Template:Example: Recurrent Events Data Non-parameteric MCF Bound Example: Difference between revisions

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'''Recurrent Events Data Non-parameteric MCF Bound Example'''
#REDIRECT [[Non-Parametric_Recurrent_Event_Data_Analysis]]
 
Using the data in [[Recurrent Events Data Non-parameteric MCF Example|Example 1]], estimate the 95% confidence bounds.
 
<br>'''Solution'''
 
Using the MCF variance equation, the following table of variance values can be obtained:
 
{| border="1" align="center"
|-
! ID
! Months
! State
! <span class="texhtml">''r''<sub>''i''</sub></span>
! <span class="texhtml">''V''''a''''r''<sub>''i''</sub></span>
|-
| 1
| 5
| F
| 5
| <math>(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.032</math>
|-
| 2
| 6
| F
| 5
| <math>0.032+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.064</math>
|-
| 1
| 10
| F
| 5
| <math>0.064+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.096</math>
|-
| 3
| 12
| F
| 5
| <math>0.096+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.128</math>
|-
| 2
| 13
| F
| 5
| <math>0.128+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.160</math>
|-
| 4
| 13
| F
| 5
| <math>0.160+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.192</math>
|-
| 1
| 15
| F
| 5
| <math>0.192+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.224</math>
|-
| 4
| 15
| F
| 5
| <math>0.224+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.256</math>
|-
| 5
| 16
| F
| 5
| <math>0.256+(\tfrac{1}{5})^2[3(0-\tfrac{1}{5})^2+2(1-\tfrac{1}{5})^2]=0.288</math>
|-
| 2
| 17
| F
| 5
| <math>0.288+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.320</math>
|-
| 1
| 17
| S
| 4
|
|-
| 2
| 19
| S
| 3
|
|-
| 3
| 20
| F
| 3
| <math>0.320+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+2(0-\tfrac{1}{5})^2]=0.394</math>
|-
| 5
| 22
| F
| 3
| <math>0.394+(\tfrac{1}{5})^2[2(0-\tfrac{1}{5})^2+4(1-\tfrac{1}{5})^2]=0.468</math>
|-
| 4
| 24
| S
| 2
|
|-
| 3
| 25
| F
| 2
| <math>0.468+(\tfrac{1}{5})^2[(1-\tfrac{1}{5})^2+4(0-\tfrac{1}{5})^2]=0.593</math>
|-
| 5
| 25
| F
| 2
| <math>0.580+(\tfrac{1}{5})^2[(0-\tfrac{1}{5})^2+4(1-\tfrac{1}{5})^2]=0.718</math>
|-
| 3
| 26
| S
| 1
|
|-
| 5
| 28
| S
| 0
|
|}
 
Using the equation for the MCF bounds and <span class="texhtml">''K''<sub>5</sub> = 1.644</span> for a 95% confidence level, the confidence bounds can be obtained as follows:
<center><math>\begin{matrix}
  ID & Months & State & MC{{F}_{i}} & Va{{r}_{i}} & MC{{F}_{{{L}_{i}}}} & MC{{F}_{{{U}_{i}}}}  \\
  \text{1} & \text{5} & \text{F} & \text{0}\text{.20} & \text{0}\text{.032} & 0.0459 & 0.8709  \\
  \text{2} & \text{6} & \text{F} & \text{0}\text{.40} & \text{0}\text{.064} & 0.1413 & 1.1320  \\
  \text{1} & \text{10} & \text{F} & \text{0}\text{.60} & \text{0}\text{.096} & 0.2566 & 1.4029  \\
  \text{3} & \text{12} & \text{F} & \text{0}\text{.80} & \text{0}\text{.128} & 0.3834 & 1.6694  \\
  \text{2} & \text{13} & \text{F} & \text{1}\text{.00} & \text{0}\text{.160} & 0.5179 & 1.9308  \\
  \text{4} & \text{13} & \text{F} & \text{1}\text{.20} & \text{0}\text{.192} & 0.6582 & 2.1879  \\
  \text{1} & \text{15} & \text{F} & \text{1}\text{.40} & \text{0}\text{.224} & 0.8028 & 2.4413  \\
  \text{4} & \text{15} & \text{F} & \text{1}\text{.60} & \text{0}\text{.256} & 0.9511 & 2.6916  \\
  \text{5} & \text{16} & \text{F} & \text{1}\text{.80} & \text{0}\text{.288} & 1.1023 & 2.9393  \\
  \text{2} & \text{17} & \text{F} & \text{2}\text{.0} & \text{0}\text{.320} & 1.2560 & 3.1848  \\
  \text{1} & \text{17} & \text{S} & {} & {} & {} & {}  \\
  \text{2} & \text{19} & \text{S} & {} & {} & {} & {}  \\
  \text{3} & \text{20} & \text{F} & \text{2}\text{.33} & \text{0}\text{.394} & 1.4990 & 3.6321  \\
  \text{5} & \text{22} & \text{F} & \text{2}\text{.66} & \text{0}\text{.468} & 1.7486 & 4.0668  \\
  \text{4} & \text{24} & \text{S} & {} & {} & {} & {}  \\
  \text{3} & \text{25} & \text{F} & \text{3}\text{.16} & \text{0}\text{.593} & 2.1226 & 4.7243  \\
  \text{5} & \text{25} & \text{F} & \text{3}\text{.66} & \text{0}\text{.718} & 2.5071 & 5.3626  \\
  \text{3} & \text{26} & \text{S} & {} & {} & {} & {}  \\
  \text{5} & \text{28} & \text{S} & {} & {} & {} & {}  \\
\end{matrix}</math></center>
The analysis presented in this example can be obtained automatically in Weibull ++ using the non-parametric RDA&nbsp;folio, as shown next.
 
[[Image:Recurrent Data Example 2 Data.png|thumb|center|400px]]
 
Note: In the&nbsp;folio above, the <span class="texhtml">''F''</span> refers to failures and <span class="texhtml">''E''</span> refers to suspensions (or censoring ages).
 
The results, with calculated MCF values and upper and lower 95% confidence limits, are shown next along with the graphical plot.
 
[[Image:Recurrent Data Example 2 Result.png|thumb|center|400px]]
 
[[Image:Recurrent Data Example 2 Plot.png|thumb|center|400px]]

Latest revision as of 02:39, 13 August 2012

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