Template:Weibull parameters Rank Regression on Y - Revision history

Lisa Hacker: Redirected page to The Weibull Distribution

2012-08-08T08:32:24Z

Redirected page to The Weibull Distribution

← Older revision		Revision as of 08:32, 8 August 2012
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	~~=== Rank Regression on Y ===~~		#REDIRECT [[The Weibull Distribution]]
	Performing rank regression on Y requires that a straight line mathematically 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. This is in essence the same methodology as the probability plotting method, except that we use the principle of least squares to determine the line through the points, as opposed to just eyeballing it. The first step is to bring our function into a linear form. For the two-parameter Weibull distribution, the (cumulative density function) is:

	~~::<math> F(t)=1-e^{-\left( \frac{t}{\eta }\right) ^{\beta }} </math>~~

	~~Taking the natural logarithm of both sides of the equation yields:~~

	~~::<math>\ln[ 1-F(t)] =-( \frac{t}{\eta }) ^{\beta } </math>~~

	~~::<math> \ln{ -\ln[ 1-F(t)]} =\beta \ln ( \frac{t}{ \eta }) </math>~~

	~~or:~~

	~~::<math> \ln \{ -\ln~~[ ~~1-F(t)]\} =-\beta \ln (\eta )+\beta \ln (t) </math>~~

	~~Now let:~~

	~~::<math> y = \ln \{ -\ln~~[ ~~1-F(t)]\} </math>~~

	~~::<math> a = − βln(\eta) </math>~~

	~~and:~~

	~~::<math> b= \beta</math>~~

	~~which results in the linear equation of:~~

	~~::<math>y=a+bx</math>~~

	The ~~least squares parameter estimation method (also known as regression analysis) was discussed in [[Parameter Estimation~~]]~~, and the following equations for regression on Y were derived:~~

	~~::<math> \hat{a}=\frac{\sum\limits_{i=1}^{N}y_{i}}{N}-\hat{b}\frac{ \sum\limits_{i=1}^{N}x_{i}}{N}=\bar{y}-\hat{b}\bar{x} </math>~~

	~~and:~~

	::<math> \hat{b}={\frac{\sum\limits_{i=1}^{N}x_{i}y_{i}-\frac{\sum \limits_{i=1}^{N}x_{i}\sum\limits_{i=1}^{N}y_{i}}{N}}{\sum \limits_{i=1}^{N}x_{i}^{2}-\frac{\left( \sum\limits_{i=1}^{N}x_{i}\right) ^{2}}{N}}} </math>

	~~In this case the equations for <span class="texhtml">''y''<sub>''i''</sub></span> and <span class="texhtml">''x''<sub>''i''</sub></span> are:~~

	~~::<math> y_{i}=\ln \left\{ -\ln [1-F(t_{i})]\right\} , </math>~~

	~~and:~~

	~~::<span class="texhtml">''x''<sub>''i''</sub> = ln(''t''<sub>''i''</sub>).</span>~~

	~~The <math> F(t_{i})s </math> are estimated from the median ranks.~~

	~~Once <math> \hat{a} </math> and <math> \hat{b} </math> are obtained, then <math> \hat{\beta } </math> and <math> \hat{\eta } </math> can easily be obtained from previous equations.~~


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

	~~The correlation coefficient is defined as follows:~~

	~~::<math> \rho ={\frac{\sigma _{xy}}{\sigma _{x}\sigma _{y}}} </math>~~

	where, <span class="texhtml">σ<sub>''x'' ''y''</sub> = </span>covariance of and , <span class="texhtml">σ<sub>''x''</sub> = </span>standard deviation of , and <span class="texhtml">σ<sub>''y''</sub> = </span>standard deviation of . The estimator of <span class="texhtml">ρ</span> is the sample ''correlation coefficient'', <math> \hat{\rho} </math>, given by:

	::<math> \hat{\rho}=\frac{\sum\limits_{i=1}^{N}(x_{i}-\overline{x})(y_{i}-\overline{y} )}{\sqrt{\sum\limits_{i=1}^{N}(x_{i}-\overline{x})^{2}\cdot \sum\limits_{i=1}^{N}(y_{i}-\overline{y})^{2}}}</math>
	~~<br>~~

	~~'''Example 3: '''~~
	~~{{Example: 2P Weibull Distribution RRY}}~~

Lisa Hacker: /* Rank Regression on Y */

2012-03-30T22:22:23Z

Rank Regression on Y

← Older revision		Revision as of 22:22, 30 March 2012
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	::<math>y=a+bx</math>		::<math>y=a+bx</math>

	The least squares parameter estimation method (also known as regression analysis) was discussed in ~~Chapter~~ [[Parameter Estimation]] and the following equations for regression on Y were derived:		The least squares parameter estimation method (also known as regression analysis) was discussed in [[Parameter Estimation]], and the following equations for regression on Y were derived:

	::<math> \hat{a}=\frac{\sum\limits_{i=1}^{N}y_{i}}{N}-\hat{b}\frac{ \sum\limits_{i=1}^{N}x_{i}}{N}=\bar{y}-\hat{b}\bar{x} </math>		::<math> \hat{a}=\frac{\sum\limits_{i=1}^{N}y_{i}}{N}-\hat{b}\frac{ \sum\limits_{i=1}^{N}x_{i}}{N}=\bar{y}-\hat{b}\bar{x} </math>

Harry Guo: /* Rank Regression on Y */

2012-02-10T22:57:28Z

Rank Regression on Y

← Older revision		Revision as of 22:57, 10 February 2012
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	::<math>y=a+bx</math>		::<math>y=a+bx</math>

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

	::<math> \hat{a}=\frac{\sum\limits_{i=1}^{N}y_{i}}{N}-\hat{b}\frac{ \sum\limits_{i=1}^{N}x_{i}}{N}=\bar{y}-\hat{b}\bar{x} </math>		::<math> \hat{a}=\frac{\sum\limits_{i=1}^{N}y_{i}}{N}-\hat{b}\frac{ \sum\limits_{i=1}^{N}x_{i}}{N}=\bar{y}-\hat{b}\bar{x} </math>

Harry Guo: /* Rank Regression on Y */

2012-02-09T22:21:40Z

Rank Regression on Y

← Older revision		Revision as of 22:21, 9 February 2012
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	Performing rank regression on Y requires that a straight line mathematically 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. This is in essence the same methodology as the probability plotting method, except that we use the principle of least squares to determine the line through the points, as opposed to just eyeballing it. The first step is to bring our function into a linear form. For the two-parameter Weibull distribution, the (cumulative density function) is:		Performing rank regression on Y requires that a straight line mathematically 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. This is in essence the same methodology as the probability plotting method, except that we use the principle of least squares to determine the line through the points, as opposed to just eyeballing it. The first step is to bring our function into a linear form. For the two-parameter Weibull distribution, the (cumulative density function) is:

	::<math> F(T)=1-e^{-\left( \frac{T}{\eta }\right) ^{\beta }} ~~(Fw)~~ </math>		::<math> F(t)=1-e^{-\left( \frac{t}{\eta }\right) ^{\beta }} </math>

	Taking the natural logarithm of both sides of the equation yields:		Taking the natural logarithm of both sides of the equation yields:

	::<math>\ln[ 1-F(T)] =-( \frac{T}{\eta }) ^{\beta } </math>		::<math>\ln[ 1-F(t)] =-( \frac{t}{\eta }) ^{\beta } </math>

	::<math> \ln{ -\ln[ 1-F(T)]} =\beta \ln ( \frac{T}{ \eta }) </math>		::<math> \ln{ -\ln[ 1-F(t)]} =\beta \ln ( \frac{t}{ \eta }) </math>

	:or:		or:

	::<math> \ln \{ -\ln[ 1-F(T)]\} =-\beta \ln (\eta )+\beta \ln (T) ~~EQNREF logw~~ </math>		::<math> \ln \{ -\ln[ 1-F(t)]\} =-\beta \ln (\eta )+\beta \ln (t) </math>

	:Now let:		Now let:

	::<math> y = \ln \{ -\ln[ 1-F(T)]\} ~~( yw )~~</math>		::<math> y = \ln \{ -\ln[ 1-F(t)]\} </math>

	::<math> a = − βln(\eta) </math> ~~(aw)~~		::<math> a = − βln(\eta) </math>

	:and:		and:

	::<math> b= \beta</math> ~~( bw )~~		::<math> b= \beta</math>

	which results in the linear equation of:		which results in the linear equation of:
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	::<math>y=a+bx</math>		::<math>y=a+bx</math>

	The least squares parameter estimation method (also known as regression analysis) was discussed in Chapter 3 and the following equations for regression on Y were derived ~~in Appendix A~~:		The least squares parameter estimation method (also known as regression analysis) was discussed in Chapter 3 and the following equations for regression on Y were derived:

	::<math> \hat{a}=\frac{\sum\limits_{i=1}^{N}y_{i}}{N}-\hat{b}\frac{ \sum\limits_{i=1}^{N}x_{i}}{N}=\bar{y}-\hat{b}\bar{x} ~~EQNREF aaw~~ </math>		::<math> \hat{a}=\frac{\sum\limits_{i=1}^{N}y_{i}}{N}-\hat{b}\frac{ \sum\limits_{i=1}^{N}x_{i}}{N}=\bar{y}-\hat{b}\bar{x} </math>

	:and:		and:

	::<math> \hat{b}={\frac{\sum\limits_{i=1}^{N}x_{i}y_{i}-\frac{\sum \limits_{i=1}^{N}x_{i}\sum\limits_{i=1}^{N}y_{i}}{N}}{\sum \limits_{i=1}^{N}x_{i}^{2}-\frac{\left( \sum\limits_{i=1}^{N}x_{i}\right) ^{2}}{N}}} ~~EQNREF bbw~~ </math>		::<math> \hat{b}={\frac{\sum\limits_{i=1}^{N}x_{i}y_{i}-\frac{\sum \limits_{i=1}^{N}x_{i}\sum\limits_{i=1}^{N}y_{i}}{N}}{\sum \limits_{i=1}^{N}x_{i}^{2}-\frac{\left( \sum\limits_{i=1}^{N}x_{i}\right) ^{2}}{N}}} </math>

	In this case the equations for <span class="texhtml">''y''<sub>''i''</sub></span> and <span class="texhtml">''x''<sub>''i''</sub></span> are:		In this case the equations for <span class="texhtml">''y''<sub>''i''</sub></span> and <span class="texhtml">''x''<sub>''i''</sub></span> are:

	::<math> y_{i}=\ln \left\{ -\ln [1-F(T_{i})]\right\} , </math>		::<math> y_{i}=\ln \left\{ -\ln [1-F(t_{i})]\right\} , </math>

	:and:		and:

	::<span class="texhtml">''x''<sub>''i''</sub> = ln(''T''<sub>''i''</sub>).</span>		::<span class="texhtml">''x''<sub>''i''</sub> = ln(''t''<sub>''i''</sub>).</span>

	The <math> F(T_{i})~~^{\prime }~~s </math> are estimated from the median ranks.		The <math> F(t_{i})s </math> are estimated from the median ranks.

			Once <math> \hat{a} </math> and <math> \hat{b} </math> are obtained, then <math> \hat{\beta } </math> and <math> \hat{\eta } </math> can easily be obtained from previous equations.

