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  We define D+ as the event a randomly selected person is positive for the disease of interest, and T+ the event the person tests positive. The events D- and T- are the corresponding complementary events of negative for disease and testing negative, respectively.

Suppose that the prevalence is given by

$$P(D+) \; = \; .25 .$$

Suppose that the sensitivity of the test is

$$P(T+\vert D+) \; = \; .8 ,$$

and the specificity is

$$P(T-\vert D-) \; = \; .4 .$$

(a) Compute the positive predictive power P(D+|T+).

Solution: Using Bayes Theorem,

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

$$\; = \; \frac{.8*.25}{.8*.25 + (1-.4)*(1-.25)} \; = \; 0.3076923$$

In spite of the great sensitivity (0.8), the test has a low specificity (0.4), so the probability of disease after test is only .308, compared with the probability of .25 before test (the prevalence).

(b) Compute the negative predictive power P(D-|T-).

Solution:

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

$$\; = \; \frac{.4*.75}{(1-.8)*.25 + .4*.75} \; = \; 0.8571429$$

Note that the probability of no disease prior to test was .75, so it has moved up to about .86 after a negative test, not a dramatic improvement (because of the low specificity, again), but still some improvement.

Problems? Questions? (I may have made a mistake, so don't hestitate to ask) dcox@stat.rice.edu