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| '''2 Parameter Exponential Distribution RRX'''
| | #REDIRECT [[The Exponential Distribution]] |
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| Using the data of [[Data in Exponential Chapter Example 2|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.
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| ''' Solution'''
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| The Table constructed in [[Data in Exponential Chapter Example 2|Example 2]] applies to this example also. Using the values from this table, we get: | |
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| ::<math>\begin{align}
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| \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}} \\
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| \\
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| \hat{b}= & \frac{-927.4899-(630)(-13.2315)/14}{22.1148-{{(-13.2315)}^{2}}/14}
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| \end{align}</math>
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| or:
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| ::<math>\hat{b}=-34.5563</math>
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| and:
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| ::<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>
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| or:
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| ::<math>\hat{a}=\frac{630}{14}-(-34.5563)\frac{(-13.2315)}{14}=12.3406</math>
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| Therefore:
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| ::<math>\hat{\lambda }=-\frac{1}{\hat{b}}=-\frac{1}{(-34.5563)}=0.0289\text{ failures/hour}</math>
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| and:
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| ::<math>\hat{\gamma }=\hat{a}=12.3406</math>
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| The correlation coefficient is found to be:
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| ::<math>\hat{\rho }=-0.9679</math>
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| 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.
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| This example can be repeated using Weibull++, choosing two-parameter exponential and rank regression on X (RRX) methods for analysis, as shown below.
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| The estimated parameters and the correlation coefficient using Weibull++ were found to be:
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| ::<math>\begin{array}{*{35}{l}}
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| \hat{\lambda }= &0.0289 \text{failures/hour} \\
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| \hat{\gamma}= & 12.3395 \text{hours} \\
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| \hat{\rho} = &-0.9679 \\
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| \end{array}</math>
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| [[Image:Exponential Distribution Example 3 Data Folio.png|thumb|center|250px|]]
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| The probability plot can be obtained simply by clicking the '''Plot''' icon.
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| [[Image:Exponential Distribution Example 3 Plot.png|thumb|center|250px|]]
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