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{{template:RGA BOOK|AB|hypothesis Tests}}
{{Template:RGA_BOOK|Appendix B|Hypothesis Tests}}
The RGA software provides two types of hypothesis tests: common beta hypothesis (CBH) and Laplace trend. Both tests are applicable to the following data types:
 
*Times-to-failure data
**Multiple Systems - Concurrent Operating Times
**Multiple Systems with Dates
**Multiple Systems with Event Codes
*Fielded data
**Repairable Systems
**Fleet


=Hypothesis Tests=
==Common Beta Hypothesis Test==
==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:
The common beta hypothesis (CBH) tests the hypothesis that all systems in the data set have similar values of beta. As shown by Crow [[RGA_References|[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>


::<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 evaluates 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>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:
where:
 
 
::<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>


*<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.


where:
<br>
• <math>K=1.</math>
<br>
• <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).
<br>
• <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:
Then for each system, assume that:


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


::<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>


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:
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 can be used to test the hypothesis <math>{{\beta }_{1}}={{\beta }_{2}}=\ldots ={{\beta }_{K}}\,\!</math>, as discussed in Crow [[RGA_References|[17]]]. Consider the following statistic:


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


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


*<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>


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


::<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>


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


===Common Beta Hypothesis Example===


::<math>D=\frac{2L}{a}</math>
Consider the data in the following table.
 
 
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>
<br>
Table B.1 - Repairable system data
 


{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|-
|colspan="4" style="text-align:center"|'''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:


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 the maximum likelihood estimates of <math>{{\tilde{\beta }}_{1}}\,\!</math> and <math>{{\tilde{\beta }}_{2}}\,\!</math>. Therefore:


::<math>\begin{align}
:<math>\begin{align}
   & {{{\tilde{\beta }}}_{1}}= & 0.3753 \\  
   & {{{\tilde{\beta }}}_{1}}= & 0.3753 \\  
  & {{{\tilde{\beta }}}_{2}}= & 0.4657   
  & {{{\tilde{\beta }}}_{2}}= & 0.4657   
\end{align}</math>
\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>
Then <math>\tfrac{{{{\tilde{\beta }}}_{2}}}{{{{\tilde{\beta }}}_{1}}}=1.2408\,\!</math>. Calculate the statistic <math>F\,\!</math> with a significance level of 0.05.


:<math>\begin{align}
F=2.0980
\end{align}\,\!</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.
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).


Now suppose that we test the hypothesis that <math>{{\beta }_{1}}={{\beta }_{2}}={{\beta }_{3}}\,\!</math>. Calculate the statistic <math>D\,\!</math>.


::<math>D=0.5260</math>
:<math>\begin{align}
D=0.5260  
\end{align}\,\!</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.


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>
==Laplace Trend Test==
==Laplace Trend Test==
<br>
The Laplace trend test evaluates the hypothesis that a trend does not exist within the data. 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:
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:
 
::<math>U=\frac{\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{X}_{i}}}{N}-\tfrac{T}{2}}{T\sqrt{\tfrac{1}{12N}}}</math>


:<math>U=\frac{\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{X}_{i}}}{N}-\tfrac{T}{2}}{T\sqrt{\tfrac{1}{12N}}}\,\!</math>


where:
where:
<br>
• <math>T</math>  = total operating time (termination time)
<br>
• <math>{{X}_{i}}</math>  = age of the system at the  <math>{{i}^{th}}</math>  successive failure
<br>
• <math>N</math>  = total number of failures
<br>
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> .
<br>
<br>
'''Example'''
<br>
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).
::<math>U=-2.6121</math>


*<math>T\,\!</math> = total operating time (termination time)
*<math>{{X}_{i}}\,\!</math> = age of the system at the <math>{{i}^{th}}\,\!</math> successive failure
*<math>N\,\!</math> = total number of failures


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>
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>.
If  <math>U>1.645</math> then a deteriorating trend would exist.  


<br>
===Laplace Trend Test Example===
==Critical Values for Cramér-von Mises Test==
<br>
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> .


<br>
Consider once again the data given in the table above. Check for a trend within System 1 assuming a significance level of 0.10. Calculate the test statistic <math>U\,\!</math> for System 1.
<br>
:::Table B.2 - Critical values for Cramér-von Mises test


