1P-Exponential MLE Solution for Interval Data

This example validates the calculations for the MLE solution, likelihood ratio bound and Fisher Matrix bound for a 1-parameter exponential distribution with interval data in Weibull++ standard folios.

Example 7.1 on page 154 in the book Statistical Methods for Reliability Data by Dr. Meeker and Dr. Escobar, John Wiley & Sons, 1998. The sample size of 200 data is used here.

The cumulative distribution function for an exponential distribution is:


 * $$F(t)=1-e^{-\left( \frac{t}{\theta }\right )}\,\!$$

The ML estimate $$\hat{\theta}\,\!$$ = 572.3, and the standard deviation is $$se_{\hat\theta}\,\!$$ = 41.72. Therefore the variance is 1740.56.

The 95% 2-sided confidence interval for $${\theta}\,\!$$ are:


 * Based on the likelihood ratio, the confidence interval is [498, 662]. The calculation is based on


 * $$-2ln\left [ \frac{L(\theta)}{L(\hat{\theta})} \right ] = X^{2}_{(0.90,1)}\,\!$$


 * The two solutions of $$\theta\,\!$$ in the above equation will be the confidence bounds for $$\theta\,\!$$.


 * Based on lognormal approximation, the confidence interval is [496, 660]. The calculation is:


 * $$\begin{alignat}{2}

[\theta_{L},\theta_{U}]&= \hat{\theta}exp\left(\pm 1.96\times \frac{se_{\hat{\theta}}}{\hat{\theta}}\right)\\ &= \left[572.3\times exp \left(-1.96\times\frac{41.72}{572.3}\right),572.3\times exp \left(1.96\times\frac{41.72}{572.3}\right)\right]\\ &= [496,660]\\ \end{alignat}$$

The ML estimator for $$\theta\,\!$$ and its variance are 572.27 and 1740.52, respectively. They are given below.



The ML estimator for $$\theta\,\!$$ and the variance are the same as the values given in the book.

The 95% 2-sided confidence interval for $$\theta\,\!$$ are:


 * Based on the likelihood ratio (Select LRB for the confidence bound), the confidence interval is:




 * Based on lognormal approximation (select FM for the bound method), the confidence bounds are: