Let X_1, \ldots , X_ n \stackrel{iid}{\sim } \mathsf{Ber}(p). We want to test
\begin{array}{rl} H_0: & p=.5 \\ H_1: & p\neq .5\\ \end{array}
at asymptotic level 5%. What is the rejection region of the Wald test? (Choose all that apply.)
\sqrt{n}\frac{\bar X_ n - 1/2}{\sqrt{\bar X_ n (1-\bar X_ n)}}>1.65
|\bar X_ n - 1/2|>1.96\sqrt{\frac{\bar X_ n (1-\bar X_ n)}{n}}
\bar X_ n > 1/2+ 1.96\sqrt{\frac{\bar X_ n (1-\bar X_ n)}{n}}
\frac{|\bar X_ n - 1/2|}{\sqrt{\textsf{Var}(\bar X_ n)}}>1.96
\sqrt{n}\frac{|\bar X_ n - p|}{\sqrt{p(1-p)}}>1.96
\mathcal{I}_{\mathrm{plug-in}}, defined to be the plug-in confidence interval at asymptotic confidence level 95% from Unit 2.
\sqrt{n}\frac{|\bar X_ n - 1/2|}{\sqrt{\bar X_ n (1-\bar X_ n)}}>1.96
The rejection region of the Wald test for testing $H_0: p = 0.5$ against $H_1: p \neq 0.5$ is given by:
$\sqrt{n}\frac{|\bar X_ n - 1/2|}{\sqrt{\bar X_ n (1-\bar X_ n)}}>1.96$
This is the correct rejection region for the test.
The rejection region of the Wald test can be found by comparing the test statistic to a critical value.
The test statistic for the Wald test is given by:
\sqrt{n}\frac{\bar X_ n - p}{\sqrt{p(1-p)}}
In this case, we are testing the null hypothesis H_0: p = 0.5 against the alternative hypothesis H_1: p ≠ 0.5.
To find the rejection region, we compare the absolute value of the test statistic to a critical value. In this case, the critical value is 1.96 because we want to test at a significance level of 5%.
Therefore, the correct rejection region is:
\sqrt{n}\frac{|\bar X_ n - 1/2|}{\sqrt{\bar X_ n (1-\bar X_ n)}} > 1.96
So the answer is:
\sqrt{n}\frac{|\bar X_ n - 1/2|}{\sqrt{\bar X_ n (1-\bar X_ n)}} > 1.96