Wald test

For an approximately normal estimator, the one-parameter Wald statisticis approximately standard normal under $H_0:\theta ={\theta }_0$. A two-sided test reports $2\Phi (-|Z|)$.

For an independent normal sample with known standard deviation $\sigma$, the signed Wald statistic for testing mean zero is $W=\sqrt{n}\bar{Y}/\sigma$. It has the standard normal distribution under the null hypothesis and distribution $N(\sqrt{n}\delta /\sigma ,1)$ under mean $\delta$. Its square has a chi-squared distribution; retaining the sign permits a one-sided test.

Independent normal arm means with variances $v_0,v_1,v_2$ give treatment-contrast Wald test statistics whose correlation coefficient is $v_0/\sqrt{(v_0+v_1)(v_0+v_2)}$. The common control mean creates positive dependence between comparisons.

For a regular $p$-parameter model, the Wald statistic for a candidate $\theta$ isUnder the true parameter it converges in distribution to ${\chi }_p^2$.

For a full-row-rank $k\times p$ matrix $R$ and the null hypothesis $R\theta =r$, putUnder the null, asymptotic normality and Slutsky theorem give $W_{n,R}\stackrel{d}{\to }{\chi }_k^2$.

In a regular scalar model under the null, Taylor expansion about the MLE givesA uniform law of large numbers for the observed information makes
$-{\ell }_n^{\prime \prime }(\tilde{\theta })/n$ and the Fisher information used in the Wald statistic converge to the same positive limit. Their ratio therefore converges in probability to one.

For $\overline{X}\sim N(\mu ,1/n)$ under $H_0:\mu =0$, the two-sided level-$\alpha$ test rejects when $|\sqrt{n}\overline{X}|>{\Phi }^{-1}(1-\alpha /2)$, and the observed p-value is