Using Bayes Theorem to calculate probability of false negative test result

Question

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?

Answer 1

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

Consider this probability tree:

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.

Answer 2

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

your calculation about false negative probability is correct. The result is about $2.29\%$

The difference between a negative (or positive) predictive value and the false negative (or positive) result is mainly that the predictive value is the probability that your disease situation is CONCORDANT with your test result

• PPV: is the probability that you actually are sick, given that the test is positive

• NPV: is the probability that you are actually healthy, given that the test is negative

What you are asked to calculate, is the probability that you actual situation is NOT CONCORDANT with test result; in particular, you correctly calculated the following probability

$$\mathbb{P}[\text{Sick}|T^-]=\frac{2.0}{87.5}$$

In these situation, a tabular representation using a contingency table will be very useful to visualize the situation