Mungkin ada baiknya untuk menjelaskan '' alasan di balik '' beberapa koreksi pengujian seperti yang dilakukan Bonferroni. Jika itu jelas maka Anda akan dapat menilai sendiri apakah Anda harus menerapkannya atau tidak.
μH0:μ=0
H1:μ≠0H0:μ=0α
H0H0
H0H0H1
Bukti palsu adalah hal yang buruk dalam sains karena kami percaya telah mendapatkan pengetahuan yang benar tentang dunia, tetapi sebenarnya kami mungkin memiliki nasib buruk dengan sampel. Kesalahan semacam ini karenanya harus dikontrol. Oleh karena itu seseorang harus meletakkan batas atas pada kemungkinan bukti semacam ini, atau seseorang harus mengendalikan kesalahan tipe I. Ini dilakukan dengan memperbaiki tingkat signifikansi yang dapat diterima di muka.
5%H05%H0H1H1
Asumsikan sekarang bahwa kita memiliki dua parameter, dan kami ingin menunjukkan bahwa setidaknya satu berbeda dari nol. Mengikuti logika '' pembuktian dengan kontradiksi '' akan kita asumsikanH0:μ1=0&μ2=0H1:μ1≠0|μ2≠0α=0.05
H(1)0:μ1=0H(1)0:μ1≠0H(2)1:μ2=0H(2)1:μ2≠0α=0.05
H(1)0 but with that same sample I may also have bad luck with the sample for the second test and erroneously reject H(1)0
Therefore, the chance that at least one of the two is an erroneous rejection is 1 minus the probability that both are not rejected, i.e. 1−(1−0.05)2=0.0975, where it was assumed that both tests are independent. In other words, the type I error has ''inflated'' to 0.0975 which is almost double α.
The important fact here is that the two tests are based on one and the sampe sample !
Note that we have assumed independence. If you can not assume independence then you can show, using the Bonferroni inequality$ that the type I error can inflate up to 0.1.
Note that Bonferroni is conservative and that Holm's stepwise procedure holds under the same assumptions as for Bonferroni, but Holm's procedure has more power.
When the variables are discrete it's better to use test statistics based on the minimum p-value and if you are ready to abandon type I error control when doing a massive number of tests then False Discovery Rate procedures may be more powerful.
EDIT :
If e.g. (see the example in the answer by @Frank Harrell)
H(1)0:μ1=0 versus H(1)1:μ1≠0 is the a test for the effect of a chemotherapy and
H(2)0:μ1=0 versus H(2)1:μ2≠0 is the test for the effect on tumor shrinkage,
then, in order to control the type I error at 5% for the hypothesis H(12)0:μ1=0&μ2=0 versus H(12)1:μ1≠0|μ2≠0 (i.e. the test that at least one of them has an effect) can be carried out by testing (on the same sample)
H(1)0 versus H(1)1 at the 2.5% level and also H(2)0 versus H(2)1 at the 2.5% level.