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==Non-parametric Analysis==
#REDIRECT [[Non-Parametric Life Data Analysis]]
 
Nonparametric analysis allows the user to analyze data without assuming an underlying distribution. This can have certain advantages as well as disadvantages. The ability to analyze data without assuming an underlying life distribution avoids the potentially large errors brought about by making incorrect assumptions about the distribution. On the other hand, the confidence bounds associated with nonparametric analysis are usually much wider than those calculated via parametric analysis, and predictions outside the range of the observations are not possible. Some practitioners recommend that any set of life data should first be subjected to a nonparametric analysis before moving on to the assumption of an underlying distribution.
There are several methods for conducting a nonparametric analysis, including the Kaplan-Meier, simple actuarial, and standard actuarial methods. A method for attaching confidence bounds to the results of these nonparametric analysis techniques can also be developed. The basis of nonparametric life data analysis is the empirical  <math>cdf</math>  function, which is given by:
 
 
::<math>\widehat{F}(t)=\frac{observations\le t}{n}</math>
 
 
Note that this is similar to the Bernard's approximation of the median ranks, as discussed in Chapter 3. The following nonparametric analysis methods are essentially variations of this concept.
 
{{kaplan-meier estimator}}
 
{{simple actuarial method}}
 
{{standard actuarial method}}
 
===Nonparametric 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>

Latest revision as of 07:34, 29 June 2012

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