Median Ranks: Difference between revisions

Latest revision as of 22:50, 21 August 2012

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Parameter Estimation#Median Ranks

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+#REDIRECT [[Parameter_Estimation#Median_Ranks]]
-Median ranks are used to obtain an estimate of the unreliability, <math>Q({{T}_{j}}),</math> for each failure at a <math>50%</math> confidence level. In the case of grouped data, the ranks are estimated for each group of failures, instead of each failure.
-For example, when using a group of 10 failures at 100 hours, 10 at 200 hours and 10 at 300 hours, Weibull++ estimates the median ranks (<math>Z</math> values) by solving the cumulative binomial equation with the appropriate values for order number and total number of test units.
-For 10 failures at 100 hours, the median rank, <math>Z,</math> is estimated by using:
-::<math>0.50=\underset{k=j}{\overset{N}{\mathop \sum }}\,\left( \begin{matrix}
-   N  \\
-   k  \\
-\end{matrix} \right){{Z}^{k}}{{\left( 1-Z \right)}^{N-k}}</math>
-with:
-::<math>N=30,\text{ }J=10</math>
-where one <math>Z</math> is obtained for the group, to represent the probability of 10 failures occurring out of 30.
-For 10 failures at 200 hours, <math>Z</math> is estimated by using:
-::<math>0.50=\underset{k=j}{\overset{N}{\mathop \sum }}\,\left( \begin{matrix}
-   N  \\
-   k  \\
-\end{matrix} \right){{Z}^{k}}{{\left( 1-Z \right)}^{N-k}}</math>
-where:
-::<math>N=30,\text{ }J=20</math>
-to represent the probability of 20 failures out of 30.
-For 10 failures at 300 hours, <math>Z</math> is estimated by using:
-::<math>0.50=\underset{k=j}{\overset{N}{\mathop \sum }}\,\left( \begin{matrix}
-   N  \\
-   k  \\
-\end{matrix} \right){{Z}^{k}}{{\left( 1-Z \right)}^{N-k}}</math>
-where:
-::<math>N=30,\text{ }J=30</math>
-to represent the probability of 30 failures out of 30.
-=Alternative Computation=
-Any rank can be computed by:
-::<math>M{{R}_{i}}=\frac{\frac{i}{n-i+1}}{{{F}_{1-\alpha ,2(n-i+1),2i}}+\frac{i}{n-i+1}}</math>
-*where F is the F-distribution and
-*<math>1-\alpha</math> is the confidence limit.
-The Median Rank is obained by setting:
-::<math>1-\alpha=0.50</math>
-or
-::<math>M{{R}_{i}}=\frac{\frac{i}{n-i+1}}{{{F}_{0.50 ,2(n-i+1),2i}}+\frac{i}{n-i+1}}</math>

Median Ranks: Difference between revisions

Latest revision as of 22:50, 21 August 2012

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