Question

2. Suppose next that we have even less knowledge of our patient, and we are only given the accuracy of the blood test and prevalence of the disease in our population. We are told that the blood test is 90 percent reliable, this means that the test will yield an accurate positive result in 90% of the cases where the disease is actually present. Gestational diabetes affects 0+1 percent of the population in our patient’s age group, and that our test has a false positive rate of 0+4 percent. Use your knowledge of Bayes’ Theorem and Conditional Probabilities to compute the following quantities based on the information given only in part 2:

1. If 100,000 people take the blood test, how many people would you expect to test positive and actually have gestational diabetes?
2. What is the probability of having the disease given that you test positive?
3. If 100,000 people take the blood test, how many people would you expect to test negative despite actually having gestational diabetes?
4. What is the probability of having the disease given that you tested negative?

Comment on what you observe in the above computations. How does the prevalence of the disease affect whether the test can be trusted?

Answer 1

The required information is given by D= Disease and ND = No Disease let TP = Test Positive TN = Test Negative Here Plopilo) - 0.9 PIINID =1-09 POTNID) = 0.1 The gestational diabetes affects I percent of the population in our patient's age group, PCD) - to PCD) = 0.01 Therefore, P(ND) = 1-0.01 P(ND) = 0.99 and that our test has a false Positive rate of 4 percent P(TPIND) = = 0.04 Therefore, P (TNIND) - 1-0.04 PITNIND = 0.96 ma) Naca first to find P(test positive and actually have gestational diabetes) - płtp and D) = 9

Now By using conditional probability P(APID)=P(1p and D) POD) e P(TP and D) 0.90 0.01 P (Tp and D) = 0.90*0.01 1. Plap and D) = 0.009 refore the number of people expected to test Positive and actually have gestational diabetes = 1,00,000 X0.009 = 900 (b) To find the probability of having the disease given that you test Positive To FP (DITP): - Now By using Bayes rule PGPIP) XP(P) PGT PID)P(D)+P(TP IND) xP (ND) P (PHP) = et - 0.90 X0.01 0.90 X0.01 + 0.04 X0.99 0.009 0.009 + 0.0396 0.009 0.0486 = 0.1852 P (DITP) = 0.1852] and despite actually having (c) Now to find P(test negative gestational diabetes

ie to find p (TN and o) Now By using conditional Probability formwa POP) 0.1 Plan and o). P(TND) - P(TN and D) 0.01 PT Nand o) = 0.1X0.01 = 0.001 Therefore the number of people expected to test negative despite actually having gestational diabetes = 1,00,000 X 0.001 = 100 (d? To find platin) :- By using Bayes rule P(TND) XP(O) Plottn) - P(TNI)*P(D) +P (TNIND) X PIND) 0.1*0.01 PlpITN) = 0.1X0.01 7 0.96 X 0.99 0.001 0.001 +0.9504 0.001 0.9514 = 0.00105 ::P (DIN) = 0.00105 Therefore the probability of having the You tested negative is 0.00105 disease given that - You tested