1
$\begingroup$

Suppose that there are 10 smokers and 590 non-smokers in a room and let $X$ be a random variable that describes the number of smokers picked out, without replacement, in $n$ random draws. What methods could be considered for determining the minimum sample size for estimating that the population contains about 10 out of 590 smokers with some level of confidence? Are there any specific methods that perform better when the ratio 10/590 is small, say, about $10^{-2}$?

$\endgroup$
7
  • 2
    $\begingroup$ Your question is a little unclear. In what manner would a sample "determine" the probability of any event? Unless the sample is a census, you can only estimate such probabilities with more or less confidence, depending on the sample size. How are the endpoints $10\pm c$ connected, if at all, to the assumed $10/590$ breakdown of the population? How do characterize a "refined" method? Compared to what? $\endgroup$
    – whuber
    Oct 5, 2022 at 20:52
  • $\begingroup$ Thanks for your comment, I updated the question. $\endgroup$
    – Steve
    Oct 5, 2022 at 21:16
  • 2
    $\begingroup$ That's getting more understandable, but the answer is zero: you have asked us to assume we know the value is $10.$ Presumably we don't know that value. If you're trying to ask about the sample size needed for the confidence to have a certain level, there's not enough information: you need to stipulate, say, the width of that confidence interval. Then we need to wonder about the meaning of "perform better:" are you looking for CI procedures that exactly achieve the intended confidence level; or are you trying to minimize expected lengths of CIs; or something else? $\endgroup$
    – whuber
    Oct 5, 2022 at 22:08
  • $\begingroup$ What questions should I ask myself when deciding on a width of a confidence interval for determining how much data to collect? Surely I want to avoid not observing any smokers. To elaborate on the "perform better" part: I was under the impression that when the estimator X/n is approximately normally distributed then I can use approximate confidence intervals based on the normal distribution. I could then fix the width and determine the number of samples approx. needed. If however there only are a few smokers then $X$ is not approx. normal? Is there a way to account for this? $\endgroup$
    – Steve
    Oct 7, 2022 at 17:51
  • 1
    $\begingroup$ Those are good questions. The current answer explains how to use the hypergeometric distribution directly. You decide on a width that meets your needs. For instance, if you want to minimize the risk of not detecting smokers who are there, you will specify the minimum number of smokers you need to detect and the maximal acceptable chance of not detecting them. $\endgroup$
    – whuber
    Oct 7, 2022 at 18:27

1 Answer 1

1
$\begingroup$

The most challenging part here is that the sampling is done without replacement, which complicates everything, if we want to stay precise.

If you know for sure that the number of smokers in the room of 600 people is 10 (a null hypothesis, which is known to be true, in a sense), you may want to first start thinking towards minimizing the probability of getting samples with zero smokers. The latter is in a sense a p-value for the lower bound of the two-sided confidence interval, if this bound is set in between 0 and 1. This probability equals: $$ \frac{590 \cdot 589 \cdot 588 \cdot ... \cdot (590-n)}{600 \cdot 599 \cdot 598 \cdot ... \cdot (600-n)} = \frac{590! \cdot (600-n)!}{600! \cdot (590-n)!}$$ You can see that, for instance, the p-value for this lower bound falls below 0.025 (if we target a 95% two-sided confidence interval) at $n=184$ (link to wolframalpha)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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