Template:Example: Lognormal Distribution RRX
Lognormal Distribution RRX Example
Using the data of Example 2 and assuming a lognormal distribution, estimate the parameters and estimate the correlation coefficient, [math]\displaystyle{ \rho }[/math] , using rank regression on X.
Solution
Table 9.2 constructed in Example 2 applies to this example as well. Using the values in this table we get:
- [math]\displaystyle{ \begin{align} & \hat{b}= & \frac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,T_{i}^{\prime }{{y}_{i}}-\tfrac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,T_{i}^{\prime }\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{y}_{i}}}{14}}{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,y_{i}^{2}-\tfrac{{{\left( \underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{y}_{i}} \right)}^{2}}}{14}} \\ & & \\ & \widehat{b}= & \frac{10.4473-(49.2220)(0)/14}{11.3646-{{(0)}^{2}}/14} \end{align} }[/math]
or:
- [math]\displaystyle{ \widehat{b}=0.9193 }[/math]
and:
- [math]\displaystyle{ \hat{a}=\overline{x}-\hat{b}\overline{y}=\frac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,T_{i}^{\prime }}{14}-\widehat{b}\frac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{y}_{i}}}{14} }[/math]
or:
- [math]\displaystyle{ \widehat{a}=\frac{49.2220}{14}-(0.9193)\frac{(0)}{14}=3.5159 }[/math]
Therefore, from Eqn. (blnx):
- [math]\displaystyle{ {{\sigma }_{{{T}'}}}=\widehat{b}=0.9193 }[/math]
and from Eqn. (alnx):
- [math]\displaystyle{ {\mu }'=\frac{\widehat{a}}{\widehat{b}}{{\sigma }_{{{T}'}}}=\frac{3.5159}{0.9193}\cdot 0.9193=3.5159 }[/math]
Using Eqns. (mean) and (sdv) we get:
- [math]\displaystyle{ \overline{T}=\mu =51.3393\text{ hours} }[/math]
and:
- [math]\displaystyle{ {{\sigma }_{T}}=59.1682\text{ hours}. }[/math]
The correlation coefficient is found using Eqn. (RHOln):
- [math]\displaystyle{ \widehat{\rho }=0.9754. }[/math]
Note that the regression on Y analysis is not necessarily the same as the regression on X. The only time when the results of the two regression types are the same (i.e. will yield the same equation for a line) is when the data lie perfectly on a line.
Using Weibull++ , with the Rank Regression on X option, the results are:
