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

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'''Recurrent Events Data Non-parameteric MCF Bound Example'''
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#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