Screening test is 98% effective in detecting a certain disease when a person has the disease. however, the test yields a false positive rate of 1% of the healthy persons tested. if 0.1% of the population have the disease, what is the probability that a person who tests positive has the diseas
Answers
Answer:
"A patient goes to see a doctor. The doctor performs a test with 99 percent reliability--that is, 99 percent of people who are sick test positive and 99 percent of the healthy people test negative. The doctor knows that only 1 percent of the people in the country are sick. Now the question is: if the patient tests positive, what are the chances the patient is sick?"
The intuitive answer is 99 percent, but the correct answer is 50 percent...."
The solution to this question can easily be calculated using Bayes's theorem. Bayes, who was a reverend who lived from 1702 to 1761 stated that the probability you test positive AND are sick is the product of the likelihood that you test positive GIVEN that you are sick and the "prior" probability that you are sick (the prevalence in the population). Bayes's theorem allows one to compute a conditional probability based on the available information.
Bayes's Theorem
equation image indicator
P(A) is the probability of event A
P(B) is the probability of event B
P(A|B) is the probability of observing event A if B is true
P(B|A) is the probability of observing event B if A is true.
Wiggins's explanation can be summarized with the help of the following table which illustrates the scenario in a hypothetical population of 10,000 people:
Diseased
Not Diseased
Test +
99
99
198
Test -
1
9,801
9,802
100
9,900
10,000
In this scenario P(A) is the unconditional probability of disease; here it is 100/10,000 = 0.01.
P(B) is the unconditional probability of a positive test; here it is 198/10,000 = 0.0198..
What we want to know is P (A | B), i.e., the probability of disease (A), given that the patient has a positive test (B). We know that prevalence of disease (the unconditional probability of disease) is 1% or 0.01; this is represented by P(A). Therefore, in a population of 10,000 there will be 100 diseased people and 9,900 non-diseased people. We also know the sensitivity of the test is 99%, i.e., P(B | A) = 0.99; therefore, among the 100 diseased people, 99 will test positive. We also know that the specificity is also 99%, or that there is a 1% error rate in non-diseased people. Therefore, among the 9,900 non-diseased people, 99 will have a positive test. And from these numbers, it follows that the unconditional probability of a positive test is 198/10,000 = 0.0198; this is P(B).
Thus, P(A | B) = (0.99 x 0.01) / 0.0198 = 0.50 = 50%.
From the table above, we can also see that given a positive test (subjects in the Test + row), the probability of disease is 99/198 = 0.05 = 50%.