Template:Normal distribution rank regression on Y - Revision history

Lisa Hacker: Redirected page to The Normal Distribution

2012-08-09T10:05:31Z

Redirected page to The Normal Distribution

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	[[~~Category: For Deletion~~]]		#REDIRECT [[The Normal Distribution]]

Lisa Hacker: Replaced content with 'Category: For Deletion'

2012-08-08T06:29:35Z

Replaced content with 'Category: For Deletion'

← Older revision		Revision as of 06:29, 8 August 2012
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	~~===Rank Regression on Y===~~		[[Category: For Deletion]]

	~~Performing 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 (regression analysis) was discussed in~~ [[~~Parameter Estimation~~]]~~, and the following equations for regression on Y were derived:~~

	~~::<math>\begin{align}\hat{a}= & \bar{b}-\hat{b}\bar{x} \\~~
	~~=& \frac{\sum_{i=1}^N y_{i}}{N}-\hat{b}\frac{\sum_{i=1}^{N}x_{i}}{N}\\~~
	~~\end{align}~~
	~~</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 the case of the normal distribution, the equations for <math>{{y}_{i}}</math> and <math>{{x}_{i}}</math> are:~~

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

	~~and:~~

	~~::<math>{{x}_{i}}={{t}_{i}}</math>~~


	where the values for <math>F({{T}_{i}})</math> are estimated from the median ranks. Once <math>\widehat{a}</math> and <math>\widehat{b}</math> are obtained, <math>\widehat{\sigma }</math> and <math>\widehat{\mu }</math> can easily be obtained from above equations.

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

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

Lisa Hacker: /* Rank Regression on Y */

2012-03-30T22:23:47Z

Rank Regression on Y

← Older revision		Revision as of 22:23, 30 March 2012
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	Performing 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 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 (regression analysis) was discussed in ~~Chapter~~ [[~~Parameter Estimation\|~~Parameter Estimation]] and the following equations for regression on Y were derived:		The least squares parameter estimation method (regression analysis) was discussed in [[Parameter Estimation]], and the following equations for regression on Y were derived:

	::<math>\begin{align}\hat{a}= & \bar{b}-\hat{b}\bar{x} \\		::<math>\begin{align}\hat{a}= & \bar{b}-\hat{b}\bar{x} \\

Harry Guo: /* Rank Regression on Y */

2012-02-10T23:32:20Z

Rank Regression on Y

← Older revision		Revision as of 23:32, 10 February 2012
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	where the values for <math>F({{T}_{i}})</math> are estimated from the median ranks. Once <math>\widehat{a}</math> and <math>\widehat{b}</math> are obtained, <math>\widehat{\sigma }</math> and <math>\widehat{\mu }</math> can easily be obtained from ~~Eqns. (an) and (bn)~~.		where the values for <math>F({{T}_{i}})</math> are estimated from the median ranks. Once <math>\widehat{a}</math> and <math>\widehat{b}</math> are obtained, <math>\widehat{\sigma }</math> and <math>\widehat{\mu }</math> can easily be obtained from above equations.

	'''The Correlation Coefficient'''		'''The Correlation Coefficient'''

Harry Guo at 18:41, 10 February 2012

2012-02-10T18:41:05Z

@@ Line 31: / Line 31: @@
 ::<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: Normal Distribution RRY}}

Harry Guo: /* Rank Regression on Y */

2012-02-10T18:26:08Z

Rank Regression on Y

← Older revision		Revision as of 18:26, 10 February 2012
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	In the case of the normal distribution, the equations for <math>{{y}_{i}}</math> and <math>{{x}_{i}}</math> are:		In the case of the normal distribution, the equations for <math>{{y}_{i}}</math> and <math>{{x}_{i}}</math> are:

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

	:and:		and:

	::<math>{{x}_{i}}={{T}_{i}}</math>		::<math>{{x}_{i}}={{t}_{i}}</math>


	where the values for <math>F({{T}_{i}})</math> are estimated from the median ranks. Once <math>\widehat{a}</math> and <math>\widehat{b}</math> are obtained, <math>\widehat{\sigma }</math> and <math>\widehat{\mu }</math> can easily be obtained from Eqns. (an) and (bn).		where the values for <math>F({{T}_{i}})</math> are estimated from the median ranks. Once <math>\widehat{a}</math> and <math>\widehat{b}</math> are obtained, <math>\widehat{\sigma }</math> and <math>\widehat{\mu }</math> can easily be obtained from Eqns. (an) and (bn).

