= Solution
Let $z=z_{1-\alpha/2}=\Phi^{-1}(1-\alpha/2)$, where $0<\alpha<1$. By the preceding <asymptotic normality> and <Slutsky's theorem>, $\sqrt n(\widehat\theta-\theta)/\widehat\theta$ is approximately standard normal. Inverting $|\sqrt n(\widehat\theta-\theta)/\widehat\theta|\leq z$ gives the <Wald confidence interval>
$$
\boxed{\left[\widehat\theta-z\frac{\widehat\theta}{\sqrt n},\quad\widehat\theta+z\frac{\widehat\theta}{\sqrt n}\right]\cap(0,\infty).}
$$
Its coverage approaches $1-\alpha$ as $n\to\infty$. The interpretation of the <confidence interval> is repeated-sample coverage of the fixed parameter; it does not assign a posterior <probability> to $\theta$.
Back to article page
