Estimating the subset of a population which are sucessful when the population itself is an estimate Let's say I have a population $S$, with an estimated size $\hat{n}$ (and standard error $\sigma_{\hat{n}}$). The way that $\hat{n}$ is estimated is through generating random samples from a larger sample space of size $m$ ($n \ll m$), and then determining how many belong to $S$. For our purposes, $n$ can't realistically be determined any other way. These samples form a Bernoulli distribution (since a sample either belongs to $S$ or doesn't), and we calculate $\sigma_\hat{n}$ through normal approximation.
I'd like to sample from $S$ and determine how many samples belong to $T$, on the basis of some arbitrary criteria for $s \in S$. Let the observed proportion of $S$ which are in $T$ be called $\hat{p}$, and let's say we also use a normal approximation. My question is: how does $\sigma_\hat{n}$ "interact" with $\sigma_\hat{p}$? (since we want to calculate $\hat{n}\hat{p}$)
Some notes:

*

*$\hat{n}$ and $\hat{p}$ are independent. There's no relationship between the two.

*Let's say we're initially sampling from $R$ (of known size $m$) to find $\hat{n}$. Why not instead determine directly how many $r \in T$? The reason is, verifying that some $r$ or $s$ is in $T$ is very complex (PSPACE-hard). The maximum number of samples I can realistically verify to be in $T$ is so small that $m\hat{q}$ (where $\hat{q}$ is the observed proportion of $R$ in $T$) would have confidence intervals much too large to mean anything useful. So instead, I can achieve a very confident estimate of $\hat{n}$, and then sample from $S$ instead.

Any guidance appreciated.
Potential answer: propagation of normally-distributed errors, in our case when multiplying some $\sigma_1$ by $\sigma_2$: notes
Other comments:
I initially asked some pretty incomprehensible questions, and really shouldn't have been given the time. Thanks for everyone's precious time, especially BruceET and whuber's.
 A: An alternative approach is to sample from the large population R of size m>n untill you have some fixed number of successes (samples from T).
The sampling is done by testing whether a sample is S and if it is S then you test whether it is T/success. (So you do not need to do all the time the costly test to see if a sample is in T)
The number of samples that you need is negative Binomial distributed and based in that you can estimate a probability $\hat{p}$ for the fraction of T and S among in R and $\hat{p}m$ will be the estimate for the size.
A: Suppose you are sampling college admission test scores from
a large high school district. Traditionally, the district
mean on this test has been 280 with a standard deviation of
25. Then an approximate 95% CI for this year's district mean would be of
the form $\bar X \pm 2(25)/\sqrt{n}.$ If you want this year's CI to have
margin of error of $\pm 10,$ then you have $50/\sqrt{n} \approx 50/\sqrt{n} = 10.$
So you need a sample size of $n \approx 25.$
Suppose you sample $n = 25$ observations at random from
$\mathsf{Norm}(\mu=280, \sigma=25),$ to get data z as below. (Sampling and computations in R.)
set.seed(2020)
x = rnorm(25, 289, 25)
summary(x);  length(x);  sd(x)
     Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
    234.9   267.9   282.3   284.0   299.3   326.0 
[1] 25         # sample size
[1] 22.52312   # sample SD

The t.test procedure in R, provides a 95% CI $(274.7,293.3)$ as part of its output,
captured below using $-notation. The margin of error for this sample is about 9.3. [Margins of error will vary from sample to sample, depending on the sample standard deviation: for example, four additional samples of size $n=25$ gave
margins of error 9.7, 12.9, 8.9, and 10.7.]
t.test(x)$conf.int
[1] 274.6643 293.2585
attr(,"conf.level")
[1] 0.95

