 Solution

Here $s=\log (0.4\times 0.7\times 0.8\times 0.9)=-1.601469743$. The maximum-likelihood estimate and estimated standard error from the observed Fisher information areThe 95% Wald confidence interval is thereforeThe whole reported confidence interval lies in $\theta >-1$, though its lower endpoint is close to the boundary. With only four observations, the normal approximation underlying this Wald confidence interval need not provide accurate 95% coverage.