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 $(*)$?

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?]
