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| '''Normal Distribution MLE Example'''
| | #REDIRECT[[The_Normal_Distribution#MLE_Example]] |
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| Using the data of [[Normal distribution example 2|Example 2]] and assuming a normal distribution, estimate the parameters using the MLE method.
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| '''Solution'''
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| 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:
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| ::<math>\begin{align}
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| \frac{\partial \Lambda }{\partial \mu }= & \frac{1}{{{\sigma }^{2}}}\underset{i=1}{\overset{14}{\mathop \sum }}\,({{t}_{i}}-\mu )=0 \\
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| \frac{\partial \Lambda }{\partial \sigma }= & \underset{i=1}{\overset{14}{\mathop \sum }}\,\left( \frac{{{t}_{i}}-\mu }{{{\sigma }^{3}}}-\frac{1}{\sigma } \right)=0
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| \end{align}</math>
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| (The derivations of these equations are presented in [[Appendix:_Log-Likelihood_Equations|an 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>
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| The Fisher matrix is:
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| ::<math>\left[ \begin{matrix}
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| \widehat{Var}\left( \widehat{\mu } \right)=62.5000 & {} & \widehat{Cov}\left( \widehat{\mu },\widehat{\sigma } \right)=0.0000 \\
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| {} & {} & {} \\
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| \widehat{Cov}\left( \widehat{\mu },\widehat{\sigma } \right)=0.0000 & {} & \widehat{Var}\left( \widehat{\sigma } \right)=31.2500 \\
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| \end{matrix} \right]</math>
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| Using Weibull++ , the MLE method can be selected from the Set Analysis page.
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| [[Image:Normal MLE Setting.png|center|550px| ]]
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| The plot of the solution for this example is shown next.
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| [[Image:Normal MLE Plot.png|center|550px| ]]
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