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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.
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 =====
'''Solution'''
 
Construct a table as shown below.  
Construct a table as shown below.  


{|align="center" border=1 cellspacing=1  
{|align="center" border=1 cellspacing=1  
|-
|-
|colspan="8" style="text-align:center"| Table 6.1 - Least Squares Analysis
|colspan="8" style="text-align:center"| Table 8.1 - Least Squares Analysis
|-  
|-  
!<math>N</math>
!<math>N</math>
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Utilizing the values from Table 6.1, calculate <math> \hat{a} </math> and <math> \hat{b} </math> using Eqns. (EQNREF aaw ) and (EQNREF bbw ):  
Utilizing the values from Table 8.1, 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> \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>  


::<math> \hat{b}=\frac{-8.0699-(23.9068)(-3.0070)/6}{97.9909-(23.9068)^{2}/6} </math>
::<math> \hat{b}=\frac{-8.0699-(23.9068)(-3.0070)/6}{97.9909-(23.9068)^{2}/6} </math>


:or
or


::<math> \hat{b}=1.4301 </math>  
::<math> \hat{b}=1.4301 </math>  


:and:  
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>  
::<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:  
:or:  
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::<math> \hat{a}=\frac{(-3.0070)}{6}-(1.4301)\frac{23.9068}{6}=-6.19935 </math>  
::<math> \hat{a}=\frac{(-3.0070)}{6}-(1.4301)\frac{23.9068}{6}=-6.19935 </math>  


Therefore, from Eqn. (EQNREF bw ):  
Therefore:


::<math> \hat{\beta }=\hat{b}=1.4301 </math>  
::<math> \hat{\beta }=\hat{b}=1.4301 </math>  


and from Eqn. (EQNREF aw ):  
and:  


::<math> \hat{\eta }=e^{-\frac{\hat{a}}{\hat{b}}}=e^{-\frac{(-6.19935)}{ 1.4301}} </math>  
::<math> \hat{\eta }=e^{-\frac{\hat{a}}{\hat{b}}}=e^{-\frac{(-6.19935)}{ 1.4301}} </math>  


:or:  
or:  


::<math> \hat{\eta }=76.318\text{ hr} </math>  
::<math> \hat{\eta }=76.318\text{ hr} </math>  


The correlation coefficient can be estimated using Eqn. (EQNREF RHOw ):  
The correlation coefficient can be estimated as:  


::<math> \hat{\rho }=0.9956 </math>  
::<math> \hat{\rho }=0.9956 </math>  
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If desired, the Weibull  <math>pdf</math> representing these data can be written as:  
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>  
::<math> f(t)={\frac{\beta }{\eta }}\left( {\frac{t}{\eta }}\right) ^{\beta -1}e^{-\left( {\frac{t}{\eta }}\right) ^{\beta }} </math>  


:or:  
: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.  
::<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]]
[[Image:weibullpdffolio16.png|thumb|center|400px]]


From this point on, different results, reports and plots can be obtained.
From this point on, different results, reports and plots can be obtained.

Revision as of 17:13, 9 February 2012

2P Weibull Distribution Example

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.

Table 8.1 - Least Squares Analysis
[math]\displaystyle{ N }[/math] [math]\displaystyle{ T_{i} }[/math] [math]\displaystyle{ ln(T_{i}) }[/math] [math]\displaystyle{ F(T_i) }[/math] [math]\displaystyle{ y_{i} }[/math] [math]\displaystyle{ (ln{T_i})^2 }[/math] [math]\displaystyle{ {y_i}^2 }[/math] [math]\displaystyle{ (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]\displaystyle{ \sum }[/math] 23.9068 -3.007 97.9909 7.1502 -8.0699


Utilizing the values from Table 8.1, calculate [math]\displaystyle{ \hat{a} }[/math] and [math]\displaystyle{ \hat{b} }[/math] using the following equations:

[math]\displaystyle{ \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]\displaystyle{ \hat{b}=\frac{-8.0699-(23.9068)(-3.0070)/6}{97.9909-(23.9068)^{2}/6} }[/math]

or

[math]\displaystyle{ \hat{b}=1.4301 }[/math]

and:

[math]\displaystyle{ \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]\displaystyle{ \hat{a}=\frac{(-3.0070)}{6}-(1.4301)\frac{23.9068}{6}=-6.19935 }[/math]

Therefore:

[math]\displaystyle{ \hat{\beta }=\hat{b}=1.4301 }[/math]

and:

[math]\displaystyle{ \hat{\eta }=e^{-\frac{\hat{a}}{\hat{b}}}=e^{-\frac{(-6.19935)}{ 1.4301}} }[/math]

or:

[math]\displaystyle{ \hat{\eta }=76.318\text{ hr} }[/math]

The correlation coefficient can be estimated as:

[math]\displaystyle{ \hat{\rho }=0.9956 }[/math]

The above example can be repeated using Weibull++. Start Weibull++ and create a new Data Folio.

Projectwizardweibull.png

Select the Times-to-failure data option.

The effect of the Weibull shape parameter on the [math]\displaystyle{ 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.

Weibullfolio16.png

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.

Parameterweibull.png

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.

Weibullanalysis.png

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,

Calculateicon.png

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.

Weibullfolio16calculate.png


You can now plot the results by clicking the Plot icon,

Ploticon.png

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.

Confidenceboundsweibull.png
Weibullconfidencebounds.png

The plot will appear as follows,

Weibullfolio16plot.png

If desired, the Weibull [math]\displaystyle{ pdf }[/math] representing these data can be written as:

[math]\displaystyle{ f(t)={\frac{\beta }{\eta }}\left( {\frac{t}{\eta }}\right) ^{\beta -1}e^{-\left( {\frac{t}{\eta }}\right) ^{\beta }} }[/math]
or:
[math]\displaystyle{ 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.
Weibullpdffolio16.png

From this point on, different results, reports and plots can be obtained.