Survival Analysis at specific time point, ltable vs. sts test logrank

[Winta Mehtsun]

I am examining the association between chemotherapy and survival among octogenarians with pancreatic cancer who have undergone surgery first - I addition to determining overall survival I want to determine the 5 year survival difference between the chemotherapy and no chemotherapy group:

Relevant code:
stset postopfu90, failure(laststatus)
xi: stcox i.anychemo AGE i.SEX i.racemm ib1.income2m i.insurancem i.hichar1 i.hlos2 i.ajcc8stage i.marginm i.radiationm i.facilitym ib2.region i.timeperiod

sts test anychemo if postopfu90<=60, logrank
ltable postopfu90, by(anychemo) interval(12)

Using the life table I am get overlapping confidence intervals at the 5 year time point, , 7.7% (95%CI [6.1%, 9.6%]) vs. 5.4% (95%CI [4.2%, 6.8%]) for patients who received adjuvant chemotherapy versus those who did not receive chemotherapy, respectively.

However when I perform the logrank test they seem to be significantly different. These results seem contradictory unless I am interpreting the ltable confidence intervals incorrectly?

[Paul Dickman]

No, these are not contradictory. The log rank test tests the equality of the two survivor functions across all values of time.

It's possible for the two curves to be statistically significantly different (considering all values of time) but the values of the curves at just one point in time to be similar.

[Winta Mehtsun]

I understand that looking at all values of time and a specific time-point would yield different results. I specifically am interested in time < 5 years (which as my command reads I restrict the log rank survivor curve to time less than five years: "sts test anychemo if postopfu90<=60, logrank. Also the life table is the conditional probability of surviving to five years and that is what I am trying to determine - is it different for the treatment vs the control group.

[Paul Dickman]

I understand that looking at all values of time and a specific time-point would yield different results.

Then what is your question? The way you phrased the original post suggested to me that you didn't understand that point.

Please know that I'm not trying to be difficult, I can think of many questions related to you initial post but it's not clear which one(s) you want answers to. Including but not limited to:

1. How does one test for a difference in 5-year survival?
2. Should one test differences in the curves (up to 5 years) or the estimates at 5 years?
3. Should one adjust for potential confounders when testing.

Note that the life table estimates are not conditional.

[Winta Mehtsun]

Your comments have been helpful - I have not found them difficult. Yes (#1) I want to determine the best way to test for a difference in the 5-year survival between the the chemotherapy and no chemotherapy group (this I believe should be the conditional probability of surviving to five years - but I welcome your feedback i.e. #2). I incorrectly interpreted that the life-table estimates are conditional. With regards to your #3 - in my main analysis of overall survival - I use Cox proportional hazards models to adjust for potential confounders and do subgroup and propensity analyses to further investigate the relationship between treatment and survival. Thus my main agenda for the 5-year survival is to start with a unadjusted analysis.

[Winta Mehtsun]

Paul Dickman Thanks very much for your comments. I took some time to review your survival content and tried to better describe my models and attempts at calculating 5 year survival between my groups of interest. If any fellow statalist users can provide further direction I would sincerely appreciate it.

For context my overall model involved a landmarked analysis (at 90 days) examining overall survival among octogenarians with pancreatic cancer who under went surgery and comparing survival between those who had adjuvant chemotherapy and those who did not
stset postopfu, failure(laststatus==1) enter(time 3)
by anychemo: stci, median
sts graph, by(anychemo) tmax(60) xtitle (Time (months)) xlabel (0 12 24 36 48 60) risktable(, size(3) order(1 "No Chemotherapy" 2 "Chemotherapy")) legend(label(1 "No Chemotherapy") label (2 "Chemotherapy") title(, size(1))) title("")
graph save "/Users/winta/Documents/Research /Wang/NCDB/KS_Final__Landmark_Overall_Cohort", replace
*Multivariate model adjusting for confounders in the relationship of chemotherapy and overall survival (I do some additional propensity and subgroup analyses to further investigate this)
xi: stcox i.anychemo AGE i.SEX i.racemm ib1.income2m i.insurancem i.hichar1 i.hlos2 i.ajcc8stage i.marginm i.radiationm i.facilitym ib2.region i.timeperiod


