# Can we use the binomial distribution instead of Bonferroni correction for large number of independent significant tests?

Supposing we are doing e.g 1000 one-sample t-tests with the null hypothesis that all population means are 0, with significance level 0.01. One common method for dealing with the number of tests is to adjust with the Bonferroni correction so we need each individual t-test to have significance level 0.01/1000 to reject the overall null hypothesis at 0.01 significance

But also under the null hypothesis the number of t-tests that give false positives is binomially distributed with mean 0.01(1000) = 10 and variance 1000(.99)(.1) = 9.9, and this is well-approximated by the Normal distribution given sample size. So can we also just run the t-tests without the Bonferroni correction and use the z-score to decide whether we can reject the null-hypothesis at 0.01 significance?

To me these seem like they should be equivalent ways to get at the same answer and they seem to use the same assumptions, but I’m curious if there are any edge-cases/situations where these methods might disagree?