Template:Common beta hypothesis test rsa - Revision history

Richard House: Redirected page to RGA Appendix B#Common Beta Hypothesis Test

2012-08-23T23:02:38Z

Redirected page to RGA Appendix B#Common Beta Hypothesis Test

← Older revision		Revision as of 23:02, 23 August 2012
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	~~==Common Beta Hypothesis Test==~~		#REDIRECT [[RGA_Appendix_B#Common_Beta_Hypothesis_Test]]
	~~The Common Beta Hypothesis (CBH) Test is applicable to the following data types: Multiple Systems-Concurrent Operating Times, Repairable and Fleet. As shown by Crow~~ [~~17], suppose that <math>K</math> number of systems are under test. Each system has an intensity function given by:~~


	~~::<math>{{u}_{q}}(t)={{\lambda }_{q}}{{\beta }_{q}}{{t}^{{{\beta }_{q}}-1}}</math>~~


	where <math>q=1,\ldots ,K</math> . You can compare the intensity functions of each of the systems by comparing the <math>{{\beta }_{q}}</math> of each system. When conducting an analysis of data consisting of multiple systems, you expect that each of the systems performed in a similar manner. In particular, you would expect the interarrival rate of the failures across the systems to be fairly consistent. Therefore, the CBH Test tests the hypothesis, <math>{{H}_{o}}</math> , such that <math>{{\beta }_{1}}={{\beta }_{2}}=\ldots ={{\beta }_{K}}</math> . Let <math>{{\tilde{\beta }}_{q}}</math> denote the conditional maximum likelihood estimate of <math>{{\beta }_{q}}</math> , which is given by:


	::<math>{{\tilde{\beta }}_{q}}=\frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{M}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{M}_{q}}}{\mathop{\sum }}}\,\ln \left( \tfrac{{{T}_{q}}}{{{X}_{iq}}} \right)}</math>


	~~where:~~
	~~<br>~~
	~~:• <math>K=1.</math>~~
	:• <math>{{M}_{q}}={{N}_{q}}</math> if data on the <math>{{q}^{th}}</math> system is time terminated or <math>{{M}_{q}}=({{N}_{q}}-1)</math> if data on the <math>{{q}^{th}}</math> system is failure terminated ( <math>{{N}_{q}}</math> is the number of failures on the <math>{{q}^{th}}</math> system).
	~~:• <math>{{X}_{iq}}</math> is the <math>{{i}^{th}}</math> time-to-failure on the <math>{{q}^{th}}</math> system.~~
	~~<br>~~
	~~Then for each system, assume that:~~


	~~::<math>\chi _{q}^{2}=\frac{2{{M}_{q}}{{\beta }_{q}}}{{{{\tilde{\beta }}}_{q}}}</math>~~


	are conditionally distributed as independent Chi-Squared random variables with <math>2{{M}_{q}}</math> degrees of freedom. When <math>K=2</math> , you can test the null hypothesis, <math>{{H}_{o}}</math> , using the following statistic:


	~~::<math>F=\frac{\tfrac{\chi _{1}^{2}}{2{{M}_{1}}}}{\tfrac{\chi _{2}^{2}}{2{{M}_{2}}}}</math>~~


	If <math>{{H}_{o}}</math> is true, then <math>F</math> equals <math>\tfrac{{{{\tilde{\beta }}}_{2}}}{{{{\tilde{\beta }}}_{1}}}</math> and conditionally has an F-distribution with <math>(2{{M}_{1}},2{{M}_{2}})</math> degrees of freedom. The critical value, <math>F</math> , can then be determined by referring to the Chi-Squared tables. Now, if <math>K\ge 2</math> , then the likelihood ratio procedure [17] ~~can be used to test the hypothesis <math>{{\beta }_{1}}={{\beta }_{2}}=\ldots ={{\beta }_{K}}</math> . Consider the following statistic:~~


	~~::<math>L=\underset{q=1}{\overset{K}{\mathop \sum }}\,{{M}_{q}}\ln ({{\tilde{\beta }}_{q}})-M\ln ({{\beta }^{*}})</math>~~


