Median Ranks - Revision history

Lisa Hacker: Redirected page to Parameter Estimation#Median Ranks

2012-08-21T22:50:13Z

Redirected page to Parameter Estimation#Median Ranks

← Older revision		Revision as of 22:50, 21 August 2012
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	~~{{InProgress\|}}~~		#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>~~

Pantelis: /* Alternative Computation */

2011-07-12T14:26:12Z

Alternative Computation

← Older revision		Revision as of 14:26, 12 July 2011
Line 52:		Line 52:

	::<math>M{{R}_{i}}=\frac{\frac{i}{N-i+1}}{{{F}_{1-\alpha ,2(N-i+1),2i}}+\frac{i}{N-i+1}}</math>		::<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		*where F is the F-distribution and
	*<math>1-\alpha</math> is the confidence limit.		*<math>1-\alpha</math> is the confidence limit.

Pantelis: /* Alternative Computation */

2011-07-12T14:26:00Z

Alternative Computation

← Older revision		Revision as of 14:26, 12 July 2011
Line 56:		Line 56:

	The Median Rank is obained by setting:		The Median Rank is obained by setting:

	::<math>1-\alpha=0.50</math>		::<math>1-\alpha=0.50</math>
	or		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>		::<math>M{{R}_{i}}=\frac{\frac{i}{N-i+1}}{{{F}_{0.50 ,2(N-i+1),2i}}+\frac{i}{N-i+1}}</math>

Pantelis: /* Alternative Computation */

2011-07-12T14:25:45Z

Alternative Computation

← Older revision		Revision as of 14:25, 12 July 2011
Line 51:		Line 51:
	Any rank can be computed by:		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>		::<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		*where F is the F-distribution and
	*<math>1-\alpha</math> is the confidence limit.		*<math>1-\alpha</math> is the confidence limit.
Line 58:		Line 58:
	::<math>1-\alpha=0.50</math>		::<math>1-\alpha=0.50</math>
	or		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>		::<math>M{{R}_{i}}=\frac{\frac{i}{N-i+1}}{{{F}_{0.50 ,2(N-i+1),2i}}+\frac{i}{N-i+1}}</math>

Pantelis at 14:24, 12 July 2011

2011-07-12T14:24:49Z

← Older revision		Revision as of 14:24, 12 July 2011
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	{{~~inprogress~~}}		{{InProgress\|}}


	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.		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.

Pantelis at 14:24, 12 July 2011

2011-07-12T14:24:02Z

← Older revision		Revision as of 14:24, 12 July 2011
Line 1:		Line 1:
			{{inprogress}}

	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.		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 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.

Pantelis: /* Alternative Computation */

2011-07-12T14:23:13Z

Alternative Computation

← Older revision		Revision as of 14:23, 12 July 2011
Line 45:		Line 45:


	==Alternative Computation==		=Alternative Computation=
	Any rank can be computed by:		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>		::<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.

	~~where F is the F-distribution and~~		The Median Rank is obained by setting:
	~~<math>1-\alpha</math> is the confidence limit.~~		::<math>1-\alpha=0.50</math>

	The ~~median~~ Rank is obained by setting:
	<math>1-\alpha=0.50</math>
	or		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>		::<math>M{{R}_{i}}=\frac{\frac{i}{n-i+1}}{{{F}_{0.50 ,2(n-i+1),2i}}+\frac{i}{n-i+1}}</math>

Pantelis at 14:22, 12 July 2011

2011-07-12T14:22:23Z

← Older revision		Revision as of 14:22, 12 July 2011
Line 43:		Line 43:

	to represent the probability of 30 failures out of 30.		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>

Pantelis: moved The Mathematics to Median Ranks

2011-07-12T14:16:58Z

moved The Mathematics to Median Ranks

← Older revision	Revision as of 14:16, 12 July 2011
(No difference)

Steve Sharp at 16:08, 29 June 2011

2011-06-29T16:08:26Z

← Older revision		Revision as of 16:08, 29 June 2011
Line 3:		Line 3:
	For 10 failures at 100 hours, the median rank, <math>Z,</math> is estimated by using:		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}		::<math>0.50=\underset{k=j}{\overset{N}{\mathop \sum }}\,\left( \begin{matrix}
	N \\		N \\
	k \\		k \\
Line 11:		Line 11:
	with:		with:

	<math>N=30,\text{ }J=10</math>		::<math>N=30,\text{ }J=10</math>


Line 17:		Line 17:
	For 10 failures at 200 hours, <math>Z</math> is estimated by using:		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}		::<math>0.50=\underset{k=j}{\overset{N}{\mathop \sum }}\,\left( \begin{matrix}
	N \\		N \\
	k \\		k \\
Line 25:		Line 25:
	where:		where:

	<math>N=30,\text{ }J=20</math>		::<math>N=30,\text{ }J=20</math>


Line 31:		Line 31:
	For 10 failures at 300 hours, <math>Z</math> is estimated by using:		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}		::<math>0.50=\underset{k=j}{\overset{N}{\mathop \sum }}\,\left( \begin{matrix}
	N \\		N \\
	k \\		k \\
Line 39:		Line 39:
	where:		where:

	<math>N=30,\text{ }J=30</math>		::<math>N=30,\text{ }J=30</math>


	to represent the probability of 30 failures out of 30.		to represent the probability of 30 failures out of 30.