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| ==Confidence Bounds==
| | #REDIRECT [[The_Lognormal_Distribution#Confidence_Bounds]] |
| The method used by the application in estimating the different types of confidence bounds for lognormally distributed data is presented in this section. Note that there are closed-form solutions for both the normal and lognormal reliability that can be obtained without the use of the Fisher information matrix. However, these closed-form solutions only apply to complete data. To achieve consistent application across all possible data types, Weibull++ always uses the Fisher matrix in computing confidence intervals. The complete derivations were presented in detail for a general function in Chapter 5. For a discussion on exact confidence bounds for the normal and lognormal, see Chapter 8.
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| ===Fisher Matrix Bounds===
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| ====Bounds on the Parameters====
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| The lower and upper bounds on the mean, <math>{\mu }'</math> , are estimated from:
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| ::<math>\begin{align}
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| & \mu _{U}^{\prime }= & {{\widehat{\mu }}^{\prime }}+{{K}_{\alpha }}\sqrt{Var({{\widehat{\mu }}^{\prime }})}\text{ (upper bound),} \\
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| & \mu _{L}^{\prime }= & {{\widehat{\mu }}^{\prime }}-{{K}_{\alpha }}\sqrt{Var({{\widehat{\mu }}^{\prime }})}\text{ (lower bound)}\text{.}
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| \end{align}</math>
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| For the standard deviation, <math>{{\widehat{\sigma }}_{{{T}'}}}</math> , <math>\ln ({{\widehat{\sigma }}_{{{T}'}}})</math> is treated as normally distributed, and the bounds are estimated from:
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| ::<math>\begin{align}
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| & {{\sigma }_{U}}= & {{\widehat{\sigma }}_{{{T}'}}}\cdot {{e}^{\tfrac{{{K}_{\alpha }}\sqrt{Var({{\widehat{\sigma }}_{{{T}'}}})}}{{{\widehat{\sigma }}_{{{T}'}}}}}}\text{ (upper bound),} \\
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| & {{\sigma }_{L}}= & \frac{{{\widehat{\sigma }}_{{{T}'}}}}{{{e}^{\tfrac{{{K}_{\alpha }}\sqrt{Var({{\widehat{\sigma }}_{{{T}'}}})}}{{{\widehat{\sigma }}_{{{T}'}}}}}}}\text{ (lower bound),}
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| \end{align}</math>
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| where <math>{{K}_{\alpha }}</math> is defined by:
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| ::<math>\alpha =\frac{1}{\sqrt{2\pi }}\int_{{{K}_{\alpha }}}^{\infty }{{e}^{-\tfrac{{{t}^{2}}}{2}}}dt=1-\Phi ({{K}_{\alpha }})</math>
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| If <math>\delta </math> is the confidence level, then <math>\alpha =\tfrac{1-\delta }{2}</math> for the two-sided bounds and <math>\alpha =1-\delta </math> for the one-sided bounds.
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| The variances and covariances of <math>{{\widehat{\mu }}^{\prime }}</math> and <math>{{\widehat{\sigma }}_{{{T}'}}}</math> are estimated as follows:
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| ::<math>\left( \begin{matrix}
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| \widehat{Var}\left( {{\widehat{\mu }}^{\prime }} \right) & \widehat{Cov}\left( {{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_{{{T}'}}} \right) \\
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| \widehat{Cov}\left( {{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_{{{T}'}}} \right) & \widehat{Var}\left( {{\widehat{\sigma }}_{{{T}'}}} \right) \\
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| \end{matrix} \right)=\left( \begin{matrix}
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| -\tfrac{{{\partial }^{2}}\Lambda }{\partial {{({\mu }')}^{2}}} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial {\mu }'\partial {{\sigma }_{{{T}'}}}} \\
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| {} & {} \\
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| -\tfrac{{{\partial }^{2}}\Lambda }{\partial {\mu }'\partial {{\sigma }_{{{T}'}}}} & -\tfrac{{{\partial }^{2}}\Lambda }{\partial \sigma _{{{T}'}}^{2}} \\
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| \end{matrix} \right)_{{\mu }'={{\widehat{\mu }}^{\prime }},{{\sigma }_{{{T}'}}}={{\widehat{\sigma }}_{{{T}'}}}}^{-1}</math>
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| where <math>\Lambda </math> is the log-likelihood function of the lognormal distribution.
