There is a test used to detect spina bifida in unborn babies, a disease which af

Question

There is a test used to detect spina bifida in unborn babies, a disease which affects 1 out of 1000 babies who are born. The test causes a positive reaction 5% of the time for a healthy baby, and 100% of the time for a baby with spina bifida.

(a) If the test produces a positive result, what is the probability that baby has spina bifida?

(b) What would the level of the test have to be (currently 0.05), in order to have a greater than 50% chance for a baby who tests positive to have spina bifida?

To answer these questions, you might want to use the Law of Total Probability P(A) = P (A|B)P (B) + P (A|Bc)P (Bc) and/or Bayes’ theorem P (A|B) = P (B|A)P (A) / P(B)

Explanation / Answer

There is a test used to detect spina bifida in unborn babies, a disease which affects 1 out of 1000 babies who are born. The test causes a positive reaction 5% of the time for a healthy baby, and 100% of the time for a baby with spina bifida.

(a) If the test produces a positive result, what is the probability that baby has spina bifida?

Let

D = has disease
P = tests positive

Hence, by Bayes' Rule,

P(P) = P(D) P(P|D) + P(D') P(P|D') = (1/1000)*(1) + (1-1/1000)*0.05 = 0.05095

Hence,

P(D|P) = P(D) P(P|D) / P(P) = (1/1000)*(1)/0.05095 = 0.019627085 [ANSWER]

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(b) What would the level of the test have to be (currently 0.05), in order to have a greater than 50% chance for a baby who tests positive to have spina bifida?

Let x = the level that we need

Hence,

P(P) = P(D) P(P|D) + P(D') P(P|D') = (1/1000)*(1) + (1-1/1000)*x = 0.001 + 0.999x

So

P(D|P) = P(D) P(P|D) / P(P)

= (1/1000)*(1)/(0.001 + 0.999x) = 0.50

= 0.001/(0.001 + 0.999x) = 0.50

Solving for x,

x = 0.001001001 [ANSWER]

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