Template:Example: Lognormal Distribution Likelihood Ratio Bound (Parameters) - Revision history

Lisa Hacker: Redirected page to The Lognormal Distribution

2012-08-13T03:36:17Z

Redirected page to The Lognormal Distribution

← Older revision		Revision as of 03:36, 13 August 2012
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	~~'''~~Lognormal Distribution ~~Likelihood Ratio Bound Example (Parameters)'''~~		#REDIRECT [[The Lognormal Distribution]]

	Five units are put on a reliability test and experience failures at 45, 60, 75, 90, and 115 hours. Assuming a lognormal distribution, the MLE parameter estimates are calculated to be <math>{{\widehat{\mu }}^{\prime }}=4.2926</math> and <math>{{\widehat{\sigma'}}}=0.32361.</math> Calculate the two-sided 75% 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 }}^{\prime }},{{\widehat{\sigma' }}})= & \underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};{{\widehat{\mu }}^{\prime }},{{\widehat{\sigma' }}}), \\~~
	= & \underset{i=1}{\overset{N}{\mathop \prod }}\,\frac{1}{{{x}_{i}}\cdot {{\widehat{\sigma' }}}\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{\text{ln}({{x}_{i}})-{{\widehat{\mu }}^{\prime }}}{{{\widehat{\sigma' }}}} \right)}^{2}}}} \\
	L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma'}}})= & \underset{i=1}{\overset{5}{\mathop \prod }}\,\frac{1}{{{x}_{i}}\cdot 0.32361\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{\text{ln}({{x}_{i}})-4.2926}{0.32361} \right)}^{2}}}} \\
	~~L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma'}}})= & 1.115256\times {{10}^{-10}}~~
	~~\end{align}</math>~~

	~~where <math>{{x}_{i}}</math> are the original time-to-failure data points. We can now rearrange the likelihod ratio equation to the form:~~

	~~::<math>L({\mu }',{{\sigma' }})-L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma' }}})\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}=0</math>~~

	Since our specified confidence level, <math>\delta </math> , is 75%, we can calculate the value of the chi-squared statistic, <math>\chi _{0.75;1}^{2}=1.323303.</math> We can now substitute this information into the equation:

	~~::<math>\begin{align}~~
	~~& L({\mu }',{{\sigma' }})-L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma' }}})\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}= & 0 \\~~
	~~& L({\mu }',{{\sigma'}})-1.115256\times {{10}^{-10}}\cdot {{e}^{\tfrac{-1.323303}{2}}}= & 0 \\~~
	~~& L({\mu }',{{\sigma'}})-5.754703\times {{10}^{-11}}= & 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>{{\sigma'}}</math> and finding the appropriate values of <math>{\mu }'</math> , and vice versa.

	~~The following table gives the values of <math>{\mu }'</math> based on given values of <math>{{\sigma'}}</math> .~~


	~~<center><math>\begin{matrix}~~
	~~{{\sigma' }} & \mu _{1}^{\prime } & \mu _{2}^{\prime } & {{\sigma' }} & \mu _{1}^{\prime } & \mu _{2}^{\prime } \\~~
	~~0.24 & 4.2421 & 4.3432 & 0.37 & 4.1145 & 4.4708 \\~~
	~~0.25 & 4.2115 & 4.3738 & 0.38 & 4.1152 & 4.4701 \\~~
	~~0.26 & 4.1909 & 4.3944 & 0.39 & 4.1170 & 4.4683 \\~~
	~~0.27 & 4.1748 & 4.4105 & 0.40 & 4.1200 & 4.4653 \\~~
	~~0.28 & 4.1618 & 4.4235 & 0.41 & 4.1244 & 4.4609 \\~~
	~~0.29 & 4.1509 & 4.4344 & 0.42 & 4.1302 & 4.4551 \\~~
	~~0.30 & 4.1419 & 4.4434 & 0.43 & 4.1377 & 4.4476 \\~~
	~~0.31 & 4.1343 & 4.4510 & 0.44 & 4.1472 & 4.4381 \\~~
	~~0.32 & 4.1281 & 4.4572 & 0.45 & 4.1591 & 4.4262 \\~~
	~~0.33 & 4.1231 & 4.4622 & 0.46 & 4.1742 & 4.4111 \\~~
	~~0.34 & 4.1193 & 4.4660 & 0.47 & 4.1939 & 4.3914 \\~~
	~~0.35 & 4.1166 & 4.4687 & 0.48 & 4.2221 & 4.3632 \\~~
	~~0.36 & 4.1150 & 4.4703 & {} & {} & {} \\~~
	~~\end{matrix}</math></center>~~