** 5-year survival**
stset postopfu, failure(laststatus==1) enter(time 3) exit(time 60)
**landmarked entry at 90 days and censorship at 5 years is this correctly set up

sts list, by(anychemo)
** These appear to be the point estimates of the specific time point 5 years (not cummulative) and the CI of the chemotherapy and no chemotherapy group overlap

ltable postopfu , by(anychemo) interval(60) noadjust
** These appear to be the point estimates of the survivial curve at 5 years, and with the noadjust I thought they would be similar to sts list but I think I misunderstand the difference in the underlying survival function for these respective commands calculations and how they treat the time interval. Again the CI of the chemotherapy and no chemotherapy group overlap in this calculation.

sts test anychemo, logrank
*here it appears the distribution of the survival curve is significantly different for the chemotherapy and no chemotherapy group when we censor at 5 years but this does not make inferences of on the point estimates of 5 year survival ** P<0.001

The confidence intervals appear to overlap for both groups by the ltable and sts list but other suggestions I found on the statalist serve recommend determining (by hand) the S(t) and the Greenwood variance for each group at the chosen time and perform a z test. It is still not clear to me what the most straightforward way using commands is to determine the 5-year survival estimate between groups.

[Paul Dickman]

Firstly, I understand now that you are estimating conditional survival. Sorry for missing that.

It seems you have two questions:
1. Why are the Kaplan-Meier estimates are life table estimates different?
2. Should I test differences in survival at 5 years or differences in the curves? Or both.

Regarding question 1, you are correct that the two estimates of 5-year survival should be identical. I can't answer why they are not without seeing the commands you used and the complete output. The FAQ recommends including the exact commands you types along with the resulting output and to present it using CODE delimiters (rather than images).

Question 2 you need to answer yourself, based on your research question and what is considered standard practice in your field. If you think the 5-year survival probabilities are the most appropriate outcome measures for your research question then you can present a test for differences in those probabilities. However, I would also present the log rank test since it is considered standard. You'll find plenty of discussion if you google "Comparing Survival Curves vs. Comparing Survival Probabilities at a Fixed Time Point".

Following is an example that I hope gives more insight into how to consider this issue. Here are two hypothetical survival curves from a hypothetical randomised trial; "A" represents an aggressive therapy and "C" represents what one might see for a conservative therapy. This is similar to your data, in that the two curves are significantly different but the 5 year survival probabilities are similar.

Which treatment is associated with the best survival? Assuming we are only interested in the period up to 5 years, I would argue treatment C. The 5 year survival probabilities are identical, but those on treatment C have a larger mean survival time. Unless you have a very good reason for doing so, I would argue that only presenting "no difference in 5-year survival" is not an accurate representation of the data.

[Paul Dickman]

On my phone, but eabted to mention that the kaplan meier and life table estimates may not be identical if time is classified differently. For exampke, if you group time in the life table but not in the kaplan meier.

[Winta Mehtsun]

Paul Dickman Thanks so much for all your comments and resources they have been very helpful. I agree very much that presenting no difference in survival will not be adequate and your examples highlights this point very well. Below is my code I have setup my follow up time in years, defined failure, restricted entry to the landmarked time of 3 months and exit time at 5 years to examine 5 year survival.

I am trying to figure out why my point estimates are so different by sts list and ltable. I agree it has something to do with how my time is binned because the estimates become more different over time.

Greatly appreciate your input.

stset postopfuyr, failure(laststatus==1) enter(time .25) exit(time 5)
sts list, by(anychemo) compare at(1 2 3 4 5)
ltable postopfuyr, by(anychemo) interval(0 1 2 3 4 5) noadjust
sts test anychemo

. sts list, by(anychemo) compare at(1 2 3 4 5 )

failure _d: laststatus == 1
analysis time _t: postopfuyr
enter on or after: time .25
exit on or before: time 5

Survivor Function
anychemo 0 1

time 1 0.6019 0.7206
2 0.3371 0.4147
3 0.2110 0.2588
4 0.1581 0.1870
5 0.1180 0.1548

. ltable postopfuyr, by(anychemo) interval(0 1 2 3 4 5) noadjust

Beg. Std.
Interval Total Deaths Lost Survival Error [95% Conf. Int.]

