Template:Non-parametric LDA Examples - Revision history

Lisa Hacker: Redirected page to Non-Parametric Life Data Analysis

2012-08-09T10:04:07Z

Redirected page to Non-Parametric Life Data Analysis

← Older revision		Revision as of 10:04, 9 August 2012
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	[[~~Category: For Deletion~~]]		#REDIRECT [[Non-Parametric_Life_Data_Analysis]]

Kate Racaza at 07:38, 24 July 2012

2012-07-24T07:38:52Z

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	~~{{Non-Parametric LDA Example (Kaplan-Meier)}}~~		[[Category: For Deletion]]

	~~{{Non-Parametric LDA Example (Simple Actuarial Method)}}~~

	~~{{Non-Parametric LDA Example (Standard Actuarial Method)}}~~

Nikki Helms at 23:33, 22 March 2012

2012-03-22T23:33:56Z

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	~~[[Non-Parametric LDA Examples]]~~
	{{Non-Parametric LDA Example (Kaplan-Meier)}}		{{Non-Parametric LDA Example (Kaplan-Meier)}}

Nikki Helms at 23:32, 22 March 2012

2012-03-22T23:32:43Z

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			[[Non-Parametric LDA Examples]]
	{{Non-Parametric LDA Example (Kaplan-Meier)}}		{{Non-Parametric LDA Example (Kaplan-Meier)}}

Harry Guo: Replaced content with '{{Non-Parametric LDA Example (Kaplan-Meier)}} {{Non-Parametric LDA Example (Simple Actuarial Method)}} {{Non-Parametric LDA Example (Standard Actuarial Method)}}'

2012-02-06T18:32:49Z

Replaced content with '{{Non-Parametric LDA Example (Kaplan-Meier)}} {{Non-Parametric LDA Example (Simple Actuarial Method)}} {{Non-Parametric LDA Example (Standard Actuarial Method)}}'

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	{{Non-Parametric LDA Example (Standard Actuarial Method)}}		{{Non-Parametric LDA Example (Standard Actuarial Method)}}

	~~====Non-Parametric LDA Example (Standard Actuarial Method)====~~
	~~=====Problem Statement(Standard Actuarial Method)=====~~

	~~Find reliability estimates for the data in Example for the Simple Actuarial Method using the standard actuarial method.~~

	~~=====Solution(Standard Actuarial Method)=====~~
	The solution to this example is similar to that of Example 10, with the exception of the inclusion of the <math>n_{i}^{\prime }</math> term, which is used in Eqn. (standact). Applying this equation to the data, we can generate the following table:

	~~<center><math>\begin{matrix}~~
	~~Start & End & Number of & Number of & Adjusted & {} & {} \\~~
	~~Time & Time & Failures, {{r}_{i}} & Suspensions, {{s}_{i}} & Units, n_{i}^{\prime } & 1-\tfrac{{{r}_{j}}}{n_{j}^{\prime }} & \mathop{}_{}^{}1-\tfrac{{{r}_{j}}}{n_{j}^{\prime }} \\~~
	~~0 & 50 & 2 & 4 & 53 & 0.962 & 0.962 \\~~
	~~50 & 100 & 0 & 5 & 46.5 & 1.000 & 0.962 \\~~
	~~100 & 150 & 2 & 2 & 43 & 0.953 & 0.918 \\~~
	~~150 & 200 & 3 & 5 & 37.5 & 0.920 & 0.844 \\~~
	~~200 & 250 & 2 & 1 & 31.5 & 0.937 & 0.791 \\~~
	~~250 & 300 & 1 & 2 & 28 & 0.964 & 0.762 \\~~
	~~300 & 350 & 2 & 1 & 25.5 & 0.922 & 0.702 \\~~
	~~350 & 400 & 3 & 3 & 21.5 & 0.860 & 0.604 \\~~
	~~400 & 450 & 3 & 4 & 15 & 0.800 & 0.484 \\~~
	~~450 & 500 & 1 & 2 & 9 & 0.889 & 0.430 \\~~
	~~500 & 550 & 2 & 1 & 6.5 & 0.692 & 0.298 \\~~
	~~550 & 600 & 1 & 0 & 4 & 0.750 & 0.223 \\~~
	~~600 & 650 & 2 & 1 & 2.5 & 0.200 & 0.045 \\~~
	~~\end{matrix}</math></center>~~


	~~As can be determined from the preceding table, the reliability estimates for the failure times are:~~

Harry Guo at 18:31, 6 February 2012

2012-02-06T18:31:45Z

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	{{Non-Parametric LDA Example (Simple Actuarial Method)}}		{{Non-Parametric LDA Example (Simple Actuarial Method)}}

			{{Non-Parametric LDA Example (Standard Actuarial Method)}}

	====Non-Parametric LDA Example (Standard Actuarial Method)====		====Non-Parametric LDA Example (Standard Actuarial Method)====

Harry Guo at 18:29, 6 February 2012

2012-02-06T18:29:39Z

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 {{Non-Parametric LDA Example (Kaplan-Meier)}}
-====Non-Parametric LDA Example (Simple Actuarial Method)====
+{{Non-Parametric LDA Example (Simple Actuarial Method)}}
 ====Non-Parametric LDA Example (Standard Actuarial Method)====

