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

Lisa Hacker: Redirected page to The Normal Distribution

2012-08-13T02:21:09Z

Redirected page to The Normal Distribution

← Older revision		Revision as of 02:21, 13 August 2012
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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> .~~

	~~<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:~~

	~~[[Image:WB.9 normal parameter contour plot.png\|center\|350px\|~~ ]]

	(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.

Richard House at 02:44, 8 August 2012

2012-08-08T02:44:08Z

← Older revision		Revision as of 02:44, 8 August 2012
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	The following table gives the values of <math>\sigma </math> based on given values of <math>\mu </math> .		The following table gives the values of <math>\sigma </math> based on given values of <math>\mu </math> .

	~~[[Image:tableofmu.png\|center\|350px\| ]]~~

	~~<math></math>~~
	~~[[Image:WB.9 normal parameter contour plot.png\|center\|350px\| ]]~~


	<center><math>\begin{matrix}		<center><math>\begin{matrix}
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	This data set is represented graphically in the following contour plot:		This data set is represented graphically in the following contour plot:

			[[Image:WB.9 normal parameter contour plot.png\|center\|350px\| ]]

	(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.		(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.

Richard House at 02:41, 8 August 2012

2012-08-08T02:41:26Z

← Older revision		Revision as of 02:41, 8 August 2012
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	::<math>L(\mu ,\sigma )-L(\widehat{\mu },\widehat{\sigma })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}=0</math>		::<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:		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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	L(\mu ,\sigma )-2.057410\times {{10}^{-9}}= & 0.		L(\mu ,\sigma )-2.057410\times {{10}^{-9}}= & 0.
	\end{align}</math>		\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.		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> .		The following table gives the values of <math>\sigma </math> based on given values of <math>\mu </math> .

	[[Image:tableofmu.png\|center\|~~250px~~\| ]]		[[Image:tableofmu.png\|center\|350px\| ]]

	<math></math>		<math></math>
	[[Image:WB.9 normal parameter contour plot.png\|center\|~~250px~~\| ]]		[[Image:WB.9 normal parameter contour plot.png\|center\|350px\| ]]

Nicolette Young at 21:13, 25 April 2012

2012-04-25T21:13:27Z

← Older revision		Revision as of 21:13, 25 April 2012
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	The following table gives the values of <math>\sigma </math> based on given values of <math>\mu </math> .		The following table gives the values of <math>\sigma </math> based on given values of <math>\mu </math> .

	[[Image:tableofmu.png\|center\|~~300px~~\| ]]		[[Image:tableofmu.png\|center\|250px\| ]]

	<math></math>		<math></math>

Nicolette Young at 19:11, 25 April 2012

2012-04-25T19:11:26Z

← Older revision		Revision as of 19:11, 25 April 2012
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	The following table gives the values of <math>\sigma </math> based on given values of <math>\mu </math> .		The following table gives the values of <math>\sigma </math> based on given values of <math>\mu </math> .

	[[Image:tableofmu.png\|center\|~~500px~~\| ]]		[[Image:tableofmu.png\|center\|300px\| ]]

	<math></math>		<math></math>
	[[Image:WB.9 normal parameter contour plot.png\|center\|~~400px~~\| ]]		[[Image:WB.9 normal parameter contour plot.png\|center\|250px\| ]]

Nikki Helms at 17:21, 3 April 2012

2012-04-03T17:21:23Z

← Older revision		Revision as of 17:21, 3 April 2012
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	The following table gives the values of <math>\sigma </math> based on given values of <math>\mu </math> .		The following table gives the values of <math>\sigma </math> based on given values of <math>\mu </math> .

	[[Image:tableofmu.png\|center\|~~400px~~\| ]]		[[Image:tableofmu.png\|center\|500px\| ]]

	<math></math>		<math></math>

Nikki Helms at 21:57, 2 April 2012

2012-04-02T21:57:12Z

← Older revision		Revision as of 21:57, 2 April 2012
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	The following table gives the values of <math>\sigma </math> based on given values of <math>\mu </math> .		The following table gives the values of <math>\sigma </math> based on given values of <math>\mu </math> .

	[[Image:tableofmu.~~gif\|thumb~~\|center\|400px\| ]]		[[Image:tableofmu.png\|center\|400px\| ]]

	<math></math>		<math></math>

Nicolette Young at 19:24, 14 March 2012

2012-03-14T19:24:31Z

← Older revision		Revision as of 19:24, 14 March 2012
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	<math></math>		<math></math>
	[[Image:~~circleplot~~.~~gif\|thumb~~\|center\|400px\| ]]		[[Image:WB.9 normal parameter contour plot.png\|center\|400px\| ]]

Harry Guo at 22:09, 10 February 2012

2012-02-10T22:09:17Z

← Older revision		Revision as of 22:09, 10 February 2012
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	\end{align}</math>		\end{align}</math>

	where <math>{{t}_{i}}</math> are the original time-to-failure data points. We can now rearrange ~~Eqn. (lratio3)~~ to the form:		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>		::<math>L(\mu ,\sigma )-L(\widehat{\mu },\widehat{\sigma })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}=0</math>

Harry Guo at 22:08, 10 February 2012

2012-02-10T22:08:35Z

← Older revision		Revision as of 22:08, 10 February 2012
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	::<math>\begin{align}		::<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{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{{{x}_{i}}-28.8}{11.2143} \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}}		L(\widehat{\mu },\widehat{\sigma })= & 4.676897\times {{10}^{-9}}
	\end{align}</math>		\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:		where <math>{{t}_{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>		::<math>L(\mu ,\sigma )-L(\widehat{\mu },\widehat{\sigma })\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}=0</math>