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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:
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| ::<math>{{u}_{q}}(t)={{\lambda }_{q}}{{\beta }_{q}}{{t}^{{{\beta }_{q}}-1}}</math>
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| 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:
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| ::<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>
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| where:
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| <br>
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| :• <math>K=1.</math>
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| :• <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).
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| :• <math>{{X}_{iq}}</math> is the <math>{{i}^{th}}</math> time-to-failure on the <math>{{q}^{th}}</math> system.
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| <br>
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| Then for each system, assume that:
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| ::<math>\chi _{q}^{2}=\frac{2{{M}_{q}}{{\beta }_{q}}}{{{{\tilde{\beta }}}_{q}}}</math>
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| 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:
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| ::<math>F=\frac{\tfrac{\chi _{1}^{2}}{2{{M}_{1}}}}{\tfrac{\chi _{2}^{2}}{2{{M}_{2}}}}</math>
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| 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:
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| ::<math>L=\underset{q=1}{\overset{K}{\mathop \sum }}\,{{M}_{q}}\ln ({{\tilde{\beta }}_{q}})-M\ln ({{\beta }^{*}})</math>
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| where:
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| <br>
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| :• <math>M=\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{M}_{q}}</math>
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| :• <math>{{\beta }^{*}}=\tfrac{M}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\tfrac{{{M}_{q}}}{{{{\tilde{\beta }}}_{q}}}}</math>
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| <br>
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| Also, let:
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| ::<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>
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| Calculate the statistic <math>D</math> , such that:
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| ::<math>D=\frac{2L}{a}</math>
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| 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.
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| <br>
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| <br>
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| '''Example'''
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| <br>
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| Consider the data in Table B.1.
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| <br>
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| {|align="center" border="1"
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| |-
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| |colspan="4" style="text-align:center"|Table B.1 - Repairable system data
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| | ||System 1||System 2||System 3
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| |-
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| |Start||0||0||0
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| |End||2000||2000||2000
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| |-
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| |Failures||1.2||1.4||0.3
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| |-
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| | ||55.6 ||35 ||32.6
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| |-
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| | ||72.7||46.8||33.4
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| |-
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| | ||111.9||65.9||241.7
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| |-
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| | ||121.9||181.1||396.2
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| |-
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| | ||303.6||712.6||444.4
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| |-
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| | ||326.9||1005.7||480.8
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| |-
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| | ||1568.4||1029.9||588.9
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| |-
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| | ||1913.5||1675.7||1043.9
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| |-
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| | || ||1787.5||1136.1
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| |-
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| | || ||1867||1288.1
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| | || || ||1408.1
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| | || || ||1439.4
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| |-
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| | || || ||1604.8
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| |}
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| 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:
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| ::<math>\begin{align}
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| & {{{\tilde{\beta }}}_{1}}= & 0.3753 \\
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| & {{{\tilde{\beta }}}_{2}}= & 0.4657
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| \end{align}</math>
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| 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.
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| ::<math>F=2.0980</math>
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| 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.
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| 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).
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| ::<math>D=0.5260</math>
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| 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.
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| <br>
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| <br>
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