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====Example 2: 2 Parameter Exponential Distribution RRY====
#REDIRECT [[The_Exponential_Distribution]]
Fourteen units were being reliability tested and the following life test data were obtained (Table 7.1):
 
 
{|align="center" border=1 cellspacing=0
|-
|colspan="2" style="text-align:center"| Table 7.1 - Life Test Data
|-
|- align="center"
!Data point index
!Time-to-failure
|- align="center"
|1 ||5
|- align="center"
|2 ||10
|- align="center"
|3 ||15
|- align="center"
|4 ||20
|- align="center"
|5 ||25
|- align="center"
|6 ||30
|- align="center"
|7 ||35
|- align="center"
|8 ||40
|- align="center"
|9 ||50
|- align="center"
|10 ||60
|- align="center"
|11 ||70
|-align="center"
|12 ||80
|- align="center"
|13 ||90
|- align="center"
|14 ||100
|}
 
Assuming that the data follow a two-parameter exponential distribution, estimate the parameters and determine the correlation coefficient, <math>\rho </math>, using rank regression on Y.
 
=====Solution to Example 2=====
 
Construct Table 7.2, as shown next.
 
<center><math>\overset{{}}{\mathop{\text{Table 7}\text{.2 - Least Squares Analysis}}}\,</math></center>
 
<center><math>\begin{matrix}
  N & T_{i} & F(t_{i}) & y_{i} & T_{i}^{2} & y_{i}^{2} & T_{i} y_{i}  \\
  \text{1} & \text{5} & \text{0}\text{.0483} & \text{-0}\text{.0495} & \text{25} & \text{0}\text{.0025} & \text{-0}\text{.2475}  \\
  \text{2} & \text{10} & \text{0}\text{.1170} & \text{-0}\text{.1244} & \text{100} & \text{0}\text{.0155} & \text{-1}\text{.2443}  \\
  \text{3} & \text{15} & \text{0}\text{.1865} & \text{-0}\text{.2064} & \text{225} & \text{0}\text{.0426} & \text{-3}\text{.0961}  \\
  \text{4} & \text{20} & \text{0}\text{.2561} & \text{-0}\text{.2958} & \text{400} & \text{0}\text{.0875} & \text{-5}\text{.9170}  \\
  \text{5} & \text{25} & \text{0}\text{.3258} & \text{-0}\text{.3942} & \text{625} & \text{0}\text{.1554} & \text{-9}\text{.8557}  \\
  \text{6} & \text{30} & \text{0}\text{.3954} & \text{-0}\text{.5032} & \text{900} & \text{0}\text{.2532} & \text{-15}\text{.0956}  \\
  \text{7} & \text{35} & \text{0}\text{.4651} & \text{-0}\text{.6257} & \text{1225} & \text{0}\text{.3915} & \text{-21}\text{.8986}  \\
  \text{8} & \text{40} & \text{0}\text{.5349} & \text{-0}\text{.7655} & \text{1600} & \text{0}\text{.5860} & \text{-30}\text{.6201}  \\
  \text{9} & \text{50} & \text{0}\text{.6046} & \text{-0}\text{.9279} & \text{2500} & \text{0}\text{.8609} & \text{-46}\text{.3929}  \\
  \text{10} & \text{60} & \text{0}\text{.6742} & \text{-1}\text{.1215} & \text{3600} & \text{1}\text{.2577} & \text{-67}\text{.2883}  \\
  \text{11} & \text{70} & \text{0}\text{.7439} & \text{-1}\text{.3622} & \text{4900} & \text{1}\text{.8456} & \text{-95}\text{.3531}  \\
  \text{12} & \text{80} & \text{0}\text{.8135} & \text{-1}\text{.6793} & \text{6400} & \text{2}\text{.8201} & \text{-134}\text{.3459}  \\
  \text{13} & \text{90} & \text{0}\text{.8830} & \text{-2}\text{.1456} & \text{8100} & \text{4}\text{.6035} & \text{-193}\text{.1023}  \\
  \text{14} & \text{100} & \text{0}\text{.9517} & \text{-3}\text{.0303} & \text{10000} & \text{9}\text{.1829} & \text{-303}\text{.0324}  \\
  \sum_{}^{} & \text{630} & {} & \text{-13}\text{.2315} & \text{40600} & \text{22}\text{.1148} & \text{-927}\text{.4899}  \\
\end{matrix}</math></center>
 
 
The median rank values ( <math>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>\hat{a}</math> and <math>\hat{b}</math>:
 
 
::<math>\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>\hat{b}=-0.02711</math>
 
and:
 
::<math>\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>\hat{a}=\frac{-13.2315}{14}-(-0.02711)\frac{630}{14}=0.2748</math>
 
 
Therefore:
 
::<math>\hat{\lambda }=-\hat{b}=-(-0.02711)=0.02711\text{ failures/hour}</math>
 
 
and:
 
::<math>\hat{\gamma }=\frac{\hat{a}}{\hat{\lambda }}=\frac{0.2748}{0.02711}</math>
 
or:
 
::<math>\hat{\gamma }=10.1365\text{ hours}</math>
 
Then:
 
::<math>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>\hat{\rho }=-0.9679</math>
 
 
This example can be repeated using Weibull++, choosing two-parameter exponential and rank regression on Y (RRY), as shown in the figure on the following page.
 
The estimated parameters and the correlation coefficient using Weibull++ were found to be:
 
::<math>\hat{\lambda }=0.0271\text{ fr/hr },\hat{\gamma }=10.1348\text{ hr },\hat{\rho }=-0.9679</math>
 
[[Image:weibullfolio1.png|thumb|center|400px|]]
 
The probability plot can be obtained simply by clicking the '''Plot''' icon.
 
[[Image:weibullfolioplot1.png|thumb|center|400px|]]

Latest revision as of 09:07, 9 August 2012

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