Tobias, 2005Econometrics 472 Down’s Syndrome Example

Tobias, 2005Econometrics 472 The Variables Consider the case of prenatal testing for Down's syndrome. Let D=1 indicate that a baby has Down's syndrome and D=0 indicate that a baby does not have the disease. In addition, suppose there is a test available (denoted T) that can help to determine if the baby has Down's Syndrome prior to birth. Let T=1 denote the event that the test suggests a positive screen for Down's syndrome, and T=0 denote a negative screen.

Tobias, 2005Econometrics 472 Information About the Disease We know the following statistics (these numbers vary according to different reports, but we will use them in this example): Pr(D=1) = 1/800 ≈.0013 Pr(D=0) = 1 - Pr(D=1) =.9987

Tobias, 2005Econometrics 472 Information About the Test In addition, we have some information regarding the accuracy of the test itself. In particular, we know that if the baby has the disease, then 80 percent of the time, the test will screen positive (T=1). This tells us that: Pr(T=1 | D=1) =.8 → Pr(T=0 | D=1) =.2 We will define the false positive rate as the probability that the test screens positive given that the baby does not have the disease. We will denote this probability abstractly as c: Pr(T=1 | D=0) = c → Pr(T=0 | D=0) = 1-c. There are a variety of screening tests, with varying degrees of accuracy. One common test (CVS) reports a false positive rate of c=.05. We will compute probabilities of interest for a variety of values of c.

Tobias, 2005Econometrics 472 What Parents Care About What parents care about is the probability of disease (or no disease) given the result of the screening test. In particular, I would be most interested in the following probabilities: Pr(D=1 | T=1) and Pr(D=0 | T=1) which are the probabilities that the baby does or does not have Down's Syndrome given that the screen was positive, and Pr(D=1 | T = 0) and Pr(D = 0 | T=0 ), the probabilities that the baby does and does not have Down's Syndrome given that the test screen is negative.

Tobias, 2005Econometrics 472 Working it Out Consider the quantity Pr(D=1 | T=1) We reduce this quantity to a functions of known quantities on the following page:

Tobias, 2005Econometrics 472

Tobias, 2005Econometrics 472 Working it out, continued… The above can be calculated for a variety of values of the false positive rate c. In particular, at c =.05, we evaluate Pr(D=1|T=1) =.02. How do we interpret this number? Does it make sense?

Tobias, 2005Econometrics 472 Interpreting the Results The results suggest that the probability of having Down's syndrome, given a positive test result is only about 2 percent! The relatively high false positive rate makes it difficult for us to actually conclude that the baby has the disease. However, the revised probability (after finding out a positive screen) of 2 percent is 16 times greater than the unconditional probability of having the disease (1/800). So, children with a positive screen are indeed at greater risk, but are still relatively unlikely to actually have the disease! (Re- calculate this by making c much smaller. Interpret your results).

Tobias, 2005Econometrics 472 Interpreting a Negative Screen What about the other quantity of interest, the probabilities of having or not having the disease if the test screen comes back negative, i.e. Pr(D=1|T=0) and Pr(D=0|T=0)? We can perform a similar calculation to evaluate these quantities:

Tobias, 2005Econometrics 472

Tobias, 2005Econometrics 472 Interpreting a Negative Screen Evaluated at c =.05, we find Pr(D=0|T=0) = Thus, if the test comes back negative, there is still a very small chance that the baby has Down's Syndrome. Also note that this number is larger than.9987, which is the overall probability of not having Down's Syndrome in the population. Finally, note that even if c=0 so that there is no possibility of a false positive, you still can not be certain that your baby will not have the disease. This is because 20 percent of the time, the test will fail to identify a baby with Down's Syndrome when the baby does, in fact, have the disease.