Using Bayes Theorem to calculate probability of false negative test result

bayes-theoremconditional probabilityprobability

I've been reading up on Bayes Theorem and thought I'd try to apply it to a hypothetical medical test, but I'm not sure I'm applying it correctly.

I contrived this scenario:

  • A medical test has a sensitivity of 60%. In other words, the false
    negative rate is 40%.
  • The test also has a specificity of 90%.
  • The prevalence of the particular disease in the population is 5%.
  • I have no symptoms of the disease, but I take the test anyway and get a
    negative result.
  • I want to know the probability that I received a false negative.

I was thinking this could be calculated as

$$ P(\text{False Negative Result}) = P(\text{Disease|Negative}) = \frac{P(\text{Negative|Disease})\cdot P(\text{Disease})}{P(\text{Negative})} = \frac{(1-\text{Sensitivity})\cdot P(\text{Disease})}{P(\text{Negative})}$$

So in this example…

$$ P(\text{False Negative Result}) = \frac{(1-0.60)(0.05)}{((1-0.60)(0.05) + (0.90)(1-0.05))} = 0.02 $$

But I'm not sure if I'm missing something?

Best Answer

Now my only confusion is how the P(False Negative Result) differs from the Negative Predictive Value.

Consider this probability tree:

https://i.stack.imgur.com/ZPmMO.png

p:  disease prevalence and other (prior) risk factors
v:  test sensitivity
f:  test specificity
D:  Diseased
H:  Healthy
+:  Positive test result
-:  Negative test result

The negative predictive value (NPV) is $$P(H|-)=\frac{P(H-)}{P(H-)+P(D-)}=\frac{(1-p)f}{(1-p)f+p(1-v)};$$

the true negative rate (specificity) is $$P(-|H)=f;$$

P(true-negative result) is $$P(H-)=(1-p)f;$$

P(false-negative result) is $$P(D-)=p(1-v);$$

the false omission rate (complement of the NPV) is $$P(D|-)=\frac{P(D-)}{P(D-)+P(H-)}=\frac{p(1-v)}{p(1-v)+(1-p)f};$$

the false discovery rate (complement of the PPV) is $$P(H|+)=\dots.$$

$P(\text{False Negative Result}) = P(\text{Disease|Negative})$

“$P(\text{False Negative Result})$” ambiguously could mean either $P(\text{False-Negative Result})$ or $P(\text{Disease|Negative}).$

The former $(2.00\%)$ is the second outcome (of four possible outcomes) in the probability tree, whereas the latter $(2.29\%)$ is conditioned on the knowledge that only outcomes 2 & 4 are possible.

Would it be correct to assume the Negative Predictive Value is associated with a test, but the P(False Negative Result) is associated with an event?

No, each term above is associated with both a test and an event. Some of them are conditional probabilities, whereas the others are not working with a reduced sample space.

Read more here: the accuracy of a medical test.

Related Question