	~~where:~~
	~~<br>~~
	~~:• <math>M=\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{M}_{q}}</math>~~
	~~:• <math>{{\beta }^{*}}=\tfrac{M}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\tfrac{{{M}_{q}}}{{{{\tilde{\beta }}}_{q}}}}</math>~~
	~~<br>~~
	~~Also, let:~~


	~~::<math>a=1+\frac{1}{6(K-1)}\left[ \underset{q=1}{\overset{K}{\mathop \sum }}\,\frac{1}{{{M}_{q}}}-\frac{1}{M} \right~~]~~</math>~~


	~~Calculate the statistic <math>D</math> , such that:~~


	~~::<math>D=\frac{2L}{a}</math>~~


	The statistic <math>D</math> is approximately distributed as a Chi-Squared random variable with <math>(K-1)</math> degrees of freedom. Then after calculating <math>D</math> , refer to the Chi-Squared tables with <math>(K-1)</math> degrees of freedom to determine the critical points. <math>{{H}_{o}}</math> is true if the statistic <math>D</math> falls between the critical points.
	~~<br>~~
	~~<br>~~
	~~'''Example'''~~
	~~<br>~~
	~~Consider the data in Table B.1.~~

	~~<br>~~
	~~{\|align="center" border="1"~~
	\|-
	~~\|colspan="4" style="text-align:center"\|Table B.1 - Repairable system data~~
	\|-
	~~\| \|\|System 1\|\|System 2\|\|System 3~~
	\|-
	~~\|Start\|\|0\|\|0\|\|0~~
	\|-
	~~\|End\|\|2000\|\|2000\|\|2000~~
	\|-
	~~\|Failures\|\|1.2\|\|1.4\|\|0.3~~
	\|-
	~~\| \|\|55.6 \|\|35 \|\|32.6~~
	\|-
	~~\| \|\|72.7\|\|46.8\|\|33.4~~
	\|-
	~~\| \|\|111.9\|\|65.9\|\|241.7~~
	\|-
	~~\| \|\|121.9\|\|181.1\|\|396.2~~
	\|-
	~~\| \|\|303.6\|\|712.6\|\|444.4~~
	\|-
	~~\| \|\|326.9\|\|1005.7\|\|480.8~~
	\|-
	~~\| \|\|1568.4\|\|1029.9\|\|588.9~~
	\|-
	~~\| \|\|1913.5\|\|1675.7\|\|1043.9~~
	\|-
	~~\| \|\| \|\|1787.5\|\|1136.1~~
	\|-
	~~\| \|\| \|\|1867\|\|1288.1~~
	\|-
	~~\| \|\| \|\| \|\|1408.1~~
	\|-
	~~\| \|\| \|\| \|\|1439.4~~
	\|-
	~~\| \|\| \|\| \|\|1604.8~~
	\|}


	Given that the intensity function for the <math>{{q}^{th}}</math> system is <math>{{u}_{q}}(t)={{\lambda }_{q}}{{\beta }_{q}}{{t}^{{{\beta }_{q}}-1}}</math> , test the hypothesis that <math>{{\beta }_{1}}={{\beta }_{2}}</math> while assuming a significance level equal to 0.05. Calculate <math>{{\tilde{\beta }}_{1}}</math> and <math>{{\tilde{\beta }}_{2}}</math> using Eqn. (CondBeta). Therefore:


	~~::<math>\begin{align}~~
	~~& {{{\tilde{\beta }}}_{1}}= & 0.3753 \\~~
	~~& {{{\tilde{\beta }}}_{2}}= & 0.4657~~
	~~\end{align}</math>~~


	~~Then <math>\tfrac{{{{\tilde{\beta }}}_{2}}}{{{{\tilde{\beta }}}_{1}}}=1.2408</math> . Using Eqn. (ftatistic) calculate the statistic <math>F</math> with a significance level of 0.05.~~


	~~::<math>F=2.0980</math>~~


	~~Since <math>1.2408<2.0980</math> we fail to reject the null hypothesis that <math>{{\beta }_{1}}={{\beta }_{2}}</math> at the 5% significance level.~~
	~~Now suppose instead it is desired to test the hypothesis that <math>{{\beta }_{1}}={{\beta }_{2}}={{\beta }_{3}}</math> . Calculate the statistic <math>D</math> using Eqn. (Dtatistic).~~


