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'''Normal Distribution RRX Example'''
#REDIRECT [[The Normal Distribution]]
 
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.
 
 
'''Solution'''
 
Table 8.2 constructed in [[Normal distribution example 2|Example 2]] applies to this example also. Using the values on 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}} \\
  \widehat{b}= & \frac{365.2711-(630)(0)/14}{11.3646-{{(0)}^{2}}/14}=32.1411 
\end{align}</math>
 
and:
 
::<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>
 
or:
 
::<math>\widehat{a}=\frac{630}{14}-(32.1411)\frac{(0)}{14}=45</math>
 
Therefore:
 
::<math>\widehat{\sigma }=\widehat{b}=32.1411</math>
 
and:
 
::<math>\widehat{\mu }=\widehat{a}=45\text{ hours}</math>
 
The correlation coefficient is obtained as:
 
::<math>\widehat{\rho }=0.979</math>
 
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.
Using Weibull++ , Rank Regression on X (RRX) can be selected from the Analysis page.
 
[[Image:weibullfolio1.png|thumb|center|400px| ]]
 
The plot of the solution for this example is shown next.
 
[[Image:weibullfolioplot1.png|thumb|center|400px| ]]
<math></math>

Latest revision as of 02:00, 13 August 2012

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