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'''Normal Distribution RRY Example'''
#REDIRECT [[The Normal Distribution]]
 
Fourteen units were reliability tested and the following life test data were obtained:
 
 
{|border="1" align="center" style="border-collapse: collapse;" cellpadding="5" cellspacing="5"
|-
|colspan="2" style="text-align:center"|The test data
|-
!Data point index
!Time-to-failure
|-
|1 ||5
|-
|2 ||10
|-
|3 ||15
|-
|4 ||20
|-
|5 ||25
|-
|6 ||30
|-
|7||35
|-
|8||40
|-
|9||50
|-
|10||60
|-
|11||70
|-
|12||80
|-
|13||90
|-
|14||100
|}
 
 
Assuming the data follow a normal distribution, estimate the parameters and determine the correlation coefficient,  <math>\rho </math> , using rank regression on Y.
 
'''Solution'''
 
Construct a table like the one shown next.
 
<center><math>\overset{{}}{\mathop{\text{Table}\text{ - Least Squares Analysis}}}\,</math></center>
<center><math>\begin{matrix}
  \text{N} & \text{T}_{i} & \text{F(T}_{i}\text{)} & \text{y}_{i} & \text{T}_{i}^{2} & \text{y}_{i}^{2} & \text{T}_{i}\text{ y}_{i}  \\
  \text{1} & \text{5} & \text{0}\text{.0483} & \text{-1}\text{.6619} & \text{25} & \text{2}\text{.7619} & \text{-8}\text{.3095}  \\
  \text{2} & \text{10} & \text{0}\text{.1170} & \text{-1}\text{.1901} & \text{100} & \text{1}\text{.4163} & \text{-11}\text{.9010}  \\
  \text{3} & \text{15} & \text{0}\text{.1865} & \text{-0}\text{.8908} & \text{225} & \text{0}\text{.7935} & \text{-13}\text{.3620}  \\
  \text{4} & \text{20} & \text{0}\text{.2561} & \text{-0}\text{.6552} & \text{400} & \text{0}\text{.4292} & \text{-13}\text{.1030}  \\
  \text{5} & \text{25} & \text{0}\text{.3258} & \text{-0}\text{.4512} & \text{625} & \text{0}\text{.2036} & \text{-11}\text{.2800}  \\
  \text{6} & \text{30} & \text{0}\text{.3954} & \text{-0}\text{.2647} & \text{900} & \text{0}\text{.0701} & \text{-7}\text{.9422}  \\
  \text{7} & \text{35} & \text{0}\text{.4651} & \text{-0}\text{.0873} & \text{1225} & \text{0}\text{.0076} & \text{-3}\text{.0542}  \\
  \text{8} & \text{40} & \text{0}\text{.5349} & \text{0}\text{.0873} & \text{1600} & \text{0}\text{.0076} & \text{3}\text{.4905}  \\
  \text{9} & \text{50} & \text{0}\text{.6046} & \text{0}\text{.2647} & \text{2500} & \text{0}\text{.0701} & \text{13}\text{.2370}  \\
  \text{10} & \text{60} & \text{0}\text{.6742} & \text{0}\text{.4512} & \text{3600} & \text{0}\text{.2036} & \text{27}\text{.0720}  \\
  \text{11} & \text{70} & \text{0}\text{.7439} & \text{0}\text{.6552} & \text{4900} & \text{0}\text{.4292} & \text{45}\text{.8605}  \\
  \text{12} & \text{80} & \text{0}\text{.8135} & \text{0}\text{.8908} & \text{6400} & \text{0}\text{.7935} & \text{71}\text{.2640}  \\
  \text{13} & \text{90} & \text{0}\text{.8830} & \text{1}\text{.1901} & \text{8100} & \text{1}\text{.4163} & \text{107}\text{.1090}  \\
  \text{14} & \text{100} & \text{0}\text{.9517} & \text{1}\text{.6619} & \text{10000} & \text{2}\text{.7619} & \text{166}\text{.1900}  \\
  \mathop{}_{}^{} & \text{630} & {} & \text{0} & \text{40600} & \text{11}\text{.3646} & \text{365}\text{.2711}  \\
\end{matrix}</math></center>
 
 
:*The median rank values ( <math>F({{t}_{i}})</math> ) can be found in rank tables, available in many statistical texts, or they can be estimated by using the Quick Statistical Reference in Weibull++.
:*The  <math>{{y}_{i}}</math>  values were obtained from standardized normal distribution's area tables by entering for  <math>F(z)</math>  and getting the corresponding  <math>z</math>  value ( <math>{{y}_{i}}</math> ).  As with the median rank values, these standard normal values can be obtained with the Quick Statistical Reference.
 
Given the values in Table 9.2, calculate  <math>\widehat{a}</math>  and  <math>\widehat{b}</math>  using:
 
::<math>\begin{align}
  & \widehat{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} \\
&  &  \\
& \widehat{b}= & \frac{365.2711-(630)(0)/14}{40,600-{{(630)}^{2}}/14}=0.02982 
\end{align}</math>
 
and:
 
::<math>\widehat{a}=\overline{y}-\widehat{b}\overline{T}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\widehat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{t}_{i}}}{N}</math>
 
or:
 
::<math>\widehat{a}=\frac{0}{14}-(0.02982)\frac{630}{14}=-1.3419</math>
 
Therefore:
 
::<math>\widehat{\sigma}=\frac{1}{\hat{b}}=\frac{1}{0.02982}=33.5367</math>
 
and:
 
::<math>\widehat{\mu }=-\widehat{a}\cdot \widehat{\sigma }=-(-1.3419)\cdot 33.5367\simeq 45</math>
 
or  <math>\widehat{\mu }=45</math>  hours <math>.</math>
 
The correlation coefficient can be estimated using:
 
::<math>\widehat{\rho }=0.979</math>
 
The preceding example can be repeated using Weibull++ .
 
:*Create a new folio for Times-to-Failure data, and enter the data given in this example.
:*Choose Normal from the Distributions list.
:*Go to the Analysis page and select Rank Regression on Y (RRY).
:*Click the Calculate icon located on the Main page.
 
[[Image:Normal RRY Setting.png|center|550px| ]]
 
The probability plot is shown next.
 
[[Image:Normal RRY Plot.png|center|550px| ]]

Latest revision as of 01:55, 13 August 2012

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