Template:Normal Distribution likelihood ratio confidence bounds - Revision history

Richard House: Redirected page to The Normal Distribution#Likelihood Ratio Confidence Bounds

2012-08-13T04:27:49Z

Redirected page to The Normal Distribution#Likelihood Ratio Confidence Bounds

← Older revision		Revision as of 04:27, 13 August 2012
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	~~===Likelihood Ratio Confidence Bounds===~~		#REDIRECT [[The_Normal_Distribution#Likelihood_Ratio_Confidence_Bounds]]
	~~====Bounds on Parameters====~~
	~~As covered in Chapter~~ [[~~Confidence Bounds~~]]~~, 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({{t}_{i}};\mu ,\sigma )=\underset{i=1}{\overset{N}{\mathop \prod }}\,\frac{1}{\sigma \cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{{{t}_{i}}-\mu }{\sigma } \right)}^{2}}}}</math>

	where the <math>{{t}_{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 the above likelihood ratio equation. 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:'''~~
	~~{{Example: Normal Distribution Likelihood Ratio Bound (Parameters)}}~~

	~~====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 the likelihood ratio equation to produce an 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{{{t}_{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. The likelihood ratio equation can be used to solve the values of interest.

	~~'''Example 6:'''~~
	~~{{Example: Normal Distribution Likelihood Ratio Bound (Time)}}~~

	~~'''Example 7:'''~~
	~~{{Example: Normal Distribution Likelihood Ratio Bound (Reliability)}}~~

Harry Guo at 22:22, 10 February 2012

2012-02-10T22:22:30Z

← Older revision		Revision as of 22:22, 10 February 2012
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	'''Example 6:'''		'''Example 6:'''
	{{Example: Normal Distribution Likelihood Ratio Bound (~~Reliability~~)}}		{{Example: Normal Distribution Likelihood Ratio Bound (Time)}}

	'''Example 7:'''		'''Example 7:'''
	{{Example: Normal Distribution Likelihood Ratio Bound (Reliability)}}		{{Example: Normal Distribution Likelihood Ratio Bound (Reliability)}}

Harry Guo at 22:21, 10 February 2012

2012-02-10T22:21:12Z

← Older revision		Revision as of 22:21, 10 February 2012
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	'''Example 6:'''		'''Example 6:'''
	{{Example: Normal Distribution Likelihood Ratio Bound (~~Time~~)}}		{{Example: Normal Distribution Likelihood Ratio Bound (Reliability)}}

	~~====~~Example 7~~====~~		'''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%.~~		{{Example: Normal Distribution Likelihood Ratio Bound (Reliability)}}

	~~=====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> .

Harry Guo at 22:15, 10 February 2012

2012-02-10T22:15:42Z

← Older revision		Revision as of 22:15, 10 February 2012
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	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. The likelihood ratio equation can be used to solve the values of interest.		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. The likelihood ratio equation can be used to solve the values of interest.

	~~====~~Example 6~~====~~		'''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.~~		{{Example: Normal Distribution Likelihood Ratio Bound (Time)}}
	~~=====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====		====Example 7====

Harry Guo: /* Bounds on Time and Reliability */

2012-02-10T22:11:59Z

Bounds on Time and Reliability

← Older revision		Revision as of 22:11, 10 February 2012
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	::<math>\mu =t-\sigma \cdot {{\Phi }^{-1}}(1-R)</math>		::<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>		where <math>{{\Phi }^{-1}}</math> is the inverse standard normal. This equation can now be substituted into the likelihood ratio equation to produce an 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{{{t}_{i}}-\left[ t-\sigma \cdot {{\Phi }^{-1}}(1-R) \right]}{\sigma } \right)}^{2}}}}</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{{{t}_{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.		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. The likelihood ratio equation can be used to solve the values of interest.

	====Example 6====		====Example 6====

Harry Guo: /* Likelihood Ratio Confidence Bounds */

2012-02-10T22:07:22Z

Likelihood Ratio Confidence Bounds

← Older revision		Revision as of 22:07, 10 February 2012
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	For complete data, the likelihood formula for the normal distribution is given by:		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>		::<math>L(\mu ,\sigma )=\underset{i=1}{\overset{N}{\mathop \prod }}\,f({{t}_{i}};\mu ,\sigma )=\underset{i=1}{\overset{N}{\mathop \prod }}\,\frac{1}{\sigma \cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{{{t}_{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 the above likelihood ratio equation. 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.		where the <math>{{t}_{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 the above likelihood ratio equation. 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:'''		'''Example 5:'''
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	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>		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>		::<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{{{t}_{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.		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.

Harry Guo: /* Likelihood Ratio Confidence Bounds */

2012-02-10T21:32:53Z

Likelihood Ratio Confidence Bounds

@@ Line 1: / Line 1: @@
 ===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:
+As covered in Chapter [[Confidence Bounds]], 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:
@@ Line 15: / Line 14: @@
 ::<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.
+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 the above likelihood ratio equation. 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.
-(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.
+'''Example 5:'''
 ====Bounds on Time and Reliability====

Harry Guo: moved Template:Nd likelihood ratio confidence bounds to Template:Normal Distribution likelihood ratio confidence bounds

2012-02-10T17:38:26Z

moved Template:Nd likelihood ratio confidence bounds to Template:Normal Distribution likelihood ratio confidence bounds

← Older revision	Revision as of 17:38, 10 February 2012
(No difference)

Nicolette Young: Created page with '===Likelihood Ratio Confidence Bounds=== ====Bounds on Parameters==== As covered in Chapter 5, the likelihood confidence bounds are calculated by finding values for ${{\the…'$

2012-01-04T17:47:00Z

Created page with '===Likelihood Ratio Confidence Bounds=== ====Bounds on Parameters==== As covered in Chapter 5, the likelihood confidence bounds are calculated by finding values for <math>{{\the…'

New page

===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> .

Template:Normal Distribution likelihood ratio confidence bounds - Revision history

Richard House: Redirected page to The Normal Distribution#Likelihood Ratio Confidence Bounds

Harry Guo at 22:22, 10 February 2012

Harry Guo at 22:21, 10 February 2012

Harry Guo at 22:15, 10 February 2012

Harry Guo: /* Bounds on Time and Reliability */

Harry Guo: /* Likelihood Ratio Confidence Bounds */

Harry Guo: /* Likelihood Ratio Confidence Bounds */

Harry Guo: moved Template:Nd likelihood ratio confidence bounds to Template:Normal Distribution likelihood ratio confidence bounds

Nicolette Young: Created page with '===Likelihood Ratio Confidence Bounds=== ====Bounds on Parameters==== As covered in Chapter 5, the likelihood confidence bounds are calculated by finding values for {{\the…'

Nicolette Young: Created page with '===Likelihood Ratio Confidence Bounds=== ====Bounds on Parameters==== As covered in Chapter 5, the likelihood confidence bounds are calculated by finding values for ${{\the…'$