You can't use the normal approximation here. The normal approximation you are likely referring to comes from the Central Limit Theorem which concerns the asymptotic distribution of sums of iid random variables. Though you are summing iid random variables in the numerator, things are complicated because of the denominator.
If you are able to make some assumptions about the distribution of weight for males and females, I suppose you could use something like the delta method. Let $w_m$ and $w_f$ be random variables for a man and woman's weights respectively. Then
$$ \sum_i^n w_{m,i} = n_m \bar{w}_m \sim \mathcal{N}(n_m\mu_m, n_m\sigma_m^2)$$
Where $\mu_m$ and $\sigma_m$ are population level mean and standard deviations for men's weights. A similar expression can be derived for women. Your statistic is then
$$ \hat{p} = \dfrac{n_m \bar{w}_m}{n_m \bar{w}_m + n_f \bar{w}_f} $$
To use the delta method, you need to know the covariance between $w_m$ and $w_f$. It might be safe to assume these are independent. If so, then the you can use the delta method to get an estimate of the variance of $\hat{p}$. I can come back later after my meeting and try to work through this some more. I also may be off in the right direction, so someone let me know if I've said something wrong.
Just for some notational clarity, let $X = n_m \bar{w}_m$ and $Y = n_f \bar{w}_f$. We need the gradient of $g(X,Y) = X/(X+Y)$ which is
$$ \nabla g = \left< \dfrac{Y}{(X+Y)^2}, \dfrac{-X}{(X+Y)^2} \right>$$
The variance of $g(X,Y)$ is
$$\nabla g^T \Sigma \nabla g$$
where $\Sigma$ is the covariance between $X$ and $Y$. Assuming they are independent, then $\Sigma$ is diagonal and the variance collapses to
$$\operatorname{Var}(g(X,Y)) = \dfrac{n_m\sigma^2_mY^2}{(X+Y)^4} + \dfrac{n_f\sigma^2_fX^2}{(X+Y)^4} $$
We can get an estimate of the variance by plugging in estimates of $X$ and $Y$ thanks to Slutsky's theorem. I think that should be correct, but let's simulate just to make sure.
nm = 20
nf = 20
weight_of_men = rnorm(nm, 80, 5)
sm = var(weight_of_men)
X = sum(weight_of_men)
weight_of_women = rnorm(nf, 70, 4)
sf = var(weight_of_women)
Y = sum(weight_of_women)
v = nm*sm*Y^2/((X+Y)^4) + nf*sf*X^2/((X+Y)^4)
ps = replicate(100000, {
nm = 20
nf = 20
weight_of_men = rnorm(nm, 80, 5)
sm = var(weight_of_men)
X = sum(weight_of_men)
weight_of_women = rnorm(nf, 70, 4)
sf = var(weight_of_women)
Y = sum(weight_of_women)
X/(X+Y)
})
#Computed variance
v
#Variance from simulation
var(ps)
Depending on your random seed, you might see different numbers but I get 2.70e-5 for my formula and 2.21e-05 from the simulation. They are at least on the same order of magnitude.
Your confidence interval is then $\hat{p} \pm 2\sigma$. Let's examine the coverage of this estimator.
library(tidyverse)
coverage = replicate(100000, {
nm = 20
nf = 20
weight_of_men = rnorm(nm, 80, 5)
sm = var(weight_of_men)
X = sum(weight_of_men)
weight_of_women = rnorm(nf, 70, 4)
sf = var(weight_of_women)
Y = sum(weight_of_women)
v = sqrt(nm*sm*Y^2/((X+Y)^4) + nf*sf*X^2/((X+Y)^4))
between(80/150, X/(X+Y) - 2*v, X/(X+Y) + 2*v)
})
mean(coverage)
>>>0.947
Bingo! 95% coverage using the derived variance. This solution heavily depends on the assumption of the random variables being independent, so have a think about that before implementing it.