anychemo = 0
0 1 1097 454 0 0.5861 0.0149 0.5564 0.6146
1 2 643 291 0 0.3209 0.0141 0.2934 0.3486
2 3 352 160 0 0.1750 0.0115 0.1532 0.1981
3 4 192 74 0 0.1076 0.0094 0.0901 0.1267
4 5 118 53 0 0.0593 0.0071 0.0463 0.0743
5 . 65 65 0 0.0000 . . .
anychemo = 1
0 1 871 251 0 0.7118 0.0153 0.6805 0.7407
1 2 620 275 0 0.3961 0.0166 0.3635 0.4284
2 3 345 150 0 0.2239 0.0141 0.1968 0.2521
3 4 195 77 0 0.1355 0.0116 0.1137 0.1591
4 5 118 39 0 0.0907 0.0097 0.0728 0.1109
5 . 79 79 0 0.0000 . . .

sts test anychemo

failure _d: laststatus == 1
analysis time _t: postopfuyr
enter on or after: time .25
exit on or before: time 5


Log-rank test for equality of survivor functions

Events Events
anychemo observed expected

0 908 819.87
1 695 783.13

Total 1603 1603.00

chi2(1) = 19.43
Pr>chi2 = 0.0000

[Paul Dickman]

ltable is not an st command so does not use the internal variables created by stset. That is, it does not restrict entry to the landmarked time of 3 months whereas your st commands use the restricted data.

You could use

ltable _t _d if _st, by(anychemo) interval(0(0.25)5) noadjust

Your ltable command doesn't include an event indicator so assumes everyone experiences the event at time postopfuyr. If some survival times are censored then this is obviously wrong. It's not possible to tell from your output if you have any censored survival times since you haven't listed them in -sts list-. When presenting results from -sts list-, please exclude the compare option so we can see the numbers at risk, events, censored, etc. I would also suggest reporting at times 0(0.25)5.

Please use CODE tags (the icon with the pound symbol) when showing commands and output.

[Winta Mehtsun]

Thanks very much have incorporated your suggestions - is it perhaps due to different handling of censoring that is resulting in the slightly different output? Forgive my inability to use the decoders properly, I have tried this time around hopefully this is now correct!

gen postopfuyr = postopfu/12
summarize postopfuyr

Variable         Obs        Mean    Std. Dev.       Min        Max
postopfuyr       1,968    2.005048    1.790747        .26    11.8525

stset postopfuyr, failure(laststatus==1) enter(time .25) exit(time 5)

failure event:  laststatus == 1
obs. time interval:  (0, postopfuyr]
enter on or after:  time .25
exit on or before:  time 5
1968  total observations
0  exclusions
1968  observations remaining, representing
1603  failures in single-record/single-failure data
3191.712  total analysis time at risk and under observation
at risk from t =         0
earliest observed entry t =       .25
last observed exit t =         5

stdescribe

failure _d:    laststatus    ==    1
analysis time _t:    postopfuyr
enter on or after:    time .25
exit on or before:    time 5
        --------------    per subject --------------
Category    total        mean    min     median        max
                
no. of subjects    1968      
no. of records    1968        1    1          1          1

(first) entry time            .25    .25        .25        .25
(final) exit time            1.871805    .26      1.425          5
subjects with gap    0      
time on gap if gap    0      
time at risk    3191.7125        1.621805    .01      1.175       4.75
failures    1603        .8145325    0          1          1

