(1) Yes.
(2) Yes. There are only $n+1$ possible outcomes for a binomial random variable, so it is possible to look at what happens for each possible outcome - in fact this is faster than simulating lots and lots of outcomes!
Let $X$ be the number of "successes" among the $n$ customers and let $\hat{p}=X/n$. The confidence interval is $\hat{p}\pm z_{\alpha/2}\sqrt{\hat{p}(1-\hat{p})/n}$, so the halfwidth is $z_{\alpha/2}\sqrt{\hat{p}(1-\hat{p})/n}$. Thus we want to compute $P(z_{\alpha/2}\sqrt{\hat{p}(1-\hat{p})/n}\leq 0.005)$. In R, we can do this as follows:
target.halfWidth<-0.005
p<-0.016 #true proportion
n.vec<-seq(from=1000, to=3000, by=100) #number of samples
# Vector to store results
prob.hw<-rep(NA,length(n.vec))
# Loop through desired sample size options
for (i in 1: length(n.vec))
{
n<-n.vec[i]
# Look at all possible outcomes
x<-0:n
p.est<-x/n
# Compute halfwidth for each option
halfWidth<-qnorm(0.95)*sqrt(p.est*(1-p.est)/n)
# What is the probability that the halfwidth is less than 0.005?
prob.hw[i]<-sum({halfWidth<=target.halfWidth}*dbinom(x,n,p))
}
# Plot results
plot(n.vec,prob.hw,type="b")
abline(0.95,0,col=2)
# Get the minimal n required
n.vec[min(which(prob.hw>=0.95))]
The answer is $n=2200$ in this case as well.
Finally, it is usually a good idea to verify that the asymptotic normal approximation interval actually gives the desired coverage. In R, we can compute the coverage probability (i.e. the actual confidence level) as:
p<-0.016
n<-2200
x<-0:n
p.est<-x/n
halfWidth<-qnorm(0.95)*sqrt(p.est*(1-p.est)/n)
# Coverage probability
sum({abs(p-p.est)<=halfWidth}*dbinom(x,n,p))
Different $p$ give different coverages. For $p$ around $0.015$, the actual confidence level of the nominal $90\%$ interval seems to be about $89\%$ in general, which I presume is fine for your purposes.
(3) When you sample from a finite population, the number of successes is not binomial but hypergeometric. If the population is large compared to your sample size, the binomial works just fine as an approximation. If you sample 1000 out of 5000, say, it does not. Have a look at confidence intervals for proportions based on the hypergeometric distribution!
Answers to additional questions:
Let $(p_L,p_U)$ be the confidence interval.
1) In that case you are no longer computing $P(p_L-p_U\leq0.01)$ but $$P\Big(p_L-p_U\leq0.01~\mbox{and}~p\in(p_L,p_U)\Big),$$ i.e. the probability that the length of intervals that actually contain $p$ is at most 0.01. This may be an interesting quantity, depending on what you're interested in...
2) Maybe, but probably not. If the population size is large compared to the sample size you don't need it, and if it's not then the binomial distribution is not appropriate to begin with!
3) Sprop seems to contain confidence intervals based on the hypergeometric intervals, so that should work just fine.
I might be able to give an answer to your question under certain conditions.
Let $x_{i}$ be your true value for the $i^{th}$ data point and $\hat{x}_{i}$ the estimated value. If we assume that the differences between the estimated and true values have
mean zero (i.e. the $\hat{x}_{i}$ are distributed around $x_{i}$)
follow a Normal distribution
and all have the same standard deviation $\sigma$
in short:
$$\hat{x}_{i}-x_{i} \sim \mathcal{N}\left(0,\sigma^{2}\right),$$
then you really want a confidence interval for $\sigma$.
If the above assumptions hold true $$\frac{n\mbox{RMSE}^{2}}{\sigma^{2}} = \frac{n\frac{1}{n}\sum_{i}\left(\hat{x_{i}}-x_{i}\right)^{2}}{\sigma^{2}}$$
follows a $\chi_{n}^{2}$ distribution with $n$ (not $n-1$) degrees of freedom.
This means
\begin{align}
P\left(\chi_{\frac{\alpha}{2},n}^{2}\le\frac{n\mbox{RMSE}^{2}}{\sigma^{2}}\le\chi_{1-\frac{\alpha}{2},n}^{2}\right) = 1-\alpha\\
\Leftrightarrow P\left(\frac{n\mbox{RMSE}^{2}}{\chi_{1-\frac{\alpha}{2},n}^{2}}\le\sigma^{2}\le\frac{n\mbox{RMSE}^{2}}{\chi_{\frac{\alpha}{2},n}^{2}}\right) = 1-\alpha\\
\Leftrightarrow P\left(\sqrt{\frac{n}{\chi_{1-\frac{\alpha}{2},n}^{2}}}\mbox{RMSE}\le\sigma\le\sqrt{\frac{n}{\chi_{\frac{\alpha}{2},n}^{2}}}\mbox{RMSE}\right) = 1-\alpha.
\end{align}
Therefore, $$\left[\sqrt{\frac{n}{\chi_{1-\frac{\alpha}{2},n}^{2}}}\mbox{RMSE},\sqrt{\frac{n}{\chi_{\frac{\alpha}{2},n}^{2}}}\mbox{RMSE}\right]$$
is your confidence interval.
Here is a python program that simulates your situation
from scipy import stats
from numpy import *
s = 3
n=10
c1,c2 = stats.chi2.ppf([0.025,1-0.025],n)
y = zeros(50000)
for i in range(len(y)):
y[i] =sqrt( mean((random.randn(n)*s)**2))
print "1-alpha=%.2f" % (mean( (sqrt(n/c2)*y < s) & (sqrt(n/c1)*y > s)),)
Hope that helps.
If you are not sure whether the assumptions apply or if you want to compare what I wrote to a different method, you could always try bootstrapping.
Best Answer
If your using linear regression I would recommend using the rms package in R. It is very easy to use and has lots of nice features.
Here's an example:
Gives this output:
Now usually you want to test the linearity assumption:
Gives this output:
And if you plot the graphs with the same code as a bove you get this picture:
If you want to make your formula more complicated just add that variable:
I don't know anything about JMP, it shouldn't be too difficult but I recommend learning R because it gives you an incredible freedom.
Hope this helped.