Question

Approximately 50% of the U. S. population has a genetic condition. Suppose to Determine the condition a rapid test is taken which has a specificity of 99.4% and a sensitivity of 100%.

23.1
Suppose a person from the U. S. population is chosen at random, given the test, and tests positive for the condition. What is the probability this person has it? What is the probability they do not have it?
23.2
Suppose a person is chosen from an "at-risk" population where the rate of the condition 10%. If that person tests positive for the condition, what is the probability they have the condition? 23.3
Using your results from (a) and (b), give an argument for or against a policy of random testing of the population. (random : implying testing someone without regard for the individual being "at-risk" or not.)

Answer 1

Step-by-step explanation:

23.1

Let's calculate the probability that a person has the genetic condition given that they tested positive, using the information provided.

Let:

P(C) = Probability of having the genetic condition (prevalence) = 0.50 (50%)

P(T|C) = Probability of testing positive given that the person has the condition (sensitivity) = 1.00 (100%)

P(T|¬C) = Probability of testing positive given that the person does not have the condition (1 - specificity) = 1 - 0.994 = 0.006

We can use Bayes' theorem to calculate the probability of having the condition given a positive test result:

P(C|T) = (P(T|C) * P(C)) / P(T)

To calculate P(T), we can use the law of total probability:

P(T) = P(T|C) * P(C) + P(T|¬C) * P(¬C)

Since P(¬C) = 1 - P(C), we have:

P(T) = P(T|C) * P(C) + P(T|¬C) * (1 - P(C))

Substituting the given values:

P(T) = (1.00 * 0.50) + (0.006 * 0.50) = 0.503

Now we can calculate P(C|T):

P(C|T) = (P(T|C) * P(C)) / P(T)

= (1.00 * 0.50) / 0.503

≈ 0.995

Therefore, the probability that a person has the genetic condition given that they tested positive is approximately 0.995 (99.5%). The probability that they do not have the condition would be 1 - 0.995 = 0.005 (0.5%).

23.2

Let's calculate the probability that a person from the "at-risk" population has the genetic condition given that they tested positive.

Given:

P(C) = Probability of having the genetic condition in the "at-risk" population = 0.10 (10%)

Using the same formula as in 23.1:

P(C|T) = (P(T|C) * P(C)) / P(T)

We already calculated P(T) as 0.503 in 23.1.

P(C|T) = (1.00 * 0.10) / 0.503

≈ 0.199

Therefore, the probability that a person from the "at-risk" population has the genetic condition given that they tested positive is approximately 0.199 (19.9%).

23.3

Given the results from 23.1 and 23.2, we can make an argument against a policy of random testing of the population.

In 23.1, we found that the probability of having the condition given a positive test result was approximately 99.5% for the general U.S. population. However, in 23.2, when considering an "at-risk" population with a known higher prevalence of the condition, the probability decreased to approximately 19.9% given a positive test result.

This shows that the context and prevalence of the condition greatly impact the interpretation of a positive test result. Random testing of the population without considering individual risk factors or prevalence could lead to a significant number of false positives, causing unnecessary stress, follow-up tests, and potential misdiagnoses.

To maximize the effectiveness and efficiency of testing, it is advisable to focus resources on targeted testing based on risk factors, symptoms, or other relevant criteria. This approach can improve the accuracy of the test results and reduce the burden on individuals and healthcare systems.