Template:Lognormal distribution rank regression on Y - Revision history

Lisa Hacker: Redirected page to The Lognormal Distribution

2012-08-13T03:22:34Z

Redirected page to The Lognormal Distribution

← Older revision		Revision as of 03:22, 13 August 2012
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	~~===Rank Regression on Y===~~		#REDIRECT [[The Lognormal Distribution]]
	~~Performing a rank regression on Y requires that a straight line be fitted to a set of data points such that the sum of the squares of the vertical deviations from the points to the line is minimized.~~

	~~The least squares parameter estimation method, or regression analysis, was discussed in [~~[~~Parameter Estimation]] Chapter and the following equations for regression on Y were derived, and are again applicable:~~

	~~::<math>\hat{a}=\bar{y}-\hat{b}\bar{x}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}}{N}</math>~~

	~~and:~~

	::<math>\hat{b}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}{{y}_{i}}-\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}}{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,x_{i}^{2}-\tfrac{{{\left( \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}} \right)}^{2}}}{N}}</math>

	~~In our case the equations for <math>{{y}_{i}}</math> and <math>x_{i}</math> are:~~

	~~::<math>{{y}_{i}}={{\Phi }^{-1}}\left~~[ ~~F(t_{i}^{\prime }) \right]</math>~~

	~~and:~~

	~~::<math>{{x}_{i}}=t_{i}^{\prime }</math>~~

	where the <math>F(t_{i}^{\prime })</math> is estimated from the median ranks. Once <math>\widehat{a}</math> and <math>\widehat{b}</math> are obtained, then <math>\widehat{\sigma }</math> and <math>\widehat{\mu }</math> can easily be obtained from the above equations.

	{{The ~~Correlation Coefficient Calculation}}~~


	~~'''Example 2:'''~~
	~~{{Example:~~ Lognormal Distribution ~~RRY}}~~

Harry Guo: /* Rank Regression on Y */

2012-02-13T21:30:48Z

Rank Regression on Y

← Older revision		Revision as of 21:30, 13 February 2012
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	where the <math>F(t_{i}^{\prime })</math> is estimated from the median ranks. Once <math>\widehat{a}</math> and <math>\widehat{b}</math> are obtained, then <math>\widehat{\sigma }</math> and <math>\widehat{\mu }</math> can easily be obtained from the above equations.		where the <math>F(t_{i}^{\prime })</math> is estimated from the median ranks. Once <math>\widehat{a}</math> and <math>\widehat{b}</math> are obtained, then <math>\widehat{\sigma }</math> and <math>\widehat{\mu }</math> can easily be obtained from the above equations.

	~~'''~~The Correlation Coefficient~~'''~~		{{The Correlation Coefficient Calculation}}

	~~The estimator of <math>\rho </math> is the sample correlation coefficient, <math>\hat{\rho }</math> , given by:~~

	::<math>\hat{\rho }=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,({{x}_{i}}-\overline{x})({{y}_{i}}-\overline{y})}{\sqrt{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{x}_{i}}-\overline{x})}^{2}}\cdot \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{y}_{i}}-\overline{y})}^{2}}}}~~</math>~~


	'''Example 2:'''		'''Example 2:'''
	{{Example: Lognormal Distribution RRY}}		{{Example: Lognormal Distribution RRY}}

Harry Guo at 21:06, 13 February 2012

2012-02-13T21:06:12Z

← Older revision		Revision as of 21:06, 13 February 2012
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	::<math>\hat{a}=\bar{y}-\hat{b}\bar{x}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}}{N}</math>		::<math>\hat{a}=\bar{y}-\hat{b}\bar{x}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}}{N}</math>

	:and:		and:

	::<math>\hat{b}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}{{y}_{i}}-\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}}{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,x_{i}^{2}-\tfrac{{{\left( \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}} \right)}^{2}}}{N}}</math>		::<math>\hat{b}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}{{y}_{i}}-\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}}{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,x_{i}^{2}-\tfrac{{{\left( \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}} \right)}^{2}}}{N}}</math>
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	In our case the equations for <math>{{y}_{i}}</math> and <math>x_{i}</math> are:		In our case the equations for <math>{{y}_{i}}</math> and <math>x_{i}</math> are:

	::<math>{{y}_{i}}={{\Phi }^{-1}}\left[ F(T_{i}^{\prime }) \right]</math>		::<math>{{y}_{i}}={{\Phi }^{-1}}\left[ F(t_{i}^{\prime }) \right]</math>

	and:		and:

	::<math>{{x}_{i}}=T_{i}^{\prime }</math>		::<math>{{x}_{i}}=t_{i}^{\prime }</math>

	where the <math>F(T_{i}^{\prime })</math> is estimated from the median ranks. Once <math>\widehat{a}</math> and <math>\widehat{b}</math> are obtained, then <math>\widehat{\sigma }</math> and <math>\widehat{\mu }</math> can easily be obtained from ~~Eqns. (aln) and (bln)~~.		where the <math>F(t_{i}^{\prime })</math> is estimated from the median ranks. Once <math>\widehat{a}</math> and <math>\widehat{b}</math> are obtained, then <math>\widehat{\sigma }</math> and <math>\widehat{\mu }</math> can easily be obtained from the above equations.

