Template:Example: 2P Weibull Distribution RRX: Difference between revisions

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'''2P Weibull Distribution RRX Example'''
#REDIRECT [[The_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:
 
::<math> \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}}}
</math>
 
::<math>\hat{b} =\frac{-8.0699-(23.9068)(-3.0070)/6}{7.1502-(-3.0070)^{2}/6} </math>
 
:or:
 
::<math> \hat{b}=0.6931 </math>
 
:and:
 
::<math> \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} </math>
 
:or:
 
::<math> \hat{a}=\frac{23.9068}{6}-(0.6931)\frac{(-3.0070)}{6}=4.3318 </math>
 
Therefore, from Eqn. (EQNREF bwx ):
 
::<math> \hat{\beta }=\frac{1}{\hat{b}}=\frac{1}{0.6931}=1.4428 </math>
 
:and from Eqn. (EQNREF awx )
 
::<math> \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} </math>
 
The correlation coefficient is found using Eqn. (EQNREF RHOw ):
 
::<math> \hat{\rho }=0.9956 </math>
 
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.
 
[[Image:onevariableplot.png|thumb|center|400px| ]]
 
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Latest revision as of 04:19, 21 August 2012

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