	~~These points are represented graphically in the following contour plot:~~

	~~[[Image:WB.10 lognormal contour plot.png\|center\|250px\|~~ ]]

	(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>{\mu }'</math> is 4.1145, while the highest is 4.4708. These represent the two-sided 75% confidence limits on this parameter. Since solutions for the equation do not exist for values of <math>{{\sigma'
	}}</math> below 0.24 or above 0.48, these can be considered the two-sided 75% confidence limits for this parameter. In order to obtain more accurate values for the confidence limits on <math>{{\sigma'}}</math> , we can perform the same procedure as before, but finding the two values of <math>\sigma </math> that correspond with a given value of <math>{\mu }'.</math> Using this method, we find that the 75% confidence limits on <math>{{\sigma'}}</math> are 0.23405 and 0.48936, which are close to the initial estimates of 0.24 and 0.48.

Richard House at 05:00, 8 August 2012

2012-08-08T05:00:14Z

← Older revision		Revision as of 05:00, 8 August 2012
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	The first step is to calculate the likelihood function for the parameter estimates:		The first step is to calculate the likelihood function for the parameter estimates:

	~~<center>~~<math>\begin{align}		::<math>\begin{align}
	L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma' }}})= & \underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};{{\widehat{\mu }}^{\prime }},{{\widehat{\sigma' }}}), \\		L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma' }}})= & \underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};{{\widehat{\mu }}^{\prime }},{{\widehat{\sigma' }}}), \\
	= & \underset{i=1}{\overset{N}{\mathop \prod }}\,\frac{1}{{{x}_{i}}\cdot {{\widehat{\sigma' }}}\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{\text{ln}({{x}_{i}})-{{\widehat{\mu }}^{\prime }}}{{{\widehat{\sigma' }}}} \right)}^{2}}}} \\		= & \underset{i=1}{\overset{N}{\mathop \prod }}\,\frac{1}{{{x}_{i}}\cdot {{\widehat{\sigma' }}}\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{\text{ln}({{x}_{i}})-{{\widehat{\mu }}^{\prime }}}{{{\widehat{\sigma' }}}} \right)}^{2}}}} \\
	L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma'}}})= & \underset{i=1}{\overset{5}{\mathop \prod }}\,\frac{1}{{{x}_{i}}\cdot 0.32361\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{\text{ln}({{x}_{i}})-4.2926}{0.32361} \right)}^{2}}}} \\		L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma'}}})= & \underset{i=1}{\overset{5}{\mathop \prod }}\,\frac{1}{{{x}_{i}}\cdot 0.32361\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{\text{ln}({{x}_{i}})-4.2926}{0.32361} \right)}^{2}}}} \\
	L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma'}}})= & 1.115256\times {{10}^{-10}}		L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma'}}})= & 1.115256\times {{10}^{-10}}
	\end{align}</math~~></center~~>		\end{align}</math>

	where <math>{{x}_{i}}</math> are the original time-to-failure data points. We can now rearrange the likelihod ratio equation to the form:		where <math>{{x}_{i}}</math> are the original time-to-failure data points. We can now rearrange the likelihod ratio equation to the form:

Nicolette Young at 21:21, 25 April 2012

2012-04-25T21:21:14Z

← Older revision		Revision as of 21:21, 25 April 2012
Line 50:		Line 50:
	These points are represented graphically in the following contour plot:		These points are represented graphically in the following contour plot:

	[[Image:WB.10 lognormal contour plot.png\|center\|~~400px~~\| ]]		[[Image:WB.10 lognormal contour plot.png\|center\|250px\| ]]