	'''The Correlation Coefficient'''		'''The Correlation Coefficient'''
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	::<math>\hat{\rho }=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,({{x}_{i}}-\overline{x})({{y}_{i}}-\overline{y})}{\sqrt{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{x}_{i}}-\overline{x})}^{2}}\cdot \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{y}_{i}}-\overline{y})}^{2}}}}</math>		::<math>\hat{\rho }=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,({{x}_{i}}-\overline{x})({{y}_{i}}-\overline{y})}{\sqrt{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{x}_{i}}-\overline{x})}^{2}}\cdot \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{y}_{i}}-\overline{y})}^{2}}}}</math>


	'''Example 2:'''		'''Example 2:'''
	{{Example: Lognormal Distribution RRY}}		{{Example: Lognormal Distribution RRY}}

Harry Guo: /* Rank Regression on Y */

2012-02-13T21:05:11Z

Rank Regression on Y

← Older revision		Revision as of 21:05, 13 February 2012
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	Performing a rank regression on Y requires that a straight line be fitted to a set of data points such that the sum of the squares of the vertical deviations from the points to the line is minimized.		Performing a rank regression on Y requires that a straight line be fitted to a set of data points such that the sum of the squares of the vertical deviations from the points to the line is minimized.

	The least squares parameter estimation method, or regression analysis, was discussed in [[Parameter ~~Estimations~~]] Chapter and the following equations for regression on Y were derived, and are again applicable:		The least squares parameter estimation method, or regression analysis, was discussed in [[Parameter Estimation]] Chapter and the following equations for regression on Y were derived, and are again applicable:

	::<math>\hat{a}=\bar{y}-\hat{b}\bar{x}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}}{N}</math>		::<math>\hat{a}=\bar{y}-\hat{b}\bar{x}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}}{N}</math>

Harry Guo: /* Rank Regression on Y */

2012-02-13T21:04:53Z

Rank Regression on Y

← Older revision		Revision as of 21:04, 13 February 2012
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	Performing a rank regression on Y requires that a straight line be fitted to a set of data points such that the sum of the squares of the vertical deviations from the points to the line is minimized.		Performing a rank regression on Y requires that a straight line be fitted to a set of data points such that the sum of the squares of the vertical deviations from the points to the line is minimized.

	The least squares parameter estimation method, or regression analysis, was discussed in Chapter 3 and the following equations for regression on Y were derived, and are again applicable:		The least squares parameter estimation method, or regression analysis, was discussed in [[Parameter Estimations]] Chapter and the following equations for regression on Y were derived, and are again applicable:

	::<math>\hat{a}=\bar{y}-\hat{b}\bar{x}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}}{N}</math>		::<math>\hat{a}=\bar{y}-\hat{b}\bar{x}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}}{N}</math>
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	::<math>{{y}_{i}}={{\Phi }^{-1}}\left[ F(T_{i}^{\prime }) \right]</math>		::<math>{{y}_{i}}={{\Phi }^{-1}}\left[ F(T_{i}^{\prime }) \right]</math>

	:and:		and:

	::<math>{{x}_{i}}=T_{i}^{\prime }</math>		::<math>{{x}_{i}}=T_{i}^{\prime }</math>

Harry Guo at 18:15, 13 February 2012

2012-02-13T18:15:00Z

@@ Line 26: / Line 26: @@
 ::<math>\hat{\rho }=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,({{x}_{i}}-\overline{x})({{y}_{i}}-\overline{y})}{\sqrt{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{x}_{i}}-\overline{x})}^{2}}\cdot \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{y}_{i}}-\overline{y})}^{2}}}}</math>
-====Example 2====
+'''Example 2:'''
-Fourteen units were reliability tested and the following life test data were obtained:
+{{Example: Lognormal Distribution RRY}}

Harry Guo: /* The Correlation Coefficient */

2012-02-13T18:13:22Z

The Correlation Coefficient

← Older revision		Revision as of 18:13, 13 February 2012
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	where the <math>F(T_{i}^{\prime })</math> is estimated from the median ranks. Once <math>\widehat{a}</math> and <math>\widehat{b}</math> are obtained, then <math>\widehat{\sigma }</math> and <math>\widehat{\mu }</math> can easily be obtained from Eqns. (aln) and (bln).		where the <math>F(T_{i}^{\prime })</math> is estimated from the median ranks. Once <math>\widehat{a}</math> and <math>\widehat{b}</math> are obtained, then <math>\widehat{\sigma }</math> and <math>\widehat{\mu }</math> can easily be obtained from Eqns. (aln) and (bln).

