9190522
51 wilson 3 10 0.3000000 0.10779127 0.6032219
52 wilson 4 10 0.4000000 0.16818033 0.6873262
53 wilson 5 10 0.5000000 0.23659309 0.7634069
54 wilson 6 10 0.6000000 0.31267377 0.8318197
55 wilson 7 10 0.7000000 0.39677815 0.8922087
I am not entirely sure what you want, but here is code to create a table that I think contains all of the parameters you are after. I dug the code out of Package binom using the Agresti-Coull method.
conf.level <- 0.95
x <- c( 4, 5, 6) # successes
n <- c(10,10,10) # trials
method <- 'ac'
# source code from package binom:
xn <- data.frame(x = x, n = n)
all.methods <- any(method == "all")
p <- x/n
alpha <- 1 - conf.level
alpha <- rep(alpha, length = length(p))
alpha2 <- 0.5 * alpha
z <- qnorm(1 - alpha2)
z2 <- z * z
res <- NULL
if(any(method %in% c("agresti-coull", "ac")) || all.methods) {
.x <- x + 0.5 * z2
.n <- n + z2
.p <- .x/.n
lcl <- .p - z * sqrt(.p * (1 - .p)/.n)
ucl <- .p + z * sqrt(.p * (1 - .p)/.n)
res.ac <- data.frame(method = rep("agresti-coull", NROW(x)),
xn, mean = p, lower = lcl, upper = ucl)
res <- res.ac
}
SE <- sqrt(.p * (1 - .p)/.n)
SE
Here is the table containing all data and parameters.
my.table <- data.frame(res, SE)
my.table
method x n mean lower upper SE
1 agresti-coull 4 10 0.4 0.1671106 0.6883959 0.1329834
2 agresti-coull 5 10 0.5 0.2365931 0.7634069 0.1343937
3 agresti-coull 6 10 0.6 0.3116041 0.8328894 0.1329834
I have not yet checked to see whether these estimates match any examples in Agresti's books. However, the first R function below from the University of Florida returns the same CI estimates as package binom. The second R function below from the University of Florida does not.
http://www.stat.ufl.edu/~aa/cda/R/one-sample/R1/
x <- 4
n <- 10
conflev <- 0.95
addz2ci <- function(x,n,conflev){
z = abs(qnorm((1-conflev)/2))
tr = z^2 #the number of trials added
suc = tr/2 #the number of successes added
ptilde = (x+suc)/(n+tr)
stderr = sqrt(ptilde * (1-ptilde)/(n+tr))
ul = ptilde + z * stderr
ll = ptilde - z * stderr
if(ll < 0) ll = 0
if(ul > 1) ul = 1
c(ll,ul)
}
# Computes the Agresti-Coull CI for x successes out of n trials
# with confidence coefficient conflev.
add4ci <- function(x,n,conflev){
ptilde = (x+2)/(n+4)
z = abs(qnorm((1-conflev)/2))
stderr = sqrt(ptilde * (1-ptilde)/(n+4))
ul = ptilde + z * stderr
ll = ptilde - z * stderr
if(ll < 0) ll = 0
if(ul > 1) ul = 1
c(ll,ul)
}
# Computes the Agresti-Coull `add 4' CI for x successes out of n trials
# with confidence coefficient conflev. Adds 2 successes and
# 4 trials.
Note also that according to the first link above the Agresti-Coull interval is not recommended when n < 40.
As for the other packages you mentioned, I rarely use them, but I am pretty sure you could include the above code in R script that calls those packages.
Here is a method for estimating multinomial 95% confidence intervals.
library(MultinomialCI)
x <- c(20,1,9,1)
multinomialCI(x,alpha=0.05,verbose=FALSE)
# [,1] [,2]
# [1,] 0.5161290 0.8322532
# [2,] 0.0000000 0.2193499
# [3,] 0.1612903 0.4774145
# [4,] 0.0000000 0.2193499
I have not yet looked at the source code to see how to obtain the SE's.
来源：https://stackoverflow.com/questions/17802320/r-proportion-confidence-interval-factor