Given a Bernoulli distribution with a success probability of p = 0.02. Let's say we have N=3000 samples. How can I compute a confidence intervals for the expected number of successes (e.g., with 5% significance level)?

  • 1
    $\begingroup$ What parameter are we computing the confidence interval for? The expected number of successes is $Np = 60$ which is known with perfect confidence regardless of what the data (the outcomes of the $3000$ trials) happen to be. The actual number of successes that have occurred is available simply by counting the successes in the data and is also known with perfect confidence. If the $3000$ trials have not occurred as yet, and the question is to find the shortest interval $[a,b]$ such that $P\{a \leq X \leq b\} \geq 0.95$, then that is a question about probability, not about confidence intervals. $\endgroup$ – Dilip Sarwate Jan 28 '13 at 15:09
  • $\begingroup$ @cbeleites No, it is no homework question but it is derived from a real scenario. Each day there are about 0.02x3000 = 60 hits. Whenever the actual number is lower than that (e.g., 50 hits), I immediately get several questions about it. I often answer that it could be natural fluctuation, but meanwhile I'm just curious about the mathematical background. $\endgroup$ – Philipp Claßen Jan 28 '13 at 16:11
  • 1
    $\begingroup$ R's pbinom() tells me that you should expect 50 or fewer hits on about 10.5% of all days. Nothing to really pester you with questions about... $\endgroup$ – S. Kolassa - Reinstate Monica Jan 28 '13 at 19:40

Your expected value is $3000\times0.02=60$, the variance is $3000\times 0.02\times(1-0.02)=58.8$. Simulating 100,000 trials and plotting a histogram, I would say you can use the normal approximation unless you have very strong requirements on accuracy (in which case the R help page says that "qbinom() uses the Cornish-Fisher Expansion to include a skewness correction to a normal approximation, followed by a search", which may help).

nn <- 3000
n.sim <- 100000
foo <- rbinom(n=n.sim,size=nn,prob=.02)

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.