# Why is the p-value given by p-value${}=P(\chi^2>\Delta D)$ rather than p-value${}=2P(\chi^2>\Delta D)$ as $(*)$?

In the Binomial Regression Models for Binary Data, we have the general Wald statistics result for the Hypothesis $$H_0: \beta_j=0, \, H_a: \beta_j \neq 0$$ that the p-value is given by

$$\text{p-value}= 2P\left( Z>\frac{\hat{\beta}_j-0}{\operatorname{se}(\hat{\beta}_j)} \right), \tag{*}$$ where $$Z\sim N(0,1)$$.

But when we do Testing Nested Models using LR/Deviance Tests, we have $$\Delta D=D_0-D_a\sim \chi^2$$. Here we have two models and $$H_0:$$ Model(1) fits the data as well as Model 2.

Why is the p-value given by $$\text{p-value}=P(\chi^2>\Delta D)$$ rather than $$\text{p-value}=2P(\chi^2>\Delta D)$$ as $$(*)$$?

• Not always a bilateral hypothesis testing give you a two sided p value, please see any test F the idea behind is to check if your statistics test is pretty large which would leads to reject H0. In this case you are wanting to verify if $\Delta D$ is very large to reject H0 this is would be it if $\Delta D>X^2(\alpha,K)$. As hypotesys testing is one sided p value does as well. Oct 26 '21 at 0:25

For these kinds of hypotheses, $$H_0$$ being false is associated with both large and small Z, while its only associated with large chi squared values, similar to if you used $$Z^2$$ rather than $$Z$$ as a test statistic - both tails of Z correspond only to the upper tail of $$Z^2$$, so if you were looking up tables for $$Z^2$$, only large values of $$Z^2$$ would lead you to reject.
[Indeed the tables you'd look up if you used $$Z^2$$ instead of Z would be $$\chi^2_1$$. Look up the 5% critical value for the $$\chi^2_1$$ and take its square root. Is that number familiar?]