Question

Consider a diagnostic test for a hypothetical disease based on measuring the amount of a certain biomarker present in blood. High levels of the biomarker are often found in individuals with the disease, but a number of non-disease conditions can also cause high levels of the biomarker. Individuals without the disease have biomarker levels that are normally distributed with mean 1.6 ng/mL (nanograms per milliliter of blood), and standard deviation 0.50 ng/mL. Individuals with the disease have biomarker levels that are normally distributed with mean 5 ng/mL and standard deviation 1.2 ng/mL. Values of 2.5 ng/mL and higher constitute a positive test result. i. Compute the accuracy of the test for those who have the disease and the accuracy of the test for those who do not have the disease. In a population where 6% of individuals are thought to have the hypothetical disease, calculate the probability that an individual who tests positive has the disease.

Answer 1

We are given here that:
P( disease ) = 0.06,

Hence, P(no disease) = 1 - 0.06 = 0.94 which is the probability of not having the disease

Note that we would be using the law of total probability and the Bayes rule later on to get the required probability here.

Now for the person without disease, we have here:

Therefore the probability of a positive test without the disease is computed here as:
P(+ | no disease) = P(X₁ > 2.5)

Converting it to a standard normal variable, we have:
P(+ | no disease) = P(Z > (2.5 - 1.6)/ 0.5 )

P(+ | no disease) = P(Z > 1.8)

Getting it from the standard normal tables, we have:
P(+ | no disease ) = 0.0359

Similarly now for the person with the disease, we have:

The positive test probability here is computed as:
P(+ | disease) = P(X₂ > 2.5)

P(+ | disease) = P(Z > (2.5 - 5) / 1.2)

P(+ | disease) = P(Z > -2.0833)

P(+ | disease) = 0.9814

Using law of total probability now, we have:
P(+) = P(+ | disease)P(disease) + P(+ | no disease)P(n0 disease)

= 0.9814*0.06 + 0.0359*0.94 = 0.0926

Therefore using Bayes theorem now, we get here:
P(disease | +) = P(+ | disease)P(disease) / P(+) = 0.9814*0.06 / 0.0926 = 0.6357

Therefore 0.6357 is the required probability here.