Dear S-News Readers,
I’ve got a model based on censorReg using the
following:
SurvCurve.2<-censorReg(censor(startTime,endTime,status)~1,data="">
Now I want to predict proportions failed with the
following:
y<-predict(SurvCurve.2,type="probability",q=endTimeSum)
where my predicted durations are as follows:
>
endTimeSum
[1]
15 24 27 36 48 60 72 96
This gives me the following predictions:
> y
$"(intercept)":
Estimate
Std.Err 95% LCL 95% UCL
15 0.03122837 0.03736268 0.0290874 0.03352149
24
0.17852421
NA
NA NA
27 0.26701616 0.02628998 0.2570531 0.27722129
36 0.61280592 0.08752239 0.5714039 0.65263977
48 0.94487842 0.27288420 0.9094281 0.96695744
60 0.99898308 0.78580469 0.9952734 0.99978185
72 0.99999915 1.87726321 0.9999665 0.99999998
96 1.00000000 0.00200296 1.0000000 1.00000000
OK so far.
Now to my questions:
(1) I
want to pick off the vector of estimated failure rates above (i.e. 0.0312,
0.1785,….., 1.0000). How can I do this?
(2) Then
I want to multiple (1) by an equivalent vector of units at risk. The number of
units by duration is as follows:
> x<-table(endTime)
> x
15
24 27
36 48
60 72
96
1
473 7
26155 2218 9285
35 13
Thus the final calculation will be the multiplication
of two vectors (0.0312, 0.1785,….., 1.0000) and (1, 473, 7, …..,
13).
Any help would be appreciated.
|
Andrew Scott
Manager - Strategic
Analysis
Tel: 02 8853 9117
Mob: 0411 555 486
Fax: 02 8853 9517
andrew.scott@icfrith.com.au
|
IC Frith & Associates (Australia’s Warranty
Provider)
Suite 1, Building B, 34-46
Brookhollow Ave, Baulkham Hills NSW 2153
Private Mail Bag 14, Castle Hill 1765
Phone: 02 8853 9100
Fax: 02 9634 2396
www.icfrith.com.au
|
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