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You 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 $[X-y , X+y].$

Thank you

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Suppose $n$ people are chosen at random from a population with $N$ members, where $n/N < 0.1.$ If $X$ of the $n$ have a certain disease, then $\hat p = X/n$ 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 $p$ is $$\hat p \pm 1.96\sqrt{\hat p(1-\hat p)/n}.$$ The 'probability factor' is changed from 1.960 (to 1.645 or 1.150) to get confidence levels 90% or 75%, respectively.

alpha = c(.05, .10, .25)
qnorm(1 - alpha/2)
## 1.959964 1.644854 1.150349

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 $\tilde n = n+4$ and $\tilde p = (X+2)/\tilde n$ to get the CI $$\tilde p \pm 1.96\sqrt{\tilde p(1-\tilde p)/\tilde n}.$$ 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 'Wilson interval for a binomial proportion', which has a somewhat more complicated formula, I refer you to the Wikipedia article for that.

Note: For more on binomial CI's see another item on this site.

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  • $\begingroup$ I like how you offered the OP multiple methods. The Wald interval is mostly used in beginning stats classes, but can also quite inaccurate at times (not necessarily in questions like the OP proposed, but other practical instances). $\endgroup$ – Brandon Dec 3 '16 at 1:30

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