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| ====Bounds on Reliability====
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| The reliability of the lognormal distribution is:
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| ::<math>\hat{R}({T}';{\mu }',{{\sigma }_{{{T}'}}})=\int_{{{T}'}}^{\infty }\frac{1}{{{\widehat{\sigma }}_{{{T}'}}}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{t-{{\widehat{\mu }}^{\prime }}}{{{\widehat{\sigma }}_{{{T}'}}}} \right)}^{2}}}}dt</math>
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| Let <math>\widehat{z}(t;{{\hat{\mu }}^{\prime }},{{\hat{\sigma }}_{{{T}'}}})=\tfrac{t-{{\widehat{\mu }}^{\prime }}}{{{\widehat{\sigma }}_{{{T}'}}}},</math> then <math>\tfrac{d\widehat{z}}{dt}=\tfrac{1}{{{\widehat{\sigma }}_{{{T}'}}}}.</math> For <math>t={T}'</math> , <math>\widehat{z}=\tfrac{{T}'-{{\widehat{\mu }}^{\prime }}}{{{\widehat{\sigma }}_{{{T}'}}}}</math> , and for <math>t=\infty ,</math> <math>\widehat{z}=\infty .</math> The above equation then becomes:
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| ::<math>\hat{R}(\widehat{z})=\int_{\widehat{z}({T}')}^{\infty }\frac{1}{\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{z}^{2}}}}dz</math>
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| The bounds on <math>z</math> are estimated from:
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| ::<math>\begin{align}
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| & {{z}_{U}}= & \widehat{z}+{{K}_{\alpha }}\sqrt{Var(\widehat{z})} \\
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| & {{z}_{L}}= & \widehat{z}-{{K}_{\alpha }}\sqrt{Var(\widehat{z})}
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| \end{align}</math>
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| :where:
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| ::<math>\begin{align}
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| & Var(\widehat{z})= & \left( \frac{\partial z}{\partial {\mu }'} \right)_{{{\widehat{\mu }}^{\prime }}}^{2}Var({{\widehat{\mu }}^{\prime }})+\left( \frac{\partial z}{\partial {{\sigma }_{{{T}'}}}} \right)_{{{\widehat{\sigma }}_{{{T}'}}}}^{2}Var({{\widehat{\sigma }}_{{{T}'}}}) \\
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| & & +2{{\left( \frac{\partial z}{\partial {\mu }'} \right)}_{{{\widehat{\mu }}^{\prime }}}}{{\left( \frac{\partial z}{\partial {{\sigma }_{{{T}'}}}} \right)}_{{{\widehat{\sigma }}_{{{T}'}}}}}Cov\left( {{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_{{{T}'}}} \right)
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| \end{align}</math>
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| :or:
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| ::<math>Var(\widehat{z})=\frac{1}{\widehat{\sigma }_{{{T}'}}^{2}}\left[ Var({{\widehat{\mu }}^{\prime }})+{{\widehat{z}}^{2}}Var({{\widehat{\sigma }}_{{{T}'}}})+2\cdot \widehat{z}\cdot Cov\left( {{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_{{{T}'}}} \right) \right]</math>
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| The upper and lower bounds on reliability are:
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| ::<math>\begin{align}
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| & {{R}_{U}}= & \int_{{{z}_{L}}}^{\infty }\frac{1}{\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{z}^{2}}}}dz\text{ (Upper bound)} \\
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| & {{R}_{L}}= & \int_{{{z}_{U}}}^{\infty }\frac{1}{\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{z}^{2}}}}dz\text{ (Lower bound)}
