Template:Example: Lognormal General Example Suspension Data

Lognormal Distribution General Example Suspension Data

From Nelson [30, p. 324]. Ninety-six locomotive controls were tested, 37 failed and 59 were suspended after running for 135,000 miles. Table 9.6 below shows their times-to-failure.

Solution

The distribution used in the publication was the base-10 lognormal. Published results (using MLE):


 * $$\begin{matrix}

{{\widehat{\mu }}^{\prime }}=2.2223 \\ =0.3064 \\ \end{matrix}$$

Published 95% confidence limits on the parameters:


 * $$\begin{matrix}

{{\widehat{\mu }}^{\prime }}=\left\{ 2.1336,2.3109 \right\} \\ =\left\{ 0.2365,0.3970 \right\} \\ \end{matrix}$$

Published variance/covariance matrix:


 * $$\left[ \begin{matrix}

\widehat{Var}\left( {{{\hat{\mu }}}^{\prime }} \right)=0.0020 & {} & \widehat{Cov}({{{\hat{\mu }}}^{\prime }},)=0.001 \\ {} & {} & {} \\   \widehat{Cov}({{{\hat{\mu }}}^{\prime }},)=0.001 & {} & \widehat{Var}\left(  \right)=0.0016  \\ \end{matrix} \right]$$

To replicate the published results (since Weibull++ uses a lognormal to the base $$e$$ ), take the base-10 logarithm of the data and estimate the parameters using the Normal distribution and MLE.


 * Weibull++ computed parameters for maximum likelihood are:


 * $$\begin{matrix}

{{\widehat{\mu }}^{\prime }}=2.2223 \\ =0.3064 \\ \end{matrix}$$


 * Weibull++ computed 95% confidence limits on the parameters:


 * $$\begin{matrix}

{{\widehat{\mu }}^{\prime }}=\left\{ 2.1364,2.3081 \right\} \\ =\left\{ 0.2395,0.3920 \right\} \\ \end{matrix}$$


 * Weibull++ computed/variance covariance matrix:


 * $$\left[ \begin{matrix}

\widehat{Var}\left( {{{\hat{\mu }}}^{\prime }} \right)=0.0019 & {} & \widehat{Cov}({{{\hat{\mu }}}^{\prime }},)=0.0009 \\ {} & {} & {} \\   \widehat{Cov}({\mu }',)=0.0009 & {} & \widehat{Var}\left(  \right)=0.0015  \\ \end{matrix} \right]$$