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

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

Harry Guo at 18:25, 10 February 2012

2012-02-10T18:25:15Z

← Older revision		Revision as of 18:25, 10 February 2012
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	Performing 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 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 (regression analysis) was discussed in Chapter[[Parameter Estimation\|~~ChapParameter~~ Estimation]] and the following equations for regression on Y were derived:		The least squares parameter estimation method (regression analysis) was discussed in Chapter [[Parameter Estimation\|Parameter Estimation]] and the following equations for regression on Y were derived:

	::<math>\begin{align}\hat{a}= & \bar{b}-\hat{b}\bar{x} \\		::<math>\begin{align}\hat{a}= & \bar{b}-\hat{b}\bar{x} \\

Harry Guo: /* Rank Regression on Y */

2012-02-10T18:24:57Z

Rank Regression on Y

← Older revision		Revision as of 18:24, 10 February 2012
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	===Rank Regression on Y===		===Rank Regression on Y===

	Performing 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 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 (regression analysis) was discussed in Chapter 3 and the following equations for regression on Y were derived:		The least squares parameter estimation method (regression analysis) was discussed in Chapter[[Parameter Estimation\|ChapParameter Estimation]] and the following equations for regression on Y were derived:

	::<math>\begin{align}\hat{a}= & \bar{b}-\hat{b}\bar{x} \\		::<math>\begin{align}\hat{a}= & \bar{b}-\hat{b}\bar{x} \\
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	</math>		</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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	::<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===
	Fourteen units were reliability tested and the following life test data were obtained:		Fourteen units were reliability tested and the following life test data were obtained:

Nicolette Young: /* Rank Regression on Y */

2012-01-04T17:40:57Z

Rank Regression on Y

← Older revision		Revision as of 17:40, 4 January 2012
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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===
			Fourteen units were reliability tested and the following life test data were obtained:


			{\|align="center" border=1 cellspacing=1
			\|-
			\|colspan="2" style="text-align:center"\|Table 8.1 -The 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 normal distribution, estimate the parameters and determine the correlation coefficient, <math>\rho </math> , using rank regression on Y.

			====Solution to Example 2====
			Construct a table like the one shown next.

			<center><math>\overset{{}}{\mathop{\text{Table 8}\text{.2 - 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 8.2, calculate <math>\widehat{a}</math> and <math>\widehat{b}</math> using Eqns. (aan) and (bbn):

			::<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, from Eqn. (bn):

			::<math>\widehat{\sigma}=\frac{1}{\hat{b}}=\frac{1}{0.02982}=33.5367</math>

			:and from Eqn. (an):

			::<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 Eqn. (RHOn):

			::<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 Table 8.1.
			:• 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:weibullfolio1.png\|thumb\|center\|400px\| ]]

			The probability plot is shown next.

			[[Image:weibullfolioplot1st.png\|thumb\|center\|400px\| ]]

Nicolette Young: Created page with '===Rank Regression on Y=== Performing 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 deviat…'

2012-01-04T17:40:22Z

Created page with '===Rank Regression on Y=== Performing 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 deviat…'

New page

===Rank Regression on Y===
Performing 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 (regression analysis) was discussed in Chapter 3 and the following equations for regression on Y were derived:

::<math>\begin{align}\hat{a}= & \bar{b}-\hat{b}\bar{x} \\
=& \frac{\sum_{i=1}^N y_{i}}{N}-\hat{b}\frac{\sum_{i=1}^{N}x_{i}}{N}\\
\end{align}
</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 the case of the normal distribution, the equations for <math>{{y}_{i}}</math> and <math>{{x}_{i}}</math> are:

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

:and:

::<math>{{x}_{i}}={{T}_{i}}</math>

where the values for <math>F({{T}_{i}})</math> are estimated from the median ranks. Once <math>\widehat{a}</math> and <math>\widehat{b}</math> are obtained, <math>\widehat{\sigma }</math> and <math>\widehat{\mu }</math> can easily be obtained from Eqns. (an) and (bn).

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