1P-Exponential MLE Solution for Interval Data - Revision history

Kate Racaza at 16:16, 28 September 2015

2015-09-28T16:16:35Z

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	{{Reference Example}}		{{Reference Example}}

	~~Compares~~ the MLE solution, likelihood ratio bound and Fisher Matrix bound for a 1-parameter exponential distribution with interval data.		This example validates the calculations for the MLE solution, likelihood ratio bound and Fisher Matrix bound for a 1-parameter exponential distribution with interval data in Weibull++ standard folios.


	{{Reference_Example_Heading1}}		{{Reference_Example_Heading1}}

Kate Racaza at 17:05, 31 July 2015

2015-07-31T17:05:41Z

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	{{Reference_Example_Heading3}}		{{Reference_Example_Heading3}}

	The cumulative distribution function for an exponential distribution is:		The cumulative distribution function for an exponential distribution is:

Kate Racaza at 21:11, 30 June 2015

2015-06-30T21:11:14Z

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Kate Racaza at 21:10, 30 June 2015

2015-06-30T21:10:13Z

← Older revision		Revision as of 21:10, 30 June 2015
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	The ML estimate <math>\hat{\theta}\,\!</math> = 572.3, and the standard deviation is <math>se_{\hat\theta}\,\!</math> = 41.72. Therefore the variance is 1740.56. ~~The 95% 2-sided confidence interval for <math>{\theta}\,\!</math> are:~~		The ML estimate <math>\hat{\theta}\,\!</math> = 572.3, and the standard deviation is <math>se_{\hat\theta}\,\!</math> = 41.72. Therefore the variance is 1740.56.


			The 95% 2-sided confidence interval for <math>{\theta}\,\!</math> are:

	* Based on the likelihood ratio, the confidence interval is [498, 662]. The calculation is based on		* Based on the likelihood ratio, the confidence interval is [498, 662]. The calculation is based on


	::<math>-2ln\left [ \frac{L(\theta)}{L(\hat{\theta})} \right ] = X^{2}_{(0.90,1)}\,\!</math>		::<math>-2ln\left [ \frac{L(\theta)}{L(\hat{\theta})} \right ] = X^{2}_{(0.90,1)}\,\!</math>
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	::<math>\begin{alignat}{2}		::<math>\begin{alignat}{2}
	[\theta_{L},\theta_{U}]&= \hat{\theta}exp(\pm 1.96\times \frac{se_{\hat{\theta}}}{\hat{\theta}})\\		[\theta_{L},\theta_{U}]&= \hat{\theta}exp\left(\pm 1.96\times \frac{se_{\hat{\theta}}}{\hat{\theta}}\right)\\
	&=\left [572.3\times exp \left(-1.96\times\~~tfrac~~{41.72}{572.3}\right),572.3\times exp \left(1.96\times\~~tfrac~~{41.72}{572.3}\right)\right]\\		&= \left[572.3\times exp \left(-1.96\times\frac{41.72}{572.3}\right),572.3\times exp \left(1.96\times\frac{41.72}{572.3}\right)\right]\\
	&= [496,660]\\		&= [496,660]\\
	\end{alignat}</math>		\end{alignat}</math>
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	* Based on the likelihood ratio (Select LRB for the confidence bound), the confidence interval is:		* Based on the likelihood ratio (Select LRB for the confidence bound), the confidence interval is:

	[[Image:~~1PE_interval_data~~.png\|center]]		[[Image:1PE_interval_data_qcp.png\|center\|550 px]]


	* Based on lognormal approximation (select FM for the bound method), the confidence bounds are:		* Based on lognormal approximation (select FM for the bound method), the confidence bounds are:

	~~::<math>\begin{alignat}{2}~~		[[Image:1PE_interval_data_qcp_FM.png\|center\|550 px]]
	[~~\theta_{L},\theta_{U}]&= \hat{\theta}exp\left(\pm 1.96 \times \frac{se_{\hat{\theta}}}{\hat{\theta}}\right)\\~~
	~~&= \left~~[~~572~~.~~3 \times exp\left(- 1.96 \times \frac{40.466}{572.3}\right), 572.3 \times exp\left(1.96 \times \frac{40.466}{572.3}\right)\right~~]\\
	~~&= [498, 657~~]\\
	~~\end{alignat}</math>~~

Kate Racaza at 20:52, 30 June 2015

2015-06-30T20:52:13Z

← Older revision		Revision as of 20:52, 30 June 2015
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	::<math>-2ln\left [ \frac{L(\theta)}{L(\hat{\theta})} \right ] = x^{2}_{(0.90,1)}\,\!</math>		::<math>-2ln\left [ \frac{L(\theta)}{L(\hat{\theta})} \right ] = X^{2}_{(0.90,1)}\,\!</math>


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	::<math>\begin{alignat}{2}		::<math>\begin{alignat}{2}
	[\theta_{L},\theta_{U}]&= \hat{\theta}exp(\pm 1.96\times \frac{se_{\hat{\theta}}}{\hat{\theta}})\\		[\theta_{L},\theta_{U}]&= \hat{\theta}exp(\pm 1.96\times \frac{se_{\hat{\theta}}}{\hat{\theta}})\\
	&=\left [572.3\times exp(-1.96\times\tfrac{41.72}{572.3}),572.3\times exp(1.96\times\tfrac{41.72}{572.3})\right]\\		&=\left [572.3\times exp \left(-1.96\times\tfrac{41.72}{572.3}\right),572.3\times exp \left(1.96\times\tfrac{41.72}{572.3}\right)\right]\\
	&= [496,660]\\		&= [496,660]\\
	\end{alignat}</math>		\end{alignat}</math>
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	The 95% 2-sided confidence interval for <math>\theta\,\!</math> are:		The 95% 2-sided confidence interval for <math>\theta\,\!</math> are:

