Determining sample size when given Confidence Interval and margin of error in a finite population I'm trying to determine n, the sample size.
We have $N=580$ and we want to estimate population proportion, $p$, at a 95%-CI and a margin of error of $0.10$, then I get 
$$n=Npq / (N-1)D + pq$$
where $D= B^2 / 4$ to be about $n=86$.
But what if instead we keep that same margin of error at $0.10$ and now only want a 90%-CI, if we use the same formula don't we get the same $n$? 
Am I using the formula wrong here? 
 A: The formulas concerning the calculation of the sample size to estimate a proportion $p$ in for finite populations are provided on this website. It also contains the derivations.
In short, the sample size necessary for estimating a population proportion $p$ of a finite population with $(1-\alpha)100\%$ confidence and a margin of error no larger than $\epsilon$ is:
$$
n = \frac{m}{1+\frac{m-1}{N}}
$$
where
$$
m=\frac{z_{1-\alpha/2}^{2}\hat{p}(1-\hat{p})}{\epsilon^{2}}.
$$
$N$ denotes the population size, $z_{1-\alpha/2}$ the $(1-\alpha/2)$-quantile of the standard normal distribution and $\hat{p}$ the estimated proportion.
For $N=580, \alpha=0.05, \epsilon=0.1, \hat{p}=0.5$ (i.e. 95% confidence), we get $n\approx 83$. If we take $\alpha = 0.1$ which corresponds to 90% confidence, we get $n\approx 61$.
The more precise we want our estimate of the popultion proportion to be, the higher our sample size needs to be.
This means that the lower $\alpha$, the higher the necessary sample size will be. The following graph illustrates this (for $N=580, \epsilon=0.1, \hat{p}=0.5$):

The necessary sample size will also increase with decreasing margin of error $\epsilon$ (note the reversed $x$-axis; graph for $N=580, \alpha=0.05, \hat{p}=0.5$):

