Chance of having covid given negative test

bayesianprobability

I was reading an article (https://www.healthline.com/health/how-accurate-are-rapid-covid-tests#at-home-tests) on the accuracy of Rapid Lateral flow tests and was interested in finding out what my chance of having covid are given that I have tested negative.

The article stated that people without symptons …..

  • If they have covid they correctly test positive 58.1%
    So from this can we assume that ..
  • If they have covid they incorrectly test negative 41.9%?

So if I want to work out p(c|neg) then this is worked out by formula p(c,neg) / (p(c,neg)+p(notC,neg)

The article didn't say what the real % of people in population who actually have covid is. I guess because this is something that we don't really know for certain. But I read one estimate being 39.42 out of 10000 people, so 0.003942

I read in another article that the specitivity of Rapid flow tests ranged from 0.924 to 1 so if I take the middle of the range as being 0.962

So the probabilities are ..
p(c) = 0.003942
p(nc) = 0.9961
p(neg|c) = 0.419
p(neg|nc) = 0.962

So does this mean that p(c|neg) = (0.003942×0.419)/(0.003942×0.419)+(0.9961×0.962)) = 0.0017
or 0.17%. I feel like I must have made a mistake in my calculation somewhere.

Also, if I take 5 tests all on the same night and they all return negative results does this mean that the chance that I have covid is 0.0017^5 or 1.419857e-14

I wanted to work out what my chance of having covid anytime over the last year would be given I have had say 8 negative test results, but this all depends on when the tests were taken i.e. if I take one test and get a negative result and then take another test and get a negative result the chance of the second test being accurate is higher than if it was taken 3 weeks later. And it would all depend on the rate of covid within population when the tests were taken?

Best Answer

$S:\quad$ symptomatic
$C:\quad$ COVID
$+:\quad$ positive test result

What you are interested in called the false omission rate $P(C|-),$ which I've written about it in some detail here and here. The following answer will be more satisfying after having returned from those two.

The article stated that people without symptoms.....

  • If they have covid they correctly test positive 58.1% So from this can we assume that ..
  • If they have covid they incorrectly test negative 41.9%?

The original passage

The researchers found that people without COVID-19 symptoms correctly tested positive in 58.1 percent of rapid tests.

is ambiguous, as the $58.1\%$ could refer to $$P(+|S^c\cap C)$$ or $$P(C\cap+|S^c)$$ or $$P(C|+\cap S^c).$$

I'm guessing the intended meaning is the first option, which agrees with your suggestion about the $41.9\%.$

So the probabilities are ..
p(c) = 0.003942
p(nc) = 0.9961
p(neg|c) = 0.419
p(neg|nc) = 0.962

So does this mean that p(c|neg) = (0.003942x0.419)/(0.003942x0.419)+(0.9961x0.962)) = 0.0017 or 0.17%. I feel like I must have made a mistake in my calculation somewhere.

Your derivation is correct, but according to our interpretation above, it is $P(-|S^c\cap C)$—not $P(-|C)$— that equals $41.9\%.$ Thus, the required value isn't $0.17\%.$

Also, if I take 5 tests all on the same night and they all return negative results does this mean that the chance that I have covid is 0.0017^5 or 1.419857e-14

No, you want $P(C|-\cap-\cap-\cap-\cap-)$ instead of $\big[P(C|-)\big]^5.$ Refer to the bottom of the first linked answer above for an explanation.

And it would all depend on the rate of covid within population when the tests were taken?

Yes. This is also addressed in the linked answers.

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