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| '''Normal Distribution Likelihood Ratio Bound Example (Parameters)'''
| | #REDIRECT [[The Normal Distribution]] |
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| 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.
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| '''Solution'''
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| The first step is to calculate the likelihood function for the parameter estimates:
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| ::<math>\begin{align}
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| 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}}}} \\
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| 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}}}} \\
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| L(\widehat{\mu },\widehat{\sigma })= & 4.676897\times {{10}^{-9}}
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| \end{align}</math>
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| where <math>{{t}_{i}}</math> are the original time-to-failure data points. We can now rearrange Eqn. (lratio3) to the form:
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| ::<math>L(\mu ,\sigma )-L(\widehat{\mu },\widehat{\sigma })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}=0</math>
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| 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:
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| ::<math>\begin{align}
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| L(\mu ,\sigma )-L(\widehat{\mu },\widehat{\sigma })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}= & 0, \\
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| \\
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| L(\mu ,\sigma )-4.676897\times {{10}^{-9}}\cdot {{e}^{\tfrac{-1.642374}{2}}}= & 0, \\
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| \\
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| L(\mu ,\sigma )-2.057410\times {{10}^{-9}}= & 0.
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| \end{align}</math>
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| 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.
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| The following table gives the values of <math>\sigma </math> based on given values of <math>\mu </math> .
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| [[Image:tableofmu.gif|thumb|center|400px| ]]
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| <math></math>
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| [[Image:circleplot.gif|thumb|center|400px| ]]
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| <center><math>\begin{matrix}
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| \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}}} \\
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| \text{22}\text{.0} & \text{12}\text{.045} & \text{14}\text{.354} & \text{29}\text{.0} & \text{7.849}& \text{19.909} \\
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| \text{22}\text{.5} & \text{11}\text{.004} & \text{15}\text{.310} & \text{29}\text{.5} & \text{7}\text{.876} & \text{17}\text{.889} \\
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| \text{23}\text{.0} & \text{10}\text{.341} & \text{15}\text{.894} & \text{30}\text{.0} & \text{7}\text{.935} & \text{17}\text{.844} \\
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| \text{23}\text{.5} & \text{9}\text{.832} & \text{16}\text{.328} & \text{30}\text{.5} & \text{8}\text{.025} & \text{17}\text{.776} \\
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| \text{24}\text{.0} & \text{9}\text{.418} & \text{16}\text{.673} & \text{31}\text{.0} & \text{8}\text{.147} & \text{17}\text{.683} \\
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| \text{24}\text{.5} & \text{9}\text{.074} & \text{16}\text{.954} & \text{31}\text{.5} & \text{8}\text{.304} & \text{17}\text{.562} \\
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| \text{25}\text{.0} & \text{8}\text{.784} & \text{17}\text{.186} & \text{32}\text{.0} & \text{8}\text{.498} & \text{17}\text{.411} \\
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| \text{25}\text{.5} & \text{8}\text{.542} & \text{17}\text{.377} & \text{32}\text{.5} & \text{8}\text{.732} & \text{17}\text{.227} \\
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| \text{26}\text{.0} & \text{8}\text{.340} & \text{17}\text{.534} & \text{33}\text{.0} & \text{9}\text{.012} & \text{17}\text{.004} \\
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| \text{26}\text{.5} & \text{8}\text{.176} & \text{17}\text{.661} & \text{33}\text{.5} & \text{9}\text{.344} & \text{16}\text{.734} \\
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| \text{27}\text{.0} & \text{8}\text{.047} & \text{17}\text{.760} & \text{34}\text{.0} & \text{9}\text{.742} & \text{16}\text{.403} \\
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| \text{27}\text{.5} & \text{7}\text{.950} & \text{17}\text{.833} & \text{34}\text{.5} & \text{10}\text{.229} & \text{15}\text{.990} \\
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| \text{28}\text{.0} & \text{7}\text{.885} & \text{17}\text{.882} & \text{35}\text{.0} & \text{10}\text{.854} & \text{15}\text{.444} \\
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| \text{28}\text{.5} & \text{7}\text{.852} & \text{17}\text{.907} & \text{35}\text{.5} & \text{11}\text{.772} & \text{14}\text{.609} \\
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| \end{matrix}</math></center>
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| This data set is represented graphically in the following contour plot:
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| (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.
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