Template:Example: Normal Distribution Likelihood Ratio Bound (Parameters): Difference between revisions

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'''Normal Distribution Likelihood Ratio Bound Example (Parameters)'''
#REDIRECT [[The Normal Distribution]]
 
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'''
 
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({{t}_{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{{{t}_{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{{{t}_{i}}-28.8}{11.2143} \right)}^{2}}}} \\
  L(\widehat{\mu },\widehat{\sigma })= & 4.676897\times {{10}^{-9}} 
\end{align}</math>
 
where  <math>{{t}_{i}}</math>  are the original time-to-failure data points. We can now rearrange the likelihood ratio equation 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.png|center|500px| ]]
 
<math></math>
[[Image:WB.9 normal parameter contour plot.png|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.

Latest revision as of 02:21, 13 August 2012

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