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{{template:RGA BOOK|AB|hypothesis Tests}}
{{template:RGA BOOK|AB|hypothesis Tests}}


=Hypothesis Tests=
{{hypothesis tests rsa}}
==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>
 
==Laplace Trend Test==
<br>
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>
 
 
where:
<br>
:• <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
<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>
 
 
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>
If  <math>U>1.645</math>  then a deteriorating trend would exist.
 
<br>
 
==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>
<br>
 
{|style= align="center" border="1"
|-
|colspan="6" style="text-align:center"|Table B.2 - Critical values for Cramér-von Mises test
|-
| ||colspan="5" style="text-align:center;"|<math>\alpha </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.

Revision as of 23:08, 10 January 2012

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Chapter AB: Hypothesis Tests


Chapter AB  
Hypothesis Tests  

Available Software:
RGA

More Resources:
RGA examples


New format available! This reference is now available in a new format that offers faster page load, improved display for calculations and images, more targeted search and the latest content available as a PDF. As of September 2023, this Reliawiki page will not continue to be updated. Please update all links and bookmarks to the latest reference at help.reliasoft.com/reference/reliability_growth_and_repairable_system_analysis

Chapter AB: Hypothesis Tests


Chapter AB  
Hypothesis Tests  

Available Software:
RGA

More Resources:
RGA examples


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