Standard error
The asymptotic formula for the standard error of maximum likelihood estimators can be used to get an approximate value for the standard error of \(\hat{\lambda} = \frac 1 {\overline{X}}\).
\[ \ell''(\lambda) \;\; = \;\; -\frac n {\lambda^2} \] \[ \se(\hat{\lambda}) \;\;\approx\;\; \sqrt {- \frac 1 {\ell''(\hat {\lambda})}} \;\;=\;\; \frac {\hat{\lambda}} {\sqrt n} \]Confidence interval
An approximate 95% confidence interval for \(\lambda\) can be found using the asymptotic normality of the maximum likelihood estimator as
\[ \begin{align} \hat{\lambda} \; \pm\; 1.96 \times \se(\hat{\lambda}) \;\;&=\;\; \frac 1 {\overline{x}} \; \pm\; 1.96 \times \frac 1 {\sqrt n \;\overline{x}} \\ &=\;\; \frac 1 {\overline{x}} \left(1 - \frac {1.96}{\sqrt n}\right) \quad\text{to}\quad \frac 1 {\overline{x}} \left(1 + \frac {1.96}{\sqrt n}\right) \end{align} \]Question: Aircraft air-conditioner failures
The mean time between failures of the \(n = 199\) air-conditioners on the previous page was \(\overline{X} = 90.92\) hours with estimated failure rate \(\hat{\lambda} = 0.0110\) failures per hour. Find an approximate 95% confidence interval for the failure rate.
(Solved in full version)