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'''2P Weibull Distribution RRY Example'''
'''2P-Weibull Distribution RRY Example'''


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.
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.

Revision as of 04:41, 23 July 2012

2P-Weibull Distribution RRY 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 - 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 the table, 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]


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.

Weibull Distribution Example 3 RRY Result.png


The following plot shows the Weibull probability plot for the data set (with 90% two-sided confidence bounds).

Weibull Distribution Example 3 RRY Confidence Plot.png

If desired, the Weibull [math]\displaystyle{ pdf }[/math] representing the data set 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 this result by selecting Pdf Plot on the Plot Type drop-down list on the control panel.

Weibull Distribution Example 3 pdf Plot.png

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