If 0.01% of a population has a disease, what sample size k is needed so that there is a 95% chance that one person with this disease falls into k?

Question

I read that this can be done using the Poisson Approximation, which I understand how to do when you are already given a sample size. How would one find the solution to this problem to find the sample size, using a Poisson approximation?

Answer 1

In this case, you can use the "exact" Binomial distribution.

We want $k$ such that $\mathbb{P}(j \geq 1) \geq 95\%$, where $j$ is the number of persons with the disease.

This is equivalent to say that $\mathbb{P}(j < 1) = \mathbb{P}(j = 0)\leq 5\%$

The probability is: $$ \mathbb{P}(j = 0) = \binom{k}{0} p^0 (1-p)^{k-0} = (1-p)^k $$ where $p = 0.0001$ is the probability of getting the disease.

Therefore, we have to solve for k $$ 0.9999^k = 0.05 $$

I.e., $$ k \log 0.9999 = \log 0.05 $$

Which gives $$ k \approx 29\ 955.8\ldots $$ So, a sample size of $29\ 956$ individuals.

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If you really want to use the Poisson approximation, then you need the same reasoning, but now the probability is: $$ \mathbb{P}(j = 0) = e^{-pk} \frac{(pk)^0}{0!} = e^{-pk} $$ where $p = 0.0001$ is the probability of getting the disease and $pk$ is the expected (average) number of people with the disease in a sample of size $k$.

Can you take it from here and get the value of $k$ in this case? The result will not be the same as before.

Edit:

We just need to plug in the value of $p$ and solve: $$ e^{-0.0001k} = 0.05 $$

So, applying natural logarithms: $$ -0.0001k = \log 0.05 $$

Which gives $$ k \approx 29\ 957.3 $$ So, a sample size of $29\ 957$ will have an approximate probability of $95\%$ of having at least one disease person. In case you need at least $95\%$ probability, $k$ will need to be at least $29\ 958$.