Question

3. You are designing an antenatal screening programme for a rare but serious disease in the foetus and you have two tests available. Test A, which is based on a blood sample has 95% sensitivity and 95% specificity, whereas test B, which is an invasive inter-uterus test , has 99% sensitivity and 99% specificity. Which test will:

a) minimize false positives?

b) minimize false negatives?

c) assume that you wish to use both screening tests–discuss the pros and cons of doing test A then B or test B then A? Remember that when using screening tests in sequence only those that are positive in the first study are then screened by the second test.

d) Work out the Net Sensitivity and Net Specificity the 2 stage sequential testing programme (A then B) assuming that the tests are applied to 100,000 pregnant women of whom 2% have a foetus with a major abnormality. Show the various stages in your calculations with appropriate tables.

Answer 1

test sensitivity is the ability of a test to correctly identify those with the disease (true positive rate), whereas test specificity is the ability of the test to correctly identify those without the disease (true negative rate)

1.

for false postives to be minimum specificity should be maximum : which is for test B

2.

for false negatives to be minimum specificity should be maximum : which is for test B

3.

This is serial testing [serial : studies are performed sequentially, with the second study dependent on the results of the first]

sensitivity = (A)sen x (B)sen = 0.95*0.99=94.05%

as we can see from the formula in serial testing for sensitivity the order doesn't make any difference

sensitivityA then B = sensitivity B then A = 94.05%

specificity:

A then B : (A)spec + [1 - (A)spec] x (B)spec = 0.95 + (1-0.95)*(0.99) = 0.9995 = 99.95%

B then A : (B)spec + [1 - (B)spec] x (A)spec = 0.99 + (0.01)*0.95 = 99.95%

A then B and B then A are same

d.

A then B :

sensitivityA then B = 94.05%

net sensitivity = (sensitivityA then B)*(P(have disease))*n

= 0.9405*0.02*100,000 = 1881

specificityA then B = 99.95%

net specificity = (specificityA then B)*(P(don't have disease))*n

= 0.9995*0.98*100,000 = 97951

P.S. (please upvote if you find the answer satisfactory)