False and true positives in research

Question

Let's suppose that there's a new apple tree disease, and there exists a test for the disease that has a false positive rate of 5%, the test produces no false negatives (if an apple tree has the disease, it's positive).

If I then tested a random number of apple trees and 10 were positive, what would be the probability that at least 75% of the positive trees were truly positive?

Could someone help me with this problem, or point me in the right direction, because I have no ide where to even begin?

I've read this article and a few others, but I couldn't apply any of them to my situation.

Thanks in advance!

Answer 1

You're interested in the quantity

$$ p = P(D+ \vert T+) $$

Which from Bayes theorem is

$$ p= {P(T+\vert D+) P(D+) \over P(T+\vert D+) P(D+) + P(T+\vert D-) P(D-) }$$

We see our false positive rate in the denominator, so we can substitute that in

$$ p={ P(T+\vert D+) P(D+) \over P(T+\vert D+) P(D+) + 0.05 (1-P(D+)) }$$

Note also that $P(T+ \vert D+) = 1 - P(T- \vert D+)$. The probability on the right hand side is the false negative rate which is 0, so

$$p = {P(D+) \over P(D+) + 0.05 (1-P(D+))} $$

So $P(D+ \vert T+)$ depends on the prevalence of this disease. Assuming the sample is iid and the sample large enough, you could use the binomial distribution to determine the probability of 3/4 or more of your positive samples having the disease. Let $X$ be the number of disease positive cases who have a positive test.

$$ P(X>7) = \sum_{k=8}^{10} \binom{10}{k} p^k(1-p)^{10-k} $$

We can compute this as a function of the prevalence

prev = seq(0, 1, 0.01)
# Result from Bayes Theorem
p = prev/(prev + 0.05*(1- prev))

# Probability you're seeking
proba = sapply(p, function(x) sum(dbinom(8:10, size = 10, x)))
plot(P_D_pos, proba, type='l', xlab = "Disease Prevalence", ylab = expression(P(X>7)))