1

Imagine that a test for a disease is 75% accurate. If someone were to take the test 2 times in a row does it affect the accuracy? i.e. is there then a 2 test accuracy?

I was thinking that the probability that the test is inaccurate is 1/4 and so for it to be wrong 2 times in a row we compute 1/16 and then [1 - (1/4)*(1/4)] = 0.9375 and so the accuracy of 2 tests is 93.75%. I just took the complement of 2 ineffective tests. I am not convinced if this is true or not though because I am not sure I am actually asking a meaningful question. Something seems subtly amiss.

Please note that I AM NOT ASKING the probability that a person actually has the disease because for that we would need the prevalence of disease in the relevant population and we could apply Bayes Theorem to find it. This is just about test accuracy.

Any clarification is appreciated

MattGeo
  • 113
  • 4
    It is unknown without more information about how the test functions. It is common for tests to look for a specific thing, for instance when testing for a particular virus testing to find the presence of antibodies. If the first time the test is taken there aren't enough antibodies to cause the test to say they are there, and the body has not created more antibodies in time for the second test... then the second time the test is taken it may well be that it returns negative again. – JMoravitz Mar 01 '22 at 00:10
  • To progress here, you have to define what the decision is if the core component test yields a T and a F. – David G. Stork Mar 01 '22 at 00:12
  • Some tests are inaccurate for reasons that are likely to be repeated on the second testing. (It's an antibody test, subject just recently infected with not enough time to make antibodies. Some unusual characteristic of patient's blood tends to trigger false positive result even in absence of virus.) In such cases, you clearly can't use anything like $(1/4)^2 1/16/$ – BruceET Mar 01 '22 at 06:33

1 Answers1

1

Let $D,H,+,-$ denote events: diseased, healthy, positive result, and negative result, respectively.

Please note that I AM NOT ASKING the probability that a person actually has the disease because for that we would need the prevalence of disease. This is just about test accuracy.

A medical test's overall accuracy is technically defined as $$P(D+)+P(H-);$$ think of this as the weighted average of its sensitivity $P(+|D)$ and specificity $P(-|H),$ with the weights assigned according to the disease prevalence $P(D);$ so, it does depend on the disease prevalence, unless its sensitivity and specificity are equal, in which case its accuracy simply equals $$P(+|D).$$

is there a 2-test accuracy?

We can coarsely analogously define two-test accuracy as $$P(D++)+P(H--);$$ if we simplistically assume—contrary to what JMoravitz has highlighted—that two tests in a row are actually independent of each other, then it equals $$P(D)P(+|D)P(+|D)+P(H)P(-|H)P(-|H);$$ if additionally the test's sensitivity and specificity are equal, then this two-test accuracy further equals $$[P(+|D)]^2,$$ which, based on these assumptions, is the probability that both tests are correct; however, a better definition of two-test accuracy might take not just $P(D++),$ but also $P(D+-)$ and $P(D-+),$ into account.

Imagine that a test for a disease is 75% accurate. I was thinking that the probability that the test is inaccurate is 1/4 and so for it to be wrong 2 times in a row we compute 1/16 and then [1 - (1/4)*(1/4)] = 0.9375 and so the accuracy of 2 tests is 93.75%. I just took the complement of 2 ineffective tests. I am not convinced if this is true or not though. Something seems subtly amiss.

You have computed $$1-[P(-|D)]^2,$$ which, based on the above assumptions, is the probability that at least one test is correct.

ryang
  • 38,879
  • 14
  • 81
  • 179