# Optimal Confidence Regions for Weibull Parameters and Quantiles under Progressive Censoring
## 5 Illustrative Applications  
A progressively censored sample studied by [PERSON] and [PERSON] ([PERSON] and [PERSON] (2008), p. 9) is considered in this section to illustrate the results developed above. This sample is based on the data reported by [PERSON] ([PERSON] and others, 2010) concerning times to breakdown (in minutes) of an insulating fluid between two electrodes subject to a voltage of 34 kV. According to engineering considerations, for a fixed voltage level, the time to breakdown, \(X\), follows a Weibull distribution. In our case, the progressive censoring scheme is \(\mathbf{r}=(0,0,3,0,3,0,0,5)\) and the sample of observed failure times is given by:  
\[\mathbf{x}=(0.19,0.78,0.96,1.31,2.78,4.85,6.50,7.35).\]  
Therefore, \(s=8\), \(n=19\) and \(w=W(\mathbf{x})=\prod_{i=1}^{s}x_{i}=120.054\). It can be shown that the maximum likelihood estimates of \(\theta\) and \(\alpha\) are \(\widehat{\theta}=9.22542\) and \(\widehat{\alpha}=0.974323\), respectively. Moreover, the value of \(K\equiv K(\mathbf{x},\mathbf{r})\) is obtained to be \(K=\exp(-34.4158)\). The posterior pdfs for \((\theta,\alpha)\), \(\theta\) and \(\alpha\) given \((\mathbf{x},\mathbf{r})\) are plotted in Figures 1-3, respectively. The HPD estimate (or posterior mode) of \((\theta,\alpha)\) is given by \((\widehat{\theta},\widehat{\alpha})=(8.28923,0.930714)\). [PERSON] and [PERSON] ([13], p. 387) derived that the joint 95% confidence region for \(\theta\) and \(\alpha\) suggested by [PERSON] [34], which is denoted by \(W_{0.95}\equiv W_{0.95}(\mathbf{x},\mathbf{r})\), is defined by  
\[W_{0.95}=\Bigg{\{}(\theta,\alpha):0.2807\leq\alpha\leq 1.9648,\quad\left(\frac{2 \upsilon_{\alpha}}{31.2070}\right)^{1/\alpha}\leq\theta\leq\left(\frac{2 \upsilon_{\alpha}}{6.0684}\right)^{1/\alpha}\Bigg{\}}.\]  
Assuming that the confidence level is \(\varepsilon=0.95\), the optimal (minimum area) 95% confidence region for \((\theta,\alpha)\), \(R_{0.95}\equiv R_{0.95}(\mathbf{x},\mathbf{r})\), proposed in this paper is given by  
\[R_{0.95}=\{(\theta,\alpha):\pi^{*}(\theta,\alpha\mid\mathbf{x},\mathbf{r}) \geq c_{0.95}\},\]  
where the 0.05-quantile of the random posterior density of \((\theta,\alpha)\) given \((\mathbf{x},\mathbf{r})\) is \(c_{0.95}=0.00268513\). This set is also the Bayesian HPD 0.95-credibility region for \((\theta,\alpha)\) in the non-informative case. For illustrative and comparative purposes, the 95% confidence regions \(W_{0.95}\) and \(R_{0.95}\) are depicted in Figure 4.  
The area of the optimal region is \(Area[R_{0.95}]=33.1901\), whereas \(Area[W_{0.95}]=148.137\). Note also that, if \(\alpha\) is small, the values of \(\theta\) such that \((\theta,\alpha)\in W_{0.95}\) could be very large. For instance, the point \((0.2807,2014.25)\) is contained in \(W_{0.95}\), which is clearly unrealistic because \(\widehat{\theta}=9.22542\) is too small compared to 2014.25. In general, our approach can greatly reduce the areas of the confidence regions for the Weibull parameters. In the above situation, \(Area[W_{0.95}]/Area[R_{0.95}]=4.46412\). Consider that a reliability engineer aims to check whether \((\theta,\alpha)=(10,1)\) is reasonable or not. Since the \(p\)-value for testing \(H_{0}:(\theta,\alpha)=(10,1)\) versus \(H_{1}:(\theta,\alpha)\
eq(10,1)\) is calculated to be \(p=0.840618\), the values \(\theta=10\) and \(\alpha=1\) are quite admissible. In contrast, \(\theta=10\) and \(\alpha=2\) are not reasonable because the \(p\)-value for testing \(H_{0}:(\theta,\alpha)=(10,2)\) against \(H_{1}:(\theta,\alpha)\
eq(10,2)\) is only \(p=0.0134487\). Suppose now that an analyst seeks to determine the smallest-size 90% confidence region for the pair of Weibull quantiles \((q_{\alpha},q_{b})\), where \(a=0.05\) and \(b=0.50\).