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 $$\begin{array}{|c|c|c|}\hline p & q & p\wedge q\\ \hline T & T & T \\ \hline T & F & F \\ \hline F & T & F \\\hline F & F & F \\\hline \end{array}.$$ 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).