Template:Non-Parametric LDA Example (Standard Actuarial Method) - Revision history

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			[[Category: For Deletion]]
	~~====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~~:

Chris Kahn: /* Solution(Standard Actuarial Method) */

2012-05-03T23:03:25Z

Solution(Standard Actuarial Method)

← Older revision		Revision as of 23:03, 3 May 2012
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	Find reliability estimates for the data in Example for the Simple Actuarial Method using the 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)=====		=====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:		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:

Chris Kahn: /* Problem Statement(Standard Actuarial Method) */

2012-05-03T23:03:16Z

Problem Statement(Standard Actuarial Method)

← Older revision		Revision as of 23:03, 3 May 2012
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	====Non-Parametric LDA Example (Standard Actuarial Method)====		====Non-Parametric LDA Example (Standard Actuarial Method)====
	=====Problem Statement(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.		Find reliability estimates for the data in Example for the Simple Actuarial Method using the standard actuarial method.

Harry Guo: Created page with ' ====Non-Parametric LDA Example (Standard Actuarial Method)==== =====Problem Statement(Standard Actuarial Method)===== Find reliability estimates for the data in Example for the…'

2012-02-06T18:32:08Z

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====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: