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===Likelihood Ratio Confidence Bounds===
#REDIRECT [[The_Normal_Distribution#Likelihood_Ratio_Confidence_Bounds]]
====Bounds on Parameters====
As covered in Chapter 5, the likelihood confidence bounds are calculated by finding values for  <math>{{\theta }_{1}}</math>  and  <math>{{\theta }_{2}}</math>  that satisfy:
 
::<math>-2\cdot \text{ln}\left( \frac{L({{\theta }_{1}},{{\theta }_{2}})}{L({{\widehat{\theta }}_{1}},{{\widehat{\theta }}_{2}})} \right)=\chi _{\alpha ;1}^{2}</math>
 
 
This equation can be rewritten as:
 
::<math>L({{\theta }_{1}},{{\theta }_{2}})=L({{\widehat{\theta }}_{1}},{{\widehat{\theta }}_{2}})\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}</math>
 
 
For complete data, the likelihood formula for the normal distribution is given by:
 
::<math>L(\mu ,\sigma )=\underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};\mu ,\sigma )=\underset{i=1}{\overset{N}{\mathop \prod }}\,\frac{1}{\sigma \cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{{{x}_{i}}-\mu }{\sigma } \right)}^{2}}}}</math>
 
where the  <math>{{x}_{i}}</math>  values represent the original time to failure data.  For a given value of  <math>\alpha </math> , values for  <math>\mu </math>  and  <math>\sigma </math>  can be found which represent the maximum and minimum values that satisfy Eqn. (lratio3). These represent the confidence bounds for the parameters at a confidence level  <math>\delta ,</math>  where  <math>\alpha =\delta </math>  for two-sided bounds and  <math>\alpha =2\delta -1</math>  for one-sided.
 
====Example 5====
Five units are put on a reliability test and experience failures at 12, 24, 28, 34, and 46 hours. Assuming a normal distribution, the MLE parameter estimates are calculated to be  <math>\widehat{\mu }=28.8</math>  and  <math>\widehat{\sigma }=11.2143.</math>  Calculate the two-sided 80% confidence bounds on these parameters using the likelihood ratio method.
=====Solution to Example 5=====
The first step is to calculate the likelihood function for the parameter estimates:
 
::<math>\begin{align}
  L(\widehat{\mu },\widehat{\sigma })= & \underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};\widehat{\mu },\widehat{\sigma })=\underset{i=1}{\overset{5}{\mathop \prod }}\,\frac{1}{\widehat{\sigma }\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{{{x}_{i}}-\widehat{\mu }}{\widehat{\sigma }} \right)}^{2}}}} \\
  L(\widehat{\mu },\widehat{\sigma })= & \underset{i=1}{\overset{5}{\mathop \prod }}\,\frac{1}{11.2143\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{{{x}_{i}}-28.8}{11.2143} \right)}^{2}}}} \\
  L(\widehat{\mu },\widehat{\sigma })= & 4.676897\times {{10}^{-9}} 
\end{align}</math>
 
where  <math>{{x}_{i}}</math>  are the original time-to-failure data points. We can now rearrange Eqn. (lratio3) to the form:
 
::<math>L(\mu ,\sigma )-L(\widehat{\mu },\widehat{\sigma })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}=0</math>
 
 
Since our specified confidence level,  <math>\delta </math> , is 80%, we can calculate the value of the chi-squared statistic,  <math>\chi _{0.8;1}^{2}=1.642374.</math>  We can now substitute this information into the equation:
 
::<math>\begin{align}
  L(\mu ,\sigma )-L(\widehat{\mu },\widehat{\sigma })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}= & 0, \\
\\
  L(\mu ,\sigma )-4.676897\times {{10}^{-9}}\cdot {{e}^{\tfrac{-1.642374}{2}}}= & 0, \\
  \\
  L(\mu ,\sigma )-2.057410\times {{10}^{-9}}= & 0. 
\end{align}</math>
 
 
It now remains to find the values of  <math>\mu </math>  and  <math>\sigma </math>  which satisfy this equation. This is an iterative process that requires setting the value of  <math>\mu </math>  and finding the appropriate values of  <math>\sigma </math> , and vice versa.
 
