1P-Exponential MLE Solution with Right Censored Data - Revision history

Kate Racaza at 16:17, 28 September 2015

2015-09-28T16:17:20Z

← Older revision		Revision as of 16:17, 28 September 2015
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	~~Compare~~ the MLE solution and Fisher Matrix bound for a 1-parameter exponential distribution with right censored and complete failure data.		This example validates the calculations for the MLE solution and Fisher Matrix bound for a 1-parameter exponential distribution with right censored and complete failure data in Weibull++ standard folios.

Kate Racaza at 15:46, 9 June 2014

2014-06-09T15:46:39Z

← Older revision		Revision as of 15:46, 9 June 2014
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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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Created page with '{{Reference Example}} Compare the MLE solution and Fisher Matrix bound for a 1-parameter exponential distribution with right censored and complete failure data. {{Reference_E…'

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

{{Reference_Example_Heading1}}
The formulas on page 166 in the book ''Statistical Methods for Reliability Data'' by Dr. Meeker and Dr. Escobar, John Wiley & Sons, 1998.

::<math>\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.

{{Reference_Example_Heading2}}
{| {{table}}
!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
|-
|}

{{Reference_Example_Heading3}}

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

::<math>se_{\hat{\theta}} = \frac{\theta}{\sqrt{6}} = \frac{198.5}{\sqrt{6}} = 81.037 \,\!</math>

So the variance of <math> \hat{\theta}\,\! </math> is 6567.04

{{Reference_Example_Heading4}}

[[Image:1PE_rcensored_data.png|center]]