Harry Guo at 18:28, 6 February 2012

2012-02-06T18:28:57Z

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	{{Non-Parametric LDA Example (Kaplan-Meier)}}		{{Non-Parametric LDA Example (Kaplan-Meier)}}
	~~====Non-Parametric LDA Example (Kaplan-Meier)====~~
	~~=====Problem Statement(Kaplan-Meier)=====~~

	~~A group of 20 units are put on a life test with the following results.~~

	~~<center><math>\begin{matrix}~~
	~~Number & State & State \\~~
	~~in State & (F or S) & End Time \\~~
	~~3 & F & 9 \\~~
	~~1 & S & 9 \\~~
	~~1 & F & 11 \\~~
	~~1 & S & 12 \\~~
	~~1 & F & 13 \\~~
	~~1 & S & 13 \\~~
	~~1 & S & 15 \\~~
	~~1 & F & 17 \\~~
	~~1 & F & 21 \\~~
	~~1 & S & 22 \\ 1 & S & 24 \\~~
	~~1 & S & 26 \\~~
	~~1 & F & 28 \\~~
	~~1 & F & 30 \\~~
	~~1 & S & 32 \\~~
	~~2 & S & 35 \\~~
	~~1 & S & 39 \\~~
	~~1 & S & 41 \\~~
	~~\end{matrix}</math></center>~~

	~~Use the Kaplan-Meier estimator to determine the reliability estimates for each failure time.~~

	~~=====Solution(Kaplan-Meier)=====~~
	~~Using the data and Eqn. (kapmeier), the following table can be constructed:~~

	~~<center><math>\begin{matrix}~~
	~~State & Number of & Number of & Available & {} & {} \\~~
	~~End Time & Failures, {{r}_{i}} & Suspensions, {{s}_{i}} & Units, {{n}_{i}} & \tfrac{{{n}_{i}}-{{r}_{i}}}{{{n}_{i}}} & \mathop{}_{}^{}\tfrac{{{n}_{i}}-{{r}_{i}}}{{{n}_{i}}} \\~~
	~~9 & 3 & 1 & 20 & 0.850 & 0.850 \\~~
	~~11 & 1 & 0 & 16 & 0.938 & 0.797 \\~~
	~~12 & 0 & 1 & 15 & 1.000 & 0.797 \\~~
	~~13 & 1 & 1 & 14 & 0.929 & 0.740 \\~~
	~~15 & 0 & 1 & 12 & 1.000 & 0.740 \\~~
	~~17 & 1 & 0 & 11 & 0.909 & 0.673 \\~~
	~~21 & 1 & 0 & 10 & 0.900 & 0.605 \\~~
	~~22 & 0 & 1 & 9 & 1.000 & 0.605 \\~~
	~~24 & 0 & 1 & 8 & 1.000 & 0.605 \\~~
	~~26 & 0 & 1 & 7 & 1.000 & 0.605 \\~~
	~~28 & 1 & 0 & 6 & 0.833 & 0.505 \\~~
	~~30 & 1 & 0 & 5 & 0.800 & 0.404 \\~~
	~~32 & 0 & 1 & 4 & 1.000 & 0.404 \\~~
	~~35 & 0 & 1 & 3 & 1.000 & 0.404 \\~~
	~~39 & 0 & 1 & 2 & 1.000 & 0.404 \\~~
	~~41 & 0 & 1 & 1 & 1.000 & 0.404 \\~~
	~~\end{matrix}</math></center>~~

	~~As can be determined from the preceding table, the reliability estimates for the failure times are:~~

	~~<center><math>\begin{matrix}~~
	~~Failure Time & Reliability Est. \\~~
	~~9 & 85.0% \\~~
	~~11 & 79.7% \\~~
	~~13 & 74.0% \\~~
	~~17 & 67.3% \\~~
	~~21 & 60.5% \\~~
	~~28 & 50.5% \\~~
	~~30 & 40.4% \\~~
	~~\end{matrix}</math></center>~~

	====Non-Parametric LDA Example (Simple Actuarial Method)====		====Non-Parametric LDA Example (Simple Actuarial Method)====

Harry Guo at 18:28, 6 February 2012

2012-02-06T18:28:35Z

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			{{Non-Parametric LDA Example (Kaplan-Meier)}}
	====Non-Parametric LDA Example (Kaplan-Meier)====		====Non-Parametric LDA Example (Kaplan-Meier)====
	=====Problem Statement(Kaplan-Meier)=====		=====Problem Statement(Kaplan-Meier)=====

Harry Guo: /* Non-Parametric LDA Example (Actuarial Method) */

2012-02-06T18:24:34Z

Non-Parametric LDA Example (Actuarial Method)

← Older revision		Revision as of 18:24, 6 February 2012
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	\end{matrix}</math></center>		\end{matrix}</math></center>

	====Non-Parametric LDA Example (Actuarial Method)====		====Non-Parametric LDA Example (Simple Actuarial Method)====
	=====Problem Statement(Actuarial Method)=====		=====Problem Statement(Simple Actuarial Method)=====

	A group of 55 units are put on a life test during which the units are evaluated every 50 hours, with the following results:		A group of 55 units are put on a life test during which the units are evaluated every 50 hours, with the following results:
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	=====Solution (Actuarial Method)=====		=====Solution (Simple Actuarial Method)=====
	The reliability estimates for the simple actuarial method can be obtained by expanding the data table to include terms used in calculation of the reliability estimates for Eqn. (simpact):		The reliability estimates for the simple actuarial method can be obtained by expanding the data table to include terms used in calculation of the reliability estimates for Eqn. (simpact):