The following table gives the values of  <math>\sigma </math>  based on given values of  <math>\mu </math> .
 
[[Image:tableofmu.gif|thumb|center|400px| ]]
 
<math></math>
[[Image:circleplot.gif|thumb|center|400px| ]]
 
 
<center><math>\begin{matrix}
  \text{ }\!\!\mu\!\!\text{ } & {{\text{ }\!\!\sigma\!\!\text{ }}_{\text{1}}} & {{\text{ }\!\!\sigma\!\!\text{ }}_{\text{2}}} & \text{ }\!\!\mu\!\!\text{ } & {{\text{ }\!\!\sigma\!\!\text{ }}_{\text{1}}} & {{\text{ }\!\!\sigma\!\!\text{ }}_{\text{2}}}  \\
  \text{22}\text{.0} & \text{12}\text{.045} & \text{14}\text{.354} & \text{29}\text{.0} & \text{7.849}& \text{19.909}    \\
  \text{22}\text{.5} & \text{11}\text{.004} & \text{15}\text{.310} & \text{29}\text{.5} & \text{7}\text{.876} & \text{17}\text{.889}  \\
  \text{23}\text{.0} & \text{10}\text{.341} & \text{15}\text{.894} & \text{30}\text{.0} & \text{7}\text{.935} & \text{17}\text{.844}  \\
  \text{23}\text{.5} & \text{9}\text{.832} & \text{16}\text{.328} & \text{30}\text{.5} & \text{8}\text{.025} & \text{17}\text{.776}  \\
  \text{24}\text{.0} & \text{9}\text{.418} & \text{16}\text{.673} & \text{31}\text{.0} & \text{8}\text{.147} & \text{17}\text{.683}  \\
  \text{24}\text{.5} & \text{9}\text{.074} & \text{16}\text{.954} & \text{31}\text{.5} & \text{8}\text{.304} & \text{17}\text{.562}  \\
  \text{25}\text{.0} & \text{8}\text{.784} & \text{17}\text{.186} & \text{32}\text{.0} & \text{8}\text{.498} & \text{17}\text{.411}  \\
  \text{25}\text{.5} & \text{8}\text{.542} & \text{17}\text{.377} & \text{32}\text{.5} & \text{8}\text{.732} & \text{17}\text{.227}  \\
  \text{26}\text{.0} & \text{8}\text{.340} & \text{17}\text{.534} & \text{33}\text{.0} & \text{9}\text{.012} & \text{17}\text{.004}  \\
  \text{26}\text{.5} & \text{8}\text{.176} & \text{17}\text{.661} & \text{33}\text{.5} & \text{9}\text{.344} & \text{16}\text{.734}  \\
  \text{27}\text{.0} & \text{8}\text{.047} & \text{17}\text{.760} & \text{34}\text{.0} & \text{9}\text{.742} & \text{16}\text{.403}  \\
  \text{27}\text{.5} & \text{7}\text{.950} & \text{17}\text{.833} & \text{34}\text{.5} & \text{10}\text{.229} & \text{15}\text{.990}  \\
  \text{28}\text{.0} & \text{7}\text{.885} & \text{17}\text{.882} & \text{35}\text{.0} & \text{10}\text{.854} & \text{15}\text{.444}  \\
  \text{28}\text{.5} & \text{7}\text{.852} & \text{17}\text{.907} & \text{35}\text{.5} & \text{11}\text{.772} & \text{14}\text{.609}  \\
\end{matrix}</math></center>
 
This data set is represented graphically in the following contour plot:
 
(Note that this plot is generated with degrees of freedom  <math>k=1</math> , as we are only determining bounds on one parameter. The contour plots generated in Weibull++ are done with degrees of freedom  <math>k=2</math> , for use in comparing both parameters simultaneously.) As can be determined from the table, the lowest calculated value for  <math>\sigma </math>  is 7.849, while the highest is 17.909.  These represent the two-sided 80% confidence limits on this parameter.  Since solutions for the equation do not exist for values of  <math>\mu </math>  below 22 or above 35.5, these can be considered the two-sided 80% confidence limits for this parameter. In order to obtain more accurate values for the confidence limits on  <math>\mu </math> , we can perform the same procedure as before, but finding the two values of  <math>\mu </math>  that correspond with a given value of  <math>\sigma .</math>  Using this method, we find that the two-sided 80% confidence limits on  <math>\mu </math>  are 21.807 and 35.793, which are close to the initial estimates of 22 and 35.5.
 
