Non Parametric RDA MCF Example - Revision history

Lisa Hacker at 18:53, 18 September 2023

2023-09-18T18:53:49Z

← Older revision		Revision as of 18:53, 18 September 2023
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	<noinclude>{{Banner Weibull Examples}}		<noinclude>{{Banner Weibull Examples}}
	''This example appears in the [~~[Non-Parametric Recurrent Event Data Analysis~~]~~] article~~.''		''This example appears in the [https://help.reliasoft.com/reference/life_data_analysis Life data analysis reference].''

	</noinclude>		</noinclude>

Nicolette Young at 17:27, 23 December 2015

2015-12-23T17:27:23Z

← Older revision		Revision as of 17:27, 23 December 2015
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	The analysis presented in this example can be performed automatically in Weibull++'s non-parametric RDA folio, as shown next.		The analysis presented in this example can be performed automatically in Weibull++'s non-parametric RDA folio, as shown next.

	[[Image:Recurrent Data Example 2 Data.png\|center\|~~600px~~]]		[[Image:Recurrent Data Example 2 Data.png\|center\|650px]]

	Note: In the folio above, the <math>F\,\!</math> refers to failures and <math>E\,\!</math> refers to suspensions (or censoring ages). The results, with calculated MCF values and upper and lower 95% confidence limits, are shown next along with the graphical plot.		Note: In the folio above, the <math>F\,\!</math> refers to failures and <math>E\,\!</math> refers to suspensions (or censoring ages). The results, with calculated MCF values and upper and lower 95% confidence limits, are shown next along with the graphical plot.

	[[Image:Recurrent Data Example 2 Result.png\|center\|~~600px~~]]		[[Image:Recurrent Data Example 2 Result.png\|center\|650px]]


	[[Image:Recurrent Data Example 2 Plot.png\|center\|550px]]		[[Image:Recurrent Data Example 2 Plot.png\|center\|550px]]

Nicolette Young at 17:26, 23 December 2015

2015-12-23T17:26:50Z

← Older revision		Revision as of 17:26, 23 December 2015
Line 197:		Line 197:
	The analysis presented in this example can be performed automatically in Weibull++'s non-parametric RDA folio, as shown next.		The analysis presented in this example can be performed automatically in Weibull++'s non-parametric RDA folio, as shown next.

	[[Image:Recurrent Data Example 2 Data.png\|center\|~~550px~~]]		[[Image:Recurrent Data Example 2 Data.png\|center\|600px]]

	Note: In the folio above, the <math>F\,\!</math> refers to failures and <math>E\,\!</math> refers to suspensions (or censoring ages). The results, with calculated MCF values and upper and lower 95% confidence limits, are shown next along with the graphical plot.		Note: In the folio above, the <math>F\,\!</math> refers to failures and <math>E\,\!</math> refers to suspensions (or censoring ages). The results, with calculated MCF values and upper and lower 95% confidence limits, are shown next along with the graphical plot.

	[[Image:Recurrent Data Example 2 Result.png\|center\|~~550px~~]]		[[Image:Recurrent Data Example 2 Result.png\|center\|600px]]


	[[Image:Recurrent Data Example 2 Plot.png\|center\|550px]]		[[Image:Recurrent Data Example 2 Plot.png\|center\|550px]]

Sharon Honecker: Added zero for consistency.

2015-08-24T19:46:33Z

Added zero for consistency.

← Older revision		Revision as of 19:46, 24 August 2015
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	\text{4} & \text{15} & \text{F} & \text{1}\text{.60} & \text{0}\text{.256} & 0.9511 & 2.6916 \\		\text{4} & \text{15} & \text{F} & \text{1}\text{.60} & \text{0}\text{.256} & 0.9511 & 2.6916 \\
	\text{5} & \text{16} & \text{F} & \text{1}\text{.80} & \text{0}\text{.288} & 1.1023 & 2.9393 \\		\text{5} & \text{16} & \text{F} & \text{1}\text{.80} & \text{0}\text{.288} & 1.1023 & 2.9393 \\
	\text{2} & \text{17} & \text{F} & \text{2}\text{.0} & \text{0}\text{.320} & 1.2560 & 3.1848 \\		\text{2} & \text{17} & \text{F} & \text{2}\text{.00} & \text{0}\text{.320} & 1.2560 & 3.1848 \\
	\text{1} & \text{17} & \text{S} & {} & {} & {} & {} \\		\text{1} & \text{17} & \text{S} & {} & {} & {} & {} \\
	\text{2} & \text{19} & \text{S} & {} & {} & {} & {} \\		\text{2} & \text{19} & \text{S} & {} & {} & {} & {} \\

Sharon Honecker: Fixed spacing in equations for consistency. Added zero to make all M*(t_i) entries have same number of digits.

2015-08-24T19:45:12Z

Fixed spacing in equations for consistency. Added zero to make all M*(t_i) entries have same number of digits.

← Older revision		Revision as of 19:45, 24 August 2015
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	\text{1} & \text{10} & \text{F} & \text{5} & \text{0}\text{.20} & \text{0}\text{.40 + 0}\text{.20 = 0}\text{.60} \\		\text{1} & \text{10} & \text{F} & \text{5} & \text{0}\text{.20} & \text{0}\text{.40 + 0}\text{.20 = 0}\text{.60} \\
	\text{3} & \text{12} & \text{F} & \text{5} & \text{0}\text{.20} & \text{0}\text{.60 + 0}\text{.20 = 0}\text{.80} \\		\text{3} & \text{12} & \text{F} & \text{5} & \text{0}\text{.20} & \text{0}\text{.60 + 0}\text{.20 = 0}\text{.80} \\
	\text{2} & \text{13} & \text{F} & \text{5} & \text{0}\text{.20} & \text{0}\text{.80+0}\text{.20 =1}\text{.00} \\		\text{2} & \text{13} & \text{F} & \text{5} & \text{0}\text{.20} & \text{0}\text{.80 + 0}\text{.20 = 1}\text{.00} \\
	\text{4} & \text{13} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.00 + 0}\text{.20 = 1}\text{.20} \\		\text{4} & \text{13} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.00 + 0}\text{.20 = 1}\text{.20} \\
	\text{1} & \text{15} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.20 + 0}\text{.20 =1}\text{.40} \\		\text{1} & \text{15} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.20 + 0}\text{.20 = 1}\text{.40} \\
	\text{4} & \text{15} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.40 + 0}\text{.20 = 1}\text{.60} \\		\text{4} & \text{15} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.40 + 0}\text{.20 = 1}\text{.60} \\
	\text{5} & \text{16} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.60 + 0}\text{.20 = 1}\text{.80} \\		\text{5} & \text{16} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.60 + 0}\text{.20 = 1}\text{.80} \\
	\text{2} & \text{17} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.80 + 0}\text{.20 = 2}\text{.0} \\		\text{2} & \text{17} & \text{F} & \text{5} & \text{0}\text{.20} & \text{1}\text{.80 + 0}\text{.20 = 2}\text{.00} \\
	\text{1} & \text{17} & \text{S} & \text{4} & {} & {} \\		\text{1} & \text{17} & \text{S} & \text{4} & {} & {} \\
	\text{2} & \text{19} & \text{S} & \text{3} & {} & {} \\		\text{2} & \text{19} & \text{S} & \text{3} & {} & {} \\

Harry Guo at 17:04, 29 January 2013

2013-01-29T17:04:28Z

← Older revision		Revision as of 17:04, 29 January 2013
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	\| F		\| F
	\| 3		\| 3
	\| <math>0.394+(\tfrac{1}{3})^2[2(0-\tfrac{1}{3})^2+2(1-\tfrac{1}{3})^2]=0.468\,\!</math>		\| <math>0.394+(\tfrac{1}{3})^2[(1-\tfrac{1}{3})^2+2(0-\tfrac{1}{3})^2]=0.468\,\!</math>
	\|-		\|-
	\| 4		\| 4
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	\| F		\| F
	\| 2		\| 2
	\| <math>0.593+(\tfrac{1}{2})^2[(0-\tfrac{1}{2})^2+(1-\tfrac{1}{2})^2]=0.718\,\!</math>		\| <math>0.593+(\tfrac{1}{2})^2[(1-\tfrac{1}{2})^2+(0-\tfrac{1}{2})^2]=0.718\,\!</math>
	\|-		\|-
	\| 3		\| 3