	~~Once <math> \hat{a} </math> and <math> \hat{b} </math> are obtained, then <math> \hat{\beta } </math> and <math> \hat{\eta } </math> can easily be obtained from Eqns. (EQNREF aw ) and (\ref {bw}).~~

	'''The Correlation Coefficient'''		'''The Correlation Coefficient'''
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	where, <span class="texhtml">σ<sub>''x'' ''y''</sub> = </span>covariance of and , <span class="texhtml">σ<sub>''x''</sub> = </span>standard deviation of , and <span class="texhtml">σ<sub>''y''</sub> = </span>standard deviation of . The estimator of <span class="texhtml">ρ</span> is the sample ''correlation coefficient'', <math> \hat{\rho} </math>, given by:		where, <span class="texhtml">σ<sub>''x'' ''y''</sub> = </span>covariance of and , <span class="texhtml">σ<sub>''x''</sub> = </span>standard deviation of , and <span class="texhtml">σ<sub>''y''</sub> = </span>standard deviation of . The estimator of <span class="texhtml">ρ</span> is the sample ''correlation coefficient'', <math> \hat{\rho} </math>, given by:

	::<math> \hat{\rho}=\frac{\sum\limits_{i=1}^{N}(x_{i}-\overline{x})(y_{i}-\overline{y} )}{\sqrt{\sum\limits_{i=1}^{N}(x_{i}-\overline{x})^{2}\cdot \sum\limits_{i=1}^{N}(y_{i}-\overline{y})^{2}}} ~~EQNREF RHOw~~ </math>		::<math> \hat{\rho}=\frac{\sum\limits_{i=1}^{N}(x_{i}-\overline{x})(y_{i}-\overline{y} )}{\sqrt{\sum\limits_{i=1}^{N}(x_{i}-\overline{x})^{2}\cdot \sum\limits_{i=1}^{N}(y_{i}-\overline{y})^{2}}}</math>
	<br>		<br>

	'''Example 3: '''		'''Example 3: '''
	{{Example: 2P Weibull Distribution RRY}}		{{Example: 2P Weibull Distribution RRY}}

Harry Guo: /* The Correlation Coefficient */

2012-02-09T21:04:14Z

The Correlation Coefficient

← Older revision		Revision as of 21:04, 9 February 2012
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	Once <math> \hat{a} </math> and <math> \hat{b} </math> are obtained, then <math> \hat{\beta } </math> and <math> \hat{\eta } </math> can easily be obtained from Eqns. (EQNREF aw ) and (\ref {bw}).		Once <math> \hat{a} </math> and <math> \hat{b} </math> are obtained, then <math> \hat{\beta } </math> and <math> \hat{\eta } </math> can easily be obtained from Eqns. (EQNREF aw ) and (\ref {bw}).

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

	The correlation coefficient is defined as follows:		The correlation coefficient is defined as follows:

Harry Guo: /* Rank Regression on Y */

2012-02-09T19:08:02Z

Rank Regression on Y

← Older revision		Revision as of 19:08, 9 February 2012
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	'''Example 3: '''		'''Example 3: '''
	{{Example:2P Weibull Distribution RRY}}		{{Example: 2P Weibull Distribution RRY}}

Harry Guo: /* Rank Regression on Y */

2012-02-09T19:06:53Z

Rank Regression on Y

← Older revision		Revision as of 19:06, 9 February 2012
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	<br>		<br>

			'''Example 3: '''
	{{Example:2P Weibull Distribution RRY}}		{{Example:2P Weibull Distribution RRY}}

Harry Guo: /* Rank Regression on Y */

2012-02-09T17:32:38Z

Rank Regression on Y

← Older revision		Revision as of 17:32, 9 February 2012
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	<br>		<br>

	{{Example:2P Weibull Distribution}}		{{Example:2P Weibull Distribution RRY}}

Nicolette Young: /* Example 3 */

2012-02-08T19:49:43Z

Example 3

← Older revision		Revision as of 19:49, 8 February 2012
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	<br>		<br>

	~~==== Example 3 ====~~		{{Example:2P Weibull Distribution}}
	Consider the data in Example 1, where six units were tested to failure and the following failure times were recorded: 16, 34, 53, 75, 93 and 120 hours. Estimate the parameters and the correlation coefficient using rank regression on Y, assuming that the data follow the two-parameter Weibull distribution.

