I'm going through the Stats Online Refresher tutorial at the moment and I've come to the estimation bit. Example question 13 says use the estimate for lambda calculated using the method of moments and the Cramer-Rao asymptotic result to obtain a 95% one sided confidence interval which provides an upper limit for lambda. The formula given is method of moments estimate + 1.6449 * square root of Cramer Rao estimate. What I don't understand is why the square root of the Cramer Rao estimate is not divided by square root of n. Can anyone explain why this isn't in the formula? Thanks.
I think you're confusing this with a typical confidence interval based on the sample mean. The asymptotic distribution (on page 23 of the Tables) is: \(\hat{\theta} \backsim N(\theta,CRLB)\) Standardising this gives: \(\frac{\hat{\theta}-\theta}{\sqrt{CRLB}} \backsim N(0,1)\) Hence a 2-sided 95% CI would be: \(\theta = \hat{\theta} ± 1.96\sqrt{CRLB}\) For a normal distribution, we have: \(\hat{\mu} = \bar{x}\) \(CRLB = \frac{\sigma^2}{n}\) Hence, we would get: \(\mu = \hat{\mu} ± 1.96\sqrt{CRLB}\) \(\mu = \bar{x} ± 1.96 \frac{\sigma}{\sqrt{n}}\)
That's exactly what I was doing. It's been a while since I've looked at confidence intervals. Thanks John.