Lognormal Distribution Examples

These examples also appear in the Life data analysis reference.

Complete Data Example
Determine the lognormal parameter estimates for the data given in the following table.

Solution

Using Weibull++, the computed parameters for maximum likelihood are:


 * $$\begin{align}

& {{{\hat{\mu }}}^{\prime }}= & 2.83 \\ & {\hat{\sigma '}}= & 1.10 \end{align}\,\!$$

For rank regression on $$X\,\!$$


 * $$\begin{align}

& {{{\hat{\mu }}}^{\prime }}= & 2.83 \\ & = & 1.24 \end{align}\,\!$$

For rank regression on $$Y:\,\!$$


 * $$\begin{align}

& {{{\hat{\mu }}}^{\prime }}= & 2.83 \\ & = & 1.36 \end{align}\,\!$$

Complete Data RRX Example
From Kececioglu [20, p. 347]. 15 identical units were tested to failure and following is a table of their failure times:

Times-to-Failure Data

$$\begin{matrix} \text{Data Point Index} & \text{Failure Times (Hr)} \\ \text{1} & \text{62}\text{.5} \\ \text{2} & \text{91}\text{.9} \\ \text{3} & \text{100}\text{.3} \\ \text{4} & \text{117}\text{.4} \\ \text{5} & \text{141}\text{.1} \\ \text{6} & \text{146}\text{.8} \\ \text{7} & \text{172}\text{.7} \\ \text{8} & \text{192}\text{.5} \\ \text{9} & \text{201}\text{.6} \\ \text{10} & \text{235}\text{.8} \\ \text{11} & \text{249}\text{.2} \\ \text{12} & \text{297}\text{.5} \\ \text{13} & \text{318}\text{.3} \\ \text{14} & \text{410}\text{.6} \\ \text{15} & \text{550}\text{.5} \\ \end{matrix}\,\!$$

Solution

Published results (using probability plotting):


 * $$\begin{matrix}

{{\widehat{\mu }}^{\prime }}=5.22575 \\ =0.62048. \\ \end{matrix}\,\!$$

Weibull++ computed parameters for rank regression on X are:


 * $$\begin{matrix}

{{\widehat{\mu }}^{\prime }}=5.2303 \\ =0.6283. \\ \end{matrix}\,\!$$

The small differences are due to the precision errors when fitting a line manually, whereas in Weibull++ the line was fitted mathematically.

Complete Data Unbiased MLE Example
From Kececioglu [19, p. 406]. 9 identical units are tested continuously to failure and failure times were recorded at 30.4, 36.7, 53.3, 58.5, 74.0, 99.3, 114.3, 140.1 and 257.9 hours.

Solution

The results published were obtained by using the unbiased model. Published Results (using MLE):


 * $$\begin{matrix}

{{\widehat{\mu }}^{\prime }}=4.3553 \\ =0.67677 \\ \end{matrix}\,\!$$

This same data set can be entered into Weibull++ by creating a data sheet capable of handling non-grouped time-to-failure data. Since the results shown above are unbiased, the Use Unbiased Std on Normal Data option in the User Setup must be selected in order to duplicate these results. Weibull++ computed parameters for maximum likelihood are:


 * $$\begin{matrix}

{{\widehat{\mu }}^{\prime }}=4.3553 \\ =0.6768 \\ \end{matrix}\,\!$$

Suspension Data Example
From Nelson [30, p. 324]. 96 locomotive controls were tested, 37 failed and 59 were suspended after running for 135,000 miles. The table below shows the failure and suspension times.

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]\,\!$$

Interval Data Example
Determine the lognormal parameter estimates for the data given in the table below.

Solution

This is a sequence of interval times-to-failure where the intervals vary substantially in length. Using Weibull++, the computed parameters for maximum likelihood are calculated to be:


 * $$\begin{align}

& {{{\hat{\mu }}}^{\prime }}= & 3.64 \\ & = & 0.18 \end{align}\,\!$$

For rank regression on $$X\ \,\!$$:


 * $$\begin{align}

& {{{\hat{\mu }}}^{\prime }}= & 3.64 \\ & = & 0.17 \end{align}\,\!$$

For rank regression on $$Y\ \,\!$$:


 * $$\begin{align}

& {{{\hat{\mu }}}^{\prime }}= & 3.64 \\ & = & 0.21 \end{align}\,\!$$