Harry Guo at 17:02, 29 January 2013

2013-01-29T17:02:36Z

← Older revision		Revision as of 17:02, 29 January 2013
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	\| F		\| F
	\| 5		\| 5
	\| <math>0.256+(\tfrac{1}{5})^2[4(0-\tfrac{1}{5})^2+1(1-\tfrac{1}{5})^2]=0.288\,\!</math>		\| <math>0.256+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.288\,\!</math>
	\|-		\|-
	\| 2		\| 2

Harry Guo at 17:02, 29 January 2013

2013-01-29T17:02:00Z

← Older revision		Revision as of 17:02, 29 January 2013
Line 149:		Line 149:
	\| F		\| F
	\| 2		\| 2
	\| <math>0.468+(\tfrac{1}{3})^2[(1-\tfrac{1}{3})^2+2(0-\tfrac{1}{3})^2]=0.593\,\!</math>		\| <math>0.468+(\tfrac{1}{2})^2[(1-\tfrac{1}{2})^2+(0-\tfrac{1}{2})^2]=0.593\,\!</math>
	\|-		\|-
	\| 5		\| 5
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	\| F		\| F
	\| 2		\| 2
	\| <math>0.593+(\tfrac{1}{3})^2[(0-\tfrac{1}{3})^2+2(1-\tfrac{1}{3})^2]=0.718\,\!</math>		\| <math>0.593+(\tfrac{1}{2})^2[(0-\tfrac{1}{2})^2+(1-\tfrac{1}{2})^2]=0.718\,\!</math>
	\|-		\|-
	\| 3		\| 3

Harry Guo at 16:56, 29 January 2013

2013-01-29T16:56:36Z

← Older revision		Revision as of 16:56, 29 January 2013
Line 155:		Line 155:
	\| F		\| F
	\| 2		\| 2
	\| <math>0.~~580~~+(\tfrac{1}{3})^2[(0-\tfrac{1}{3})^2+2(1-\tfrac{1}{3})^2]=0.718\,\!</math>		\| <math>0.593+(\tfrac{1}{3})^2[(0-\tfrac{1}{3})^2+2(1-\tfrac{1}{3})^2]=0.718\,\!</math>
	\|-		\|-
	\| 3		\| 3

Harry Guo at 16:55, 29 January 2013

2013-01-29T16:55:36Z

← Older revision		Revision as of 16:55, 29 January 2013
Line 137:		Line 137:
	\| F		\| F
	\| 3		\| 3
	\| <math>0.394+(\tfrac{1}{3})^2[2(0-\tfrac{1}{3})^2+4(1-\tfrac{1}{3})^2]=0.468\,\!</math>		\| <math>0.394+(\tfrac{1}{3})^2[2(0-\tfrac{1}{3})^2+2(1-\tfrac{1}{3})^2]=0.468\,\!</math>
	\|-		\|-
	\| 4		\| 4
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	\| F		\| F
	\| 2		\| 2
	\| <math>0.468+(\tfrac{1}{3})^2[(1-\tfrac{1}{3})^2+4(0-\tfrac{1}{3})^2]=0.593\,\!</math>		\| <math>0.468+(\tfrac{1}{3})^2[(1-\tfrac{1}{3})^2+2(0-\tfrac{1}{3})^2]=0.593\,\!</math>
	\|-		\|-
	\| 5		\| 5
Line 155:		Line 155:
	\| F		\| F
	\| 2		\| 2
	\| <math>0.580+(\tfrac{1}{3})^2[(0-\tfrac{1}{3})^2+4(1-\tfrac{1}{3})^2]=0.718\,\!</math>		\| <math>0.580+(\tfrac{1}{3})^2[(0-\tfrac{1}{3})^2+2(1-\tfrac{1}{3})^2]=0.718\,\!</math>
	\|-		\|-
	\| 3		\| 3