Question

The sensitivity of a diagnostic test is the probability that a true positive (e.g. someone with the disease under question) is correctly identified as such. The specificity, on the other hand, is the probability that a true negative (e.g. a healthy person) is not reported as a positive. You have a COVID-19 antibody test with specificity α and sensitivity β. There are N people in a particular population, and suppose we already know that n of them have the disease.

(a) What is the probability that someone picked out of this population will be correctly diagnosed (either as a positive or a negative)?

(b) What is the probability that a positive diagnosis means someone actually has the disease? Or that a negative diagnosis means they don’t have it?

(c) Only ten people out of a sampled population of 10,000 actually have the disease. Let α = 0.97 and β = 0.95. How many false positives and false negatives are expected?

(d) Now let’s say 9,000 actually have the disease. How many false positives and false negatives are expected this time?

Answer 1

sensitivity of a diagnostic test = P (a true positive is correctly identified) = β

specificity of a diagnostic test = P (a true negative is not reported as a positive) = α

& given that n out of N have the disease.

a) P (someone picked out of this population will be correctly diagnosed)

= P (the person is choosen from n affected persons & identified correctly) + P (the person is choosen from (N-n) healthy persons & not reported as a positive)

=

b) P (a diagnosis results positive given that someone actually has the disease)

  

similarly, P (a diagnosis results negative given that someone don't have the disease)

c)Only 10 people out of a sampled population of 10,000 actually have the disease

=> P (false positive) = P(the person is choosen from (N-n) healthy persons & reported as a positive)

= 0.02997

=> expected no. of false positive = 10000*0.02997 = 299.7

and, P (false negative) = P (the person is choosen from n affected persons & identified negative)

= 0.00005

=> expected no. of false negative = 10000*0.00005 = 0.5

d) If 9,000 actually have the disease,

P (false positive) = 0.003

=> expected no. of false positive = 10000*0.003 = 30

P (false negative) = 0.045

=> expected no. of false negative = 10000*0.045 = 450