Template:Non-Parametric LDA Example (Kaplan-Meier) - Revision history

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

2012-08-09T10:03:38Z

Redirected page to Non-Parametric Life Data Analysis

← Older revision		Revision as of 10:03, 9 August 2012
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	[[~~Category: For Deletion~~]]		#REDIRECT [[Non-Parametric_Life_Data_Analysis]]

Kate Racaza: Replaced content with 'Category: For Deletion'

2012-07-24T07:39:47Z

Replaced content with 'Category: For Deletion'

← Older revision		Revision as of 07:39, 24 July 2012
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	~~====Non-Parametric LDA Example (Kaplan-Meier)====~~		[[Category: For Deletion]]
	~~=====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>~~

Chris Kahn: /* Solution(Kaplan-Meier) */

2012-05-03T23:00:15Z

Solution(Kaplan-Meier)

← Older revision		Revision as of 23:00, 3 May 2012
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	Use the Kaplan-Meier estimator to determine the reliability estimates for each failure time.		Use the Kaplan-Meier estimator to determine the reliability estimates for each failure time.

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

Chris Kahn: /* Problem Statement(Kaplan-Meier) */

2012-05-03T22:58:39Z

Problem Statement(Kaplan-Meier)

← Older revision		Revision as of 22:58, 3 May 2012
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	====Non-Parametric LDA Example (Kaplan-Meier)====		====Non-Parametric LDA Example (Kaplan-Meier)====
	=====Problem Statement(Kaplan-Meier)=====		=====Problem Statement (Kaplan-Meier)=====

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

Harry Guo: Created page with '====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.
2012-02-06T18:27:23Z

Created page with '====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…'

New page
====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>