Confidence interval for population proportionYou have a population of 10M people. Suppose you want to determine the share of people with a certain disease. To do so you randomly choose X people and check if they have the disease. You determine that a share P of those people have the disease.How do I compute the 75% / 90% / 95% confidence intervals for the disease prevalence for the whole population?e.g. find y so that there is a 95% chance the disease rate in the whole population is in the range [X−y,X+y].

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Denise Daniel · Accepted Answer

Suppose n people are chosen at random from a population with N members, where nN&amp;lt;0.1. If X of the n have a certain disease, then p^=Xn is an estimate of the proportion p of people in the population who have the disease.Wald. Then a traditional Wald 95% confidence interval (CI) for p isp^±1.96p^1−p^n.The &#039;probability factor&#039; is changed from 1.960 (to 1.645 or 1.150) to get confidence levels 90% or 75%, respectively.Agresti. When the confidence level is 95% and n is small (some say less than 100), then it is better to use the Agresti (CI): Let n~=n+4 and p~=X+2n~ to get the CIp~±1.96p~1−p~n~.This form of the Agrest CI is intended for use only at the 95% confidence level.Wilson. In general, a slightly more accurate form of CI than either of these is the &#039;Wilson interval for a binomial proportion&#039;, which has a somewhat more complicated formula, I refer you to the Wikipedia article for that.

Confidence interval for population proportion You have a population

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