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| ==Grouped Data== <!-- THIS SECTION HEADER IS LINKED FROM: Operational Mission Profile Testing. IF YOU RENAME THE SECTION, YOU MUST UPDATE THE LINK(S). -->
| | #REDIRECT [[Crow-AMSAA - NHPP]] |
| For analyzing grouped data, we follow the same logic described previously for the [[Duane Model|Duane model]]. If Eqn. (amsaa2a) is linearized:
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| ::<math>\ln [E(N(T))]=\ln \lambda +\beta \ln T</math>
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| According to Crow [9], the likelihood function for the grouped data case, (where <math>{{n}_{1}},</math> <math>{{n}_{2}},</math> <math>{{n}_{3}},\ldots ,</math> <math>{{n}_{k}}</math> failures are observed and <math>k</math> is the number of groups), is:
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| ::<math>\underset{i=1}{\overset{k}{\mathop \prod }}\,\underset{}{\overset{}{\mathop{\Pr }}}\,({{N}_{i}}={{n}_{i}})=\underset{i=1}{\overset{k}{\mathop \prod }}\,\frac{{{(\lambda T_{i}^{\beta }-\lambda T_{i-1}^{\beta })}^{{{n}_{i}}}}\cdot {{e}^{-(\lambda T_{i}^{\beta }-\lambda T_{i-1}^{\beta })}}}{{{n}_{i}}!}</math>
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| And the MLE of <math>\lambda </math> based on this relationship is:
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| ::<math>\widehat{\lambda }=\frac{n}{T_{k}^{\widehat{\beta }}}</math>
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| And the estimate of <math>\beta </math> is the value <math>\widehat{\beta }</math> that satisfies:
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| ::<math>\underset{i=1}{\overset{k}{\mathop \sum }}\,{{n}_{i}}\left[ \frac{T_{i}^{\widehat{\beta }}\ln {{T}_{i}}-T_{i-1}^{\widehat{\beta }}\ln {{T}_{i-1}}}{T_{i}^{\widehat{\beta }}-T_{i-1}^{\widehat{\beta }}}-\ln {{T}_{k}} \right]=0</math>
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| '''Example 4'''
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| <br>
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| Consider the grouped failure times data given in Table 5.2. Solve for the Crow-AMSAA parameters using MLE.
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| <br>
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| {|style= align="center" border="1"
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| |+'''Table 5.2 - Grouped failure times data'''
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| !Run Number
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| !Cumulative Failures
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| !End Time(hr)
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| !<math>\ln{(T_i)}</math>
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| !<math>\ln{(T_i)^2}</math>
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| !<math>\ln{(\theta_i)}</math>
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| !<math>\ln{(T_i)}\cdot\ln{(\theta_i)}</math>
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| |-
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| |1|| 2|| 200|| 5.298|| 28.072|| 0.693|| 3.673
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| |-
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| |2|| 3|| 400|| 5.991 ||35.898|| 1.099|| 6.582
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| |-
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| |3|| 4|| 600|| 6.397|| 40.921|| 1.386|| 8.868
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| |-
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| |4|| 11|| 3000|| 8.006|| 64.102|| 2.398 ||19.198
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| |-
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| | || ||<span style="color:blue">Sum =</span>|| <span style="color:blue">25.693</span> ||<span style="color:blue">168.992</span>||<span style="color:blue">5.576</span>||<span style="color:blue">38.321</span>
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| |}
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| '''Solution'''
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| To obtain the estimator of <math>\beta </math> , Eqn. (vv) must be solved numerically for <math>\beta </math> . Using RGA, the value of <math>\widehat{\beta }</math> is <math>0.6315</math> . Now plugging this value into Eqn. (vv1), the estimator of <math>\lambda </math> is:
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| ::<math>\begin{align}
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| & \widehat{\lambda }= & \frac{11}{3,{{000}^{0.6315}}} \\
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| & = & 0.0701
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| \end{align}</math>
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| Therefore, the intensity function becomes:
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| ::<math>\widehat{\rho }(T)=0.0701\cdot 0.6315\cdot {{T}^{-0.3685}}</math>
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