Template:Example: Weibull MLE

Maximum Likelihood Estimation Example

Repeat Example 1 using maximum likelihood estimation.

Solution

In this case, we have non-grouped data with no suspensions or intervals, i.e. complete data. The equations for the partial derivatives of the log-likelihood function are derived in Appendix and given next:

[math]\displaystyle{ \frac{\partial \Lambda }{\partial \beta }=\frac{6}{\beta } +\sum_{i=1}^{6}\ln \left( \frac{T_{i}}{\eta }\right) -\sum_{i=1}^{6}\left( \frac{T_{i}}{\eta }\right) ^{\beta }\ln \left( \frac{T_{i}}{\eta }\right) =0 }[/math]

and:

[math]\displaystyle{ \frac{\partial \Lambda }{\partial \eta }=\frac{-\beta }{\eta }\cdot 6+\frac{ \beta }{\eta }\sum\limits_{i=1}^{6}\left( \frac{T_{i}}{\eta }\right) ^{\beta }=0 }[/math]

Solving the above equations simultaneously we get:

[math]\displaystyle{ \hat{\beta }=1.933, }[/math] [math]\displaystyle{ \hat{\eta }=73.526 }[/math]

The variance/covariance matrix is found to be,

[math]\displaystyle{ \left[ \begin{array}{ccc} \hat{Var}\left( \hat{\beta }\right) =0.4211 & \hat{Cov}( \hat{\beta },\hat{\eta })=3.272 \\ \hat{Cov}(\hat{\beta },\hat{\eta })=3.272 & \hat{Var} \left( \hat{\eta }\right) =266.646 \end{array} \right] }[/math]

The results and the associated graph using Weibull++ (MLE) are shown next.

You can view the variance/covariance matrix directly by clicking the Analysis Summary at the Main Panel.

Note that the decimal accuracy displayed and used is based on your individual User Setup.