{|style= align="center" border="1"
:<math>\begin{align}
|-
U=-2.6121
| || |colsplan="5" style="text-align:center;"|<math>\alpha </math>
\end{align}\,\!</math>
|-
|<math>M</math>|| 0.20|| 0.15|| 0.10|| 0.05|| 0.01
|-
|2|| 0.138|| 0.149|| 0.162|| 0.175|| 0.186
|-
|3|| 0.121|| 0.135|| 0.154|| 0.184||0.23
|-
|4|| 0.121|| 0.134|| 0.155|| 0.191||0.28
|-
|5|| 0.121|| 0.137|| 0.160|| 0.199||0.30
|-
|6|| 0.123|| 0.139|| 0.162|| 0.204||0.31
|-
|7|| 0.124|| 0.140|| 0.165|| 0.208||0.32
|-
|8|| 0.124|| 0.141|| 0.165|| 0.210||0.32
|-
|9|| 0.125|| 0.142|| 0.167|| 0.212||0.32
|-
|10|| 0.125|| 0.142|| 0.167|| 0.212||0.32
|-
|11|| 0.126|| 0.143|| 0.169|| 0.214||0.32
|-
|12|| 0.126|| 0.144|| 0.169|| 0.214||0.32
|-
|13|| 0.126|| 0.144|| 0.169|| 0.214||0.33
|-
|14|| 0.126|| 0.144|| 0.169|| 0.214||0.33
|-
|15|| 0.126|| 0.144|| 0.169|| 0.215||0.33
|-
|16|| 0.127|| 0.145|| 0.171|| 0.216|| 0.33
|-
|17|| 0.127|| 0.145|| 0.171|| 0.217|| 0.33
|-
|18|| 0.127|| 0.146|| 0.171|| 0.217|| 0.33
|-
|19|| 0.127|| 0.146|| 0.171|| 0.217|| 0.33
|-
|20|| 0.128|| 0.146|| 0.172|| 0.217|| 0.33
|-
|30|| 0.128|| 0.146|| 0.172|| 0.218|| 0.33
|-
|60|| 0.128|| 0.147|| 0.173|| 0.220|| 0.33
|-
|100|| 0.129|| 0.147|| 0.173|| 0.220|| 0.34
|}


For application of the Cramér-von Mises critical values, refer to Sections 5.5.1 and 10.1.6.1.
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. If <math>U>1.645\,\!</math> then a deteriorating trend would exist.

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Chapter Appendix B: Hypothesis Tests


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Chapter Appendix B  
Hypothesis Tests  

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Available Software:
RGA

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More Resources:
RGA examples

The RGA software provides two types of hypothesis tests: common beta hypothesis (CBH) and Laplace trend. Both tests are applicable to the following data types:

  • Times-to-failure data
    • Multiple Systems - Concurrent Operating Times
    • Multiple Systems with Dates
    • Multiple Systems with Event Codes
  • Fielded data
    • Repairable Systems
    • Fleet

Common Beta Hypothesis Test

The common beta hypothesis (CBH) tests the hypothesis that all systems in the data set have similar values of beta. As shown by Crow [17], suppose that [math]\displaystyle{ K\,\! }[/math] number of systems are under test. Each system has an intensity function given by:

[math]\displaystyle{ {{u}_{q}}(t)={{\lambda }_{q}}{{\beta }_{q}}{{t}^{{{\beta }_{q}}-1}}\,\! }[/math]

where [math]\displaystyle{ q=1,\ldots ,K\,\! }[/math]. You can compare the intensity functions of each of the systems by comparing the [math]\displaystyle{ {{\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 evaluates the hypothesis, [math]\displaystyle{ {{H}_{o}}\,\! }[/math], such that [math]\displaystyle{ {{\beta }_{1}}={{\beta }_{2}}=\ldots ={{\beta }_{K}}\,\! }[/math]. Let [math]\displaystyle{ {{\tilde{\beta }}_{q}}\,\! }[/math] denote the conditional maximum likelihood estimate of [math]\displaystyle{ {{\beta }_{q}}\,\! }[/math], which is given by:

[math]\displaystyle{ {{\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]\displaystyle{ K=1.\,\! }[/math]
  • [math]\displaystyle{ {{M}_{q}}={{N}_{q}}\,\! }[/math] if data on the [math]\displaystyle{ {{q}^{th}}\,\! }[/math] system is time terminated or [math]\displaystyle{ {{M}_{q}}=({{N}_{q}}-1)\,\! }[/math] if data on the [math]\displaystyle{ {{q}^{th}}\,\! }[/math] system is failure terminated ( [math]\displaystyle{ {{N}_{q}}\,\! }[/math] is the number of failures on the [math]\displaystyle{ {{q}^{th}}\,\! }[/math] system).
  • [math]\displaystyle{ {{X}_{iq}}\,\! }[/math] is the [math]\displaystyle{ {{i}^{th}}\,\! }[/math] time-to-failure on the [math]\displaystyle{ {{q}^{th}}\,\! }[/math] system.

Then for each system, assume that:

[math]\displaystyle{ \chi _{q}^{2}=\frac{2{{M}_{q}}{{\beta }_{q}}}{{{{\tilde{\beta }}}_{q}}}\,\! }[/math]

are conditionally distributed as independent chi-squared random variables with [math]\displaystyle{ 2{{M}_{q}}\,\! }[/math] degrees of freedom. When [math]\displaystyle{ K=2\,\! }[/math], you can test the null hypothesis, [math]\displaystyle{ {{H}_{o}}\,\! }[/math], using the following statistic:

[math]\displaystyle{ F=\frac{\tfrac{\chi _{1}^{2}}{2{{M}_{1}}}}{\tfrac{\chi _{2}^{2}}{2{{M}_{2}}}}\,\! }[/math]

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

[math]\displaystyle{ L=\underset{q=1}{\overset{K}{\mathop \sum }}\,{{M}_{q}}\ln ({{\tilde{\beta }}_{q}})-M\ln ({{\beta }^{*}})\,\! }[/math]

where:

  • [math]\displaystyle{ M=\underset{q=1}{\overset{K}{\mathop{\sum }}}\,{{M}_{q}}\,\! }[/math]
  • [math]\displaystyle{ {{\beta }^{*}}=\tfrac{M}{\underset{q=1}{\overset{K}{\mathop{\sum }}}\,\tfrac{{{M}_{q}}}{{{{\tilde{\beta }}}_{q}}}}\,\! }[/math]

Also, let:

[math]\displaystyle{ 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]\displaystyle{ D\,\! }[/math], such that:

[math]\displaystyle{ D=\frac{2L}{a}\,\! }[/math]

The statistic [math]\displaystyle{ D\,\! }[/math] is approximately distributed as a chi-squared random variable with [math]\displaystyle{ (K-1)\,\! }[/math] degrees of freedom. Then after calculating [math]\displaystyle{ D\,\! }[/math], refer to the chi-squared tables with [math]\displaystyle{ (K-1)\,\! }[/math] degrees of freedom to determine the critical points. [math]\displaystyle{ {{H}_{o}}\,\! }[/math] is true if the statistic [math]\displaystyle{ D\,\! }[/math] falls between the critical points.

Common Beta Hypothesis Example

Consider the data in the following table.

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]\displaystyle{ {{q}^{th}}\,\! }[/math] system is [math]\displaystyle{ {{u}_{q}}(t)={{\lambda }_{q}}{{\beta }_{q}}{{t}^{{{\beta }_{q}}-1}}\,\! }[/math], test the hypothesis that [math]\displaystyle{ {{\beta }_{1}}={{\beta }_{2}}\,\! }[/math] while assuming a significance level equal to 0.05. Calculate the maximum likelihood estimates of [math]\displaystyle{ {{\tilde{\beta }}_{1}}\,\! }[/math] and [math]\displaystyle{ {{\tilde{\beta }}_{2}}\,\! }[/math]. Therefore:

[math]\displaystyle{ \begin{align} & {{{\tilde{\beta }}}_{1}}= & 0.3753 \\ & {{{\tilde{\beta }}}_{2}}= & 0.4657 \end{align}\,\! }[/math]

Then [math]\displaystyle{ \tfrac{{{{\tilde{\beta }}}_{2}}}{{{{\tilde{\beta }}}_{1}}}=1.2408\,\! }[/math]. Calculate the statistic [math]\displaystyle{ F\,\! }[/math] with a significance level of 0.05.

[math]\displaystyle{ \begin{align} F=2.0980 \end{align}\,\! }[/math]

Since [math]\displaystyle{ 1.2408\lt 2.0980\,\! }[/math] we fail to reject the null hypothesis that [math]\displaystyle{ {{\beta }_{1}}={{\beta }_{2}}\,\! }[/math] at the 5% significance level.

Now suppose that we test the hypothesis that [math]\displaystyle{ {{\beta }_{1}}={{\beta }_{2}}={{\beta }_{3}}\,\! }[/math]. Calculate the statistic [math]\displaystyle{ D\,\! }[/math].

[math]\displaystyle{ \begin{align} D=0.5260 \end{align}\,\! }[/math]

Using the chi-square tables with [math]\displaystyle{ 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]\displaystyle{ 0.1026\lt D\lt 5.9915\,\! }[/math], we fail to reject the null hypothesis that [math]\displaystyle{ {{\beta }_{1}}={{\beta }_{2}}={{\beta }_{3}}\,\! }[/math] at the 5% significance level.

Laplace Trend Test

The Laplace trend test evaluates the hypothesis that a trend does not exist within the data. 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]\displaystyle{ U\,\! }[/math], using the following equation:

[math]\displaystyle{ U=\frac{\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{X}_{i}}}{N}-\tfrac{T}{2}}{T\sqrt{\tfrac{1}{12N}}}\,\! }[/math]

where:

  • [math]\displaystyle{ T\,\! }[/math] = total operating time (termination time)
  • [math]\displaystyle{ {{X}_{i}}\,\! }[/math] = age of the system at the [math]\displaystyle{ {{i}^{th}}\,\! }[/math] successive failure
  • [math]\displaystyle{ N\,\! }[/math] = total number of failures

The test statistic [math]\displaystyle{ 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]\displaystyle{ \alpha \,\! }[/math].

Laplace Trend Test Example

Consider once again the data given in the table above. Check for a trend within System 1 assuming a significance level of 0.10. Calculate the test statistic [math]\displaystyle{ U\,\! }[/math] for System 1.

[math]\displaystyle{ \begin{align} U=-2.6121 \end{align}\,\! }[/math]

From the standard normal tables with a significance level of 0.10, the critical value is equal to 1.645. If [math]\displaystyle{ -1.645\lt U\lt 1.645\,\! }[/math] then we would fail to reject the hypothesis of no trend. However, since [math]\displaystyle{ U\lt -1.645\,\! }[/math] then an improving trend exists within System 1. If [math]\displaystyle{ U\gt 1.645\,\! }[/math] then a deteriorating trend would exist.

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