	* Based on the likelihood ratio (Select LRB for the confidence bound), the confidence interval is		* Based on the likelihood ratio (Select LRB for the confidence bound), the confidence interval is:

			[[Image:1PE_interval_data.png\|center]]

	~~::<math>F(t)= 1-e^{-(\frac{t}{\theta})}</math>~~

			* Based on lognormal approximation (select FM for the bound method), the confidence bounds are:

	~~The ML estimate~~ <math>\hat{\theta}\,\~~!</math> = 572~~.~~3, and the standard deviation is <math>~~se_{\hat{\theta}}\,\~~!</math> = 41.72. Therefore the variance is 1740.56.~~		::<math>\begin{alignat}{2}
			[\theta_{L},\theta_{U}]&= \hat{\theta}exp\left(\pm 1.96 \times \frac{se_{\hat{\theta}}}{\hat{\theta}}\right)\\
			&= \left[572.3 \times exp\left(- 1.96 \times \frac{40.466}{572.3}\right), 572.3 \times exp\left(1.96 \times \frac{40.466}{572.3}\right)\right]\\
	~~The 95% 2-sided confidence interval for <math>~~\~~theta~~\,\~~!</math> are:~~		&= [498, 657]\\
			\end{alignat}</math>
	* Based on the likelihood ratio, the confidence interval is [~~498, 662]~~. ~~The calculation is based on~~

	~~::<math>~~-~~2ln~~ \frac{L(\~~theta~~)~~}{L~~(\~~hat{~~\~~theta})} = X^~~{2}_{(0.~~90,1~~)~~}</math>~~


	~~The two solutions of <math>~~\~~theta~~\,\~~!</math> in the above equation will be the confidence bounds for <math>\theta~~\,\!</math>.

Kate Racaza at 18:54, 30 June 2015

2015-06-30T18:54:57Z

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	Compares the MLE solution, likelihood ratio bound and Fisher Matrix bound for a 1-parameter exponential distribution with interval data.		Compares the MLE solution, likelihood ratio bound and Fisher Matrix bound for a 1-parameter exponential distribution with interval data.

	~~<!---~~
	{{Reference_Example_Heading1}}		{{Reference_Example_Heading1}}
	Example 7.1 on page 154 in book ''Statistical Methods for Reliability Data'' by Dr. Meeker and Dr. Escobar, John Wiley & Sons, 1998. The sample size of 200 data is used here.		Example 7.1 on page 154 in the book ''Statistical Methods for Reliability Data'' by Dr. Meeker and Dr. Escobar, John Wiley & Sons, 1998. The sample size of 200 data is used here.


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	The ML estimator for <math>\theta\,\!</math> and its variance are 572.27 and ~~1637~~.~~489~~, respectively. They are given below.		The ML estimator for <math>\theta\,\!</math> and its variance are 572.27 and 1740.52, respectively. They are given below.

	[[Image:1PE_interval_data.png\|center]]		[[Image:1PE_interval_data.png\|center]]

	The ML estimator for <math>\theta\,\!</math> and the variance are the same as the values given in the book~~. The following calculation provides the step by step analytical results for calculating the variance of <math>\theta\,\!</math>~~.		The ML estimator for <math>\theta\,\!</math> and the variance are the same as the values given in the book.


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	* Based on the likelihood ratio (Select LRB for the confidence bound), the confidence interval is		* Based on the likelihood ratio (Select LRB for the confidence bound), the confidence interval is
	--->

			::<math>F(t)= 1-e^{-(\frac{t}{\theta})}</math>


			The ML estimate <math>\hat{\theta}\,\!</math> = 572.3, and the standard deviation is <math>se_{\hat{\theta}}\,\!</math> = 41.72. Therefore the variance is 1740.56.


			The 95% 2-sided confidence interval for <math>\theta\,\!</math> are:

			* Based on the likelihood ratio, the confidence interval is [498, 662]. The calculation is based on

			::<math>-2ln \frac{L(\theta)}{L(\hat{\theta})} = X^{2}_{(0.90,1)}</math>


			The two solutions of <math>\theta\,\!</math> in the above equation will be the confidence bounds for <math>\theta\,\!</math>.

Kate Racaza at 15:40, 11 November 2014

2014-11-11T15:40:35Z

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Kate Racaza at 22:19, 29 August 2014

2014-08-29T22:19:37Z

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Kate Racaza at 22:18, 29 August 2014

2014-08-29T22:18:49Z

← Older revision		Revision as of 22:18, 29 August 2014
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	Compares the MLE solution, likelihood ratio bound and Fisher Matrix bound for a 1-parameter exponential distribution with interval data.		Compares the MLE solution, likelihood ratio bound and Fisher Matrix bound for a 1-parameter exponential distribution with interval data.

			<!---
	{{Reference_Example_Heading1}}		{{Reference_Example_Heading1}}
	Example 7.1 on page 154 in book ''Statistical Methods for Reliability Data'' by Dr. Meeker and Dr. Escobar, John Wiley & Sons, 1998. The sample size of 200 data is used here.		Example 7.1 on page 154 in book ''Statistical Methods for Reliability Data'' by Dr. Meeker and Dr. Escobar, John Wiley & Sons, 1998. The sample size of 200 data is used here.
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	* Based on the likelihood ratio (Select LRB for the confidence bound), the confidence interval is		* Based on the likelihood ratio (Select LRB for the confidence bound), the confidence interval is
			--->

Kate Racaza at 18:38, 4 June 2014

2014-06-04T18:38:43Z

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