Question

In an article that appears on the web site of the American Statistical Association (www.amstat.org), Carlton Gunn, a public defender in Seattle, Washington, wrote about how he uses statistics in his work as an attorney. He states: I personally have used statistics in trying to challenge the reliability of drug testing results. Suppose the chance of a mistake in the taking and processing of a urine sample for a drug test is just 1 in 100 . And your client has a "dirty" (i.e., positive) test result. Only a 1 in 100 chance that it could be wrong? Not necessarily. If the vast majority of all tests given - say 99 in 100 \- are truly clean, then you get one false dirty and one true dirty in every 100 tests, so that half of the dirty tests are false. Define the following events as \(T D=\) event that the test result is dirty, \(T C=\) event that the test result is clean, \(D=\) event that the person tested is actually dirty, and \(C=\) event that the person tested is actually clean. a. Using the information in the quote, what are the values of \mathbf{i} . ~ \(P(T D \mid D)\) iii. \(P(C)\) ii. \(P(T D \mid C) \quad\) iv. \(P(D)\) b. Use the law of total probability to find \(P(T D)\). c. Use Bayes' rule to evaluate \(P(C \mid T D)\). Is this value consistent with the argument given in the quote? Explain.

Short answer

The values of i. \(P(T D | D) = 1\), ii. \(P(T D | C) = 0.01\), iii. \(P(C) = 0.99\), and iv. \(P(D) = 0.01\). The total probability \(P(T D) = 0.02\). The Bayesian probability \(P(C | T D) = 0.495\). This result is consistent with the argument in the quote indicating that about half of the positive tests could be false positives.

Step-by-step solution

Assign Values

From the statement, we know that:\n- \(P(T D | D) = 1\) (if person is dirty, the test is always dirty), \n- \(P(T D | C) = 0.01\) (there is a 1% chance that a clean person will test dirty), \n- \(P(C) = 0.99\) (99% of people tested are clean), \n- \(P(D) = 0.01\) (1% of people tested are dirty).

Calculate Total Probability

We can use the law of total probability to calculate \(P(T D)\). The total probability law states that the probability of event A is equal to the sum of the probability of A intersecting with each event Bi in a partition of the sample space.\nSo, \(P(T D) = P(T D | D) * P(D) + P(T D | C) * P(C) = 1 * 0.01 + 0.01 * 0.99 = 0.02.\)

Use Bayes' Rule

Bayes' Rule is used to reverse conditional probabilities. The formula is: \n\(P(A | B) = P(B | A) * P(A) / P(B)\)\nWe need to find \(P(C | T D)\), the probability that a person is clean given a dirty test.\nSo, \(P(C | T D) = P(T D | C) * P(C) / P(T D) = 0.01 * 0.99 / 0.02 = 0.495\)\nThis indicates that there is a 49.5% chance that a person is clean even if the test is dirty. This result is consistent with the public defender's statement that 'half of the dirty tests are false'.

Interpretation

The public defender's statistical argument checks out mathematically. For a test that only makes a mistake 1% of the time, if 99% of the people are clean, then about 50% of the positive tests could be false positives. It is important to consider the base rates, in this case the fact that a very large majority of the tested people are clean.