I know that I need mean and s.d to find the interval, however, what if the question is:
For a survey of 1,000 randomly chosen workers, 520 of them are female. Create a 95% confidence interval for the proportion of workers who are female based on the survey.
How do I find mean and s.d for that?
You can also use prop.test from package stats, or binom.test
prop.test(x, n, conf.level=0.95, correct = FALSE)
1-sample proportions test without continuity correction
data: x out of n, null probability 0.5
X-squared = 1.6, df = 1, p-value = 0.2059
alternative hypothesis: true p is not equal to 0.5
95 percent confidence interval:
0.4890177 0.5508292
sample estimates:
p
0.52
You may find interesting this article, where in Table 1 on page 861 are given different confidence intervals, for a single proportion, calculated using seven methods (for selected combinations of n and r). Using prop.test you can get the results found in rows 3 and 4 of the table, while binom.test returns what you see in row 5.
In this case, you have binomial distribution, so you will be calculating binomial proportion confidence interval.
In R, you can use binconf() from package Hmisc
> binconf(x=520, n=1000)
PointEst Lower Upper
0.52 0.4890177 0.5508292
Or you can calculate it yourself:
> p <- 520/1000
> p + c(-qnorm(0.975),qnorm(0.975))*sqrt((1/1000)*p*(1-p))
[1] 0.4890345 0.5509655
Alternatively, use function propCI from the prevalence package, to get the five most commonly used binomial confidence intervals:
> library(prevalence)
> propCI(x = 520, n = 1000)
x n p method level lower upper
1 520 1000 0.52 agresti.coull 0.95 0.4890176 0.5508293
2 520 1000 0.52 exact 0.95 0.4885149 0.5513671
3 520 1000 0.52 jeffreys 0.95 0.4890147 0.5508698
4 520 1000 0.52 wald 0.95 0.4890351 0.5509649
5 520 1000 0.52 wilson 0.95 0.4890177 0.5508292
