正規近似に基づく信頼区間
n <- 100
x <- 20
r <- x/n
se <- (r * (1 - r)/n)^0.5
ci <- r + qnorm(0.975) * se * c(-1, 1)
prop.test(x, n)
##
## 1-sample proportions test with continuity correction
##
## data: x out of n, null probability 0.5
## X-squared = 34.81, df = 1, p-value = 3.635e-09
## alternative hypothesis: true p is not equal to 0.5
## 95 percent confidence interval:
## 0.1292 0.2943
## sample estimates:
## p
## 0.2
##
binom.test(x, n)
##
## Exact binomial test
##
## data: x and n
## number of successes = 20, number of trials = 100, p-value =
## 1.116e-09
## alternative hypothesis: true probability of success is not equal to 0.5
## 95 percent confidence interval:
## 0.1267 0.2918
## sample estimates:
## probability of success
## 0.2
##
library(binom)
## Loading required package: lattice
binom.confint(x, n) # 11種類の信頼区間
## method x n mean lower upper
## 1 agresti-coull 20 100 0.200 0.1326 0.2896
## 2 asymptotic 20 100 0.200 0.1216 0.2784
## 3 bayes 20 100 0.203 0.1308 0.2863
## 4 cloglog 20 100 0.200 0.1283 0.2832
## 5 exact 20 100 0.200 0.1267 0.2918
## 6 logit 20 100 0.200 0.1328 0.2898
## 7 probit 20 100 0.200 0.1310 0.2872
## 8 profile 20 100 0.200 0.1298 0.2854
## 9 lrt 20 100 0.200 0.1297 0.2854
## 10 prop.test 20 100 0.200 0.1292 0.2943
## 11 wilson 20 100 0.200 0.1334 0.2888
二項分布に基づく母比率の信頼区間のグラフ
n <- 100
x <- 0:n
ci <- sapply(x, function(x) {
c(x/n, binom.test(x, n)$conf.int[1:2])
})
library(plotrix)
plotCI(x/n, ci[1, ], ci[3, ] - ci[1, ], ci[1, ] - ci[2, ])