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| \end{align}</math>
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| ====Bounds on Time====
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| The bounds around time for a given lognormal percentile, or unreliability, are estimated by first solving the reliability equation with respect to time, as follows:
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| ::<math>{T}'({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_{{{T}'}}})={{\widehat{\mu }}^{\prime }}+z\cdot {{\widehat{\sigma }}_{{{T}'}}}</math>
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| :where:
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| ::<math>z={{\Phi }^{-1}}\left[ F({T}') \right]</math>
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| :and:
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| ::<math>\Phi (z)=\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{z({T}')}{{e}^{-\tfrac{1}{2}{{z}^{2}}}}dz</math>
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| The next step is to calculate the variance of <math>{T}'({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_{{{T}'}}}):</math>
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| ::<math>\begin{align}
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| & Var({{{\hat{T}}}^{\prime }})= & {{\left( \frac{\partial {T}'}{\partial {\mu }'} \right)}^{2}}Var({{\widehat{\mu }}^{\prime }})+{{\left( \frac{\partial {T}'}{\partial {{\sigma }_{{{T}'}}}} \right)}^{2}}Var({{\widehat{\sigma }}_{{{T}'}}}) \\
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| & & +2\left( \frac{\partial {T}'}{\partial {\mu }'} \right)\left( \frac{\partial {T}'}{\partial {{\sigma }_{{{T}'}}}} \right)Cov\left( {{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_{{{T}'}}} \right) \\
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| & & \\
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| & Var({{{\hat{T}}}^{\prime }})= & Var({{\widehat{\mu }}^{\prime }})+{{\widehat{z}}^{2}}Var({{\widehat{\sigma }}_{{{T}'}}})+2\cdot \widehat{z}\cdot Cov\left( {{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_{{{T}'}}} \right)
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| \end{align}</math>
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| The upper and lower bounds are then found by:
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| ::<math>\begin{align}
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| & T_{U}^{\prime }= & \ln {{T}_{U}}={{{\hat{T}}}^{\prime }}+{{K}_{\alpha }}\sqrt{Var({{{\hat{T}}}^{\prime }})} \\
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| & T_{L}^{\prime }= & \ln {{T}_{L}}={{{\hat{T}}}^{\prime }}-{{K}_{\alpha }}\sqrt{Var({{{\hat{T}}}^{\prime }})}
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| \end{align}</math>
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| Solving for <math>{{T}_{U}}</math> and <math>{{T}_{L}}</math> we get:
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| ::<math>\begin{align}
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| & {{T}_{U}}= & {{e}^{T_{U}^{\prime }}}\text{ (upper bound),} \\
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| & {{T}_{L}}= & {{e}^{T_{L}^{\prime }}}\text{ (lower bound)}\text{.}
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| \end{align}</math>
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| ====Example 4====
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| Using the data of Example 2 and assuming a lognormal distribution, estimate the parameters using the MLE method.
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| =====Solution to Example 4=====
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| In this example we have only complete data. Thus, the partials reduce to:
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| ::<math>\begin{align}
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| & \frac{\partial \Lambda }{\partial {\mu }'}= & \frac{1}{\sigma _{{{T}'}}^{2}}\cdot \underset{i=1}{\overset{14}{\mathop \sum }}\,\ln ({{T}_{i}})-{\mu }'=0 \\
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| & \frac{\partial \Lambda }{\partial {{\sigma }_{{{T}'}}}}= & \underset{i=1}{\overset{14}{\mathop \sum }}\,\left( \frac{\ln ({{T}_{i}})-{\mu }'}{\sigma _{{{T}'}}^{3}}-\frac{1}{{{\sigma }_{{{T}'}}}} \right)=0
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| \end{align}</math>
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| Substituting the values of <math>{{T}_{i}}</math> and solving the above system simultaneously, we get:
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| ::<math>\begin{align}
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| & {{{\hat{\sigma }}}_{{{T}'}}}= & 0.849 \\
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| & {{{\hat{\mu }}}^{\prime }}= & 3.516
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| \end{align}</math>
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| Using Eqns. (mean) and (sdv) we get:
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| ::<math>\overline{T}=\hat{\mu }=48.25\text{ hours}</math>
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| :and:
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| ::<math>{{\hat{\sigma }}_{{{T}'}}}=49.61\text{ hours}.</math>
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| The variance/covariance matrix is given by:
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| ::<math>\left[ \begin{matrix}
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| \widehat{Var}\left( {{{\hat{\mu }}}^{\prime }} \right)=0.0515 & {} & \widehat{Cov}\left( {{{\hat{\mu }}}^{\prime }},{{{\hat{\sigma }}}_{{{T}'}}} \right)=0.0000 \\
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| {} & {} & {} \\
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| \widehat{Cov}\left( {{{\hat{\mu }}}^{\prime }},{{{\hat{\sigma }}}_{{{T}'}}} \right)=0.0000 & {} & \widehat{Var}\left( {{{\hat{\sigma }}}_{{{T}'}}} \right)=0.0258 \\
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| \end{matrix} \right]</math>
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| ====Note About Bias====
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| See the discussion regarding bias with the normal distribution in Chapter 8 for information regarding parameter bias in the lognormal distribution.
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| ===Likelihood Ratio Confidence Bounds===
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| ====Bounds on Parameters====
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| 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:
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| ::<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>
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| This equation can be rewritten as:
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| ::<math>L({{\theta }_{1}},{{\theta }_{2}})=L({{\widehat{\theta }}_{1}},{{\widehat{\theta }}_{2}})\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}</math>
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| For complete data, the likelihood formula for the normal distribution is given by:
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| ::<math>L({\mu }',{{\sigma }_{{{T}'}}})=\underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};{\mu }',{{\sigma }_{{{T}'}}})=\underset{i=1}{\overset{N}{\mathop \prod }}\,\frac{1}{{{x}_{i}}\cdot {{\sigma }_{{{T}'}}}\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{\text{ln}({{x}_{i}})-{\mu }'}{{{\sigma }_{{{T}'}}}} \right)}^{2}}}}</math>
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| 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 }_{{{T}'}}}</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.
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| ====Example 5====
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| 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.
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| =====Solution to Example 5=====
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| The first step is to calculate the likelihood function for the parameter estimates:
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| <center><math>\begin{align}
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| L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_{{{T}'}}})= & \underset{i=1}{\overset{N}{\mathop \prod }}\,f({{x}_{i}};{{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_{{{T}'}}}), \\
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| = & \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}}}} \\
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| 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}}}} \\
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| L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_{{{T}'}}})= & 1.115256\times {{10}^{-10}}
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| \end{align}</math></center>
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| where <math>{{x}_{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 }_{{{T}'}}})-L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_{{{T}'}}})\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}=0</math>
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| 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:
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| ::<math>\begin{align}
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| & L({\mu }',{{\sigma }_{{{T}'}}})-L({{\widehat{\mu }}^{\prime }},{{\widehat{\sigma }}_{{{T}'}}})\cdot {{e}^{\tfrac{-\chi _{\alpha ;1}^{2}}{2}}}= & 0 \\
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| & L({\mu }',{{\sigma }_{{{T}'}}})-1.115256\times {{10}^{-10}}\cdot {{e}^{\tfrac{-1.323303}{2}}}= & 0 \\
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| & L({\mu }',{{\sigma }_{{{T}'}}})-5.754703\times {{10}^{-11}}= & 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 }_{{{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.
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| The following table gives the values of <math>{\mu }'</math> based on given values of <math>{{\sigma }_{{{T}'}}}</math> .
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| <center><math>\begin{matrix}
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| {{\sigma }_{{{T}'}}} & \mu _{1}^{\prime } & \mu _{2}^{\prime } & {{\sigma }_{{{T}'}}} & \mu _{1}^{\prime } & \mu _{2}^{\prime } \\
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| 0.24 & 4.2421 & 4.3432 & 0.37 & 4.1145 & 4.4708 \\
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| 0.25 & 4.2115 & 4.3738 & 0.38 & 4.1152 & 4.4701 \\
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| 0.26 & 4.1909 & 4.3944 & 0.39 & 4.1170 & 4.4683 \\
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| 0.27 & 4.1748 & 4.4105 & 0.40 & 4.1200 & 4.4653 \\
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| 0.28 & 4.1618 & 4.4235 & 0.41 & 4.1244 & 4.4609 \\
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| 0.29 & 4.1509 & 4.4344 & 0.42 & 4.1302 & 4.4551 \\
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| 0.30 & 4.1419 & 4.4434 & 0.43 & 4.1377 & 4.4476 \\
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| 0.31 & 4.1343 & 4.4510 & 0.44 & 4.1472 & 4.4381 \\
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| 0.32 & 4.1281 & 4.4572 & 0.45 & 4.1591 & 4.4262 \\
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| 0.33 & 4.1231 & 4.4622 & 0.46 & 4.1742 & 4.4111 \\
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| 0.34 & 4.1193 & 4.4660 & 0.47 & 4.1939 & 4.3914 \\
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| 0.35 & 4.1166 & 4.4687 & 0.48 & 4.2221 & 4.3632 \\
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| 0.36 & 4.1150 & 4.4703 & {} & {} & {} \\
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| \end{matrix}</math></center>
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| These points are represented graphically in the following contour plot:
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| [[Image:ldachp9ex5.gif|thumb|center|400px| ]]
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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>{\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.
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| ====Bounds on Time and Reliability====
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| 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:
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| ::<math>R=1-\Phi \left( \frac{\text{ln}(t)-{\mu }'}{{{\sigma }_{{{T}'}}}} \right)</math>
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| This can be rearranged to the form:
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| ::<math>{\mu }'=\text{ln}(t)-{{\sigma }_{{{T}'}}}\cdot {{\Phi }^{-1}}(1-R)</math>
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| where <math>{{\Phi }^{-1}}</math> is the inverse standard normal. This equation can now be substituted into Eqn. (lognormlikelihood) to produce a likelihood equation in terms of <math>{{\sigma }_{{{T}'}}},</math> <math>t</math> and <math>R\ \ :</math>
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| ::<math>L({{\sigma }_{{{T}'}}},t/R)=\underset{i=1}{\overset{N}{\mathop \prod }}\,\frac{1}{{{x}_{i}}\cdot {{\sigma }_{{{T}'}}}\cdot \sqrt{2\pi }}\cdot {{e}^{-\tfrac{1}{2}{{\left( \tfrac{\text{ln}({{x}_{i}})-\left( \text{ln}(t)-{{\sigma }_{{{T}'}}}\cdot {{\Phi }^{-1}}(1-R) \right)}{{{\sigma }_{{{T}'}}}} \right)}^{2}}}}</math>
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| The unknown variable <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 variable. 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 variable. Either way, Eqn. (lognormliketr) can be used to solve Eqn. (lratio3) for the values of interest.
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| ====Example 6====
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| For the data given in Example 5, determine the two-sided 75% confidence bounds on the time estimate for a reliability of 80%. The ML estimate for the time at <math>R(t)=80%</math> is 55.718.
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| =====Solution to Example 6=====
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| In this example, we are trying to determine the two-sided 75% confidence bounds on the time estimate of 55.718. This is accomplished by substituting <math>R=0.80</math> and <math>\alpha =0.75</math> into Eqn. (lognormliketr), and varying <math>{{\sigma }_{{{T}'}}}</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 }_{{{T}'}}}</math> .
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| <center><math>\begin{matrix}
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| {{\sigma }_{{{T}'}}} & {{t}_{1}} & {{t}_{2}} & {{\sigma }_{{{T}'}}} & {{t}_{1}} & {{t}_{2}} \\
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| 0.24 & 56.832 & 62.879 & 0.37 & 44.841 & 64.031 \\
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| 0.25 & 54.660 & 64.287 & 0.38 & 44.494 & 63.454 \\
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| 0.26 & 53.093 & 65.079 & 0.39 & 44.200 & 62.809 \\
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| 0.27 & 51.811 & 65.576 & 0.40 & 43.963 & 62.093 \\
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| 0.28 & 50.711 & 65.881 & 0.41 & 43.786 & 61.304 \\
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| 0.29 & 49.743 & 66.041 & 0.42 & 43.674 & 60.436 \\
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| 0.30 & 48.881 & 66.085 & 0.43 & 43.634 & 59.481 \\
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| 0.31 & 48.106 & 66.028 & 0.44 & 43.681 & 58.426 \\
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| 0.32 & 47.408 & 65.883 & 0.45 & 43.832 & 57.252 \\
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| 0.33 & 46.777 & 65.657 & 0.46 & 44.124 & 55.924 \\
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| 0.34 & 46.208 & 65.355 & 0.47 & 44.625 & 54.373 \\
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| 0.35 & 45.697 & 64.983 & 0.48 & 45.517 & 52.418 \\
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| 0.36 & 45.242 & 64.541 & {} & {} & {} \\
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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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| [[Image:ldachp9ex6.gif|thumb|center|400px| ]]
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| As can be determined from the table, the lowest calculated value for <math>t</math> is 43.634, while the highest is 66.085. These represent the two-sided 75% confidence limits on the time at which reliability is equal to 80%.
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| ====Example 7====
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| For the data given in Example 5, determine the two-sided 75% confidence bounds on the reliability estimate for <math>t=65</math> . The ML estimate for the reliability at <math>t=65</math> is 64.261%.
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| =====Solution to Example 7=====
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| In this example, we are trying to determine the two-sided 75% confidence bounds on the reliability estimate of 64.261%. This is accomplished by substituting <math>t=65</math> and <math>\alpha =0.75</math> into Eqn. (lognormliketr), and varying <math>{{\sigma }_{{{T}'}}}</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 }_{{{T}'}}}</math> .
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| <center><math>\begin{matrix}
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| {{\sigma }_{{{T}'}}} & {{R}_{1}} & {{R}_{2}} & {{\sigma }_{{{T}'}}} & {{R}_{1}} & {{R}_{2}} \\
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| 0.24 & 61.107% & 75.910% & 0.37 & 43.573% & 78.845% \\
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| 0.25 & 55.906% & 78.742% & 0.38 & 43.807% & 78.180% \\
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| 0.26 & 55.528% & 80.131% & 0.39 & 44.147% & 77.448% \\
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| 0.27 & 50.067% & 80.903% & 0.40 & 44.593% & 76.646% \\
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| 0.28 & 48.206% & 81.319% & 0.41 & 45.146% & 75.767% \\
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| 0.29 & 46.779% & 81.499% & 0.42 & 45.813% & 74.802% \\
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| 0.30 & 45.685% & 81.508% & 0.43 & 46.604% & 73.737% \\
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| 0.31 & 44.857% & 81.387% & 0.44 & 47.538% & 72.551% \\
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| 0.32 & 44.250% & 81.159% & 0.45 & 48.645% & 71.212% \\
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| 0.33 & 43.827% & 80.842% & 0.46 & 49.980% & 69.661% \\
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| 0.34 & 43.565% & 80.446% & 0.47 & 51.652% & 67.789% \\
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| 0.35 & 43.444% & 79.979% & 0.48 & 53.956% & 65.299% \\
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| 0.36 & 43.450% & 79.444% & {} & {} & {} \\
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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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| [[Image:ldachp9ex7.gif|thumb|center|400px| ]] | |
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| As can be determined from the table, the lowest calculated value for <math>R</math> is 43.444%, while the highest is 81.508%. These represent the two-sided 75% confidence limits on the reliability at <math>t=65</math> .
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