Question

A genetic test is used to determine if people have a predisposition for thrombosis, which is the formation of a blood clot inside a blood vessel that obstructs the flow of blood through the circulatory system. It is believed that 3% of people have this predisposition. The test is 95% accurate for those who have the predisposition, and 97% accurate for those who do not have the predisposition. Simulate the results for administering this test to a population of 100,000 individuals.

a) How many individuals in this hypothetical population are expected to test positive for the predisposition?

b) Estimate the probability that an individual who tests positive has the predisposition.

c) Suppose that two new tests have been developed. Test A improves accuracy for those who have the predisposition to 98% (while accuracy for those who do not have the predisposition remains at 97%), while Test B improves accuracy for those who do not have the predisposition to 99% (while accuracy for those who do have the predisposition remains at 95%). Which test offers a higher increase in the probability that a person who tests positive actually has the predisposition? Explain the reasoning behind your answer. Limit your answer to at most seven sentences.

Answer 1

Answer:-

Given That:-

A genetic test is used to determine if people have a predisposition for thrombosis, which is the formation of a blood clot inside a blood vessel that obstructs the flow of blood through the circulatory system. It is believed that 3% of people have this predisposition. The test is 95% accurate for those who have the predisposition, and 97% accurate for those who do not have the predisposition. Simulate the results for administering this test to a population of 100,000 individuals.

a) How many individuals in this hypothetical population are expected to test positive for the predisposition?

Given,

We know

P(A|B) = P(B|A) x P(A)/P(B)

P(A) = P(A|B).P(B) + P(A|B') . P(B')

Here

P(Positive) = P(positive/disease).P(disease) + P(tests positive).P(No disease)

Therefore,

P(Positive) = (0.95) (0.03) + (0.03)(0.97)

= 0.0576

= 5.76 %

We know

P(Positive|disease) = 0.95

P(disease) = 0.03 (3%)

P(Positive |no disease) = 0.03

= 1 - 0.97

P(No disease) = 0.97

here disease is predisposition.

Therefore,

Individuals from hypothetical situation that are expected to test are

= 5.76/100 * 100000

= 5760

5760 individuals are expected to test positive .

b) Estimate the probability that an individual who tests positive has the predisposition.

P(Disease|positive)

P(actually has the disease given he was tested positive) = P(tested positive given the disease) * {P(disease/P(tested positive)

P(disease|positive) = P(positive | disease) x P(disease)/P(positive)

P(disease|positive) = 0.95 * 0.03/0.0576

= 0.4947916

49.479 % of people who tested positive have the disease.

P (Predisposition / tested positive) =0.4947

c) Suppose that two new tests have been developed. Test A improves accuracy for those who have the predisposition to 98% (while accuracy for those who do not have the predisposition remains at 97%), while Test B improves accuracy for those who do not have the predisposition to 99% (while accuracy for those who do have the predisposition remains at 95%). Which test offers a higher increase in the probability that a person who tests positive actually has the predisposition? Explain the reasoning behind your answer. Limit your answer to at most seven sentences.

Now new tests are present

(i) Test - A

P(Positive| predisposition.) = 0.98

P(positive|No  predisposition) = 0.03

i.e, 1 - 0.97

P( predisposition) = 0.03

P(no disease) = 0.97

a) P(positive) = (0.98 * 0.03) + (0.03 * 0.97)

= 0.0294 + 0.0291

= 0.0585 i.e, 5.85%

b) P(Actually has  predisposition given positive) = P(disease |positive)

= 0.98 * 0.03/0.0585

= 0.502564

= 50.2%

disease is  predisposition.

ii) Test - B

P(Positive| predisposition.) = 0.95

P(positive|No  predisposition) = 0.01

i.e, 1 - 0.99

P( predisposition) = 0.03

P(no disease) = 0.97

a) P(positive) = (0.95 * 0.03) + (0.01 * 0.97)

= 0.0285 + 0.0097

= 0.0382

i.e, 3.82%

b) P(disease|positive) = P(positive|disease) x P(disease)/P(positive)

= 0.95 * 0.03/0.0382

= 0.74607

= 74.6 %

Therefore Test B has best predisposition.

Test B has the best probability as 74.6% > 50.2%

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