I have a sample with 1000 people which 382 agree, 578 disagree and 40 can't decide. I want to find the 99% confidence interval of the proportion of people who agree.

I know how how to discover using R the standard error, and then use it to find the 95% confidence interval. It's just make this calculation: (0.382-2SE,0.382+2SE).

My question is how to find the 99% confidence interval. Do I need R programming or I can do as a simple calculation?

  • 1
    $\begingroup$ en.wikipedia.org/wiki/68%E2%80%9395%E2%80%9399.7_rule $\endgroup$ Sep 4, 2020 at 8:10
  • $\begingroup$ @ericperkerson How can I find other values like 96.54% for example? do we have a way to do this using R programming? $\endgroup$
    – user45523
    Sep 4, 2020 at 8:13
  • 1
    $\begingroup$ Yes, since you are using the normal approximation to find the 95% confidence level with ±2SE, you can run -qnorm((1-x)/2) in R to find how many standard deviations are required to get probability x. For example, -qnorm((1-0.99)/2) = 2.575829 tells you that ±2.57829SE will contain 99% of the area/probability under the standard normal distribution. $\endgroup$ Sep 4, 2020 at 9:23

2 Answers 2


If you choose a significance level $\alpha$, you get a confidence interval with confidence level $1-\alpha$: $$C_{1-\alpha}=\left(X-z_{1-\frac{\alpha}{2}}\text{se},X+z_{1-\frac{\alpha}{2}}\text{se}\right)$$ where $z_{1-\frac{\alpha}{2}}$ is the $(1-\frac{\alpha}{2})$-quantile of a standard normal variable $Z$.

In your confidence interval, (0.382-2SE,0.382+2SE), that '2' actually is:

> alpha <- 0.05                     # significance level
> qnorm(1-alpha/2)
[1] 1.959964

If you wish a $0.99=1-0.01$ confidence level, you replace '2' with:

> alpha <- 0.01
> qnorm(1-alpha/2)
[1] 2.575829

If you are happy with R, you can use the binom.test function. E.g:

binom.test(382, 382+578+40, conf.level= 0.9654)

    Exact binomial test

data:  382 and 382 + 578 + 40
number of successes = 382, number of trials = 1000, p-value = 8.291e-14
alternative hypothesis: true probability of success is not equal to 0.5
96.54 percent confidence interval:
 0.3494766 0.415319
sample estimates:
probability of success 

(although I'm not sure how the 40 undecided should be treated)

  • $\begingroup$ Thank you for your answer. $\endgroup$
    – user45523
    Sep 4, 2020 at 8:30
  • $\begingroup$ I'm trying to solve it in a broader way. The answer should be (0.382-zSE,0.382+zSE). But how to find this z? $\endgroup$
    – user45523
    Sep 4, 2020 at 8:33

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