5 ms·
Well, given that 90/15000 = 0.6% in the control group developed the disease, you can consider the vaccine group as a Bernouili trial of n = 15000 with probabili
by roland_nilsson 6y ago
Well, given that 90/15000 = 0.6% in the control group developed the disease, you can consider the vaccine group as a Bernouili trial of n = 15000 with probability p = 0.6%. Then the probability of observing 5 or fewer cases is 3.4*10^-32, from the tail probability of the binomial distribution.
https://en.wikipedia.org/wiki/Bernoulli_trial https://en.wikipedia.org/wiki/Bernoulli_trial
Of course, that's assuming that five guys from the vaccine group didn't get infected at the same after-ski party, or any funny business that violates statistical independence ...
- whimsicalism 6y agoNaive question: If the null hypothesis is that there is no difference, wouldn't that imply 95/30,000, p = 0.0031, putting the probability of observing less than 5 cases at a much more reasonable 1*10^(-14).
- subroutine 6y agoI think Fisher's exact test [1] is most commonly used in these types of trials. But the P-value roland (parent comment) provided also make sense to me. For the Fisher's test R says... Covid NoCovid _____ _______ Vaccine 5 1495 NoVaccine 90 1410 9.0e-22 one tailed 4.5e-22 two tailed [1] https://en.wikipedia.org/wiki/Fisher%27s_exact_test https://en.wikipedia.org/wiki/Fisher%27s_exact_test
- flobosg 6y agoShouldn't the NoCovid values be 14995 and 14910?
- subroutine 6y agoAh yeah good catch (2.7e-21 for 2-tailed, mutatis mutandis)
- roland_nilsson 6y agoNo, I think you must estimate p only using the cohort where people were not treated, otherwise you will underestimate the population fraction. In order to test if the null hypothesis is true, we can't assume that's its true when constructing the test.