Suppose we have some population of size $N$ about which we are interested in which individuals have some particular property. In particular, we would like to know the fraction of the whole population which has that property. This will be the population proportion $p=a/N$, where $a$ is the number of individuals with the desired property. ($p$ is of course a parameter, since it is a number computed from the entire population.)
If $N$ is large, it may be impractical simply to go out and check every individual for the property of interest, so it will be hard to find $p$.
Instead, we might take a sample of size $n$ and compute the sample proportion, $\widehat{p}=b/n$, where $b$ is the number of individuals in the sample which have the desired property.
If $n$ is fairly large and if we think the sample is well (very randomly!) chosen, we might guess that $p\approx\widehat{p}$.
We will again make a Confidence Interval, now for the proportion, which will be precise enough to say what is the probability that $p$ is a certain distance from $\widehat{p}$. That is: for any confidence level $L$ (which is a number from $0$ to $1$, often expressed as a percentage), we will make an interval of real numbers centered at $\widehat{p}$ with the probability of $p$ being in that interval equalling (at least) $L$.
In confidence intervals for the population mean, the CLT was a vital tool. Here, there is a substite bit of mathematical technology (the Normal approximation to the binomial) which tells us that for large $n$, $\widehat{p}$ is approximately Normally distributed with mean $p$ and standard deviation $\sqrt{\frac{p (1-p)}{n}}$. Then the same sort of argument as in our discussion of CIs for means will tell us that
| The confidence interval at confidence level $L$ for the population proportion $p$ having some property of interest has endpoints $\widehat{p}\pm z^*_L \sqrt{\frac{\widehat{p} (1-\widehat{p})}{n}}$, where the sample size $n$ is fairly large (typically, large enough so that there are at least 15 individuals in the sample which have the desired property and 15 which do not), $\widehat{p}$ is the sample proportion, and $z^*_L$ is the usual Normal critical value at confidene level $L$. |
The same table of critical values as before can still be used:
| confidence level $L$ | .5 | .8 | .9 | .95 | .99 | .999 |
|---|---|---|---|---|---|---|
| critical value $z^*_L$ | .674 | 1.282 | 1.645 | 1.960 | 2.576 | 3.291 |
Note that the condition on how big the sample must be can also be expressed by saying that $n\widehat{p}$ and $n(1-\widehat{p})$ must both be at least 15. (This is because $(1-\widehat{p})$ is the proportion of the sample which does not have the property we are investigating.
We interpret the confidence for a population proportion in much the same way that we did for the population mean. In particular, the confidence level is a statement about the method, not the validity of any particular CI that we have constructed. So, e.g., if the newspaper says that Mickey Mouse will win the election for president this year by gaining 73% of the vote, plus or minus 4%, and if on Election Day Mickey only gets 62% of the vote, we should not cancel our newpaper subscription in anger. Instead, we should conclude that if we had done 100 surveys of the same size as the newpspaper researchers had done, then 95 or more of them would have yielded a CI for the population proportion which did include the true value of 62%. Apparently, the survey which was actually done for that newspaper story was one of the 5 out of 100 which yielded incorrect results, in the sense that the true population proportion of Mickey supporters, 62%, did not lie in the ranges predicted (such as the one in the newspaper, which was 69% to 77%). That particular interval in the newspaper was not a good one, but the method that produced it was likely with probability .95 of producing an interval which did contain the true value. (Note the 95% in this example is because when the confidence level is not mentioned in coverage of a CI in the press, they are usually implicitly using confidence level $L=.95$.)