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)