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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>

Latest revision as of 23:02, 23 August 2012

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