	~~===== Solution to Example 3 =====~~
	~~Construct a table as shown below.~~

	{~~\|align="center" border=1 cellspacing=1~~
	\|-
	~~\|colspan="8" style="text-align:center"\| Table 6.1 - Least Squares Analysis~~
	\|-
	~~!<math>N</math>~~
	~~!<math>T_~~{~~i}</math>~~
	~~!<math>ln(T_{i})</math>~~
	~~!<math>F(T_i)</math>~~
	~~!<math>y_{i}</math>~~
	~~!<math>(ln{T_i})^2</math>~~
	~~!<math>{y_i}^2</math>~~
	~~!<math>(ln{T_i})y_i</math>~~
	\|-
	~~\|1 \|\|16\|\|2.7726\|\|0.1091\|\|-2.1583\|\|7.6873\|\|4.6582\|\|-5.9840~~
	\|-
	~~\|2 \|\|34\|\|3.5264\|\|0.2645\|\|-1.1802\|\|12.4352\|\|1.393\|\|-4.1620~~
	\|-
	~~\|3 \|\|53\|\|3.9703\|\|0.4214\|\|-0.6030\|\|15.7632\|\|0.3637\|\|-2.3943~~
	\|-
	~~\|4 \|\|75\|\|4.3175\|\|0.5786\|\|-0.146\|\|18.6407\|\|0.0213\|\|-0.6303~~
	\|-
	~~\|5 \|\|93\|\|4.5326\|\|0.7355\|\|0.2851\|\|20.5445\|\|0.0813\|\|1.2923~~
	\|-
	~~\|6 \|\|120\|\|4.7875\|\|0.8909\|\|0.7955\|\|22.9201\|\|0.6328\|\|3.8083~~
	\|-
	~~\|<math>\sum</math>\|\| \|\|23.9068\|\| \|\|-3.007\|\|97.9909\|\|7.1502\|\|-8.0699~~
	\|}


	~~Utilizing the values from Table 6.1, calculate <math> \hat{a} </math> and <math> \hat{b} </math> using Eqns. (EQNREF aaw ) and (EQNREF bbw ):~~
	::<math> \hat{b} =\frac{\sum\limits_{i=1}^{6}(\ln T_{i})y_{i}-(\sum\limits_{i=1}^{6}\ln T_{i})(\sum\limits_{i=1}^{6}y_{i})/6}{ \sum\limits_{i=1}^{6}(\ln T_{i})^{2}-(\sum\limits_{i=1}^{6}\ln T_{i})^{2}/6}
	~~</math>~~

	~~::<math> \hat{b}=\frac{-8.0699-(23.9068)(-3.0070)/6}{97.9909-(23.9068)^{2}/6} </math>~~

	~~:or~~

	~~::<math> \hat{b}=1.4301 </math>~~

	~~:and:~~

	~~::<math> \hat{a}=\overline{y}-\hat{b}\overline{T}=\frac{\sum \limits_{i=1}^{N}y_{i}}{N}-\hat{b}\frac{\sum\limits_{i=1}^{N}\ln T_{i}}{N } </math>~~

	~~:or:~~

	~~::<math> \hat{a}=\frac{(-3.0070)}{6}-(1.4301)\frac{23.9068}{6}=-6.19935 </math>~~

	~~Therefore, from Eqn. (EQNREF bw ):~~

	~~::<math> \hat{\beta }=\hat{b}=1.4301 </math>~~

	~~and from Eqn. (EQNREF aw ):~~

	~~::<math> \hat{\eta }=e^{-\frac{\hat{a}}{\hat{b}}}=e^{-\frac{(-6.19935)}{ 1.4301}} </math>~~

	~~:or:~~

	~~::<math> \hat{\eta }=76.318\text{ hr} </math>~~

	~~The correlation coefficient can be estimated using Eqn. (EQNREF RHOw ):~~

	~~::<math> \hat{\rho }=0.9956 </math>~~

	~~The above example can be repeated using Weibull++. Start Weibull++ and create a new ''Data Folio''.~~