	~~====~~The Correlation Coefficient~~====~~		'''The Correlation Coefficient'''

	The estimator of <math>\rho </math> is the sample correlation coefficient, <math>\hat{\rho }</math> , given by:		The estimator of <math>\rho </math> is the sample correlation coefficient, <math>\hat{\rho }</math> , given by:

	::<math>\hat{\rho }=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,({{x}_{i}}-\overline{x})({{y}_{i}}-\overline{y})}{\sqrt{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{x}_{i}}-\overline{x})}^{2}}\cdot \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{y}_{i}}-\overline{y})}^{2}}}}</math>		::<math>\hat{\rho }=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,({{x}_{i}}-\overline{x})({{y}_{i}}-\overline{y})}{\sqrt{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{x}_{i}}-\overline{x})}^{2}}\cdot \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{y}_{i}}-\overline{y})}^{2}}}}</math>


	====Example 2====		====Example 2====

Harry Guo: moved Template:Ld rank regression on Y to Template:Lognormal distribution rank regression on Y

2012-02-13T17:27:58Z

moved Template:Ld rank regression on Y to Template:Lognormal distribution rank regression on Y

← Older revision	Revision as of 17:27, 13 February 2012
(No difference)

Nicolette Young: Created page with '===Rank Regression on Y=== Performing a rank regression on Y requires that a straight line be fitted to a set of data points such that the sum of the squares of the vertical devi…'

2012-01-04T18:31:55Z

Created page with '===Rank Regression on Y=== Performing a rank regression on Y requires that a straight line be fitted to a set of data points such that the sum of the squares of the vertical devi…'

New page

===Rank Regression on Y===
Performing a rank regression on Y requires that a straight line be fitted to a set of data points such that the sum of the squares of the vertical deviations from the points to the line is minimized.

The least squares parameter estimation method, or regression analysis, was discussed in Chapter 3 and the following equations for regression on Y were derived, and are again applicable:

::<math>\hat{a}=\bar{y}-\hat{b}\bar{x}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}}{N}</math>

:and:

::<math>\hat{b}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}{{y}_{i}}-\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}}{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,x_{i}^{2}-\tfrac{{{\left( \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}} \right)}^{2}}}{N}}</math>

In our case the equations for <math>{{y}_{i}}</math> and <math>x_{i}</math> are:

::<math>{{y}_{i}}={{\Phi }^{-1}}\left[ F(T_{i}^{\prime }) \right]</math>

:and:

::<math>{{x}_{i}}=T_{i}^{\prime }</math>

where the <math>F(T_{i}^{\prime })</math> is estimated from the median ranks. Once <math>\widehat{a}</math> and <math>\widehat{b}</math> are obtained, then <math>\widehat{\sigma }</math> and <math>\widehat{\mu }</math> can easily be obtained from Eqns. (aln) and (bln).

====The Correlation Coefficient====
The estimator of <math>\rho </math> is the sample correlation coefficient, <math>\hat{\rho }</math> , given by:

::<math>\hat{\rho }=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,({{x}_{i}}-\overline{x})({{y}_{i}}-\overline{y})}{\sqrt{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{x}_{i}}-\overline{x})}^{2}}\cdot \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{y}_{i}}-\overline{y})}^{2}}}}</math>

====Example 2====
Fourteen units were reliability tested and the following life test data were obtained:

{|align="center" border=1 cellspacing=0
|-
|colspan="2" style="text-align:center"| Table 9.1 - Life Test Data for Example 2
|-
!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 lognormal distribution, estimate the parameters and the correlation coefficient, <math>\rho </math> , using rank regression on Y.

=====Solution to Example 2=====
Construct Table 9.2, as shown next.

<center><math>\overset{{}}{\mathop{\text{Table 9}\text{.2 - Least Squares Analysis}}}\,</math></center>

