Template:Non-parametric LDA confidence bounds - Revision history

Kate Racaza: Redirected page to Non-Parametric Life Data Analysis

2012-06-29T07:35:56Z

Redirected page to Non-Parametric Life Data Analysis

← Older revision		Revision as of 07:35, 29 June 2012
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	== Non-Parametric ~~Confidence Bounds ==~~		#REDIRECT [[Non-Parametric Life Data Analysis]]

	Confidence bounds for non-parametric reliability estimates can be calculated using a method similar to that of parametric confidence bounds. The difficulty in dealing with nonparametric data lies in the estimation of the variance. To estimate the variance for non-parametric data, Weibull++ uses Greenwood's formula [[Appendix: Weibull References\|[27]]]:

	~~::<math>\widehat{Var}(\widehat{R}({{t}_{i}}))={{\left[ \widehat{R}({{t}_{i}}) \right~~]~~}^{2}}\cdot \underset{j=1}{\overset{i}{\mathop \sum }}\,\frac{\tfrac{{{r}_{j}}}{{{n}_{j}}}}{{{n}_{j}}\cdot \left( 1-\tfrac{{{r}_{j}}}{{{n}_{j}}} \right)}</math>~~

	~~where:~~

	~~::<math>\begin{align}~~
	~~& m= \text{ the total number of intervals} \\~~
	~~& n= \text{ the total number of units}~~
	~~\end{align}</math>~~

	~~The variable <span class="texhtml">''n''<sub>''i''</sub></span> is defined by:~~

	~~::<math>{{n}_{i}}=n-\underset{j=0}{\overset{i-1}{\mathop \sum }}\,{{s}_{j}}-\underset{j=0}{\overset{i-1}{\mathop \sum }}\,{{r}_{j,}}\text{ }i=1,...,m</math>~~

	~~where:~~

	~~::<math>\begin{align}~~
	~~& {{r}_{j}}= \text{the number of failures in interval }j \\~~
	~~& {{s}_{j}}= \text{the number of suspensions in interval }j~~
	~~\end{align}</math>~~

	~~Once the variance has been calculated, the standard error can be determined by taking the square root of the variance:~~

	~~::<math>{{\widehat{se}}_{\widehat{R}}}=\sqrt{\widehat{Var}(\widehat{R}({{t}_{i}}))}</math>~~

	~~This information can then be applied to determine the confidence bounds:~~

	::<math>\left[ LC{{B}_{\widehat{R}}},\text{ }UC{{B}_{\widehat{R}}} \right]=\left[ \frac{\widehat{R}}{\widehat{R}+(1-\widehat{R})\cdot w},\text{ }\frac{\widehat{R}}{\widehat{R}+(1-\widehat{R})/w} \right]</math>

	~~where:~~

	~~::<math>w={{e}^{{{z}_{\alpha }}\cdot \tfrac{{{\widehat{se}}_{\widehat{R}}}}{\left[ \widehat{R}\cdot (1-\widehat{R}) \right]}}}</math>~~

	~~and <span class="texhtml">α</span> is the desired confidence level for the 1-sided confidence bounds.~~

	~~<br>'''Example 4:''' {{Example: Non-parametric LDA Confidence Bounds Example}}~~

Kate Racaza at 22:24, 9 March 2012

2012-03-09T22:24:16Z

← Older revision		Revision as of 22:24, 9 March 2012
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	==Non-~~parametric~~ Confidence Bounds==		== Non-Parametric Confidence Bounds ==
	Confidence bounds for ~~nonparametric~~ reliability estimates can be calculated using a method similar to that of parametric confidence bounds. The difficulty in dealing with nonparametric data lies in the estimation of the variance. To estimate the variance for ~~nonparametric~~ data, Weibull++ uses Greenwood's formula [[Appendix: Weibull References\|[27]]]:
			Confidence bounds for non-parametric reliability estimates can be calculated using a method similar to that of parametric confidence bounds. The difficulty in dealing with nonparametric data lies in the estimation of the variance. To estimate the variance for non-parametric data, Weibull++ uses Greenwood's formula [[Appendix: Weibull References\|[27]]]:

	::<math>\widehat{Var}(\widehat{R}({{t}_{i}}))={{\left[ \widehat{R}({{t}_{i}}) \right]}^{2}}\cdot \underset{j=1}{\overset{i}{\mathop \sum }}\,\frac{\tfrac{{{r}_{j}}}{{{n}_{j}}}}{{{n}_{j}}\cdot \left( 1-\tfrac{{{r}_{j}}}{{{n}_{j}}} \right)}</math>		::<math>\widehat{Var}(\widehat{R}({{t}_{i}}))={{\left[ \widehat{R}({{t}_{i}}) \right]}^{2}}\cdot \underset{j=1}{\overset{i}{\mathop \sum }}\,\frac{\tfrac{{{r}_{j}}}{{{n}_{j}}}}{{{n}_{j}}\cdot \left( 1-\tfrac{{{r}_{j}}}{{{n}_{j}}} \right)}</math>
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	\end{align}</math>		\end{align}</math>

	The variable <~~math~~>{{n~~}_{~~i}}</~~math~~> is defined by:		The variable <span class="texhtml">''n''<sub>''i''</sub></span> is defined by:

	::<math>{{n}_{i}}=n-\underset{j=0}{\overset{i-1}{\mathop \sum }}\,{{s}_{j}}-\underset{j=0}{\overset{i-1}{\mathop \sum }}\,{{r}_{j,}}\text{ }i=1,...,m</math>		::<math>{{n}_{i}}=n-\underset{j=0}{\overset{i-1}{\mathop \sum }}\,{{s}_{j}}-\underset{j=0}{\overset{i-1}{\mathop \sum }}\,{{r}_{j,}}\text{ }i=1,...,m</math>
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	::<math>w={{e}^{{{z}_{\alpha }}\cdot \tfrac{{{\widehat{se}}_{\widehat{R}}}}{\left[ \widehat{R}\cdot (1-\widehat{R}) \right]}}}</math>		::<math>w={{e}^{{{z}_{\alpha }}\cdot \tfrac{{{\widehat{se}}_{\widehat{R}}}}{\left[ \widehat{R}\cdot (1-\widehat{R}) \right]}}}</math>

	and <~~math~~>~~\alpha~~ </~~math~~> is the desired confidence level for the 1-sided confidence bounds.		and <span class="texhtml">α</span> is the desired confidence level for the 1-sided confidence bounds.


	'''Example 4:'''		<br>'''Example 4:''' {{Example: Non-parametric LDA Confidence Bounds Example}}
	{{Example: Non-parametric LDA Confidence Bounds Example}}

Harry Guo: /* Non-parametric Confidence Bounds */

2012-02-20T23:54:24Z

Non-parametric Confidence Bounds

← Older revision		Revision as of 23:54, 20 February 2012
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	::<math>\begin{align}		::<math>\begin{align}
	& m= & \text{ the total number of intervals} \\		& m= \text{ the total number of intervals} \\
	& n= & \text{ the total number of units}		& n= \text{ the total number of units}
	\end{align}</math>		\end{align}</math>

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	::<math>\begin{align}		::<math>\begin{align}
	& {{r}_{j}}= & \text{the number of failures in interval }j \\		& {{r}_{j}}= \text{the number of failures in interval }j \\
	& {{s}_{j}}= & \text{the number of suspensions in interval }j		& {{s}_{j}}= \text{the number of suspensions in interval }j
	\end{align}</math>		\end{align}</math>

Harry Guo: /* Non-parametric Confidence Bounds */

2012-02-20T19:09:19Z

Non-parametric Confidence Bounds

← Older revision		Revision as of 19:09, 20 February 2012
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	==Non-parametric Confidence Bounds==		==Non-parametric Confidence Bounds==
	Confidence bounds for nonparametric reliability estimates can be calculated using a method similar to that of parametric confidence bounds. The difficulty in dealing with nonparametric data lies in the estimation of the variance. To estimate the variance for nonparametric data, Weibull++ uses Greenwood's formula [27]:		Confidence bounds for nonparametric reliability estimates can be calculated using a method similar to that of parametric confidence bounds. The difficulty in dealing with nonparametric data lies in the estimation of the variance. To estimate the variance for nonparametric data, Weibull++ uses Greenwood's formula [[Appendix: Weibull References\|[27]]]:

	::<math>\widehat{Var}(\widehat{R}({{t}_{i}}))={{\left[ \widehat{R}({{t}_{i}}) \right]}^{2}}\cdot \underset{j=1}{\overset{i}{\mathop \sum }}\,\frac{\tfrac{{{r}_{j}}}{{{n}_{j}}}}{{{n}_{j}}\cdot \left( 1-\tfrac{{{r}_{j}}}{{{n}_{j}}} \right)}</math>		::<math>\widehat{Var}(\widehat{R}({{t}_{i}}))={{\left[ \widehat{R}({{t}_{i}}) \right]}^{2}}\cdot \underset{j=1}{\overset{i}{\mathop \sum }}\,\frac{\tfrac{{{r}_{j}}}{{{n}_{j}}}}{{{n}_{j}}\cdot \left( 1-\tfrac{{{r}_{j}}}{{{n}_{j}}} \right)}</math>

	:where:		where:

	::<math>\begin{align}		::<math>\begin{align}
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	::<math>{{n}_{i}}=n-\underset{j=0}{\overset{i-1}{\mathop \sum }}\,{{s}_{j}}-\underset{j=0}{\overset{i-1}{\mathop \sum }}\,{{r}_{j,}}\text{ }i=1,...,m</math>		::<math>{{n}_{i}}=n-\underset{j=0}{\overset{i-1}{\mathop \sum }}\,{{s}_{j}}-\underset{j=0}{\overset{i-1}{\mathop \sum }}\,{{r}_{j,}}\text{ }i=1,...,m</math>

	:where:		where:

	::<math>\begin{align}		::<math>\begin{align}
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	::<math>\left[ LC{{B}_{\widehat{R}}},\text{ }UC{{B}_{\widehat{R}}} \right]=\left[ \frac{\widehat{R}}{\widehat{R}+(1-\widehat{R})\cdot w},\text{ }\frac{\widehat{R}}{\widehat{R}+(1-\widehat{R})/w} \right]</math>		::<math>\left[ LC{{B}_{\widehat{R}}},\text{ }UC{{B}_{\widehat{R}}} \right]=\left[ \frac{\widehat{R}}{\widehat{R}+(1-\widehat{R})\cdot w},\text{ }\frac{\widehat{R}}{\widehat{R}+(1-\widehat{R})/w} \right]</math>

	:where:		where:

	::<math>w={{e}^{{{z}_{\alpha }}\cdot \tfrac{{{\widehat{se}}_{\widehat{R}}}}{\left[ \widehat{R}\cdot (1-\widehat{R}) \right]}}}</math>		::<math>w={{e}^{{{z}_{\alpha }}\cdot \tfrac{{{\widehat{se}}_{\widehat{R}}}}{\left[ \widehat{R}\cdot (1-\widehat{R}) \right]}}}</math>
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	and <math>\alpha </math> is the desired confidence level for the 1-sided confidence bounds.		and <math>\alpha </math> is the desired confidence level for the 1-sided confidence bounds.

	~~==== Example 12====~~
	~~Determine the 1-sided confidence bounds for the reliability estimates in Example 11, with a 95% confidence level.~~
	~~===== Solution to Example 12=====~~
	~~Once again, this type of problem is most readily solved by constructing a table similar to the following:~~

	~~[[Image~~:~~lda22.1.gif\|thumb\|center\|600px\| ]]~~		'''Example 4:'''
			{{Example: Non-parametric LDA Confidence Bounds Example}}
	~~The following plot illustrates these results graphically~~:

	~~[[Image:lda22.2.gif\|thumb\|center\|400px\| ]]~~
	~~<br>~~

Harry Guo at 18:48, 20 February 2012

2012-02-20T18:48:28Z

← Older revision		Revision as of 18:48, 20 February 2012
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	===Non-parametric Confidence Bounds===		==Non-parametric Confidence Bounds==
	Confidence bounds for nonparametric reliability estimates can be calculated using a method similar to that of parametric confidence bounds. The difficulty in dealing with nonparametric data lies in the estimation of the variance. To estimate the variance for nonparametric data, Weibull++ uses Greenwood's formula [27]:		Confidence bounds for nonparametric reliability estimates can be calculated using a method similar to that of parametric confidence bounds. The difficulty in dealing with nonparametric data lies in the estimation of the variance. To estimate the variance for nonparametric data, Weibull++ uses Greenwood's formula [27]:

Harry Guo: moved Template:Non-parametric confidence bounds to Template:Non-parametric LDA confidence bounds

2012-02-20T18:46:13Z

moved Template:Non-parametric confidence bounds to Template:Non-parametric LDA confidence bounds

← Older revision	Revision as of 18:46, 20 February 2012
(No difference)

Nicolette Young: Created page with '===Non-parametric Confidence Bounds=== Confidence bounds for nonparametric reliability estimates can be calculated using a method similar to that of parametric confidence bounds.…'

2012-01-05T00:03:27Z

Created page with '===Non-parametric Confidence Bounds=== Confidence bounds for nonparametric reliability estimates can be calculated using a method similar to that of parametric confidence bounds.…'

New page

===Non-parametric Confidence Bounds===
Confidence bounds for nonparametric reliability estimates can be calculated using a method similar to that of parametric confidence bounds. The difficulty in dealing with nonparametric data lies in the estimation of the variance. To estimate the variance for nonparametric data, Weibull++ uses Greenwood's formula [27]:

::<math>\widehat{Var}(\widehat{R}({{t}_{i}}))={{\left[ \widehat{R}({{t}_{i}}) \right]}^{2}}\cdot \underset{j=1}{\overset{i}{\mathop \sum }}\,\frac{\tfrac{{{r}_{j}}}{{{n}_{j}}}}{{{n}_{j}}\cdot \left( 1-\tfrac{{{r}_{j}}}{{{n}_{j}}} \right)}</math>

:where:

::<math>\begin{align}
& m= & \text{ the total number of intervals} \\
& n= & \text{ the total number of units}
\end{align}</math>

The variable <math>{{n}_{i}}</math> is defined by:

::<math>{{n}_{i}}=n-\underset{j=0}{\overset{i-1}{\mathop \sum }}\,{{s}_{j}}-\underset{j=0}{\overset{i-1}{\mathop \sum }}\,{{r}_{j,}}\text{ }i=1,...,m</math>

:where:

::<math>\begin{align}
& {{r}_{j}}= & \text{the number of failures in interval }j \\
& {{s}_{j}}= & \text{the number of suspensions in interval }j
\end{align}</math>

Once the variance has been calculated, the standard error can be determined by taking the square root of the variance:

::<math>{{\widehat{se}}_{\widehat{R}}}=\sqrt{\widehat{Var}(\widehat{R}({{t}_{i}}))}</math>

This information can then be applied to determine the confidence bounds:

::<math>\left[ LC{{B}_{\widehat{R}}},\text{ }UC{{B}_{\widehat{R}}} \right]=\left[ \frac{\widehat{R}}{\widehat{R}+(1-\widehat{R})\cdot w},\text{ }\frac{\widehat{R}}{\widehat{R}+(1-\widehat{R})/w} \right]</math>

:where:

::<math>w={{e}^{{{z}_{\alpha }}\cdot \tfrac{{{\widehat{se}}_{\widehat{R}}}}{\left[ \widehat{R}\cdot (1-\widehat{R}) \right]}}}</math>

and <math>\alpha </math> is the desired confidence level for the 1-sided confidence bounds.

==== Example 12====
Determine the 1-sided confidence bounds for the reliability estimates in Example 11, with a 95% confidence level.
===== Solution to Example 12=====
Once again, this type of problem is most readily solved by constructing a table similar to the following:

[[Image:lda22.1.gif|thumb|center|600px| ]]

The following plot illustrates these results graphically:

[[Image:lda22.2.gif|thumb|center|400px| ]]
<br>