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

	~~Select the '' Times-to-failure data'' option.~~

	~~[[Image:times--to-failure.png\|thumb\|center\|400px\| The effect of the Weibull shape parameter on the <math>pdf</math>. ]]~~

	~~Enter the times-to-failure in the datasheet (ignore the Subset ID column), as shown next. The times-to-failure need not be sorted, Weibull++ will automatically sort the data.~~
	~~<br>~~
	~~[[Image:weibullfolio16.png\|thumb\|center\|400px\| ]]~~

	Select the desired method of analysis. Note that we are assuming that the underlying distribution is the Weibull, so make sure that the Weibull distribution is selected. Under ''Parameters/Type'' on the Main page, select ''2''.
	~~<br>~~
	~~[[Image:parameterweibull.png\|thumb\|center\|400px\|]]~~

	Also, so that you get the same results as this example, switch to the ''Analysis'' page and make sure you are using the ''Rank Regression on Y (RRY)'' calculation method with this example, as shown next.

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

	Note that this can also be done from the ''Main'' page by clicking the left bottom box under the Results area. Each time you click that box you will see the method switch between MLE, RRX, and RRY. Click the ''Calculate'' icon,

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

	~~or select ''Calculate'' from the ''Data'' menu. The results will appear in the Data Folio's ''Results area''. The next figure shows the results for this example.~~

	~~[[Image:weibullfolio16calculate.png\|thumb\|center\|400px\| ]]~~

	~~<br>~~
	~~You can now plot the results by clicking the ''Plot'' icon,~~

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

	~~or by selecting ''Plot Probability'' from the ''Data'' menu.~~

	~~The Weibull probability plot for these data is shown next.~~

	The confidence bounds, as determined from the Fisher matrix, can also be plotted. Select ''Confidence Bounds'' from the ''Plot'' menu, choose ''Two-Sided'' under ''Sides,'' ''Reliability (Type II)'' under ''Type'' and enter ''90'' for the ''Confidence level.

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

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

	~~The plot will appear as follows,~~

	~~[[Image~~:~~weibullfolio16plot.png\|thumb\|center\|400px\| ]]~~

	~~If desired, the~~ Weibull ~~<math>pdf</math> representing these data can be written as:~~

	~~::<math> f(T)={\frac{\beta }{\eta }~~}~~\left( {\frac{T~~}~~{\eta }}\right) ^{\beta -1}e^{-\left( {\frac{T}{\eta }}\right) ^{\beta }} </math>~~

	~~:or:~~

	::<math> f(T)={\frac{1.4302}{76.317}}\left( {\frac{T}{76.317}}\right) ^{0.4302}e^{-\left( {\frac{T}{76.317}}\right) ^{1.4302}} </math> You can also plot the Weibull by selecting ''Pdf Plot'' from the ''Plot Type'' drop-down menu on the control panel to the right of the plot area.

	~~[[Image:weibullpdffolio16.png\|thumb\|center\|400px]]~~

	~~From this point on, different results, reports and plots can be obtained.~~

Nicolette Young: /* Example 3 */

2012-02-08T19:48:57Z

Example 3

← Older revision		Revision as of 19:48, 8 February 2012
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	<br>		<br>

	===== Example 3 =====		==== Example 3 ====
	Consider the data in Example 1, where six units were tested to failure and the following failure times were recorded: 16, 34, 53, 75, 93 and 120 hours. Estimate the parameters and the correlation coefficient using rank regression on Y, assuming that the data follow the two-parameter Weibull distribution.		Consider the data in Example 1, where six units were tested to failure and the following failure times were recorded: 16, 34, 53, 75, 93 and 120 hours. Estimate the parameters and the correlation coefficient using rank regression on Y, assuming that the data follow the two-parameter Weibull distribution.

	===== Solution to Example 3 =====		===== Solution to Example 3 =====