Binomial confidence intervals of means with R

Question

I have got 4 different data.frames that have observations that follow a binomial distribution and I need to calculate, for each one, the confidence intervals related to the means of a second column ( Flow ).

The number of successes are reported in the column Success and the total number of trials = 85 .

How can I calculate confidence intervals? How can I do it with R?

Here an example of my data.frames:

df1 <- read.table(text = 'Flow Success
725.661   4
25.54     4
318.481   4
230.556   4
2.823     3
12.6      3
9.891     3
11.553    1', header = TRUE)

> mean(df1$Flow)
[1] 167.1381


df2 <- read.table(text = 'Flow Success
725.661    3
25.54      3
318.481    3
230.556    2
2.823      2
12.6       1', header = TRUE)

> mean(df2$Flow)
[1] 219.2768

df3 <- read.table(text = 'Flow Success
725.661     2
25.54       2
318.481     1', header = TRUE)

> mean(df3$Flow)
[1] 356.5607

df4 <- read.table(text = 'Flow Success
725.661    2
25.54      2', header = TRUE)

> mean(df4$Flow)
[1] 375.6005

I need to calculate the confidence intervals of the above means.

I can give you more info about the data if needed.

Thanks for anyone who will help me.

Answer 1

The package binom provides methods for calculating binomial confidence intervals. One can choose to use all available methods, or specify a single method.

x gives the number of successes, and n the number of Bernouli trials.

library(binom)

binom.confint(x = 5, n = 10)
          method x  n mean     lower     upper
1  agresti-coull 5 10  0.5 0.2365931 0.7634069
2     asymptotic 5 10  0.5 0.1901025 0.8098975
3          bayes 5 10  0.5 0.2235287 0.7764713
4        cloglog 5 10  0.5 0.1836056 0.7531741
5          exact 5 10  0.5 0.1870860 0.8129140
6          logit 5 10  0.5 0.2245073 0.7754927
7         probit 5 10  0.5 0.2186390 0.7813610
8        profile 5 10  0.5 0.2176597 0.7823403
9            lrt 5 10  0.5 0.2176212 0.7823788
10     prop.test 5 10  0.5 0.2365931 0.7634069
11        wilson 5 10  0.5 0.2365931 0.7634069

binom.confint(x = 5, n = 10, method = "exact")
  method x  n mean    lower    upper
1  exact 5 10  0.5 0.187086 0.812914