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From |
Allen Buxton <[email protected]> |

To |
[email protected] |

Subject |
Re: st: test survival curve differences at 5, 10, 15 year cut points |

Date |
Wed, 27 Jul 2005 08:36:32 -0700 (PDT) |

Here is a crude little program to do this. (There is an old package we sometimes use called Epilog that provides point-wise KM curve comparisons). Hence the name of the program epiwald because this is a standard test of the form of an estimate divided by a standard error. This is the way that I have seen this done. Informally, it is like saying that two population standard errors are known. H0 mu1 = mu2 vs mu1 ~= mu2. -Allen Buxton capture program drop epiwald program define epiwald version 7.0 args m1 s1 m2 s2 local d = abs(`m2' - `m1') local s = sqrt(`s1'^2 + `s2'^2) local z = `d'/`s' local p = 2*(1-norm(`z')) local p1 = 1*(1-norm(`z')) foreach Number in m1 s1 m2 s2 s z p p1 { local txt = `"``Number''"' local `Number'=(round(`txt'*10000)/10000) } di `"mu1=`m1', mu2=`m2', std1=`s1', std2=`s2'"' di `"difference=`d', std=`s'"' di `"z-value=`z'"' di `"p-value=`p'"' di `"p-1side=`p1'"' di `""' end Example: . sts list , by(contx_g0) at(6,12,48) failure _d: event_i == 1 2 analysis time _t: event_d/30.4375 Beg. Survivor Std. Time Total Fail Function Error [95% Conf. Int.] ------------------------------------------------------------------------------- bmt 6 12 5 0.6875 0.1159 0.4046 0.8563 12 10 2 0.5625 0.1240 0.2954 0.7622 48 4 1 0.4922 0.1269 0.2355 0.7064 chm 6 6 1 0.8333 0.1521 0.2731 0.9747 12 4 2 0.5000 0.2041 0.1109 0.8037 48 2 1 0.3333 0.1925 0.0461 0.6756 ------------------------------------------------------------------------------- Note: Survivor function is calculated over full data and evaluated at indicated times; it is not calculated from aggregates shown at left. . epiwald .6875 .1159 .8333 .1521 mu1=.6875, mu2=.8333, std1=.1159, std2=.1521 difference=.1458, std=.1912 z-value=.7625 p-value=.4458 p-1side=.2229 . epiwald .5625 .1240 .5000 .2041 mu1=.5625, mu2=.5, std1=.124, std2=.2041 difference=.0625, std=.2388 z-value=.2617 p-value=.7935 p-1side=.3968 --- "Wagner, Joseph" <[email protected]> wrote: > > I have been asked to test the difference between two > lines on a KM graph > at 5, 10, and 15 year points. I had originally > intended to use -sts > test- but there isn't the option -at(5 10 15)- like > there is with -sts > list- (but the option -test- isn't allowed when the > -at- option is used) > and I prefer to not create new variables for 5, 10, > 15 year endpoints. > I am at a loss how to do this. Ideally, I want a > single graph with the > test coefficient at each 5 year point. (I am using > version 8) > > Here are my -stset- and -stcox- statements: > > stset datelr, id(patkey) failure(death_yn) > origin(dxdate) scale(365.25) > stcox type agedxfr race sex, nolog > sts list, by(type) adjustfor(agedxfr race sex) > compare at(5 10 15) > > > I have considered using different -exit- options > with -stset- but I > don't think you can specify an exit of 5 years > post-entrance. Correct > me if I am wrong. > > > > > > * > * For searches and help try: > * > http://www.stata.com/support/faqs/res/findit.html > * http://www.stata.com/support/statalist/faq > * http://www.ats.ucla.edu/stat/stata/ > * * For searches and help try: * http://www.stata.com/support/faqs/res/findit.html * http://www.stata.com/support/statalist/faq * http://www.ats.ucla.edu/stat/stata/

**References**:**st: test survival curve differences at 5, 10, 15 year cut points***From:*"Wagner, Joseph" <[email protected]>

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