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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 $(*)$?

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  • $\begingroup$ 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. $\endgroup$ Oct 26 '21 at 0:25
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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.

That is you'd consider upper tail area only, no doubling.

[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?]

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