NBME 4 Block 4 Q17 - vitf

[ArchivalUser]

screening test for colon cancer is administered to 1000 people with biopsy proven colon cancer and to 1000 people without cancer. The test results are positive for 250 of proven cases and 100 of those without colon cancer. The screening test is now to be usedon a population of 100,000 with a known prevelance rate of colon cancer of 80/100,00. what is expected no of TP and Fp in population of 100,000.

Ans is TP=20
FP=9992.

I am able to make the 2 by 2 table , but dont know how to implement the prevelance part..
99hope do explain in detail..
thankx..

[ArchivalUser]

find the true positive rate of the test: 250/1000 = 25%...
the number true negative rate of test : 100/1000 = 10%

80 ppl out of 100000 have the disease...so
True positives are the ppl out of those 80 ... so how much ppl as TRUE POSITIVES the test can find??

25% x 80 = 20

HERE how much population is not diseased(HEALTHY )?


100000- 80 = 99920

The false positive rate is 10% so 10% of this HEALTHY population will be found positive according to the test............................which is the definition of being false positive..
10% x 99920 = 9992

[ArchivalUser]

God bless you...
you are a life saver..
thankx