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| ===Non-parametric Confidence Bounds===
| | #REDIRECT [[Non-Parametric Life Data Analysis]] |
| 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]:
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| ::<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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| :where:
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| ::<math>\begin{align}
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| & m= & \text{ the total number of intervals} \\
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| & n= & \text{ the total number of units}
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| \end{align}</math>
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| The variable <math>{{n}_{i}}</math> is defined by:
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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>
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| :where:
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| ::<math>\begin{align}
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| & {{r}_{j}}= & \text{the number of failures in interval }j \\
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| & {{s}_{j}}= & \text{the number of suspensions in interval }j
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| \end{align}</math>
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| Once the variance has been calculated, the standard error can be determined by taking the square root of the variance:
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| ::<math>{{\widehat{se}}_{\widehat{R}}}=\sqrt{\widehat{Var}(\widehat{R}({{t}_{i}}))}</math>
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| This information can then be applied to determine the confidence bounds:
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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>
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| :where:
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| ::<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.
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| ==== Example 12====
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| Determine the 1-sided confidence bounds for the reliability estimates in Example 11, with a 95% confidence level.
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| ===== Solution to Example 12=====
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| Once again, this type of problem is most readily solved by constructing a table similar to the following:
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| [[Image:lda22.1.gif|thumb|center|600px| ]]
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| The following plot illustrates these results graphically:
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| [[Image:lda22.2.gif|thumb|center|400px| ]]
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| <br>
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