Template:Example: 2P Weibull Distribution RRY

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.

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

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.

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