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:
 * $$ \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

$$


 * and:


 * $$ \frac{\partial \Lambda }{\partial \eta }=\frac{-\beta }{\eta }\cdot 6+\frac{ \beta }{\eta }\sum\limits_{i=1}^{6}\left( \frac{T_{i}}{\eta }\right) ^{\beta }=0 $$

Solving the above equations simultaneously we get:


 * $$ \hat{\beta }=1.933,$$ $$\hat{\eta }=73.526 $$

The variance/covariance matrix is found to be,


 * $$ \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] $$

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



You can view the variance/covariance matrix directly by clicking the Quick Calculation Pad (QCP) icon





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