One of the application of Student's t-test is to test the location of one sequence of independent and identically distributed random variables. If we want to test the locations of multiple sequences of such variables, Šidák correction should be applied in order to calibrate the level of the Student's t-test. Moreover, if we want to test the locations of nearly infinitely many sequences of variables, then Šidák correction should be used, but with caution. More specifically, the validity of Šidák correction depends on how fast the number of sequences goes to infinity.
Suppose we are interested in different hypotheses,
H1,...,Hm
Hnull
Hi
Halternative
Hi
Let
\alpha
Hnull
We aim to design a test with certain level
\alpha
Suppose when testing each hypothesis
Hi
ti
If these
ti
Hnull
Step 1, we test each of null hypotheses at level
| ||||
1-(1-\alpha) |
Step 2, if any of these null hypotheses is rejected, we reject
Hnull
For finitely many t-tests,suppose
Yij=\mui+\epsilonij,i=1,...,N,j=1,...,n,
\epsiloni1,...,\epsilonin
\epsilon1j,...,\epsilonNj
\epsilonij
Our goal is to design a test for
Hnull:\mui=0,\foralli=1,...,N
ti=
\bar{Y | |
i |
where:
\bar{Y}i=
1 | |
n |
n | |
\sum | |
j=1 |
Yij,
2 | ||
S | = | |
i |
1 | |
n |
n | |
\sum | |
j=1 |
(Yij-\bar{Y}i)2.
Using Šidák correction, we reject
Hnull
| ||||
1-(1-\alpha) |
.
Hnull
\existsi\in\{1,\ldots,N\}:|ti|>\zeta\alpha,N,
where
P(|Z|>\zeta\alpha,N
| ||||
)=1-(1-\alpha) |
, Z\simN(0,1)
The test defined above has asymptotic level, because
\begin{align} level&=Pnull\left(rejectHnull\right)\\ &=Pnull\left(\existsi\in\{1,\ldots,N\}:|ti|>\zeta\alpha,N\right)\\ &=1-Pnull\left(\foralli\in\{1,\ldots,N\}:|ti|\leq\zeta\alpha,N\right)
N | |
\\ &=1-\prod | |
i=1 |
Pnull\left(|ti|\leq\zeta\alpha,N\right)\\ &\to
N | |
1-\prod | |
i=1 |
P\left(|Zi|\leq\zeta\alpha,N\right)&&Zi\simN(0,1)\\ &=\alpha\end{align}
In some cases, the number of sequences,
N
n
N(n) → inftyasn → infty
To design a test, Šidák correction may be applied, as in the case of finitely many t-test. However, when
N(n) → inftyasn → infty
\alpha
\alpha
N(n) → infty
\epsilonij
logN=o(n1/3)
\alpha
\epsilonij
logN=o(n1/2)
\alpha
logN=o(n1/3)
\epsilonij
The results above are based on Central Limit Theorem. According to Central Limit Theorem, each of our t-statistics
ti
ti
ti
When we have finitely many
ti
ti
N(n) → infty
\alpha