Template:Grouped data camsaa

Grouped Data

For analyzing grouped data, we follow the same logic described in Chapter 4 for the Duane model. If Eqn. (amsaa2a) is linearized:

[math]\displaystyle{ \ln [E(N(T))]=\ln \lambda +\beta \ln T }[/math]

According to Crow [9], the likelihood function for the grouped data case, (where [math]\displaystyle{ {{n}_{1}}, }[/math] [math]\displaystyle{ {{n}_{2}}, }[/math] [math]\displaystyle{ {{n}_{3}},\ldots , }[/math] [math]\displaystyle{ {{n}_{k}} }[/math] failures are observed and [math]\displaystyle{ k }[/math] is the number of groups), is:

[math]\displaystyle{ \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]

And the MLE of [math]\displaystyle{ \lambda }[/math] based on this relationship is:

[math]\displaystyle{ \widehat{\lambda }=\frac{n}{T_{k}^{\widehat{\beta }}} }[/math]

And the estimate of [math]\displaystyle{ \beta }[/math] is the value [math]\displaystyle{ \widehat{\beta } }[/math] that satisfies:

[math]\displaystyle{ \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]

Example 4
Consider the grouped failure times data given in Table 5.2. Solve for the Crow-AMSAA parameters using MLE.

Table 5.2 - Grouped failure times data

Run Number	Cumulative Failures	End Time(hr)	[math]\displaystyle{ \ln{(T_i)} }[/math]	[math]\displaystyle{ \ln{(T_i)^2} }[/math]	[math]\displaystyle{ \ln{(\theta_i)} }[/math]	[math]\displaystyle{ \ln{(T_i)}\cdot\ln{(\theta_i)} }[/math]
1	2	200	5.298	28.072	0.693	3.673
2	3	400	5.991	35.898	1.099	6.582
3	4	600	6.397	40.921	1.386	8.868
4	11	3000	8.006	64.102	2.398	19.198
		Sum =	25.693	168.992	5.576	38.321

Solution To obtain the estimator of [math]\displaystyle{ \beta }[/math] , Eqn. (vv) must be solved numerically for [math]\displaystyle{ \beta }[/math] . Using RGA, the value of [math]\displaystyle{ \widehat{\beta } }[/math] is [math]\displaystyle{ 0.6315 }[/math] . Now plugging this value into Eqn. (vv1), the estimator of [math]\displaystyle{ \lambda }[/math] is:

[math]\displaystyle{ \begin{align} & \widehat{\lambda }= & \frac{11}{3,{{000}^{0.6315}}} \\ & = & 0.0701 \end{align} }[/math]

Therefore, the intensity function becomes:

[math]\displaystyle{ \widehat{\rho }(T)=0.0701\cdot 0.6315\cdot {{T}^{-0.3685}} }[/math]