# Standard error for the sum of weighted means

I am looking at a two-stage process of diagnosis.

Stage 1 is a screen that identifies people with the diagnosis with some error, Stage 2 is a detailed examination (a gold standard)

In Stage 1, N people were screened, and S were screened positively.

When the S people screened positively were examined in detail it was discovered that G actually had the disease

A 10% random sample of the those screened negative were also examined in detail, and it was found that J actually had the disease (the false negatives from the first stage)

The prevalence of the disease is the sum of the true positives and the false negatives from the first stage, or (G/S)(S/N) + (J/(N-S))((N-S)/S)) = (G+J)/N

My question is, what is the standard error of this prevalence estimate?

• Don't forget that the $J$ is based on a 10% sample, not all of $N-S$. Presumably, there are more false negatives in the other 90% of $N-S$. – jbowman Jan 20 '12 at 20:43

## 1 Answer

It seems like you have:

Estimate/variance of true positive rate P is estimate of success probability given G successes of S Bernoulli trials. Call these values PE and PV

Estimate/variance of false negative rate Q is estimate of success probability given J successes of 0.1(N-S) Bernoulli trials. Call these values QE and QV.

So we want the estimate/variance of P(S/N) + Q((N-S)/N).

Estimate of P(S/N) is PE(S/N) with variance PV((S/N)^2)

Estimate of Q((N-S)/N) is QE((N-S)/N) with variance QV(((N-S)/N)^2).

So you add these together to get an estimate of PE(S/N) + QE((N-S)/N) with variance PV((S/N)^2) + QV(((N-S)/N)^2). Remember that when you add two independent random variables the variance of the sum is equal to the sum of the variances. Then you can take the square root to get the standard error.

There are a couple things I haven't taken into account though:

• There is variance in S as well. For instance, let's say the screen was 100 percent accurate so P=PE=1, Q=QE=0, PV=QV=0. Then the formula above would give zero variance but clearly there still is variance (just the variance in estimated success probability given S successes out of N Bernoulli trials). I'm not sure exactly how you incorporate that in the formula. One thing you could do is rearrange terms to get the estimated prevalence being Q + (P-Q)(S/N) and then use the formula for variance of the product of two independent random variables (P-Q) and (S/N), except that...

• I don't think that (P-Q) and (S/N) are independent. In fact I'm not even sure if P and Q are independent. And certainly Q and (P-Q)(S/N) are not independent so the addition of independent variables formula wouldn't work there.

• I found an article that gives the answer to this, Shrout and Newman, Biometrica 45, 1989 I tried to copy their answer and explanation into this comment but the formatting got all screwed up. It is along the lines of Alex319's answer – Dan Jan 23 '12 at 13:28