Finite Population Correction on sample size Estimate of percent of an infected species is accurate to within $\pm 0.04%$ with $95\%$ CI
Between $15\%$ and $35\%$ of the population are infected
Size of the population is between $1100$ and $2300$
Calculate the sample size.
My calculations:
For $15\%$ , $n = 306$
For $35\%$, $n = 546$
Since population is between $1100$ and $2300$ , finite population correction is required.
How can I determine the sample size adjusted by finite population correction?
Do I have to take the weighted average of the smaller and greater $n$? 
Or should I simply use the larger $n$?
 A: n is the sample size, not the population or something else. You do not need to know the true porpotion of the infected people.  The equation for the deviation is $$\large{z_{0.975}} \normalsize{\cdot \sqrt{\frac{0.2 \cdot 0.8}{n} }=0.04}$$
$z_{0.975}$ is the value for z, where the cdf of the standard normal distribution is equal to $0.975$
Remark:
$0.25$, the mean of $0.35$ and $0.15$, is the best estimation, since you have no other information.
A: The issue is what we should use for the variance. The variance for any individual trial is $p(1-p)$, where $p$ is the probability of success. What should we use for $p$? The conservative choice  is $p=0.35$. There is a very informal case for $0.25$. This part depends to an uncomfortable degree on the preferences of your instructor.
If $N$ is the population size, the small population version of the variance is 
$$\frac{p(1-p)}{n}\cdot \frac{N-n}{N-1}.$$
So we end up wanting
$$1.96\sqrt{\frac{p(1-p)}{n}}\sqrt{\frac{N-n}{N-1}}\approx 0.04.$$
Square both sides and manipulate a little. We end up with a linear equation for $n$. Solve. It is not clear what we should use for $N$. The conservative approach is to use $2300$, since that gives the highest value of variance.  But again, your instructor may have different preferences.
