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'''Bayesian Test Design with Prior Information from Expert Opinion'''
#REDIRECT [[Non-Parametric Bayesian - Expert Opinion]]
 
Suppose you wanted to know the reliability of a system and you had the following prior knowledge of the system:
 
:*Lowest possible reliability: ''a'' = 0.8
 
:*Most likely reliability: ''b'' = 0.85
 
:*Highest possible reliability: ''c'' = 0.97
 
This information can be used to approximate the expected value and the variance of the prior system reliability.
 
::<math> E\left(R_{0}\right)=\frac{a+4b+c}{6}=0.861667 </math>
 
::<math> Var\left(R_{0}\right)=\frac{c-a}{6}=0.028333 </math>
 
<br> These approximations of the expected value and variance of the prior system reliability can then be used to estimate <math>\alpha_{0}</math> and <math>\beta_{0}</math>, as given next:
 
::<math> \alpha_{0}=E\left(R_{0}\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right]=2.763331 </math>
 
::<math> \beta_{0}=\left(1-E\left(R_{0}\right)\right)\left[\frac{E\left(R_{0}\right)-E^{2}\left(R_{0}\right)}{Var\left(R_{0}\right)}-1\right]=0.44363 </math>
 
With <math>\alpha\,\!_{0}</math> and <math>\beta\,\!_{0}</math> known, any single value of the 4 quantities system reliability ''R'', confidence level ''CL'', number of units ''n'', or number of failures ''r'' can be calculated from the other 3.
 
System reliability ''R'' can be found if confidence level ''CL'', number of units ''n'', and number of failures ''r'' are known. Given the following data
<center>
''CL'' = 0.8
 
''n'' = 20
 
''r'' = 1
</center>
the number of successes ''s'' is
 
::<math> s = n – r = 19 </math>
 
and the posterior distribution is calculated as
 
::<math> \alpha\,\!=\alpha\,\!_{0}+s=2.763331+19=21.76333 </math>
 
::<math> \beta\,\!=\beta\,\!_{0}+r=0.44363+1=1.44363 </math>
 
::<math> R=\text{BetaINV}\left(1-CL,\alpha\,\!,\beta\,\!\right)=0.902996 </math>
 
The confidence level ''CL'' can be found if system reliability ''R'', number of units ''n'', and number of failures ''r'' are known. Given the following data
<center>
''R'' = 0.9
 
''n'' = 20
 
''r'' = 1
</center>
the number of successes ''s'' is
 
::<math> s = n – r = 19 </math>
 
and the posterior distribution is calculated as
 
::<math> \alpha\,\!=\alpha\,\!_{0}+s=2.763331+19=21.76333 </math>
 
::<math> \beta\,\!=\beta\,\!_{0}+r=0.44363+1=1.44363 </math>
 
::<math> CL=\text{BetaDist}\left(R,\alpha\,\!,\beta\,\!\right)=0.812164 </math>
 
The number of units ''n'' can be found if system reliability ''R'', confidence level ''CL'', and number of failures ''r'' are known. Given the following data
<center>
''R'' = 0.9
 
''CL'' = 0.8
 
''r'' = 1
</center>
the '''Number of Units''' utility in the '''Non-Parametric Binomial''' tab of the '''Design a Reliability Demonstration Test''' window can be used to solve for ''n''.
 
[[Image:drt8.jpg|thumb|center|500px| ]]  
 
The figure above shows that, in this case, ''n'' = 28.925085. The posterior distribution can now be calculated as <math> s=n-r=27.925085 </math> <math> \alpha\,\!=\alpha\,\!_{0}+s=2.763331+27.925085=30.688416 </math> <math> \beta\,\!=\beta\,\!_{0}+r=0.44363+1=1.44363 </math>
 
which results in a confidence level of <math> CL=1-1-\text{BetaDist}\left(R,\alpha\,\!,\beta\,\!\right)=0.91775 </math> Since the confidence level (0.91775) is greater than that which is required (0.8), we can reduce the number of units ''n'' until the calculated confidence level is close to the required value of 0.8. This results in the number of units ''n'' = 19.31.

Latest revision as of 01:28, 15 August 2012

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