We know for a sample from a population we can use the normal approximation to get a confidence interval around $\hat{p}$.

If we know the population completely then we know $p$ completely.

The normal approximation for C.I. is $\hat{p} \pm Z_{\alpha/2}\sqrt{\hat{p}(1-\hat{p})/n}$.

What is the population size $m$ such that a C.I. of $p$ is useful given a sample size $n$?


We have a known population of 400 people and 100 of them go to the movies on the weekend.

Thus, we know the proportion of people going to the movies on the weekend is p = $0.25$.

We don't need create a C.I. because we know the population completely. If we assumed this was a sample however we would get the following:

> 0.25 + c(-1,1) * qnorm(0.975) * sqrt(0.25*(1-0.25)/400)
[1] 0.2075655 0.2924345

What if the population was 401? We have a sample $n$ of 400. We could use the C.I. now, but would it wouldn't really be useful at this ratio of $n/m$.

  • $\begingroup$ I am afraid that "useful" is a very imprecise concept. What makes the estimate "useful"? $\endgroup$
    – Tim
    Apr 3 '17 at 14:01
  • $\begingroup$ @Tim Useful as in practical. A C.I. for a proportion of a known population isn't useful it adds uncertainty when there is none. Hence, isn't "useful". If I take a large sample that contains every item of the population except one the C.I. for the proportion isn't useful because it would be more useful to use the estimated proportion less one sample that is missing to know the entire population. Of course there would be some uncertainty, but could assume it as negligible. $\endgroup$
    – Matt L.
    Apr 3 '17 at 14:34
  • $\begingroup$ Why do you think the normal approximation can always be used? $\endgroup$ Apr 3 '17 at 14:54

You are talking about binomial distribution. Binomial distribution is a distribution of draws with replacement, so you can even draw a sample of size equal to the population and end up, by chance, with not sampling each of the individuals in the population, but rather sampling some of them multiple times. So no matter of size of your samples, there will be variability in your results.

But you may be confusing binomial distribution with hypergeometric distribution that describes sampling without replacement from a finite population. As described by Mansuri (2012), it also has a normal approximation, but it is different then what you've described. Notice that since drawing without replacement, each time you draw new value, the probability of what you could draw next changes, so there is no fixed $p$.

Moreover, even in case of hypergeometric distribution the notion of "usefulness" does not hold if you consider the extreme case: Imagine that you have drawn sample of size 499 from a population of size 500 and observed that 499 members of the population are right handed. Can you be certain that the probability of finding right handed person in the population is equal to unity? You cannot, since there is still $1/500$ chance that the remaining individual is left-handed. Obviously, if you have a whole-population data, then deriving confidence intervals does not make any sense since you are certain about it's properties.

Mansuri, S. B. (2012). A note on negative hypergeometric distribution and its approximation. World Academy of Science, Engineering & Technology, 6(71), 1082-1084.

  • $\begingroup$ Okay, that makes more sense to me know. I think I was over thinking it. I wasn't thinking about this as samples from a specific distribution. After reading this and my notes I've made. It makes a lot more sense. That this approximation would work when taking samples with replacement from a binomial distribution. Like lab trials or drug test trials where my $m$ value could be infinite because I could keep sampling forever until I ran out of money for instance. Thanks Tim! $\endgroup$
    – Matt L.
    Apr 3 '17 at 14:47

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