1P-Exponential MLE Solution with Right Censored Data: Difference between revisions

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::<math>\hat{\theta} = \frac{TTT}{r} = \frac{16+34+53+75+93+120+4\times 200}{6} = \frac{1191}{6} = 198.5\,\!</math>
::<math>\begin{align}
 
\hat{\theta} =& \frac{TTT}{r} = \frac{16+34+53+75+93+120+4\times 200}{6} = \frac{1191}{6} = 198.5 \\
 
\\
::<math>se_{\hat{\theta}} = \frac{\theta}{\sqrt{6}} = \frac{198.5}{\sqrt{6}} = 81.037 \,\!</math>
se_{\hat{\theta}} =& \frac{\theta}{\sqrt{r}} = \frac{198.5}{\sqrt{6}} = 81.037 \\
\end{align}\,\!</math>




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1P-Exponential MLE Solution with Right Censored Data

Compare the MLE solution and Fisher Matrix bound for a 1-parameter exponential distribution with right censored and complete failure data.


Reference Case

The formulas on page 166 in the book Statistical Methods for Reliability Data by Dr. Meeker and Dr. Escobar, John Wiley & Sons, 1998.

[math]\displaystyle{ \hat{\theta}=\frac{TTT}{r}\ \ and\ \ se_{\hat{\theta}} = \frac{\hat{\theta}}{\sqrt{r}}\,\! }[/math]


where TTT is the total test time and r is the number of failures.


Data

Number in State State F or S Time to Failure
1 F 16
1 F 34
1 F 53
1 F 75
1 F 93
1 F 120
4 S 200


Result

[math]\displaystyle{ \begin{align} \hat{\theta} =& \frac{TTT}{r} = \frac{16+34+53+75+93+120+4\times 200}{6} = \frac{1191}{6} = 198.5 \\ \\ se_{\hat{\theta}} =& \frac{\theta}{\sqrt{r}} = \frac{198.5}{\sqrt{6}} = 81.037 \\ \end{align}\,\! }[/math]


So the variance of [math]\displaystyle{ \hat{\theta}\,\! }[/math] is 6567.04


Results in Weibull++


1PE rcensored data.png
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