Lognormal MLE Solution with Right Censored Data: Difference between revisions
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Kate Racaza (talk | contribs) (Created page with '{{Reference Example}} This example compares the Lognormal MLE solution with Fisher matrix bound for right censored data. {{Reference_Example_Heading1}} The data on page 199 of…') |
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{{Reference Example}} | {{Reference Example}} | ||
This example | This example validates the calculations for the Lognormal MLE solution with Fisher matrix bound for right censored data in Weibull++ standard folios. | ||
{{Reference_Example_Heading1}} | {{Reference_Example_Heading1}} | ||
The data on page 199 of the book ''Statistical Methods for Reliability Data'' by Dr. Meeker and Dr. Escobar, John Wiley & Sons, 1998 is used. | 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. | ||
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* The MLE solution is <math>\hat{\mu} = 6.56, \ \hat{\sigma} = 0.543\,\!</math>. | * The MLE solution is <math>\hat{\mu} = 6.56, \ \hat{\sigma} = 0.543\,\!</math>. | ||
* The variance and covariance matrix is | * The variance and covariance matrix is | ||
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* The MLE solution and the variance/covariance matrix: | * The following picture shows the MLE solution and the variance/covariance matrix: | ||
[[Image:Lg_right_censored_matrix.png|center]] | [[Image:Lg_right_censored_matrix.png|center]] | ||
* The Fisher matrix bound for parameters: | * The Fisher matrix bound for parameters are: | ||
:For Ln-mu (using normal approximation of Eqn. 8.7 on page 187): | :For Ln-mu (using normal approximation of Eqn. 8.7 on page 187): | ||
::<math>[\hat{\mu}'_{L}, \hat{\mu}'_{U}] = \hat{\mu}' \pm z_{(1-\alpha | ::<math>[\hat{\mu}'_{L}, \hat{\mu}'_{U}] = \hat{\mu}' \pm z_{(1-\alpha / 2)}se_{\hat{\mu}'}\,\!</math> | ||
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::<math>\begin{alignat}{2} | ::<math>\begin{alignat}{2} | ||
[\hat{\mu}'_{L}, \hat{\mu}'_{U}] =& \hat{\mu}' \pm z_{(1-\alpha | [\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}\\ | =& 6.564256 \pm 1.96 \times (0.0581)^{0.5}\\ | ||
=& [6.0918, 7.0366933]\\ | =& [6.0918, 7.0366933]\\ | ||
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:For Ln-Std (using log-normal approximation of Eqn. 8.8 on page 188): | :For Ln-Std (using log-normal approximation of Eqn. 8.8 on page 188): | ||
::<math>[\hat{\sigma}'_{L}, \hat{\sigma}'_{U}] = \hat{\sigma}'exp (\pm \frac{z_{(1-\alpha | ::<math>[\hat{\sigma}'_{L}, \hat{\sigma}'_{U}] = \hat{\sigma}'exp (\pm \frac{z_{(1-\alpha /2)}se_{\hat{\sigma}'}}{\hat{\sigma}'})\,\!</math> | ||
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::<math>\begin{alignat}{2} | ::<math>\begin{alignat}{2} | ||
[\hat{\sigma}'_{L}, \hat{\sigma}'_{U}] =& \hat{\sigma}'exp (\pm \frac{z_{(1-\alpha | [\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})\\ | =& 0.534049 \times exp(\pm \tfrac{1.96\times 0.040562^{0.5}}{0.534049})\\ | ||
=& [0.255, 1.118]\\ | =& [0.255, 1.118]\\ | ||
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:The results in Weibull++ | :The following picture shows the results in Weibull++: | ||
[[Image:Lg_right_censored_bounds.png|center]] | [[Image:Lg_right_censored_bounds.png|center]] | ||
Latest revision as of 16:18, 28 September 2015
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