Template:Example: 2P Weibull Distribution RRX

Example 4: 2P Weibull Distribution RRX Example
Repeat Example 1 using rank regression on X.

Solution

Solution to Example 4 Table 8.1, constructed in Example 3, can also be applied to this example.

Using the values from this table we get:


 * $$ \hat{b} ={\frac{\sum\limits_{i=1}^{6}(\ln T_{i})y_{i}-\frac{ \sum\limits_{i=1}^{6}\ln T_{i}\sum\limits_{i=1}^{6}y_{i}}{6}}{ \sum\limits_{i=1}^{6}y_{i}^{2}-\frac{\left( \sum\limits_{i=1}^{6}y_{i}\right) ^{2}}{6}}}

$$


 * $$\hat{b} =\frac{-8.0699-(23.9068)(-3.0070)/6}{7.1502-(-3.0070)^{2}/6} $$


 * or:


 * $$ \hat{b}=0.6931 $$


 * and:


 * $$ \hat{a}=\overline{x}-\hat{b}\overline{y}=\frac{\sum\limits_{i=1}^{6}\ln T_{i} }{6}-\hat{b}\frac{\sum\limits_{i=1}^{6}y_{i}}{6} $$


 * or:


 * $$ \hat{a}=\frac{23.9068}{6}-(0.6931)\frac{(-3.0070)}{6}=4.3318 $$

Therefore, from Eqn. (EQNREF bwx ):


 * $$ \hat{\beta }=\frac{1}{\hat{b}}=\frac{1}{0.6931}=1.4428 $$


 * and from Eqn. (EQNREF awx )


 * $$ \hat{\eta }=e^{\frac{\hat{a}}{\hat{b}}\cdot \frac{1}{\hat{ \beta }}}=e^{\frac{4.3318}{0.6931}\cdot \frac{1}{1.4428}}=76.0811\text{ hr} $$

The correlation coefficient is found using Eqn. (EQNREF RHOw ):


 * $$ \hat{\rho }=0.9956 $$

The results and the associated graph using Weibull++ are given next. Note that the slight variation in the results is due to the number of significant figures used in the estimation of the median ranks. Weibull++ by default uses double precision accuracy when computing the median ranks.