Past exam of the options I course of the University of Oxford / 2013 / O / 1 / c / 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, $\sqrt{n}(\hat{\theta }-\theta )/\hat{\theta }$ is approximately standard normal. Inverting $|\sqrt{n}(\hat{\theta }-\theta )/\hat{\theta }|\le z$ gives the Wald confidence intervalIts 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 $\theta$.