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| '''Normal Distribution RRX Example'''
| | #REDIRECT [[The Normal Distribution]] |
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| Using the data of [[Normal distribution example 2|Example 2]] and assuming a normal distribution, estimate the parameters and determine the correlation coefficient, <math>\rho </math> , using rank regression on X.
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| '''Solution'''
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| The table constructed in [[Normal distribution example 2|Example 2]] applies to this example also. Using the values on 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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| \widehat{b}= & \frac{365.2711-(630)(0)/14}{11.3646-{{(0)}^{2}}/14}=32.1411
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| \end{align}</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}-\widehat{b}\frac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{y}_{i}}}{14}</math>
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| or:
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| ::<math>\widehat{a}=\frac{630}{14}-(32.1411)\frac{(0)}{14}=45</math>
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| Therefore:
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| ::<math>\widehat{\sigma }=\widehat{b}=32.1411</math>
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| and:
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| ::<math>\widehat{\mu }=\widehat{a}=45\text{ hours}</math>
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| The correlation coefficient is obtained as:
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| ::<math>\widehat{\rho }=0.979</math>
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| Note that the results for regression on X are not necessarily the same as the results for regression on Y. The only time when the two regressions are the same (i.e. will yield the same equation for a line) is when the data lie perfectly on a straight line.
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| Using Weibull++ , Rank Regression on X (RRX) can be selected from the Analysis page.
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| [[Image:Normal RRX Setting.png|thumb|center|400px| ]]
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| The plot of the solution for this example is shown next.
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| [[Image:Normal RRX Plot.png|thumb|center|400px| ]]
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| <math></math>
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