I am trying to estimate a confidence interval containing the true number of people who would vote for Trump. So far, I have calculated the mean number of men who would vote for Trump and the mean number of women who would vote for Trump. I have also rejected the null hypothesis that the means are equal.
How would I go about constructing a 95%-confidence interval for the true number of people (men and women) who would vote for him GIVEN my estimates and rejection of equality between the means?
I don't know what your data looks like, but I will provide an example of what you need to do. Since you want a 95% confidence interval for the proportion that favors Trump (for men and women), and NOT the difference between men and women who favor trump, then you can do this in a couple ways. One way being by hand or one way using computer software R. I will use the method by hand first. At the end I supply you with code for R.
Lets say of 200 men, 100 favor trump and of 100 women, 40 favor trump.
Thus, the proportions of men and women who favor trump is $$p1 = .50$$
$$p2 = .40$$
We know $$n1 = 200$$
$$n2 = 100$$
Now we need to compute the Standard Error (SE)
$$\text{SE}=\sqrt{\frac{p (1-p)}{n}}$$
Thus,
$$SE1 = .035355$$ $$SE2 = .0489898$$
Now we need to find the Z-Value for a 95% confidence interval, which is $$Z = 1.96$$
We can now compute the Margin of Error(ME) by using the formula $$Z*SE$$
Thus, $$ME1 = 1.96* .0353553 = .0692964$$ $$ME2 = 1.96*.0489898 = .09602$$
Now we can compute the 95% Confidence Interval (CI) for the proportions of men and women who favor Trump, using $p1$ and $p2$, and $ME1$ and $ME2$
Thus, $$CI1= (.431,.569)$$ $$CI2 = (.304,.496)$$
Alternatively computer software can do this much faster for you
prop.test(100,200,conf.level = .95, correct = TRUE)
$$0.4313609, 0.5686391$$
prop.test(40,100,conf.level = .95, correct = TRUE)
$$0.3047801, 0.5029964$$