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==== 2P Weibull Distribution Example ====
#REDIRECT [[The Weibull Distribution]]
Consider the data in 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 6.1 - 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 Table 6.1, calculate <math> \hat{a} </math> and <math> \hat{b} </math> using Eqns. (EQNREF aaw ) and (EQNREF bbw ):
::<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, from Eqn. (EQNREF bw ):
 
::<math> \hat{\beta }=\hat{b}=1.4301 </math>
 
and from Eqn. (EQNREF aw ):
 
::<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 using Eqn. (EQNREF RHOw ):
 
::<math> \hat{\rho }=0.9956 </math>
 
The above example can be repeated using Weibull++. Start Weibull++ and create a new ''Data Folio''.
 
[[Image:projectwizardweibull.png|thumb|center|400px|]]
 
Select the '' Times-to-failure data'' option.
 
[[Image:times--to-failure.png|thumb|center|400px| The effect of the Weibull shape parameter on the <math>pdf</math>. ]]
 
Enter the times-to-failure in the datasheet (ignore the Subset ID column), as shown next. The times-to-failure need not be sorted, Weibull++ will automatically sort the data.
<br>
[[Image:weibullfolio16.png|thumb|center|400px| ]]
 
Select the desired method of analysis. Note that we are assuming that the underlying distribution is the Weibull, so make sure that the Weibull distribution is selected. Under ''Parameters/Type'' on the Main page, select ''2''.
<br>
[[Image:parameterweibull.png|thumb|center|400px|]]
 
Also, so that you get the same results as this example, switch to the ''Analysis'' page and make sure you are using the ''Rank Regression on Y (RRY)'' calculation method with this example, as shown next.
 
[[Image:weibullanalysis.png|thumb|center|400px| ]]
 
Note that this can also be done from the ''Main'' page by clicking the left bottom box under the Results area. Each time you click that box you will see the method switch between MLE, RRX, and RRY. Click the ''Calculate'' icon,
 
[[Image:calculateicon.png|thumb|center|400px|]]
or select ''Calculate'' from the ''Data'' menu. The results will appear in the Data Folio's ''Results area''. The next figure shows the results for this example.
 
[[Image:weibullfolio16calculate.png|thumb|center|400px| ]]
 
<br>
You can now plot the results by clicking the ''Plot'' icon,
 
[[Image:ploticon.png|thumb|center|400px| ]]
 
or by selecting ''Plot Probability'' from the ''Data'' menu.
 
The Weibull probability plot for these data is shown next.
 
The confidence bounds, as determined from the Fisher matrix, can also be plotted. Select ''Confidence Bounds'' from the ''Plot'' menu, choose ''Two-Sided'' under ''Sides,'' ''Reliability (Type II)'' under ''Type'' and enter ''90'' for the ''Confidence level.
 
[[Image:confidenceboundsweibull.png|thumb|center|400px| ]]
 
[[Image:weibullconfidencebounds.png|thumb|center|400px| ]]
The plot will appear as follows,
 
[[Image:weibullfolio16plot.png|thumb|center|400px| ]]
 
If desired, the Weibull  <math>pdf</math> representing these data 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 the Weibull  by selecting ''Pdf Plot'' from the ''Plot Type'' drop-down menu on the control panel to the right of the plot area.
 
[[Image:weibullpdffolio16.png|thumb|center|400px]]
 
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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