Question

Two tests (A and B) for osteoporosis have the following characteristics:

• Test A has a sensitivity of 60% and a specificity of 80%.
• Test B has a sensitivity of 90% and a specificity of 90%.

If the nurse reversed the order of the testing (i.e. used Test B first and the administered Test A to those who were positive on Test B), what would happen to the net specificity?

a. decrease

b. stay the same

c. increase

d. not enough information to tell

Answer 1

For the purpose for contingency table assuming that 50% of patients have osteoporosis

Scenerio 1:

IF Test A is administered first and then Test B is administed on Test positive cases of Test A

Test A
No disease	Disease
P(D)=0.5	P(D)=0.5
0.5	0.5
Test positive	Test negative	Test positive	Test negative
20.00%	80.00%	60.00%	40.00%
False Positive	True Negative	True positive	False negative
0.1	0.4	0.3	0.2

P(Test A positive) = 0.3

Sensitivity is 90% for Test B.

Hence P(testing positve in Test B) = 0.3 *0.9 = 0.270

Net sensitivity = 0.27 if Test A is administered first and then Test B is administed on test positive cases.

Scenerio 2:

IF Test B is administered first and then Test A is administed on Test positive cases of Test B

Test B
No disease	Tested positive
P(D)=0.5	P(D)=0.5
0.5	0.5
Test positive	Test negative	Test positive	Test negative
10.00%	90.00%	90.00%	10.00%
False Positive	True Negative	True positive	False negative
0.05	0.45	0.45	0.05

P(Test B positive) = 0.45

Sensitivity is 60% for Test A.

Hence P(testing positve in Test A) = 0.45 *0.6 = 0.270

Net sensitivity = 0.27 if Test A is administered first and then Test B is administed on test positive cases.

ANS: Option B : Net specificity stays the same.