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%.

  • 2
    $\begingroup$ Assuming A and B are uncorrelated, use the variance propagation formula on Wikipedia at en.wikipedia.org/wiki/Fieller%27s_theorem#Case_1 . Because your CIs are so small, we know the sampling distributions are approximately normal and the sample sizes are fairly large. Therefore you can back-calculate Var(A) and Var(B) from your CIs in the usual way, for input in the variance propagation formula. (BTW, your edit is mysterious. Perhaps you are confusing the coverage of a CI with the interval itself?) $\endgroup$ – whuber Feb 1 '11 at 16:23
  • $\begingroup$ You also might want to think about measurement error. e.g., if we're talking about proportion of people who agree with "A," and 2% of everyone says the opposite of their true feelings, then most of the people who say "A," don't really agree with "A." $\endgroup$ – Michael Bishop Mar 22 '11 at 21:50

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.


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