Solution to Exercise 2.64.

  Problem Statement: In Example 2.30, suppose that the incidence rate for disease is 1 in 25 rather than 1 in 1000. What then is the probability of a positive test result? Given that the test result is positive, what is the probability that the individual has the disease? Given a negative test result, what is the probability that the individual does not have the disease?

Solution: The author is changing the prevalence (percentage of persons in the population who have the disease) from .001 to .04 = 1/25. (``Incidence'' actually means something else, and shouldn't have been used here.) We shall use the notations given in class, i.e. D+ means the event that a person chosen at random is positive for the disease, and T+ means that such a person is positive for the test. The same notations with the minus signs denote the complementary events; e.g. D- means negative for disease. We are given that the sensitivity and specificity are

$$P(T+\vert D+) \; = \; .99 , \quad P(T-\vert D-) \; = \; 1- .02 \; = \; .98 ,$$

respectively. The prevalence is P(D+) = .04, so $P(D^{\prime}) = .96$. Applying Bayes Theorem, the positive predictive power of the test is

$$P(D+\vert T+) \; = \; \frac{P(T+\vert D+) P(D+)} {P(T+\vert D+) P(D+) + P(T+\vert D-) P(D-)}$$

$$\; = \; \frac{.99 \times .04}{.99 \times .04 + (1 -.98) * .96} \; = \; .6735 .$$

The negative predictive power is

$$P(D-\vert T-) \; = \; \frac{P(T-\vert D-) P(D-)} {P(T-\vert D+) P(D+) + P(T-\vert D-) P(D-)}$$

$$\frac{.98 \times (1 - .04)} { (1-.99) \times .04 \, + \, .98 \times (1 - .04)} \; = \; .9996$$

Because the disease is so rare, the negative predictive power is very high.