Template:Example: Normal Distribution MLE: Difference between revisions

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'''Normal Distribution MLE Example'''
#REDIRECT[[The_Normal_Distribution#MLE_Example]]
 
Using the data of [[Normal distribution example 2|Example 2]] and assuming a normal distribution, estimate the parameters using the MLE method.
 
'''Solution'''
 
In this example we have non-grouped data without suspensions and without interval data. The partial derivatives of the normal log-likelihood function,  <math>\Lambda ,</math>  are given by:
 
 
::<math>\begin{align}
  \frac{\partial \Lambda }{\partial \mu }= & \frac{1}{{{\sigma }^{2}}}\underset{i=1}{\overset{14}{\mathop \sum }}\,({{t}_{i}}-\mu )=0 \\
  \frac{\partial \Lambda }{\partial \sigma }= & \underset{i=1}{\overset{14}{\mathop \sum }}\,\left( \frac{{{t}_{i}}-\mu }{{{\sigma }^{3}}}-\frac{1}{\sigma } \right)=0 
\end{align}</math>
 
 
(The derivations of these equations are presented in [[Appendix: Distribution Log-Likelihood Equations|Appendix]].) Substituting the values of  <math>{{t}_{i}}</math>  and solving the above system simultaneously, we get  <math>\widehat{\sigma }=29.58</math>  hours <math>,</math>  <math>\widehat{\mu }=45</math>  hours <math>.</math>
 
The Fisher matrix is:
 
::<math>\left[ \begin{matrix}
  \widehat{Var}\left( \widehat{\mu } \right)=62.5000 & {} & \widehat{Cov}\left( \widehat{\mu },\widehat{\sigma } \right)=0.0000  \\
  {} & {} & {}  \\
  \widehat{Cov}\left( \widehat{\mu },\widehat{\sigma } \right)=0.0000 & {} & \widehat{Var}\left( \widehat{\sigma } \right)=31.2500  \\
\end{matrix} \right]</math>
 
 
Using Weibull++ , the MLE method can be selected from the Set Analysis page.
 
[[Image:Normal MLE Setting.png|thumb|center|250px| ]]
 
The plot of the solution for this example is shown next.
 
[[Image:Normal MLE Plot.png|thumb|center|250px| ]]

Latest revision as of 05:59, 14 August 2012

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