Consider a likelihood ratio test for the hypotheses
This is based on the log-likelihood when H0 holds, \(\ell(\theta_0)\), and its maximum possible value under HA, \(\ell(\hat{\theta})\), where \(\hat{\theta}\) is the maximum likelihood estimate of \(\theta\). For a hypothesis test with 5% significance level, we reject H0 if
\[ 2\left(\ell(\hat{\theta}) - \ell(\theta_0)\right) \;\;\gt\;\; K \]where the constant \(K\) is the 95th percentile of the \(\ChiSqrDistn(1 \text{ df})\) distribution, \(K = 3.841\).
Inverting the likelihood ratio test
A 95% confidence interval for \(\theta\) can therefore be found as the values for which
\[ \begin{align} 2\left(\ell(\hat{\theta}) - \ell(\theta_0)\right) \;\;&\le\;\; 3.841 \\[0.4em] \ell(\theta_0)\;\;&\gt\;\; \ell(\hat{\theta}) - \frac{3.841}{2} \\[0.5em] \ell(\theta_0)\;\;&\gt\;\; \ell(\hat{\theta}) - 1.921 \end{align} \]It is therefore the values of \(\theta\) for which the log-likelihood is within 1.921 of its maximum.
Question
Clinical records give the survival time in months from diagnosis of 30 sufferers from a certain disease as
9.73 5.56 4.28 4.87 |
1.55 6.20 1.08 7.17 |
28.65 6.10 16.16 9.92 |
2.40 6.19 7.67 1.11 |
4.66 4.35 7.31 3.28 |
13.38 3.08 0.41 4.33 |
2.16 4.49 0.75 |
4.45 10.29 0.90 |
If the survival times are exponentially distributed with a death rate of \(\lambda\), find a 95% confidence interval for \(\lambda\).
(Solved in full version)