# Finding a statistically different proportion

I have the following problem: given a proportion $p_1=\frac{X_1}{n}$, where $X_1$ is the number of succesful outcomes in $n$ tries ($X_1$ and $n$ are both > 100), I have to find the smallest proportion $p_2$ that is greater than and statistically different from $p_1$ at a given confidence level (e.g. 95%). An application of this would be to find out a success rate (i.e., $X_2$ successes in $n$ tries) that would ensure that a given enhancement to a system would actually be beneficial.

I could find the confidence interval for $p_1$ but merely having $p_2$ outside this interval wouldn't be enough to assure these proportions would be different. Doubling the interval didn't seem right either.

I also tried using the formulas for the confidence interval for the difference of proportions

$c = p\mp z\cdot s_p$

where $p = p_1-p_2$ is the difference between the proportions and $s_p$ is given by

$s_p = \sqrt{\frac{p_1(1-p_1)+p_2(1-p_2)}{n}}$

but I couldn't figure out how to get $p_2$ from them. Any hints on how I might proceed?

For what it's worth, this is not homework, but a slightly reworded problem from a book I'm using for self-study.

EDIT: Glen's answer put me in the right direction. It also occurred to me to brute force a solution, posted below.

• "a slightly reworded problem from a book I'm using for self-study" ... which sounds like it might still count as worthy of the self-study tag. (take a look and decide for yourself) Commented Apr 14, 2014 at 4:05
• Your 'brute-force' answer at the end is a legitimate way of solving the problem. In practice, often all that's needed in practice is a fast numerical answer in this case, not an algebraic answer in every case. I also use such approaches to double check my algebra, or to guide my intuition. I encourage you to post it below as an answer, so that your question and answer stand on their own, as question and answer. Commented Apr 14, 2014 at 20:50

This is pretty straightforward.

Consider that framed in terms of a confidence interval you need an interval for $p_2-p_1$ to be all positive:

In particular, you need the lower limit of the interval to be positive (that the rest is positive then follows automatically).

If we assume $n$ is large enough to apply the normal approximation, then the lower limit is $p_2 - p_1 - z_{1-\alpha/2}s_p$ and you need

$p_2 - p_1 - z_{1-\alpha/2}s_p >0\,$, or

$p_2 - p_1 - z_{1-\alpha/2}\sqrt{\frac{p_1(1-p_1)+p_2(1-p_2)}{n}}>0\,$.

Here $p_1$, $n$ and $z_{1-\alpha/2}$ are all known, so $p_2$ is the only variable.

Can you do it from there?

The algebra is doable, but if it gets too hard, one can simply use simple numerical calculation over a grid (and then find the zeroes) to identify the region that satisfies the conditions $p_2>p_1$ and $p_2 - p_1 - z_{1-\alpha/2}s_p >0\,$.

In addition to Glen's analytical approach, I also developed an iterative solution (kind of a brute force solution, actually). R snippet follows:

prop2 <- function(x1, n1, n2=n1, ascending=TRUE, conf.level=0.95) {
p1 <- x1/n1
start <- end <- 0
if (ascending) {
start <- round(p1*n2)+1
end <- n2
} else {
start <- round(p1*n2)-1
end <- 1
}
alpha <- 1-conf.level
z <- qnorm(1-alpha/2)
found <- FALSE
for (x2 in start:end) {
p2 <- x2/n2
p <- p1-p2
sp <- sqrt(p1*(1-p1)/n1 + p2*(1-p2)/n2)
ci <- p + c(-1,1)*z*sp
if ((ci[1] < 0 & ci[2] < 0) | (ci[1] > 0 & ci[2] > 0)) {
cat(sprintf("x2=%d, p2=%f, ci=[%f ; %f]\n", x2, p2, ci[1], ci[2]))
found <- TRUE
break
}
}
if (!found)

So, prop2(x1, n1) will show the values for $x_2$ and $p_2$ that solve the problem. With ascending=FALSE, it finds the largest $p_2$ that is less than and statistically different from $p_1$. It also works for $n_2\not= n_1$, and for any confidence level (just set conf.level).