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'''
+
'''Recurrent Events Data Non-parameteric MCF Bound Example'''  
  
Using the data in [[Recurrent Events Data Non-parameteric MCF Example|Example 1]], estimate the 95% confidence bounds.
+
Using the data in [[Recurrent Events Data Non-parameteric MCF Example|Example 1]], estimate the 95% confidence bounds.  
  
 +
<br>'''Solution'''
  
'''Solution'''
+
Using the MCF variance equation, the following table of variance values can be obtained:
  
Using MCF variance equation the following table of variance values can be obtained:
+
{| border="1" align="center"
 
 
{|align="center" border="1"
 
 
|-
 
|-
!ID
+
! ID  
!Months
+
! Months  
!State
+
! State  
!<math>r_i</math>
+
! <span class="texhtml">''r''<sub>''i''</sub></span>  
!<math>Var_i</math>
+
! <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>
+
| 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>
+
| 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>
+
| 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>
+
| 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>
+
| 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>
+
| 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>
+
| 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>
+
| 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>
+
| 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>
+
| 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 ||  
+
| 1  
 +
| 17  
 +
| S  
 +
| 4  
 +
|  
 
|-
 
|-
| 2||19 ||S ||3 ||  
+
| 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>
+
| 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>
+
| 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 ||  
+
| 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>
+
| 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>
+
| 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 ||
+
| 3  
 +
| 26  
 +
| S  
 +
| 1  
 +
|  
 
|-
 
|-
| 5||28 ||S ||0 ||
+
| 5  
|}
+
| 28  
 
+
| S  
Using equation for the MCF bounds and  <math>{{K}_{5}}=1.644</math>  for a 95% confidence level, the confidence bounds can be obtained as follows:
+
| 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}
 
<center><math>\begin{matrix}
 
   ID & Months & State & MC{{F}_{i}} & Va{{r}_{i}} & MC{{F}_{{{L}_{i}}}} & MC{{F}_{{{U}_{i}}}}  \\
 
   ID & Months & State & MC{{F}_{i}} & Va{{r}_{i}} & MC{{F}_{{{L}_{i}}}} & MC{{F}_{{{U}_{i}}}}  \\
Line 80: Line 153:
 
   \text{5} & \text{28} & \text{S} & {} & {} & {} & {}  \\
 
   \text{5} & \text{28} & \text{S} & {} & {} & {} & {}  \\
 
\end{matrix}</math></center>
 
\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.
  
The analysis presented in this example can be obtained automatically in Weibull ++ using the Non-Parametric RDA Specialized Folio, as shown next.
+
[[Image:Recurrent Data Example 2 Data.png|thumb|center|400px]]  
 
 
[[Image:Recurrent Data Example 2 Data.png|thumb|center|400px| ]]  
 
  
Note: In the above Folio, the <math>F</math> refers to failures and <math>E</math> refers to suspensions (or censoring ages).
+
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.
+
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 Result.png|thumb|center|400px]]  
  
[[Image:Recurrent Data Example 2 Plot.png|thumb|center|400px| ]]
+
[[Image:Recurrent Data Example 2 Plot.png|thumb|center|400px]]

Revision as of 16:53, 8 March 2012

Recurrent Events Data Non-parameteric MCF Bound Example

Using the data in Example 1, estimate the 95% confidence bounds.


Solution

Using the MCF variance equation, the following table of variance values can be obtained:

ID Months State ri V'a'ri
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 K5 = 1.644 for a 95% confidence level, the confidence bounds can be obtained as follows:

[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]

The analysis presented in this example can be obtained automatically in Weibull ++ using the non-parametric RDA folio, as shown next.

Recurrent Data Example 2 Data.png

Note: In the folio above, the F refers to failures and E 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.

Recurrent Data Example 2 Result.png
Recurrent Data Example 2 Plot.png