<center><math>\begin{matrix}
N & T_{i} & F(T_{i}) & {T_{i}}'& y_{i} & {{T_{i}}'}^{2} & y_{i}^{2} & T_{i} y_{i} \\
\text{1} & \text{5} & \text{0}\text{.0483} & \text{1}\text{.6094}& \text{-1}\text{.6619} & \text{2}\text{.5903} & \text{2}\text{.7619} & \text{-2}\text{.6747} \\
\text{2} & \text{10} & \text{0}\text{.1170} & \text{2.3026}& \text{-1.1901} & \text{5.3019} & \text{1.4163} & \text{-2.7403} \\
\text{3} & \text{15} & \text{0}\text{.1865} & \text{2.7080}&\text{-0.8908} & \text{7.3335} & \text{0.7935} & \text{-2.4123} \\
\text{4} & \text{20} & \text{0}\text{.2561} & \text{2.9957} &\text{-0.6552} & \text{8.9744} & \text{0.4292} & \text{-1.9627} \\
\text{5} & \text{25} & \text{0}\text{.3258} & \text{3.2189}& \text{-0.4512} & \text{10.3612} & \text{0.2036} & \text{-1.4524} \\
\text{6} & \text{30} & \text{0}\text{.3954} & \text{3.4012}& \text{-0.2647} & \text{11.5681} & \text{0.0701} & \text{-0.9004} \\
\text{7} & \text{35} & \text{0}\text{.4651} & \text{3.5553} & \text{-0.0873} & \text{12.6405} & \text{-0.0076}& \text{-0.3102} \\
\text{8} & \text{40} & \text{0}\text{.5349} & \text{3.6889}& \text{0.0873} & \text{13.6078} & \text{0.0076} & \text{0.3219} \\
\text{9} & \text{50} & \text{0}\text{.6046} & \text{3.912} & \text{0.2647} & \text{15.3039} & \text{0.0701} &\text{1.0357} \\
\text{10} & \text{60} & \text{0}\text{.6742} & \text{4.0943} & \text{0.4512} & \text{16.7637} & \text{0.2036}&\text{1.8474} \\
\text{11} & \text{70} & \text{0}\text{.7439} & \text{4.2485} & \text{0.6552} & \text{18.0497}& \text{0.4292} & \text{2.7834} \\
\text{12} & \text{80} & \text{0}\text{.8135} & \text{4.382} & \text{0.8908} & \text{19.2022} & \text{0.7935} & \text{3.9035} \\
\text{13} & \text{90} & \text{0}\text{.8830} & \text{4.4998} & \text{1.1901} & \text{20.2483}&\text{1.4163} & \text{5.3552} \\
\text{14} & \text{100}& \text{1.9517} & \text{4.6052} & \text{1.6619} & \text{21.2076} &\text{2.7619} & \text{7.6533} \\
\sum_{}^{} & \text{ } & \text{ } & \text{49.222} & \text{0} & \text{183.1531} & \text{11.3646} & \text{10.4473} \\

\end{matrix}</math></center>

The median rank values ( <math>F({{T}_{i}})</math> ) can be found in rank tables or by using the Quick Statistical Reference in Weibull++ .

The <math>{{y}_{i}}</math> values were obtained from the 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> ).

Given the values in the table above, calculate <math>\widehat{a}</math> and <math>\widehat{b}</math> using Eqns. (aaln) and (bbln):

::<math>\begin{align}
& \widehat{b}= & \frac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,T_{i}^{\prime }{{y}_{i}}-(\underset{i=1}{\overset{14}{\mathop{\sum }}}\,T_{i}^{\prime })(\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{y}_{i}})/14}{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,T_{i}^{\prime 2}-{{(\underset{i=1}{\overset{14}{\mathop{\sum }}}\,T_{i}^{\prime })}^{2}}/14} \\
& & \\
& \widehat{b}= & \frac{10.4473-(49.2220)(0)/14}{183.1530-{{(49.2220)}^{2}}/14}
\end{align}</math>

:or:

::<math>\widehat{b}=1.0349</math>

:and:

::<math>\widehat{a}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}-\widehat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,T_{i}^{\prime }}{N}</math>

:or:

::<math>\widehat{a}=\frac{0}{14}-(1.0349)\frac{49.2220}{14}=-3.6386</math>

:Therefore, from Eqn. (bln):

::<math>{{\sigma }_{{{T}'}}}=\frac{1}{\widehat{b}}=\frac{1}{1.0349}=0.9663</math>

:and from Eqn. (aln):

::<math>{\mu }'=-\widehat{a}\cdot {{\sigma }_{{{T}'}}}=-(-3.6386)\cdot 0.9663</math>

:or:

::<math>{\mu }'=3.516</math>

The mean and the standard deviation of the lognormal distribution are obtained using Eqns. (mean) and (sdv):

::<math>\overline{T}=\mu ={{e}^{3.516+\tfrac{1}{2}{{0.9663}^{2}}}}=53.6707\text{ hours}</math>

:and:

::<math>{{\sigma }_{T}}=\sqrt{({{e}^{2\cdot 3.516+{{0.9663}^{2}}}})({{e}^{{{0.9663}^{2}}}}-1)}=66.69\text{ hours}</math>

The correlation coefficient can be estimated using Eqn. (RHOln):

::<math>\widehat{\rho }=0.9754</math>

The above example can be repeated using Weibull++ , using RRY.

[[Image:5folio.png|thumb|center|400px| ]]

The mean can be obtained from the QCP and both the mean and the standard deviation can be obtained from the Function Wizard.