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Lognormal MLE Solution with Right Censored Data
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This example validates the calculations for the Lognormal MLE solution with Fisher matrix bound for right censored data in Weibull++ standard folios.
Reference Case
The data set on page 199 of the book Statistical Methods for Reliability Data by Dr. Meeker and Dr. Escobar, John Wiley & Sons, 1998 is used.
Data
| Number in State
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State F or S
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Time to Failure
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| 1 |
F |
252
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| 1 |
F |
315
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| 1 |
F |
369
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| 1 |
F |
403
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| 1 |
F |
474
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| 15 |
S |
500
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Result
- The MLE solution is [math]\displaystyle{ \hat{\mu} = 6.56, \ \hat{\sigma} = 0.543\,\! }[/math].
- The variance and covariance matrix is
- [math]\displaystyle{
\sum =\begin{bmatrix}
0.0581 & 0.0374 \\
0.0374 & 0.0405
\end{bmatrix}\,\! }[/math]
Results in Weibull++
- The following picture shows the MLE solution and the variance/covariance matrix:
- The Fisher matrix bound for parameters are:
- For Ln-mu (using normal approximation of Eqn. 8.7 on page 187):
- [math]\displaystyle{ [\hat{\mu}'_{L}, \hat{\mu}'_{U}] = \hat{\mu}' \pm z_{(1-\alpha / 2)}se_{\hat{\mu}'}\,\! }[/math]
- For a confidence level of 0.95, it is:
- [math]\displaystyle{ \begin{alignat}{2}
[\hat{\mu}'_{L}, \hat{\mu}'_{U}] =& \hat{\mu}' \pm z_{(1-\alpha /2)}se_{\hat{\mu}'}\\
=& 6.564256 \pm 1.96 \times (0.0581)^{0.5}\\
=& [6.0918, 7.0366933]\\
\end{alignat}\,\! }[/math]
- For Ln-Std (using log-normal approximation of Eqn. 8.8 on page 188):
- [math]\displaystyle{ [\hat{\sigma}'_{L}, \hat{\sigma}'_{U}] = \hat{\sigma}'exp (\pm \frac{z_{(1-\alpha /2)}se_{\hat{\sigma}'}}{\hat{\sigma}'})\,\! }[/math]
- For confidence level of 0.95, it is:
- [math]\displaystyle{ \begin{alignat}{2}
[\hat{\sigma}'_{L}, \hat{\sigma}'_{U}] =& \hat{\sigma}'exp (\pm \frac{z_{(1-\alpha /2)}se_{\hat{\sigma}'}}{\hat{\sigma}'})\\
=& 0.534049 \times exp(\pm \tfrac{1.96\times 0.040562^{0.5}}{0.534049})\\
=& [0.255, 1.118]\\
\end{alignat} \,\! }[/math]
- The following picture shows the results in Weibull++:
