Template:Example: Recurrent Events Data Non-parameteric MCF Bound Example

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	$({\tfrac {1}{5}})^{2}[(1-{\tfrac {1}{5}})^{2}+4(0-{\tfrac {1}{5}})^{2}]=0.032$
2	6	F	5	$0.032+({\tfrac {1}{5}})^{2}[(1-{\tfrac {1}{5}})^{2}+4(0-{\tfrac {1}{5}})^{2}]=0.064$
1	10	F	5	$0.064+({\tfrac {1}{5}})^{2}[(1-{\tfrac {1}{5}})^{2}+4(0-{\tfrac {1}{5}})^{2}]=0.096$
3	12	F	5	$0.096+({\tfrac {1}{5}})^{2}[(1-{\tfrac {1}{5}})^{2}+4(0-{\tfrac {1}{5}})^{2}]=0.128$
2	13	F	5	$0.128+({\tfrac {1}{5}})^{2}[(1-{\tfrac {1}{5}})^{2}+4(0-{\tfrac {1}{5}})^{2}]=0.160$
4	13	F	5	$0.160+({\tfrac {1}{5}})^{2}[(1-{\tfrac {1}{5}})^{2}+4(0-{\tfrac {1}{5}})^{2}]=0.192$
1	15	F	5	$0.192+({\tfrac {1}{5}})^{2}[(1-{\tfrac {1}{5}})^{2}+4(0-{\tfrac {1}{5}})^{2}]=0.224$
4	15	F	5	$0.224+({\tfrac {1}{5}})^{2}[(1-{\tfrac {1}{5}})^{2}+4(0-{\tfrac {1}{5}})^{2}]=0.256$
5	16	F	5	$0.256+({\tfrac {1}{5}})^{2}[3(0-{\tfrac {1}{5}})^{2}+2(1-{\tfrac {1}{5}})^{2}]=0.288$
2	17	F	5	$0.288+({\tfrac {1}{5}})^{2}[(1-{\tfrac {1}{5}})^{2}+4(0-{\tfrac {1}{5}})^{2}]=0.320$
1	17	S	4
2	19	S	3
3	20	F	3	$0.320+({\tfrac {1}{5}})^{2}[(1-{\tfrac {1}{5}})^{2}+2(0-{\tfrac {1}{5}})^{2}]=0.394$
5	22	F	3	$0.394+({\tfrac {1}{5}})^{2}[2(0-{\tfrac {1}{5}})^{2}+4(1-{\tfrac {1}{5}})^{2}]=0.468$
4	24	S	2
3	25	F	2	$0.468+({\tfrac {1}{5}})^{2}[(1-{\tfrac {1}{5}})^{2}+4(0-{\tfrac {1}{5}})^{2}]=0.593$
5	25	F	2	$0.580+({\tfrac {1}{5}})^{2}[(0-{\tfrac {1}{5}})^{2}+4(1-{\tfrac {1}{5}})^{2}]=0.718$
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:

${\begin{matrix}ID&Months&State&MC{{F}_{i}}&Va{{r}_{i}}&MC{{F}_{{L}_{i}}}&MC{{F}_{{U}_{i}}}\\{\text{1}}&{\text{5}}&{\text{F}}&{\text{0}}{\text{.20}}&{\text{0}}{\text{.032}}&0.0459&0.8709\\{\text{2}}&{\text{6}}&{\text{F}}&{\text{0}}{\text{.40}}&{\text{0}}{\text{.064}}&0.1413&1.1320\\{\text{1}}&{\text{10}}&{\text{F}}&{\text{0}}{\text{.60}}&{\text{0}}{\text{.096}}&0.2566&1.4029\\{\text{3}}&{\text{12}}&{\text{F}}&{\text{0}}{\text{.80}}&{\text{0}}{\text{.128}}&0.3834&1.6694\\{\text{2}}&{\text{13}}&{\text{F}}&{\text{1}}{\text{.00}}&{\text{0}}{\text{.160}}&0.5179&1.9308\\{\text{4}}&{\text{13}}&{\text{F}}&{\text{1}}{\text{.20}}&{\text{0}}{\text{.192}}&0.6582&2.1879\\{\text{1}}&{\text{15}}&{\text{F}}&{\text{1}}{\text{.40}}&{\text{0}}{\text{.224}}&0.8028&2.4413\\{\text{4}}&{\text{15}}&{\text{F}}&{\text{1}}{\text{.60}}&{\text{0}}{\text{.256}}&0.9511&2.6916\\{\text{5}}&{\text{16}}&{\text{F}}&{\text{1}}{\text{.80}}&{\text{0}}{\text{.288}}&1.1023&2.9393\\{\text{2}}&{\text{17}}&{\text{F}}&{\text{2}}{\text{.0}}&{\text{0}}{\text{.320}}&1.2560&3.1848\\{\text{1}}&{\text{17}}&{\text{S}}&{}&{}&{}&{}\\{\text{2}}&{\text{19}}&{\text{S}}&{}&{}&{}&{}\\{\text{3}}&{\text{20}}&{\text{F}}&{\text{2}}{\text{.33}}&{\text{0}}{\text{.394}}&1.4990&3.6321\\{\text{5}}&{\text{22}}&{\text{F}}&{\text{2}}{\text{.66}}&{\text{0}}{\text{.468}}&1.7486&4.0668\\{\text{4}}&{\text{24}}&{\text{S}}&{}&{}&{}&{}\\{\text{3}}&{\text{25}}&{\text{F}}&{\text{3}}{\text{.16}}&{\text{0}}{\text{.593}}&2.1226&4.7243\\{\text{5}}&{\text{25}}&{\text{F}}&{\text{3}}{\text{.66}}&{\text{0}}{\text{.718}}&2.5071&5.3626\\{\text{3}}&{\text{26}}&{\text{S}}&{}&{}&{}&{}\\{\text{5}}&{\text{28}}&{\text{S}}&{}&{}&{}&{}\\\end{matrix}}$

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

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.