How to compute Confidence Interval associated to a Binomial proportion's increase? Not sure if you can help me but here's the problem.
I have two binomial proportions A and B (95% CI) -
$A = 2\% \pm 0.2\%,\quad B = 3\% \pm 0.2\%.$
In other words, B's proportion is 50% higher than A's.
Business people tend to prefer speaking about this difference in terms of a percentage increase. In other words, B's proportion is 50% higher than A's. How do I go about calculating a Confidence Interval for the 50% increase of B's proportion over A.
I've done some research into Fieller’s formula which is said to provide exact percent effect. However, besides not understanding Fieller's theorem, I don't know how I would go about calculating this in R.
Can anyone help? 
Edit: E.g., if the CI for those intervals is 95%, the CI for the 50% increase = 98.52%.
 A: Following whuber's link to Wikipedia you have 

Assume that $a$ and $b$ are jointly
  normally distributed, and that $b$ is
  not too near zero (i.e. more
  specifically, that the standard error
  of $b$ is small compared to $b$)
$$\operatorname{Var} \left( \frac{a}{b} \right) = \left(
 \frac{a}{b} \right)^{2} \left( \frac{\operatorname{Var}(a)}{a^2} +
 \frac{\operatorname{Var}(b)}{b^2}\right).$$

though in fact you want $\operatorname{Var} \left( \frac{B}{A} \right)$.
If your 95% CI is $\pm 0.002$ then your variances for $A$ and $B$ are $(0.002/1.96)^2 \approx 0.00000104$, so $\operatorname{Var} \left( \frac{B}{A} \right) \approx 0.00846$. Taking the square root and multiplying by 1.96 you get $$\frac{B}{A} \approx 1.5 \pm 0.18$$
If you must turn this into percentages (I think it confuses more than it enlightens) then it becomes

B's proportion is 50% higher than A's, plus or minus 18%, i.e. between 32% higher and 68% higher.

In R you could simulate this by something like
> n <- 1000000
> A <- 0.02 + (0.002 / qnorm(0.975)) * rnorm(n)
> B <- 0.03 + (0.002 / qnorm(0.975)) * rnorm(n)
> C <- B / A
> quantile(C,  probs = c(0.025, 0.5, 0.975))
    2.5%      50%    97.5% 
1.333514 1.499955 1.697418 

which is reasonably close.
