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'''2-Parameter Weibull Distribution RRY Example'''
#REDIRECT [[The Weibull Distribution]]
 
Consider the data in [[Weibull Example 1 Data|Example 1]], where six units were tested to failure and the following failure times were recorded: 16, 34, 53, 75, 93 and 120 hours. Estimate the parameters and the correlation coefficient using rank regression on Y, assuming that the data follow the two-parameter Weibull distribution.
 
 
'''Solution'''
 
Construct a table as shown below.
 
{|align="center" border=1 cellspacing=1
|-
|colspan="8" style="text-align:center"| Table - Least Squares Analysis
|-
!<math>N</math>
!<math>T_{i}</math>
!<math>ln(T_{i})</math>
!<math>F(T_i)</math>
!<math>y_{i}</math>
!<math>(ln{T_i})^2</math>
!<math>{y_i}^2</math>
!<math>(ln{T_i})y_i</math>
|-
|1 ||16||2.7726||0.1091||-2.1583||7.6873||4.6582||-5.9840
|-
|2 ||34||3.5264||0.2645||-1.1802||12.4352||1.393||-4.1620
|-
|3 ||53||3.9703||0.4214||-0.6030||15.7632||0.3637||-2.3943
|-
|4 ||75||4.3175||0.5786||-0.146||18.6407||0.0213||-0.6303
|-
|5 ||93||4.5326||0.7355||0.2851||20.5445||0.0813||1.2923
|-
|6 ||120||4.7875||0.8909||0.7955||22.9201||0.6328||3.8083
|-
|<math>\sum</math>||  ||23.9068|| ||-3.007||97.9909||7.1502||-8.0699
|}
 
 
Utilizing the values from the table, calculate <math> \hat{a} </math> and <math> \hat{b} </math> using the following equations:
::<math> \hat{b} =\frac{\sum\limits_{i=1}^{6}(\ln t_{i})y_{i}-(\sum\limits_{i=1}^{6}\ln t_{i})(\sum\limits_{i=1}^{6}y_{i})/6}{ \sum\limits_{i=1}^{6}(\ln t_{i})^{2}-(\sum\limits_{i=1}^{6}\ln t_{i})^{2}/6}
</math>
 
::<math> \hat{b}=\frac{-8.0699-(23.9068)(-3.0070)/6}{97.9909-(23.9068)^{2}/6} </math>
 
or
 
::<math> \hat{b}=1.4301 </math>
 
and:
 
::<math> \hat{a}=\overline{y}-\hat{b}\overline{T}=\frac{\sum \limits_{i=1}^{N}y_{i}}{N}-\hat{b}\frac{\sum\limits_{i=1}^{N}\ln t_{i}}{N } </math>
 
or:
 
::<math> \hat{a}=\frac{(-3.0070)}{6}-(1.4301)\frac{23.9068}{6}=-6.19935 </math>
 
Therefore:
 
::<math> \hat{\beta }=\hat{b}=1.4301 </math>
 
and:
 
::<math> \hat{\eta }=e^{-\frac{\hat{a}}{\hat{b}}}=e^{-\frac{(-6.19935)}{ 1.4301}} </math>
 
or:
 
::<math> \hat{\eta }=76.318\text{ hr} </math>
 
The correlation coefficient can be estimated as:
 
::<math> \hat{\rho }=0.9956 </math>
<br>
This example can be repeated in the Weibull++ software, as shown next. The following picture shows a Weibull++ standard folio data sheet calculated with the 2P-Weibull distribution and rank regression on Y.
 
[[Image:Weibull Distribution Example 3 RRY Result.png|thumb|center|250px| ]]
 
 
The following plot shows the Weibull probability plot for the data set (with 90% two-sided confidence bounds).
 
[[Image:Weibull Distribution Example 3 RRY Confidence Plot.png|thumb|center|250px| ]]
 
If desired, the Weibull  <math>pdf</math> representing the data set can be written as:
 
::<math> f(t)={\frac{\beta }{\eta }}\left( {\frac{t}{\eta }}\right) ^{\beta -1}e^{-\left( {\frac{t}{\eta }}\right) ^{\beta }} </math>
 
or:
 
::<math> f(t)={\frac{1.4302}{76.317}}\left( {\frac{t}{76.317}}\right) ^{0.4302}e^{-\left( {\frac{t}{76.317}}\right) ^{1.4302}} </math>
 
You can also plot this result by selecting '''Pdf Plot''' on the Plot Type drop-down list on the control panel.
 
[[Image:Weibull Distribution Example 3 pdf Plot.png|thumb|center|250px]]
 
From this point on, different results, reports and plots can be obtained.

Latest revision as of 08:35, 8 August 2012

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