2P Exponential Example: Difference between revisions
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<noinclude>{{Banner Weibull Examples}} | <noinclude>{{Banner Weibull Examples}} | ||
''This example also appears in the [[The_Exponential_Distribution#Probability_Plotting|Life Data Analysis Reference book]]''. | ''This example also appears in the [[The_Exponential_Distribution#Probability_Plotting|Life Data Analysis Reference book]]''. | ||
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Revision as of 05:06, 15 August 2012
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This example also appears in the Life Data Analysis Reference book.
2-Parameter Exponential RRY Example
14 units were being reliability tested and the following life test data were obtained:
Life Test Data | |
---|---|
Data point index | Time-to-failure |
1 | 5 |
2 | 10 |
3 | 15 |
4 | 20 |
5 | 25 |
6 | 30 |
7 | 35 |
8 | 40 |
9 | 50 |
10 | 60 |
11 | 70 |
12 | 80 |
13 | 90 |
14 | 100 |
Assuming that the data follow a two-parameter exponential distribution, estimate the parameters and determine the correlation coefficient, [math]\displaystyle{ \rho }[/math], using rank regression on Y.
Solution
Construct the following table, as shown next.
The median rank values ( [math]\displaystyle{ F({{t}_{i}}) }[/math] ) can be found in rank tables or they can be estimated using the Quick Statistical Reference in Weibull++.
Given the values in the table above, calculate [math]\displaystyle{ \hat{a} }[/math] and [math]\displaystyle{ \hat{b} }[/math]:
- [math]\displaystyle{ \begin{align} \hat{b}= & \frac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{t}_{i}}{{y}_{i}}-(\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{t}_{i}})(\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{y}_{i}})/14}{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,t_{i}^{2}-{{(\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{t}_{i}})}^{2}}/14} \\ \\ \hat{b}= & \frac{-927.4899-(630)(-13.2315)/14}{40,600-{{(630)}^{2}}/14} \end{align} }[/math]
or:
- [math]\displaystyle{ \hat{b}=-0.02711 }[/math]
and:
- [math]\displaystyle{ \hat{a}=\overline{y}-\hat{b}\overline{t}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{t}_{i}}}{N} }[/math]
or:
- [math]\displaystyle{ \hat{a}=\frac{-13.2315}{14}-(-0.02711)\frac{630}{14}=0.2748 }[/math]
Therefore:
- [math]\displaystyle{ \hat{\lambda }=-\hat{b}=-(-0.02711)=0.02711\text{ failures/hour} }[/math]
and:
- [math]\displaystyle{ \hat{\gamma }=\frac{\hat{a}}{\hat{\lambda }}=\frac{0.2748}{0.02711} }[/math]
or:
- [math]\displaystyle{ \hat{\gamma }=10.1365\text{ hours} }[/math]
Then:
- [math]\displaystyle{ f(t)=(0.02711)\cdot {{e}^{-0.02711(T-10.136)}} }[/math]
The correlation coefficient can be estimated using equation for calculating the correlation coefficient:
- [math]\displaystyle{ \hat{\rho }=-0.9679 }[/math]
This example can be repeated using Weibull++, choosing two-parameter exponential and rank regression on Y (RRY), as shown next.
The estimated parameters and the correlation coefficient using Weibull++ were found to be:
- [math]\displaystyle{ \hat{\lambda }=0.0271\text{ fr/hr },\hat{\gamma }=10.1348\text{ hr },\hat{\rho }=-0.9679 }[/math]
Please note that the user must deselect the Reset if location parameter > T1 on Exponential option on the Calculations page of the Application Setup window.
The probability plot can be obtained simply by clicking the Plot icon.