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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 $H_0^1, H_0^2, H_0^3$?

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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).

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  • $\begingroup$ Thanks. Hence, just to be sure about my final question: I reject $H_0$ if at least one among $H_0^1, H_0^2, H_0^3$ is false. Correct? $\endgroup$ – TEX May 10 '17 at 15:21
  • $\begingroup$ Yes, indeed. You are right. $\endgroup$ – m_gnacik May 10 '17 at 15:23
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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.

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