## Multiple Hypothesis Testing

It's a bit messy to try and fit all of the pairwise hypothesis tests you have to do into one statement like this, but to overthink everything like I usually do :stuck_out_tongue_winking_eye: here is the full-on logic statement for the hypotheses you would have. In words, from what I understand:

* The **null hypothesis** $H_0$ is that all the artists are "the same", in terms of the Spotify features that can be obtained for all their songs, and then
* The **alternative hypothesis** $H_A$ is the logical negation of this: that it is not true that all the artists are "the same" in this sense.

Mathematically, since in this class we have only learned pairwise tests (and going beyond pairwise tests can actually be quite tricky, because of **corrections** you need to make when using the **same dataset** to test **multiple hypotheses**), you'd want to write out the individual hypotheses as follows:

Let $A = \{\text{Weeknd},\text{TaySwift},\text{Bad Bunny},\text{BTS},\text{Drake}\}$ be the full set of artists you want to consider, let $F$ be the full set of features you want to consider, and let $\mu_{f,a}$ represent the mean of feature $f$ for artist $a$. Then the null hypothesis is

$$
H_0: \forall f \in F \left[ \forall a, b \in A, a \neq b \left[ \mu_{f,a} = \mu_{f,b} \right] \right]
$$

Which can be read as "for all features $f$, and all pairs of artists $a$ and $b$ such that $a \neq b$, the mean of feature $f$ for artist $a$ is the same as the mean of feature $f$ for artist $b$". Another way to write this would be (remembering that $\wedge$ is the logical symbol for **and**: $p \wedge q$ is true if both $p$ and $q$ are true):

$$
\bigwedge_{f \in F}\bigwedge_{\{a, b \, \in \, A \, \mid \, a \neq b\}}\mu_{f,a} = \mu_{f,b}
$$

but that notation is somewhat less common.

The alternative hypothesis is the logical negation:

$$
H_A: \exists f \in F \left[ \exists a, b \in A, a \neq b \left[ \mu_{f,a} \neq \mu_{f,b} \right] \right]
$$

Which could be rewritten similarly to the null hypothesis as (remembering that $\vee$ is the logical symbol for **inclusive or**: $p \vee q$ is true if $p$ is true or $q$ is true or both are true):

$$
\bigvee_{f \in F}\bigvee_{\{a, b \, \in \, A \, \mid \, a \neq b\}}\mu_{f,a} \neq \mu_{f,b}
$$