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| '''Normal Distribution RRY Example'''
| | #REDIRECT [[The Normal Distribution]] |
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| Fourteen units were reliability tested and the following life test data were obtained:
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| {|align="center" border=1 cellspacing=1
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| |-
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| |colspan="2" style="text-align:center"|Table -The test data
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| |-
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| !Data point index
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| !Time-to-failure
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| |-
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| |1 ||5
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| |-
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| |2 ||10
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| |-
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| |3 ||15
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| |-
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| |4 ||20
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| |-
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| |5 ||25
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| |-
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| |6 ||30
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| |-
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| |7||35
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| |-
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| |8||40
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| |-
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| |9||50
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| |-
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| |10||60
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| |-
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| |11||70
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| |-
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| |12||80
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| |-
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| |13||90
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| |-
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| |14||100
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| |}
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| Assuming the data follow a normal distribution, estimate the parameters and determine the correlation coefficient, <math>\rho </math> , using rank regression on Y.
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| '''Solution'''
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| Construct a table like the one shown next.
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| <center><math>\overset{{}}{\mathop{\text{Table}\text{ - Least Squares Analysis}}}\,</math></center>
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| <center><math>\begin{matrix}
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| \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} \\
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| \text{1} & \text{5} & \text{0}\text{.0483} & \text{-1}\text{.6619} & \text{25} & \text{2}\text{.7619} & \text{-8}\text{.3095} \\
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| \text{2} & \text{10} & \text{0}\text{.1170} & \text{-1}\text{.1901} & \text{100} & \text{1}\text{.4163} & \text{-11}\text{.9010} \\
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| \text{3} & \text{15} & \text{0}\text{.1865} & \text{-0}\text{.8908} & \text{225} & \text{0}\text{.7935} & \text{-13}\text{.3620} \\
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| \text{4} & \text{20} & \text{0}\text{.2561} & \text{-0}\text{.6552} & \text{400} & \text{0}\text{.4292} & \text{-13}\text{.1030} \\
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| \text{5} & \text{25} & \text{0}\text{.3258} & \text{-0}\text{.4512} & \text{625} & \text{0}\text{.2036} & \text{-11}\text{.2800} \\
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| \text{6} & \text{30} & \text{0}\text{.3954} & \text{-0}\text{.2647} & \text{900} & \text{0}\text{.0701} & \text{-7}\text{.9422} \\
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| \text{7} & \text{35} & \text{0}\text{.4651} & \text{-0}\text{.0873} & \text{1225} & \text{0}\text{.0076} & \text{-3}\text{.0542} \\
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| \text{8} & \text{40} & \text{0}\text{.5349} & \text{0}\text{.0873} & \text{1600} & \text{0}\text{.0076} & \text{3}\text{.4905} \\
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| \text{9} & \text{50} & \text{0}\text{.6046} & \text{0}\text{.2647} & \text{2500} & \text{0}\text{.0701} & \text{13}\text{.2370} \\
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| \text{10} & \text{60} & \text{0}\text{.6742} & \text{0}\text{.4512} & \text{3600} & \text{0}\text{.2036} & \text{27}\text{.0720} \\
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| \text{11} & \text{70} & \text{0}\text{.7439} & \text{0}\text{.6552} & \text{4900} & \text{0}\text{.4292} & \text{45}\text{.8605} \\
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| \text{12} & \text{80} & \text{0}\text{.8135} & \text{0}\text{.8908} & \text{6400} & \text{0}\text{.7935} & \text{71}\text{.2640} \\
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| \text{13} & \text{90} & \text{0}\text{.8830} & \text{1}\text{.1901} & \text{8100} & \text{1}\text{.4163} & \text{107}\text{.1090} \\
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| \text{14} & \text{100} & \text{0}\text{.9517} & \text{1}\text{.6619} & \text{10000} & \text{2}\text{.7619} & \text{166}\text{.1900} \\
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| \mathop{}_{}^{} & \text{630} & {} & \text{0} & \text{40600} & \text{11}\text{.3646} & \text{365}\text{.2711} \\
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| \end{matrix}</math></center>
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| :*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++.
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| :*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.
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| Given the values in Table 9.2, calculate <math>\widehat{a}</math> and <math>\widehat{b}</math> using:
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| ::<math>\begin{align}
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| & \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} \\
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| & & \\
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| & \widehat{b}= & \frac{365.2711-(630)(0)/14}{40,600-{{(630)}^{2}}/14}=0.02982
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| \end{align}</math>
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| and:
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| ::<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>
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| or:
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| ::<math>\widehat{a}=\frac{0}{14}-(0.02982)\frac{630}{14}=-1.3419</math>
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| Therefore:
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| ::<math>\widehat{\sigma}=\frac{1}{\hat{b}}=\frac{1}{0.02982}=33.5367</math>
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| and:
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| ::<math>\widehat{\mu }=-\widehat{a}\cdot \widehat{\sigma }=-(-1.3419)\cdot 33.5367\simeq 45</math>
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| or <math>\widehat{\mu }=45</math> hours <math>.</math>
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| The correlation coefficient can be estimated using:
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| ::<math>\widehat{\rho }=0.979</math>
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| The preceding example can be repeated using Weibull++ .
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| :*Create a new folio for Times-to-Failure data, and enter the data given in this example.
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| :*Choose Normal from the Distributions list.
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| :*Go to the Analysis page and select Rank Regression on Y (RRY).
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| :*Click the Calculate icon located on the Main page.
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| [[Image:Normal RRY Setting.png|thumb|center|250px| ]] | |
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| The probability plot is shown next. | |
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| [[Image:Normal RRY Plot.png|thumb|center|250px| ]]
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