Cfrgtkky

Could you help me to clarify how to use Bonferroni correction in hypothesis testing? Suppose I want to test
$$
H_0: X perp Y text{ , } Z perp Y text{ , } X perp Z hspace{1cm} text{at level $alpha=5%$}
$$
where $perp$ denotes independence.

One way to do this is to test
$$
H_0^1: X perp Y hspace{1cm} text{at level $alpha=frac{5}{3}%$}
$$
$$
H_0^2: Z perp Y hspace{1cm} text{at level $alpha=frac{5}{3}%$}
$$
$$
H_0^3: X perp Z hspace{1cm} text{at level $alpha=frac{5}{3}%$}
$$
What I am not sure about is the following: once I have the results of these three tests, what should I conclude about $H_0$? Should I reject $H_0$ if I reject at least one among $H_0^1, H_0^2, H_0^3$? Should I reject $H_0$ if I reject all of $H_0^1, H_0^2, H_0^3$?

The solution here is quite simple, however I thought that I might be able to explain the logic in here in more formal way.

The comma is often used as the replacement for AND ($wedge$ in logic).
Let me write your problem in an equivalent form using truth tables. Recall that the truth table for logical conjunction is
$$begin{array}{|c|c|c|}hline
p & q & pwedge q\
hline
T & T & T \ hline
T & F & F \ hline
F & T & F \hline
F & F & F \hline
end{array}.$$
The hypothesis $H_0$ will hold if the last column of the below table returns the value true ($T$). We obtain

$$begin{array}{|c|c|c|c|c|}hline H_0^1 & H_0^1 & H_0^3 & H_0^1 wedge H_0^2 & (H_0^1 wedge H_0^2) wedge H_0^3 \
hline T & T & T & T& T\
hline F & T & T & F &F \
hline F & F & T & F& F\
hline F & F & F & F&F \
hline T & F & F & F&F \
hline T & T & F & T&F \
hline F & T & F & F&F \
hline T & F & T & F&F \ hline
end{array}$$
Therefore, you can see that $H_0$ holds only in one case. If your comma (AND) was replaced by OR, then you could translate this problem into logic tables and use the one for disjunction (OR).

Usually there are more powerful ways to test a global hypothesis than the Bonferroni test. The advantage of using Bonferroni (and its variants) is that, not only can you reject the global hypotheses $H_0$ when there is at least one rejection among the tests at the $alpha/3$ level, you can also reject every individual component hypothesis $H_0^j$ that was rejected at that level.

For most tests of global hypotheses (e.g., the $F$ test), rejection of the global hypothesis $H_0$ does not allow you to make any statements about any of the component hypotheses $H_0^j$. More specifically, you can't say anything about the components following a global rejection if you wish you to control the Familywise Error Rate (FWER) over all components, when you use global tests such as the $F$ test. The benefit of the Bonferroni test of the global intersection hypothesis (unlike the $F$ global test), is that you can make statements about the components of the intersection, with complete FWER control.