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It is known that 5% of the members of a population have
Chapter 3, Problem 215SE(choose chapter or problem)
Problem 215SE
It is known that 5% of the members of a population have disease A, which can be discovered by a blood test. Suppose that N (a large number) people are to be tested. This can be done in two ways: (1) Each person is tested separately, or (2) the blood samples of k people are pooled together and analyzed. (Assume that N = nk, with n an integer.) If the test is negative, all of them are healthy (that is, just this one test is needed). If the test is positive, each of the k persons must be tested separately (that is, a total of k + 1 tests are needed).
a For fixed k, what is the expected number of tests needed in option 2?
b Find the k that will minimize the expected number of tests in option 2.
c If k is selected as in part (b), on the average how many tests does option 2 save in comparison with option 1?
Questions & Answers
QUESTION:
Problem 215SE
It is known that 5% of the members of a population have disease A, which can be discovered by a blood test. Suppose that N (a large number) people are to be tested. This can be done in two ways: (1) Each person is tested separately, or (2) the blood samples of k people are pooled together and analyzed. (Assume that N = nk, with n an integer.) If the test is negative, all of them are healthy (that is, just this one test is needed). If the test is positive, each of the k persons must be tested separately (that is, a total of k + 1 tests are needed).
a For fixed k, what is the expected number of tests needed in option 2?
b Find the k that will minimize the expected number of tests in option 2.
c If k is selected as in part (b), on the average how many tests does option 2 save in comparison with option 1?
ANSWER:
Solution
Step 1 of 3
a) We have to find expected no.of tests needed in option 2
Given that 5% of the population having disease
Then 95% of the population are healthy
Here k people are pooled together
Let N be the no.of tests
If the test is negative
If k people are healthy that is N=1
The probability of k people are healthy is (0.95)k
If the test is positive
The k people must be tested separately that is N=k+1
The probability of k people are having disease is 1-(0.95)k
The expected no.of tests for one group E(N)=1(0.95)k+(k+1) (1-(0.95k))
=(0.95)k+k-k(0.95)k+1-(0.95)k
=1+k(1-0.95k)
Hence the expected no.of tests for ‘n’ groups is E(N)=n[1+ k(1-0.95k)]