I'm solving a question regarding the independence of two criteria and I already found the right value for the likelihood ratio test (which in this case is 0.482, by summing 21 $ln$ (0.21) + ... + 14 $ln$ (0.14) - 34$ln$(0.34) ... - 43$ln$(0.43) ). Now here's the two parts i don't get it. First although mechanically i've solved a few questions like these and i got them right, i don't know what this 0.482 means, by letting $Q_{0} = -2ln(\lambda) = 2(0.241)= 0.482$. Secondly, and this is my biggest doubt, why in the solution manual shows a graph of a right tail of 2.5% but in the question it asks for a $\alpha$ = 5%? (7.38 value using a $\chi{2}$ table with 2 degrees in this case since (a-1)(b-1)=(3-1)(2-1)=2) This question of when to use the given alpha versus to use it divided by two or other value gets me everytime, i guess it's related to the alternative hypothesis being different or bigger / smaller than a value. Is there any guideline on how to set the proper alpha? Thanks! Here's the table used in the question. N = 100 subscribers, of $A_{1},A_{2},A_{3}$ magazine and B meaning different academic levels of the readers.
The value of $\alpha$, the significance level, is predetermined depending on how likely the tester wants a Type I error to be. The typical values used are $0.1, 0.05,$ or $0.01$. Then, we calculate the likelihood ratio test statistic $\Lambda$, which is a random variable. Now, by the Neyman-Pearson lemma, for the most powerful test, we will always have $\Lambda \leq k$ for some $k \in \mathbb{R}$ as the rejection region, where $k$ is determined by $\alpha$ in the following way: as I have said, $\alpha$ is the probability of a Type I error, or the error of rejecting the null hypothesis when it is in fact true. So, in order to determine $k$, we solve the equation $$P(\Lambda \leq k \, | \, H_0) = \alpha \,\,,$$ where $H_0$ is the null hypothesis of the test. Note that this chi-square test for independence that you are using is derived from this setup, and so it is actually a likelihood ratio test - in other words, it is what this most powerful test reduces to when we compute $\Lambda$. The demonstration of this equivalence can be found in most college-level textbooks on statistics. Thus, the choice between $\chi_{n}^{2}(\frac{\alpha}{2})$ and $\chi_{n}^{2}(\alpha)$, for some $n$ degrees of freedom that depends on the sample size, as the critical value for rejection comes from this construction of the test, fundamentally.