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| =Hypothesis Tests=
| | #REDIRECT [[RGA_Appendix_B]] |
| ==Common Beta Hypothesis Test==
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| 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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| |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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| |Start||0||0||0
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| |-
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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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| | || || ||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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| ==Laplace Trend Test==
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| <br>
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| The Laplace Trend Test tests the hypothesis that a trend does not exist within the data. The Laplace Trend test is applicable to the following data types: Multiple Systems-Concurrent Operating Times, Repairable and Fleet. The Laplace Trend Test can determine whether the system is deteriorating, improving, or if there is no trend at all. Calculate the test statistic, <math>U</math> , using the following equation:
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| ::<math>U=\frac{\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{X}_{i}}}{N}-\tfrac{T}{2}}{T\sqrt{\tfrac{1}{12N}}}</math>
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| where:
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| <br>
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| :• <math>T</math> = total operating time (termination time)
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| :• <math>{{X}_{i}}</math> = age of the system at the <math>{{i}^{th}}</math> successive failure
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| :• <math>N</math> = total number of failures
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| <br>
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| The test statistic <math>U</math> is approximately a standard normal random variable. The critical value is read from the Standard Normal tables with a given significance level, <math>\alpha </math> .
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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 once again the data in Table B.1. Check for a trend within System 1 assuming a significance level of 0.10. Calculate the test statistic <math>U</math> for System 1 using Eqn. (Utatistic).
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| ::<math>U=-2.6121</math>
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| From the Standard Normal tables with a significance level of 0.10, the critical value is equal to 1.645. If <math>-1.645<U<1.645</math> then we would fail to reject the hypothesis of no trend. However, since <math>U<-1.645</math> then an improving trend exists within System 1. <br>
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| If <math>U>1.645</math> then a deteriorating trend would exist.
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| <br>
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| ==Critical Values for Cramér-von Mises Test==
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| <br>
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| Table B.2 displays the critical values for the Cramér-von Mises goodness-of-fit test given the sample size, <math>M</math> , and the significance level, <math>\alpha </math> .
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| <br>
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| <br>
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| {|style= align="center" border="1"
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| |colspan="6" style="text-align:center"|Table B.2 - Critical values for Cramér-von Mises test
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| | ||colspan="5" style="text-align:center;"|<math>\alpha </math>
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| |<math>M</math>|| 0.20|| 0.15|| 0.10|| 0.05|| 0.01
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| |2|| 0.138|| 0.149|| 0.162|| 0.175|| 0.186
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| |3|| 0.121|| 0.135|| 0.154|| 0.184||0.23
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| |4|| 0.121|| 0.134|| 0.155|| 0.191||0.28
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| |5|| 0.121|| 0.137|| 0.160|| 0.199||0.30
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| |6|| 0.123|| 0.139|| 0.162|| 0.204||0.31
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| |7|| 0.124|| 0.140|| 0.165|| 0.208||0.32
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| |8|| 0.124|| 0.141|| 0.165|| 0.210||0.32
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| |9|| 0.125|| 0.142|| 0.167|| 0.212||0.32
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| |10|| 0.125|| 0.142|| 0.167|| 0.212||0.32
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| |11|| 0.126|| 0.143|| 0.169|| 0.214||0.32
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| |12|| 0.126|| 0.144|| 0.169|| 0.214||0.32
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| |13|| 0.126|| 0.144|| 0.169|| 0.214||0.33
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| |14|| 0.126|| 0.144|| 0.169|| 0.214||0.33
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| |15|| 0.126|| 0.144|| 0.169|| 0.215||0.33
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| |16|| 0.127|| 0.145|| 0.171|| 0.216|| 0.33
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| |17|| 0.127|| 0.145|| 0.171|| 0.217|| 0.33
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| |18|| 0.127|| 0.146|| 0.171|| 0.217|| 0.33
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| |19|| 0.127|| 0.146|| 0.171|| 0.217|| 0.33
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| |20|| 0.128|| 0.146|| 0.172|| 0.217|| 0.33
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| |30|| 0.128|| 0.146|| 0.172|| 0.218|| 0.33
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| |60|| 0.128|| 0.147|| 0.173|| 0.220|| 0.33
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| |100|| 0.129|| 0.147|| 0.173|| 0.220|| 0.34
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| |}
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| For application of the Cramér-von Mises critical values, refer to Sections 5.5.1 and 10.1.6.1.
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