Template:Normal distribution estimation of the parameters - Revision history

Lisa Hacker: Redirected page to The Normal Distribution

2012-08-03T09:06:04Z

Redirected page to The Normal Distribution

← Older revision		Revision as of 09:06, 3 August 2012
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	~~==Estimation of the Parameters==~~		#REDIRECT [[The_Normal_Distribution]]

	~~{{normal distribution probability plotting}}~~

	~~{{normal distribution rank regression on Y}}~~

	~~{{normal distribution rank regression on X}}~~

	~~{{normal distribution maximum likelihood estimation}}~~

Nicolette Young: Replaced content with '==Estimation of the Parameters== {{normal distribution probability plotting}} {{normal distribution rank regression on Y}} {{normal distribution rank regression on X}} {{…'

2012-01-04T17:42:51Z

Replaced content with '==Estimation of the Parameters== {{normal distribution probability plotting}} {{normal distribution rank regression on Y}} {{normal distribution rank regression on X}} {{…'

← Older revision		Revision as of 17:42, 4 January 2012
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	{{normal distribution rank regression on X}}		{{normal distribution rank regression on X}}

	~~===Maximum Likelihood Estimation===~~		{{normal distribution maximum likelihood estimation}}
	As it was outlined in Chapter 3, maximum likelihood estimation works by developing a likelihood function based on the available data and finding the values of the parameter estimates that maximize the likelihood function. This can be achieved by using iterative methods to determine the parameter estimate values that maximize the likelihood function. This can be rather difficult and time-consuming, particularly when dealing with the three-parameter distribution. Another method of finding the parameter estimates involves taking the partial derivatives of the likelihood function with respect to the parameters, setting the resulting equations equal to zero, and solving simultaneously to determine the values of the parameter estimates. The log-likelihood functions and associated partial derivatives used to determine maximum likelihood estimates for the normal distribution are covered in Appendix C.

	~~===Special Note About Bias===~~
	Estimators (i.e. parameter estimates) have properties such as unbiasedness, minimum variance, sufficiency, consistency, squared error constancy, efficiency and completeness [7][5]. Numerous books and papers deal with these properties and this coverage is beyond the scope of this reference.

	However, we would like to briefly address one of these properties, unbiasedness. An estimator is said to be unbiased if the estimator <math>\widehat{\theta }=d({{X}_{1,}}{{X}_{2,}}...,{{X}_{n)}}</math> satisfies the condition <math>E\left[ \widehat{\theta } \right]</math> <math>=\theta </math> for all <math>\theta \in \Omega .</math>
	~~Note that <math>E\left[ X \right]</math> denotes the expected value of X and is defined (for continuous distributions) by:~~

	~~::<math>\begin{align}~~
	~~E\left[ X \right]= \int_~~{~~\varpi }x\cdot f(x)dx \\~~
	~~X\in & \varpi .~~
	~~\end~~{~~align}</math>~~


	~~It can be shown [7][5] that the MLE estimator for the mean of the~~ normal ~~(and lognormal)~~ distribution does satisfy the unbiasedness criteria, or <math>E\left[ \widehat{\mu } \right]</math> <math>=\mu .</math> The same is not true for the estimate of the variance <math>\hat{\sigma }_{T}^{2}</math> . The maximum likelihood ~~estimate for the variance for the normal distribution is given by:~~

	~~::<math>\hat{\sigma }_{T}^{2}=\frac{1}{N}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{T}_{i}}-\bar{T})}^{2}}</math>~~

	~~with a standard deviation of:~~

	~~::<math>{{\hat{\sigma }}_{T}}=\sqrt{\frac{1}{N}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{T}_{i}}-\bar{T})}^{2}}}</math>~~


	~~These estimates, however, have been shown to be biased. It can be shown [7][5] that the unbiased estimate of the variance and standard deviation for complete data is given by:~~


	~~::<math>\begin{align}~~
	\hat{\sigma }_{T}^{2}= & \left[ \frac{N}{N-1} \right]\cdot \left[ \frac{1}{N}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{T}_{i}}-\bar{T})}^{2}} \right]=\frac{1}{N-1}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{T}_{i}}-\bar{T})}^{2}} \\
	~~{{{\hat{\sigma }}}_{T~~}}~~= & \sqrt{\left[ \frac{N}{N-1} \right]\cdot \left[ \frac{1}{N}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{T}_{i}}-\bar{T})}^{2}} \right]} \\~~
	~~= & \sqrt{\frac{1}{N-1}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{T}_{i}}-\bar{T})}^{2}}}~~
	~~\end{align}</math>~~


	~~Note that for larger values of <math>N</math> , <math>\sqrt{\left[ N/(N-1) \right]}</math> tends to 1.~~

	~~Weibull++ by default returns the standard deviation as defined by Eqn. (NormSt2).~~
	~~The Use Unbiased Std on Normal Data option in the User Setup under the Calculations tab allows biasing to be considered when estimating the parameters.~~

	When this option is selected, Weibull++ returns the standard deviation as defined by Eqn. (NormSt2). This is only true for complete data sets. For all other data types, Weibull++ by default returns the standard deviation as defined by Eqn. (normbias2) regardless of the selection status of this option. The next figure shows this setting in Weibull++.


	~~<math></math>~~
	~~[[Image:ldachp8fig3.gif\|thumb\|center\|300px\| ]]~~

Nicolette Young at 17:41, 4 January 2012

2012-01-04T17:41:54Z

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 {{normal distribution rank regression on Y}}
-===Rank Regression on X===
+{{normal distribution rank regression on X}}
 ===Maximum Likelihood Estimation===

Nicolette Young: /* Example 2 */

2012-01-04T17:40:39Z

Example 2

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	{{normal distribution rank regression on Y}}		{{normal distribution rank regression on Y}}

	~~===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\| ]]~~

	===Rank Regression on X===		===Rank Regression on X===

Nicolette Young: /* Rank Regression on Y */

2012-01-04T17:40:13Z

Rank Regression on Y

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 {{normal distribution probability plotting}}
-===Rank Regression on Y===
+{{normal distribution rank regression on Y}}
 ===Example 2===

Nicolette Young: /* Example 1 */

2012-01-04T17:38:04Z

Example 1

← Older revision		Revision as of 17:38, 4 January 2012
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	{{normal distribution probability plotting}}		{{normal distribution probability plotting}}

	~~=== Example 1===~~
	Seven units are put on a life test and run until failure. The failure times are 85, 90, 95, 100, 105, 110, and 115 hours. Assuming a normal distribution, estimate the parameters using probability plotting.

	~~==== Solution to Example 1====~~
	In order to plot the points for the probability plot, the appropriate unreliability estimate values must be obtained. These will be estimated through the use of median ranks, which can be obtained from statistical tables or the Quick Statistical Reference in Weibull++. The following table shows the times-to-failure and the appropriate median rank values for this example:


	~~<center><math>\begin{matrix}~~
	~~\text{Time-to-} & \text{Median} \\~~
	~~\text{Failure (hr)} & \text{Rank ( }\!\!%\!\!\text{ )} \\~~
	~~\text{85} & \text{ 9}\text{.43 }\!\!%\!\!\text{ } \\~~
	~~\text{90} & \text{22}\text{.85 }\!\!%\!\!\text{ } \\~~
	~~\text{95} & \text{36}\text{.41 }\!\!%\!\!\text{ } \\~~
	~~\text{100} & \text{50}\text{.00 }\!\!%\!\!\text{ } \\~~
	~~\text{105} & \text{63}\text{.59 }\!\!%\!\!\text{ } \\~~
	~~\text{110} & \text{77}\text{.15 }\!\!%\!\!\text{ } \\~~
	~~\text{115} & \text{90}\text{.57 }\!\!%\!\!\text{ } \\~~
	~~\end{matrix}</math></center>~~



	~~These points can now be plotted on normal probability plotting paper as shown in the next figure.~~

	~~[[Image:chp8PP.gif\|thumb\|center\|300px\| ]]~~


	Draw the best possible line through the plot points. The time values where this line intersects the 15.85%, 50%, and 84.15% unreliability values should be projected down to the abscissa, as shown in the following plot.

	~~[[Image:chp8PP2.gif\|thumb\|center\|300px\| ]]~~


	The estimate of <math>\mu </math> is determined from the time value at the 50% unreliability level, which in this case is 100 hours. The value of the estimator of <math>\sigma </math> is determined by Eqn. (sigplot):

	~~::<math>\begin{align}~~
	~~\widehat{\sigma }= & \frac{t(Q=84.15%)-t(Q=15.85%)}{2} \\~~
	~~\widehat{\sigma }= & \frac{112-88}{2}=\frac{24}{2} \\~~
	~~\widehat{\sigma }= & 12\text{ hours}~~
	~~\end{align}</math>~~

	Alternately, <math>\widehat{\sigma }</math> could be determined by measuring the distance from <math>t(Q=15.85%)</math> to <math>t(Q=50%)</math> , or <math>t(Q=50%)</math> to <math>t(Q=84.15%)</math> , as either of these two distances is equal to the value of one standard deviation.

	===Rank Regression on Y===		===Rank Regression on Y===

Nicolette Young: /* Probability Plotting */

2012-01-04T17:37:44Z

Probability Plotting

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 ==Estimation of the Parameters==
-====Probability Plotting====
+{{normal distribution probability plotting}}
 ===  Example 1===

Nicolette Young: Created page with '==Estimation of the Parameters== ====Probability Plotting==== As described before, probability plotting involves plotting the failure times and associated unreliability estimate…'

2012-01-04T17:37:00Z

Created page with '==Estimation of the Parameters== ====Probability Plotting==== As described before, probability plotting involves plotting the failure times and associated unreliability estimate…'

New page

==Estimation of the Parameters==

====Probability Plotting====
As described before, probability plotting involves plotting the failure times and associated unreliability estimates on specially constructed probability plotting paper. The form of this paper is based on a linearization of the <math>cdf</math> of the specific distribution. For the normal distribution, the cumulative density function can be written as:

::<math>F(T)=\Phi \left( \frac{T-\mu }{{{\sigma }_{T}}} \right)</math>

:or:

::<math>{{\Phi }^{-1}}\left[ F(T) \right]=-\frac{\mu}{\sigma}+\frac{1}{\sigma}T</math>

:where:

::<math>\Phi (x)=\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{x}{{e}^{-\tfrac{{{t}^{2}}}{2}}}dt</math>

:Now, let:

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

::<math>a=-\frac{\mu }{\sigma }</math>

:and:

::<math>b=\frac{1}{\sigma }</math>

which results in the linear equation of:

::<math>y=a+bT</math>

The normal probability paper resulting from this linearized <math>cdf</math> function is shown next.

[[Image:normalPP.gif|thumb|center|300px| ]]

Since the normal distribution is symmetrical, the area under the <math>pdf</math> curve from <math>-\infty </math> to <math>\mu </math> is <math>0.5</math> , as is the area from <math>\mu </math> to <math>+\infty </math> . Consequently, the value of <math>\mu </math> is said to be the point where <math>R(t)=Q(t)=50%</math> . This means that the estimate of <math>\mu </math> can be read from the point where the plotted line crosses the 50% unreliability line.

To determine the value of <math>\sigma </math> from the probability plot, it is first necessary to understand that the area under the <math>pdf</math> curve that lies between one standard deviation in either direction from the mean (or two standard deviations total) represents 68.3% of the area under the curve. This is represented graphically in the following figure.

[[Image:68.3.gif|thumb|center|300px| ]]

Consequently, the interval between <math>Q(t)=84.15%</math> and <math>Q(t)=15.85%</math> represents two standard deviations, since this is an interval of 68.3% ( <math>84.15-15.85=68.3</math> ), and is centered on the mean at 50%. As a result, the standard deviation can be estimated from:

::<math>\widehat{\sigma }=\frac{t(Q=84.15%)-t(Q=15.85%)}{2}</math>

That is: the value of <math>\widehat{\sigma }</math> is obtained by subtracting the time value where the plotted line crosses the 84.15% unreliability line from the time value where the plotted line crosses the 15.85% unreliability line and dividing the result by two. This process is illustrated in the following example.

=== Example 1===
Seven units are put on a life test and run until failure. The failure times are 85, 90, 95, 100, 105, 110, and 115 hours. Assuming a normal distribution, estimate the parameters using probability plotting.

==== Solution to Example 1====
In order to plot the points for the probability plot, the appropriate unreliability estimate values must be obtained. These will be estimated through the use of median ranks, which can be obtained from statistical tables or the Quick Statistical Reference in Weibull++. The following table shows the times-to-failure and the appropriate median rank values for this example:

<center><math>\begin{matrix}
\text{Time-to-} & \text{Median} \\
\text{Failure (hr)} & \text{Rank ( }\!\!%\!\!\text{ )} \\
\text{85} & \text{ 9}\text{.43 }\!\!%\!\!\text{ } \\
\text{90} & \text{22}\text{.85 }\!\!%\!\!\text{ } \\
\text{95} & \text{36}\text{.41 }\!\!%\!\!\text{ } \\
\text{100} & \text{50}\text{.00 }\!\!%\!\!\text{ } \\
\text{105} & \text{63}\text{.59 }\!\!%\!\!\text{ } \\
\text{110} & \text{77}\text{.15 }\!\!%\!\!\text{ } \\
\text{115} & \text{90}\text{.57 }\!\!%\!\!\text{ } \\
\end{matrix}</math></center>

These points can now be plotted on normal probability plotting paper as shown in the next figure.

[[Image:chp8PP.gif|thumb|center|300px| ]]

Draw the best possible line through the plot points. The time values where this line intersects the 15.85%, 50%, and 84.15% unreliability values should be projected down to the abscissa, as shown in the following plot.

[[Image:chp8PP2.gif|thumb|center|300px| ]]

The estimate of <math>\mu </math> is determined from the time value at the 50% unreliability level, which in this case is 100 hours. The value of the estimator of <math>\sigma </math> is determined by Eqn. (sigplot):

::<math>\begin{align}
\widehat{\sigma }= & \frac{t(Q=84.15%)-t(Q=15.85%)}{2} \\
\widehat{\sigma }= & \frac{112-88}{2}=\frac{24}{2} \\
\widehat{\sigma }= & 12\text{ hours}
\end{align}</math>

Alternately, <math>\widehat{\sigma }</math> could be determined by measuring the distance from <math>t(Q=15.85%)</math> to <math>t(Q=50%)</math> , or <math>t(Q=50%)</math> to <math>t(Q=84.15%)</math> , as either of these two distances is equal to the value of one standard deviation.

===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>

===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| ]]

===Rank Regression on X===
As was mentioned previously, performing a rank regression on X requires that a straight line be fitted to a set of data points such that the sum of the squares of the horizontal deviations from the points to the fitted line is minimized.

Again, the first task is to bring our function, Eqn. (Fnorm), into a linear form. This step is exactly the same as in regression on Y analysis and Eqns. (norm), (yn), (an), and (bn) apply in this case as they did for the regression on Y. The deviation from the previous analysis begins on the least squares fit step where: in this case, we treat <math>x</math> as the dependent variable and <math>y</math> as the independent variable. The best-fitting straight line for the data, for regression on X, is the straight line:

::<math>x=\widehat{a}+\widehat{b}y</math>

The corresponding equations for <math>\widehat{a}</math> and <math>\widehat{b}</math> are:

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

:where:

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

:and:

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

and the <math>F({{T}_{i}})</math> values are estimated from the median ranks. Once <math>\widehat{a}</math> and <math>\widehat{b}</math> are obtained, solve Eqn. (xlinen) for the unknown value of <math>y</math> which corresponds to:

::<math>y=-\frac{\widehat{a}}{\widehat{b}}+\frac{1}{\widehat{b}}x</math>

Solving for the parameters from Eqns. (an) and (bn), we get:

::<math>a=-\frac{\widehat{a}}{\widehat{b}}=-\frac{\mu }{\sigma }\Rightarrow \mu =\widehat{a}</math>

:and:

::<math>b=\frac{1}{\widehat{b}}=\frac{1}{\sigma }\Rightarrow \sigma =\widehat{b}</math>

The correlation coefficient is evaluated as before using Eqn. (RHOn).

===Example 3===
Using the data of Example 2 and assuming a normal distribution, estimate the parameters and determine the correlation coefficient, <math>\rho </math> , using rank regression on X.

====Solution to Example 3====
Table 8.2 constructed in Example 2 applies to this example also. Using the values on this table, we get:

::<math>\begin{align}
\hat{b}= & \frac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{T}_{i}}{{y}_{i}}-\tfrac{\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 }}}\,y_{i}^{2}-\tfrac{{{\left( \underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{y}_{i}} \right)}^{2}}}{14}} \\
\widehat{b}= & \frac{365.2711-(630)(0)/14}{11.3646-{{(0)}^{2}}/14}=32.1411
\end{align}</math>

:and:

::<math>\hat{a}=\overline{x}-\hat{b}\overline{y}=\frac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{T}_{i}}}{14}-\widehat{b}\frac{\underset{i=1}{\overset{14}{\mathop{\sum }}}\,{{y}_{i}}}{14}</math>

:or:

::<math>\widehat{a}=\frac{630}{14}-(32.1411)\frac{(0)}{14}=45</math>

Therefore, from Eqn. (bnx):

::<math>\widehat{\sigma }=\widehat{b}=32.1411</math>

:and from Eqn. (anx):

::<math>\widehat{\mu }=\widehat{a}=45\text{ hours}</math>

The correlation coefficient is found using Eqn. (RHOn):

::<math>\widehat{\rho }=0.979</math>

Note that the results for regression on X are not necessarily the same as the results for regression on Y. The only time when the two regressions are the same (i.e. will yield the same equation for a line) is when the data lie perfectly on a straight line.
Using Weibull++ , Rank Regression on X (RRX) can be selected from the Analysis page.

[[Image:weibullfolio1.png|thumb|center|400px| ]]

The plot of the solution for this example is shown next.

[[Image:weibullfolioplot1.png|thumb|center|400px| ]]
<math></math>

===Maximum Likelihood Estimation===
As it was outlined in Chapter 3, maximum likelihood estimation works by developing a likelihood function based on the available data and finding the values of the parameter estimates that maximize the likelihood function. This can be achieved by using iterative methods to determine the parameter estimate values that maximize the likelihood function. This can be rather difficult and time-consuming, particularly when dealing with the three-parameter distribution. Another method of finding the parameter estimates involves taking the partial derivatives of the likelihood function with respect to the parameters, setting the resulting equations equal to zero, and solving simultaneously to determine the values of the parameter estimates. The log-likelihood functions and associated partial derivatives used to determine maximum likelihood estimates for the normal distribution are covered in Appendix C.

===Special Note About Bias===
Estimators (i.e. parameter estimates) have properties such as unbiasedness, minimum variance, sufficiency, consistency, squared error constancy, efficiency and completeness [7][5]. Numerous books and papers deal with these properties and this coverage is beyond the scope of this reference.

However, we would like to briefly address one of these properties, unbiasedness. An estimator is said to be unbiased if the estimator <math>\widehat{\theta }=d({{X}_{1,}}{{X}_{2,}}...,{{X}_{n)}}</math> satisfies the condition <math>E\left[ \widehat{\theta } \right]</math> <math>=\theta </math> for all <math>\theta \in \Omega .</math>
Note that <math>E\left[ X \right]</math> denotes the expected value of X and is defined (for continuous distributions) by:

::<math>\begin{align}
E\left[ X \right]= \int_{\varpi }x\cdot f(x)dx \\
X\in & \varpi .
\end{align}</math>

It can be shown [7][5] that the MLE estimator for the mean of the normal (and lognormal) distribution does satisfy the unbiasedness criteria, or <math>E\left[ \widehat{\mu } \right]</math> <math>=\mu .</math> The same is not true for the estimate of the variance <math>\hat{\sigma }_{T}^{2}</math> . The maximum likelihood estimate for the variance for the normal distribution is given by:

::<math>\hat{\sigma }_{T}^{2}=\frac{1}{N}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{T}_{i}}-\bar{T})}^{2}}</math>

with a standard deviation of:

::<math>{{\hat{\sigma }}_{T}}=\sqrt{\frac{1}{N}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{T}_{i}}-\bar{T})}^{2}}}</math>

These estimates, however, have been shown to be biased. It can be shown [7][5] that the unbiased estimate of the variance and standard deviation for complete data is given by:

::<math>\begin{align}
\hat{\sigma }_{T}^{2}= & \left[ \frac{N}{N-1} \right]\cdot \left[ \frac{1}{N}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{T}_{i}}-\bar{T})}^{2}} \right]=\frac{1}{N-1}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{T}_{i}}-\bar{T})}^{2}} \\
{{{\hat{\sigma }}}_{T}}= & \sqrt{\left[ \frac{N}{N-1} \right]\cdot \left[ \frac{1}{N}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{T}_{i}}-\bar{T})}^{2}} \right]} \\
= & \sqrt{\frac{1}{N-1}\underset{i=1}{\overset{N}{\mathop \sum }}\,{{({{T}_{i}}-\bar{T})}^{2}}}
\end{align}</math>

Note that for larger values of <math>N</math> , <math>\sqrt{\left[ N/(N-1) \right]}</math> tends to 1.

Weibull++ by default returns the standard deviation as defined by Eqn. (NormSt2).
The Use Unbiased Std on Normal Data option in the User Setup under the Calculations tab allows biasing to be considered when estimating the parameters.

When this option is selected, Weibull++ returns the standard deviation as defined by Eqn. (NormSt2). This is only true for complete data sets. For all other data types, Weibull++ by default returns the standard deviation as defined by Eqn. (normbias2) regardless of the selection status of this option. The next figure shows this setting in Weibull++.

<math></math>
[[Image:ldachp8fig3.gif|thumb|center|300px| ]]