Alternative to 2 dimensional χ² test which assumes one dependent and one independent variable A 2 dimensional χ² test can 'prove' that nominal variable A is associated with nominal variable B, and can express the strength of that association using φ[0,1].
But what test can I use to test the hypothesis that nominal variable A predicts nominal variable B?
I'll show why this is an important difference using a contrived example: I'm investigating the relationship between eating lots of crisps and being almost unbelievably spotty, using a sample of 222 teenagers. I want to test the hypothesis that almost unbelievably spotty teenagers are more likely to eat lots of crisps than other teenagers. I build a 2-by-2 contingency table:





Not so many crisps
Lots of crisps




Not especially spotty
100
100


Almost unbelievably spotty
1
11




(Please ignore the fact that these variables are ordinal and built by collapsing continuous variables. It's not a great example)
A χ² test of the above data is sufficient to reject the null hypothesis (assume α = 0.05), giving a χ² statistic of 6.297.
I can use my χ² statistic to calculate φ:
> sqrt(6.297 / 222)
#[1] 0.1684187

That's a pretty low φ, which makes sense because it's describing the strength of the 2 way association between being almost unbelievably spotty and eating lots of crisps. There is an association, but it's not mega-strong in both directions:

*

*only ~10% of teenagers who eat a lot of crisps are almost unbelievably spotty

*~92% of teenagers who are almost unbelievably spotty eat lots of crisps.

Assuming I had good reason, a priori, to be testing the hypothesis that teenagers who are almost unbelievably spotty are more likely than other teenagers to eat lots of crisps, what test could I use to find out if it's true?
For clarity this is an actual problem I'm having at work with data that I have good reason to treat as dependent/independent, but are weakly associated because one row/column of the contingency table is much more populated than the other.

THEY ARE SYNONYMS
Throughout, assume the following null and alternative hypotheses, where $p$ is the probability of eating lots of crisps.
$$
H_0: p_{\text{spotty}} = p_{\text{not spotty}}\\
H_a: p_{\text{spotty}} \ne p_{\text{not spotty}}\\
$$
If you use the hypothesis test to infer that the two levels of spottiness differ on crisp consumption, then if you know the level of spottiness, you can make a better prediction of the probability of eating lots of crisps than you would by guessing the overall probability (pooled across both spottiness groups), regardless of spottiness
If you can make good predictions of the probability of eating lots of crisps, given the level of spottiness (good predictions in the sense of being better than just guessing based the baseline probability of eating lots of crisps, regardless of spottiness), then it must be that the two levels of spottiness differ in their probability of eating many crisps.
The $\chi^2$ test is a score test, and the score test has a mathematical setup that does not, at least to me, make it clear that this is the case. For a likelihood ratio test (another of the big three classical tests, along with Wald tests), however, you literally calculate the performance, measured by likelihood in the technical sense in statistics, of each model (predict the baseline for one model, predict the probability by spottiness level for the other), and compare how much better the latter model does. If the latter model does much better, you reject the null hypothesis that the spottiness level have equal probabilities of eating many crisps.