	(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>{\mu }'</math> is 4.1145, while the highest is 4.4708. These represent the two-sided 75% confidence limits on this parameter. Since solutions for the equation do not exist for values of <math>{{\sigma'		(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>{\mu }'</math> is 4.1145, while the highest is 4.4708. These represent the two-sided 75% confidence limits on this parameter. Since solutions for the equation do not exist for values of <math>{{\sigma'
	}}</math> below 0.24 or above 0.48, these can be considered the two-sided 75% confidence limits for this parameter. In order to obtain more accurate values for the confidence limits on <math>{{\sigma'}}</math> , we can perform the same procedure as before, but finding the two values of <math>\sigma </math> that correspond with a given value of <math>{\mu }'.</math> Using this method, we find that the 75% confidence limits on <math>{{\sigma'}}</math> are 0.23405 and 0.48936, which are close to the initial estimates of 0.24 and 0.48.		}}</math> below 0.24 or above 0.48, these can be considered the two-sided 75% confidence limits for this parameter. In order to obtain more accurate values for the confidence limits on <math>{{\sigma'}}</math> , we can perform the same procedure as before, but finding the two values of <math>\sigma </math> that correspond with a given value of <math>{\mu }'.</math> Using this method, we find that the 75% confidence limits on <math>{{\sigma'}}</math> are 0.23405 and 0.48936, which are close to the initial estimates of 0.24 and 0.48.

Nicolette Young at 19:12, 14 March 2012

2012-03-14T19:12:30Z

← Older revision		Revision as of 19:12, 14 March 2012
Line 50:		Line 50:
	These points are represented graphically in the following contour plot:		These points are represented graphically in the following contour plot:

	[[Image:~~ldachp9ex5~~.~~gif\|thumb~~\|center\|400px\| ]]		[[Image:WB.10 lognormal contour plot.png\|center\|400px\| ]]

	(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>{\mu }'</math> is 4.1145, while the highest is 4.4708. These represent the two-sided 75% confidence limits on this parameter. Since solutions for the equation do not exist for values of <math>{{\sigma'		(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>{\mu }'</math> is 4.1145, while the highest is 4.4708. These represent the two-sided 75% confidence limits on this parameter. Since solutions for the equation do not exist for values of <math>{{\sigma'
	}}</math> below 0.24 or above 0.48, these can be considered the two-sided 75% confidence limits for this parameter. In order to obtain more accurate values for the confidence limits on <math>{{\sigma'}}</math> , we can perform the same procedure as before, but finding the two values of <math>\sigma </math> that correspond with a given value of <math>{\mu }'.</math> Using this method, we find that the 75% confidence limits on <math>{{\sigma'}}</math> are 0.23405 and 0.48936, which are close to the initial estimates of 0.24 and 0.48.		}}</math> below 0.24 or above 0.48, these can be considered the two-sided 75% confidence limits for this parameter. In order to obtain more accurate values for the confidence limits on <math>{{\sigma'}}</math> , we can perform the same procedure as before, but finding the two values of <math>\sigma </math> that correspond with a given value of <math>{\mu }'.</math> Using this method, we find that the 75% confidence limits on <math>{{\sigma'}}</math> are 0.23405 and 0.48936, which are close to the initial estimates of 0.24 and 0.48.

Harry Guo at 23:18, 13 February 2012

2012-02-13T23:18:26Z

Harry Guo: Created page with ''''Lognormal Distribution Likelihood Ratio Bound Example (Parameters)''' Five units are put on a reliability test and experience failures at 45, 60, 75, 90, and 115 hours. Assum…'

2012-02-13T18:49:22Z

Created page with ''''Lognormal Distribution Likelihood Ratio Bound Example (Parameters)''' Five units are put on a reliability test and experience failures at 45, 60, 75, 90, and 115 hours. Assum…'

New page

'''Lognormal Distribution Likelihood Ratio Bound Example (Parameters)'''

Five units are put on a reliability test and experience failures at 45, 60, 75, 90, and 115 hours. Assuming a lognormal distribution, the MLE parameter estimates are calculated to be <math>{{\widehat{\mu }}^{\prime }}=4.2926</math> and <math>{{\widehat{\sigma }}_{{{T}'}}}=0.32361.</math> Calculate the two-sided 75% confidence bounds on these parameters using the likelihood ratio method.

'''Solution'''

The first step is to calculate the likelihood function for the parameter estimates:

<center><math>\begin{align}
L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_{{{T}'}}})= & \underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};{{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_{{{T}'}}}), \\
= & \underset{i=1}{\overset{N}{\mathop \prod }}\,\frac{1}{{{x}_{i}}\cdot {{\widehat{\sigma }}_{{{T}'}}}\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{\text{ln}({{x}_{i}})-{{\widehat{\mu }}^{\prime }}}{{{\widehat{\sigma }}_{{{T}'}}}} \right)}^{2}}}} \\
L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_{{{T}'}}})= & \underset{i=1}{\overset{5}{\mathop \prod }}\,\frac{1}{{{x}_{i}}\cdot 0.32361\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{\text{ln}({{x}_{i}})-4.2926}{0.32361} \right)}^{2}}}} \\
L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_{{{T}'}}})= & 1.115256\times {{10}^{-10}}
\end{align}</math></center>

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 }_{{{T}'}}})-L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_{{{T}'}}})\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}=0</math>

Since our specified confidence level, <math>\delta </math> , is 75%, we can calculate the value of the chi-squared statistic, <math>\chi _{0.75;1}^{2}=1.323303.</math> We can now substitute this information into the equation:

::<math>\begin{align}
& L({\mu }',{{\sigma }_{{{T}'}}})-L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_{{{T}'}}})\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}= & 0 \\
& L({\mu }',{{\sigma }_{{{T}'}}})-1.115256\times {{10}^{-10}}\cdot {{e}^{\tfrac{-1.323303}{2}}}= & 0 \\
& L({\mu }',{{\sigma }_{{{T}'}}})-5.754703\times {{10}^{-11}}= & 0
\end{align}</math>

It now remains to find the values of <math>{\mu }'</math> and <math>{{\sigma }_{{{T}'}}}</math> which satisfy this equation. This is an iterative process that requires setting the value of <math>{{\sigma }_{{{T}'}}}</math> and finding the appropriate values of <math>{\mu }'</math> , and vice versa.

The following table gives the values of <math>{\mu }'</math> based on given values of <math>{{\sigma }_{{{T}'}}}</math> .

<center><math>\begin{matrix}
{{\sigma }_{{{T}'}}} & \mu _{1}^{\prime } & \mu _{2}^{\prime } & {{\sigma }_{{{T}'}}} & \mu _{1}^{\prime } & \mu _{2}^{\prime } \\
0.24 & 4.2421 & 4.3432 & 0.37 & 4.1145 & 4.4708 \\
0.25 & 4.2115 & 4.3738 & 0.38 & 4.1152 & 4.4701 \\
0.26 & 4.1909 & 4.3944 & 0.39 & 4.1170 & 4.4683 \\
0.27 & 4.1748 & 4.4105 & 0.40 & 4.1200 & 4.4653 \\
0.28 & 4.1618 & 4.4235 & 0.41 & 4.1244 & 4.4609 \\
0.29 & 4.1509 & 4.4344 & 0.42 & 4.1302 & 4.4551 \\
0.30 & 4.1419 & 4.4434 & 0.43 & 4.1377 & 4.4476 \\
0.31 & 4.1343 & 4.4510 & 0.44 & 4.1472 & 4.4381 \\
0.32 & 4.1281 & 4.4572 & 0.45 & 4.1591 & 4.4262 \\
0.33 & 4.1231 & 4.4622 & 0.46 & 4.1742 & 4.4111 \\
0.34 & 4.1193 & 4.4660 & 0.47 & 4.1939 & 4.3914 \\
0.35 & 4.1166 & 4.4687 & 0.48 & 4.2221 & 4.3632 \\
0.36 & 4.1150 & 4.4703 & {} & {} & {} \\
\end{matrix}</math></center>

These points are represented graphically in the following contour plot:

[[Image:ldachp9ex5.gif|thumb|center|400px| ]]

(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>{\mu }'</math> is 4.1145, while the highest is 4.4708. These represent the two-sided 75% confidence limits on this parameter. Since solutions for the equation do not exist for values of <math>{{\sigma }_{{{T}'}}}</math> below 0.24 or above 0.48, these can be considered the two-sided 75% confidence limits for this parameter. In order to obtain more accurate values for the confidence limits on <math>{{\sigma }_{{{T}'}}}</math> , we can perform the same procedure as before, but finding the two values of <math>\sigma </math> that correspond with a given value of <math>{\mu }'.</math> Using this method, we find that the 75% confidence limits on <math>{{\sigma }_{{{T}'}}}</math> are 0.23405 and 0.48936, which are close to the initial estimates of 0.24 and 0.48.

← Older revision		Revision as of 23:18, 13 February 2012
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	'''Lognormal Distribution Likelihood Ratio Bound Example (Parameters)'''		'''Lognormal Distribution Likelihood Ratio Bound Example (Parameters)'''

	Five units are put on a reliability test and experience failures at 45, 60, 75, 90, and 115 hours. Assuming a lognormal distribution, the MLE parameter estimates are calculated to be <math>{{\widehat{\mu }}^{\prime }}=4.2926</math> and <math>{{\widehat{\sigma ~~}}_{{{T}~~'}}}=0.32361.</math> Calculate the two-sided 75% confidence bounds on these parameters using the likelihood ratio method.		Five units are put on a reliability test and experience failures at 45, 60, 75, 90, and 115 hours. Assuming a lognormal distribution, the MLE parameter estimates are calculated to be <math>{{\widehat{\mu }}^{\prime }}=4.2926</math> and <math>{{\widehat{\sigma'}}}=0.32361.</math> Calculate the two-sided 75% confidence bounds on these parameters using the likelihood ratio method.

	'''Solution'''		'''Solution'''
Line 8:		Line 8:

	<center><math>\begin{align}		<center><math>\begin{align}
	L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma ~~}}_{{{T}~~'}}})= & \underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};{{\widehat{\mu }}^{\prime }},{{\widehat{\sigma ~~}}_{{{T}~~'}}}), \\		L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma' }}})= & \underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};{{\widehat{\mu }}^{\prime }},{{\widehat{\sigma' }}}), \\
	= & \underset{i=1}{\overset{N}{\mathop \prod }}\,\frac{1}{{{x}_{i}}\cdot {{\widehat{\sigma ~~}}_{{{T}~~'}}}\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{\text{ln}({{x}_{i}})-{{\widehat{\mu }}^{\prime }}}{{{\widehat{\sigma ~~}}_{{{T}~~'}}}} \right)}^{2}}}} \\		= & \underset{i=1}{\overset{N}{\mathop \prod }}\,\frac{1}{{{x}_{i}}\cdot {{\widehat{\sigma' }}}\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{\text{ln}({{x}_{i}})-{{\widehat{\mu }}^{\prime }}}{{{\widehat{\sigma' }}}} \right)}^{2}}}} \\
	L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma ~~}}_{{{T}~~'}}})= & \underset{i=1}{\overset{5}{\mathop \prod }}\,\frac{1}{{{x}_{i}}\cdot 0.32361\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{\text{ln}({{x}_{i}})-4.2926}{0.32361} \right)}^{2}}}} \\		L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma'}}})= & \underset{i=1}{\overset{5}{\mathop \prod }}\,\frac{1}{{{x}_{i}}\cdot 0.32361\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{\text{ln}({{x}_{i}})-4.2926}{0.32361} \right)}^{2}}}} \\
	L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma ~~}}_{{{T}~~'}}})= & 1.115256\times {{10}^{-10}}		L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma'}}})= & 1.115256\times {{10}^{-10}}
	\end{align}</math></center>		\end{align}</math></center>

	where <math>{{x}_{i}}</math> are the original time-to-failure data points. We can now rearrange ~~Eqn. (lratio3)~~ to the form:		where <math>{{x}_{i}}</math> are the original time-to-failure data points. We can now rearrange the likelihod ratio equation to the form:

	::<math>L({\mu }',{{\sigma ~~}_{{{T}~~'}}})-L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma ~~}}_{{{T}~~'}}})\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}=0</math>		::<math>L({\mu }',{{\sigma' }})-L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma' }}})\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}=0</math>

	Since our specified confidence level, <math>\delta </math> , is 75%, we can calculate the value of the chi-squared statistic, <math>\chi _{0.75;1}^{2}=1.323303.</math> We can now substitute this information into the equation:		Since our specified confidence level, <math>\delta </math> , is 75%, we can calculate the value of the chi-squared statistic, <math>\chi _{0.75;1}^{2}=1.323303.</math> We can now substitute this information into the equation:

	::<math>\begin{align}		::<math>\begin{align}
	& L({\mu }',{{\sigma ~~}_{{{T}~~'}}})-L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma ~~}}_{{{T}~~'}}})\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}= & 0 \\		& L({\mu }',{{\sigma' }})-L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma' }}})\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}= & 0 \\
	& L({\mu }',{{\sigma ~~}_{{{T}~~'}}})-1.115256\times {{10}^{-10}}\cdot {{e}^{\tfrac{-1.323303}{2}}}= & 0 \\		& L({\mu }',{{\sigma'}})-1.115256\times {{10}^{-10}}\cdot {{e}^{\tfrac{-1.323303}{2}}}= & 0 \\
	& L({\mu }',{{\sigma ~~}_{{{T}~~'}}})-5.754703\times {{10}^{-11}}= & 0		& L({\mu }',{{\sigma'}})-5.754703\times {{10}^{-11}}= & 0
	\end{align}</math>		\end{align}</math>

	It now remains to find the values of <math>{\mu }'</math> and <math>{{\sigma ~~}_{{{T}~~'}}}</math> which satisfy this equation. This is an iterative process that requires setting the value of <math>{{\sigma ~~}_{{{T}~~'}}}</math> and finding the appropriate values of <math>{\mu }'</math> , and vice versa.		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>{{\sigma'}}</math> and finding the appropriate values of <math>{\mu }'</math> , and vice versa.

	The following table gives the values of <math>{\mu }'</math> based on given values of <math>{{\sigma ~~}_{{{T}~~'}}}</math> .		The following table gives the values of <math>{\mu }'</math> based on given values of <math>{{\sigma'}}</math> .


	<center><math>\begin{matrix}		<center><math>\begin{matrix}
	{{\sigma ~~}_{{{T}~~'}}} & \mu _{1}^{\prime } & \mu _{2}^{\prime } & {{\sigma ~~}_{{{T}~~'}}} & \mu _{1}^{\prime } & \mu _{2}^{\prime } \\		{{\sigma' }} & \mu _{1}^{\prime } & \mu _{2}^{\prime } & {{\sigma' }} & \mu _{1}^{\prime } & \mu _{2}^{\prime } \\
	0.24 & 4.2421 & 4.3432 & 0.37 & 4.1145 & 4.4708 \\		0.24 & 4.2421 & 4.3432 & 0.37 & 4.1145 & 4.4708 \\
	0.25 & 4.2115 & 4.3738 & 0.38 & 4.1152 & 4.4701 \\		0.25 & 4.2115 & 4.3738 & 0.38 & 4.1152 & 4.4701 \\
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	[[Image:ldachp9ex5.gif\|thumb\|center\|400px\| ]]		[[Image:ldachp9ex5.gif\|thumb\|center\|400px\| ]]

	(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>{\mu }'</math> is 4.1145, while the highest is 4.4708. These represent the two-sided 75% confidence limits on this parameter. Since solutions for the equation do not exist for values of <math>{{\sigma ~~}_{{{T}~~'}}}</math> below 0.24 or above 0.48, these can be considered the two-sided 75% confidence limits for this parameter. In order to obtain more accurate values for the confidence limits on <math>{{\sigma ~~}_{{{T}~~'}}}</math> , we can perform the same procedure as before, but finding the two values of <math>\sigma </math> that correspond with a given value of <math>{\mu }'.</math> Using this method, we find that the 75% confidence limits on <math>{{\sigma ~~}_{{{T}~~'}}}</math> are 0.23405 and 0.48936, which are close to the initial estimates of 0.24 and 0.48.		(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>{\mu }'</math> is 4.1145, while the highest is 4.4708. These represent the two-sided 75% confidence limits on this parameter. Since solutions for the equation do not exist for values of <math>{{\sigma'
			}}</math> below 0.24 or above 0.48, these can be considered the two-sided 75% confidence limits for this parameter. In order to obtain more accurate values for the confidence limits on <math>{{\sigma'}}</math> , we can perform the same procedure as before, but finding the two values of <math>\sigma </math> that correspond with a given value of <math>{\mu }'.</math> Using this method, we find that the 75% confidence limits on <math>{{\sigma'}}</math> are 0.23405 and 0.48936, which are close to the initial estimates of 0.24 and 0.48.