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This should be simple I hope. In lecture notes from a physics course, the professor writes:

$\text{Prob(success)}=\dfrac{k}{N}$. So query $q$ times, see some number $\mathcal{l}$ successes and estimate probability $\dfrac{k}{N}=\dfrac{\mathcal{l}}{q}$. i.e. estimate $\tilde{k}$ of $k$ is $\tilde{k}=\dfrac{\mathcal{l}N}{q}$

Everything up to this point makes sense to me. But...

Now by properties of binomial distribution (with probability $\dfrac{k}{N}$), the standard deviation of $\mathcal{l}$ is $\sqrt{\dfrac{k(N-k)}{q}}$.

From what I recollect, the standard deviation of a binomial distribution $\text{bin}(n,p)$ is $\sqrt{npq}\equiv\sqrt{np(1-p)}$. With respect to the lecture notes, if we let $n\rightarrow q$ and $p\rightarrow \dfrac{k}{N}$, then that gives us stddev $=\sqrt{\dfrac{qk(N-k)}{N^2}}\neq\sqrt{\dfrac{k(N-k)}{q}}$.

Where am I messing up in the calculation?

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You are estimating the variance of the statistic, $p$ in this case (proportion), this is a Bernoulli trial, which has variance $p(1-p)$.

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  • $\begingroup$ Well this is for the standard deviation (or variance) of $\mathcal{l}$, not $p$ (where we are using $\dfrac{\mathcal{l}}{q}$ to estimate $p$). $\mathcal{l}$ is a regular sample from a binomial distribution and should have the regular variance $np(1-p)$. I do agree that $\dfrac{\mathcal{l}}{q}$ should have variance $p(1-p)=\dfrac{k(N-k)}{N^2}$, but that still does not match up with the above conclusion. $\endgroup$ – Dan Jan 18 '19 at 9:05
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I realized that the professor had a typo, and put down the standard deviation for $\tilde{k}$ where he was instead writing about the standard deviation for $l$.

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