====Bounds on Time and Reliability====
In order to calculate the bounds on a time estimate for a given reliability, or on a reliability estimate for a given time, the likelihood function needs to be rewritten in terms of one parameter and time/reliability, so that the maximum and minimum values of the time can be observed as the parameter is varied. This can be accomplished by substituting a form of the normal reliability equation into the likelihood function. The normal reliability equation can be written as:
 
::<math>R=1-\Phi \left( \frac{t-\mu }{\sigma } \right)</math>
 
This can be rearranged to the form:
 
::<math>\mu =t-\sigma \cdot {{\Phi }^{-1}}(1-R)</math>
 
where  <math>{{\Phi }^{-1}}</math>  is the inverse standard normal. This equation can now be substituted into Eqn. (normlikelihood), to produce a likelihood equation in terms of  <math>\sigma ,</math>  <math>t</math>  and  <math>R\ \ :</math> 
 
::<math>L(\sigma ,t/R)=\underset{i=1}{\overset{N}{\mathop \prod }}\,\frac{1}{\sigma \cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{{{x}_{i}}-\left[ t-\sigma \cdot {{\Phi }^{-1}}(1-R) \right]}{\sigma } \right)}^{2}}}}</math>
 
The unknown parameter  <math>t/R</math>  depends on what type of bounds are being determined.  If one is trying to determine the bounds on time for a given reliability, then  <math>R</math>  is a known constant and  <math>t</math>  is the unknown parameter. Conversely, if one is trying to determine the bounds on reliability for a given time, then  <math>t</math>  is a known constant and  <math>R</math>  is the unknown parameter. Either way, Eqn. (normliketr) can be used to solve Eqn. (lratio3) for the values of interest.
 
====Example 6====
For the data given in Example 5, determine the two-sided 80% confidence bounds on the time estimate for a reliability of 40%.  The ML estimate for the time at  <math>R(t)=40%</math>  is 31.637.
=====Solution to Example 6=====
In this example, we are trying to determine the two-sided 80% confidence bounds on the time estimate of 31.637. This is accomplished by substituting  <math>R=0.40</math>  and  <math>\alpha =0.8</math>  into Eqn. (normliketr), and varying  <math>\sigma </math>  until the maximum and minimum values of  <math>t</math>  are found. The following table gives the values of  <math>t</math>  based on given values of  <math>\sigma </math> .
 
[[Image:tabletbasedonsigma.gif|thumb|center|400px| ]]
<math></math>
 
This data set is represented graphically in the following contour plot:
 
[[Image:ovalplot.gif|thumb|center|400px| ]]
 
 
As can be determined from the table, the lowest calculated value for  <math>t</math>  is 25.046, while the highest is 39.250. These represent the 80% confidence limits on the time at which reliability is equal to 40%.
 
====Example 7====
For the data given in Example 5, determine the two-sided 80% confidence bounds on the reliability estimate for  <math>t=30</math> .  The ML estimate for the reliability at  <math>t=30</math>  is 45.739%.
 
=====Solution to Example 7=====
In this example, we are trying to determine the two-sided 80% confidence bounds on the reliability estimate of 45.739%. This is accomplished by substituting  <math>t=30</math>  and  <math>\alpha =0.8</math>  into Eqn. (normliketr), and varying  <math>\sigma </math>  until the maximum and minimum values of  <math>R</math>  are found. The following table gives the values of  <math>R</math>  based on given values of  <math>\sigma </math> .
 
[[Image:tablerbasedonsigma.gif|thumb|center|400px| ]]
 
This data set is represented graphically in the following contour plot:
 
[[Image:crazyoplot.gif|thumb|center|400px| ]]
 
 
As can be determined from the table, the lowest calculated value for  <math>R</math>  is 24.776%, while the highest is 68.000%. These represent the 80% two-sided confidence limits on the reliability at  <math>t=30</math> .

Latest revision as of 04:27, 13 August 2012

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