	~~::<math>D=0.5260</math>~~


	Using the Chi-Square tables with <math>K-1=2</math> degrees of freedom, the critical values at the 2.5 and 97.5 percentiles are 0.1026 and 5.9915, respectively. Since <math>0.1026<D<5.9915</math> , we fail to reject the null hypothesis that <math>{{\beta }_{1}}={{\beta }_{2}}={{\beta }_{3}}</math> at the 5% significance level.
	~~<br>~~
	~~<br>~~

Nicolette Young: Created page with '==Common Beta Hypothesis Test== The Common Beta Hypothesis (CBH) Test is applicable to the following data types: Multiple Systems-Concurrent Operating Times, Repairable and Fleet…'

2012-01-10T23:10:01Z

Created page with '==Common Beta Hypothesis Test== The Common Beta Hypothesis (CBH) Test is applicable to the following data types: Multiple Systems-Concurrent Operating Times, Repairable and Fleet…'

New page

==Common Beta Hypothesis Test==
The Common Beta Hypothesis (CBH) Test is applicable to the following data types: Multiple Systems-Concurrent Operating Times, Repairable and Fleet. As shown by Crow [17], suppose that <math>K</math> number of systems are under test. Each system has an intensity function given by:

::<math>{{u}_{q}}(t)={{\lambda }_{q}}{{\beta }_{q}}{{t}^{{{\beta }_{q}}-1}}</math>

where <math>q=1,\ldots ,K</math> . You can compare the intensity functions of each of the systems by comparing the <math>{{\beta }_{q}}</math> of each system. When conducting an analysis of data consisting of multiple systems, you expect that each of the systems performed in a similar manner. In particular, you would expect the interarrival rate of the failures across the systems to be fairly consistent. Therefore, the CBH Test tests the hypothesis, <math>{{H}_{o}}</math> , such that <math>{{\beta }_{1}}={{\beta }_{2}}=\ldots ={{\beta }_{K}}</math> . Let <math>{{\tilde{\beta }}_{q}}</math> denote the conditional maximum likelihood estimate of <math>{{\beta }_{q}}</math> , which is given by:

::<math>{{\tilde{\beta }}_{q}}=\frac{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{M}_{q}}}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\underset{i=1}{\overset{{{M}_{q}}}{\mathop{\sum }}}\,\ln \left( \tfrac{{{T}_{q}}}{{{X}_{iq}}} \right)}</math>

where:
 
:• <math>K=1.</math>
:• <math>{{M}_{q}}={{N}_{q}}</math> if data on the <math>{{q}^{th}}</math> system is time terminated or <math>{{M}_{q}}=({{N}_{q}}-1)</math> if data on the <math>{{q}^{th}}</math> system is failure terminated ( <math>{{N}_{q}}</math> is the number of failures on the <math>{{q}^{th}}</math> system).
:• <math>{{X}_{iq}}</math> is the <math>{{i}^{th}}</math> time-to-failure on the <math>{{q}^{th}}</math> system.
 
Then for each system, assume that:

::<math>\chi _{q}^{2}=\frac{2{{M}_{q}}{{\beta }_{q}}}{{{{\tilde{\beta }}}_{q}}}</math>

are conditionally distributed as independent Chi-Squared random variables with <math>2{{M}_{q}}</math> degrees of freedom. When <math>K=2</math> , you can test the null hypothesis, <math>{{H}_{o}}</math> , using the following statistic:

::<math>F=\frac{\tfrac{\chi _{1}^{2}}{2{{M}_{1}}}}{\tfrac{\chi _{2}^{2}}{2{{M}_{2}}}}</math>

If <math>{{H}_{o}}</math> is true, then <math>F</math> equals <math>\tfrac{{{{\tilde{\beta }}}_{2}}}{{{{\tilde{\beta }}}_{1}}}</math> and conditionally has an F-distribution with <math>(2{{M}_{1}},2{{M}_{2}})</math> degrees of freedom. The critical value, <math>F</math> , can then be determined by referring to the Chi-Squared tables. Now, if <math>K\ge 2</math> , then the likelihood ratio procedure [17] can be used to test the hypothesis <math>{{\beta }_{1}}={{\beta }_{2}}=\ldots ={{\beta }_{K}}</math> . Consider the following statistic:

::<math>L=\underset{q=1}{\overset{K}{\mathop \sum }}\,{{M}_{q}}\ln ({{\tilde{\beta }}_{q}})-M\ln ({{\beta }^{*}})</math>

where:
 
:• <math>M=\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{M}_{q}}</math>
:• <math>{{\beta }^{*}}=\tfrac{M}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\tfrac{{{M}_{q}}}{{{{\tilde{\beta }}}_{q}}}}</math>
 
Also, let:

::<math>a=1+\frac{1}{6(K-1)}\left[ \underset{q=1}{\overset{K}{\mathop \sum }}\,\frac{1}{{{M}_{q}}}-\frac{1}{M} \right]</math>

Calculate the statistic <math>D</math> , such that:

::<math>D=\frac{2L}{a}</math>

The statistic <math>D</math> is approximately distributed as a Chi-Squared random variable with <math>(K-1)</math> degrees of freedom. Then after calculating <math>D</math> , refer to the Chi-Squared tables with <math>(K-1)</math> degrees of freedom to determine the critical points. <math>{{H}_{o}}</math> is true if the statistic <math>D</math> falls between the critical points.
 
 
'''Example'''
 
Consider the data in Table B.1.

 
{|align="center" border="1"
|-
|colspan="4" style="text-align:center"|Table B.1 - Repairable system data
|-
| ||System 1||System 2||System 3
|-
|Start||0||0||0
|-
|End||2000||2000||2000
|-
|Failures||1.2||1.4||0.3
|-
| ||55.6 ||35 ||32.6
|-
| ||72.7||46.8||33.4
|-
| ||111.9||65.9||241.7
|-
| ||121.9||181.1||396.2
|-
| ||303.6||712.6||444.4
|-
| ||326.9||1005.7||480.8
|-
| ||1568.4||1029.9||588.9
|-
| ||1913.5||1675.7||1043.9
|-
| || ||1787.5||1136.1
|-
| || ||1867||1288.1
|-
| || || ||1408.1
|-
| || || ||1439.4
|-
| || || ||1604.8
|}

Given that the intensity function for the <math>{{q}^{th}}</math> system is <math>{{u}_{q}}(t)={{\lambda }_{q}}{{\beta }_{q}}{{t}^{{{\beta }_{q}}-1}}</math> , test the hypothesis that <math>{{\beta }_{1}}={{\beta }_{2}}</math> while assuming a significance level equal to 0.05. Calculate <math>{{\tilde{\beta }}_{1}}</math> and <math>{{\tilde{\beta }}_{2}}</math> using Eqn. (CondBeta). Therefore:

::<math>\begin{align}
& {{{\tilde{\beta }}}_{1}}= & 0.3753 \\
& {{{\tilde{\beta }}}_{2}}= & 0.4657
\end{align}</math>

Then <math>\tfrac{{{{\tilde{\beta }}}_{2}}}{{{{\tilde{\beta }}}_{1}}}=1.2408</math> . Using Eqn. (ftatistic) calculate the statistic <math>F</math> with a significance level of 0.05.

::<math>F=2.0980</math>

Since <math>1.2408<2.0980</math> we fail to reject the null hypothesis that <math>{{\beta }_{1}}={{\beta }_{2}}</math> at the 5% significance level.
Now suppose instead it is desired to test the hypothesis that <math>{{\beta }_{1}}={{\beta }_{2}}={{\beta }_{3}}</math> . Calculate the statistic <math>D</math> using Eqn. (Dtatistic).

::<math>D=0.5260</math>

Using the Chi-Square tables with <math>K-1=2</math> degrees of freedom, the critical values at the 2.5 and 97.5 percentiles are 0.1026 and 5.9915, respectively. Since <math>0.1026<D<5.9915</math> , we fail to reject the null hypothesis that <math>{{\beta }_{1}}={{\beta }_{2}}={{\beta }_{3}}</math> at the 5% significance level.