Template:Example: 2 Parameter Exponential Distribution RRX: Difference between revisions

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'''2 Parameter Exponential Distribution RRX'''
#REDIRECT [[The Exponential Distribution]]
 
Using the data of Example 2 and assuming a two-parameter exponential distribution, estimate the parameters and determine the correlation coefficient estimate, <math>\hat{\rho }</math>, using rank regression on X.
 
''' Solution to Example 3'''
 
Table 7.2 constructed in Example 2 applies to this example also. Using the values from this table, we get:
 
::<math>\begin{align}
  \hat{b}= & \frac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{t}_{i}}{{y}_{i}}-\tfrac{\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 }}}\,y_{i}^{2}-\tfrac{{{\left( \underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{y}_{i}} \right)}^{2}}}{14}} \\
  \\
  \hat{b}= & \frac{-927.4899-(630)(-13.2315)/14}{22.1148-{{(-13.2315)}^{2}}/14} 
\end{align}</math>
 
or:
 
::<math>\hat{b}=-34.5563</math>
 
and:
 
::<math>\hat{a}=\overline{x}-\hat{b}\overline{y}=\frac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{t}_{i}}}{14}-\hat{b}\frac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{y}_{i}}}{14}</math>
 
or:
 
::<math>\hat{a}=\frac{630}{14}-(-34.5563)\frac{(-13.2315)}{14}=12.3406</math>
 
Therefore:
 
::<math>\hat{\lambda }=-\frac{1}{\hat{b}}=-\frac{1}{(-34.5563)}=0.0289\text{ failures/hour}</math>
 
and:
 
::<math>\hat{\gamma }=\hat{a}=12.3406</math>
 
The correlation coefficient is found to be:
 
::<math>\hat{\rho }=-0.9679</math>
 
Note that the equation for regression on Y is not necessarily the same as that for the regression on X. The only time when the two regression methods yield identical results is when the data lie perfectly on a line. If this were the case, the correlation coefficient would be <math>-1</math>. The negative value of the correlation coefficient is due to the fact that the slope of the exponential probability plot is negative.
 
This example can be repeated using Weibull++, choosing two-parameter exponential and rank regression on X (RRX) methods for analysis, as shown below.
The estimated parameters and the correlation coefficient using Weibull++ were found to be:
 
::<math>\begin{array}{*{35}{l}}
  \hat{\lambda }= &0.0289 \text{failures/hour} \\
  \hat{\gamma}= & 12.3395 \text{hours} \\
  \hat{\rho} = &-0.9679  \\
\end{array}</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 08:32, 10 August 2012

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