sts list, by(anychemo) enter

    failure _d:  laststatus    ==    1
    analysis time _t:  postopfuyr
    enter on or after:  time .25
    exit on or before:  time 5
Beg.                Survivor    Std.
Time    Total    Fail    Lost    Enter    Function    Error    [95% Conf. Int.]                       
anychemo=0 
.25        0    0    0    1097    1.0000    .    .         .
.26     1097    3    0    0    0.9973    0.0016    0.9915    0.9991
.2683     1094    1    0    0    0.9964    0.0018    0.9903    0.9986
.2708     1093    1    0    0    0.9954    0.0020    0.9891    0.9981
.2767     1092    3    0    0    0.9927    0.0026    0.9855    0.9963
.2792     1089    3    0    0    0.9900    0.0030    0.9820    0.9944
.2817     1086    1    0    0    0.9891    0.0031    0.9808    0.9938
.285     1085    1    0    0    0.9881    0.0033    0.9797    0.9931
.2875     1084    1    0    0    0.9872    0.0034    0.9785    0.9924
.29     1083    0    1    0    0.9872    0.0034    0.9785    0.9924
.2933     1082    3    0    0    0.9845    0.0037    0.9752    0.9903
.2958     1079    1    0    0    0.9836    0.0038    0.9741    0.9896
.2983     1078    2    0    0    0.9818    0.0040    0.9719    0.9882
.3008     1076    2    0    0    0.9799    0.0042    0.9697    0.9867
.3042     1074    2    0    0    0.9781    0.0044    0.9675    0.9853
.3092     1072    3    0    0    0.9754    0.0047    0.9643    0.9830
.3125     1069    1    0    0    0.9745    0.0048    0.9632    0.9823
.3175     1068    2    0    0    0.9726    0.0049    0.9611    0.9808

It appears I do have censored data ( I trunctaed the output above), when I specify at below - it does not show the censored columns

sts list, at(0(0.25)5) by(anychemo) enter
failure _d:  laststatus == 1
analysis time _t:  postopfuyr enter on or after:  time .25
exit on or before:  time 5
Beg.        Survivor    Std.
Time    Total    Fail    Function    Error    [95% Conf. Int.]
anychemo=0 
0    0    0    1.0000    .    .         .
.25    0    0    1.0000    .    .         .
.5    930    159    0.8544    0.0107    0.8321    0.8740
.75    771    149    0.7166    0.0137    0.6888    0.7424
1    644    123    0.6019    0.0149    0.5720    0.6304
1.25    559    83    0.5240    0.0152    0.4937    0.5533
1.5    474    83    0.4457    0.0152    0.4157    0.4752
1.75    407    63    0.3858    0.0149    0.3566    0.4149
2    353    51    0.3371    0.0145    0.3088    0.3656
2.25    304    45    0.2938    0.0140    0.2666    0.3214
2.5    268    29    0.2655    0.0136    0.2392    0.2925
2.75    223    33    0.2322    0.0131    0.2070    0.2582
3    193    20    0.2110    0.0127    0.1866    0.2364
3.25    173    13    0.1964    0.0125    0.1726    0.2214
3.5    154    11    0.1836    0.0122    0.1603    0.2082
3.75    136    9    0.1724    0.0120    0.1496    0.1967
4    118    11    0.1581    0.0118    0.1358    0.1820
4.25    105    7    0.1485    0.0116    0.1265    0.1720
4.5    89    6    0.1395    0.0115    0.1180    0.1629
4.75    75    10    0.1234    0.0112    0.1025    0.1464
5    65    3    0.1180    0.0112    0.0972    0.1410
anychemo=1 
0    0    0    1.0000    .    .         .
.25    0    0    1.0000    .    .         .
.5    821    51    0.9414    0.0080    0.9236    0.9552
.75    727    90    0.8378    0.0125    0.8116    0.8607
1    620    101    0.7206    0.0153    0.6895    0.7493
1.25    541    76    0.6321    0.0164    0.5990    0.6633
1.5    469    69    0.5512    0.0170    0.5173    0.5837
1.75    395    70    0.4685    0.0171    0.4347    0.5015
2    345    45    0.4147    0.0169    0.3815    0.4476
2.25    302    37    0.3698    0.0166    0.3373    0.4023
2.5    268    30    0.3326    0.0162    0.3010    0.3646
2.75    229    28    0.2970    0.0158    0.2663    0.3283
3    196    29    0.2588    0.0153    0.2292    0.2892
3.25    173    15    0.2386    0.0150    0.2098    0.2685
3.5    152    14    0.2187    0.0146    0.1907    0.2480
3.75    134    12    0.2011    0.0143    0.1738    0.2299
4    119    9    0.1870    0.0141    0.1603    0.2154
4.25    103    9    0.1724    0.0138    0.1463    0.2003
4.5    93    6    0.1621    0.0136    0.1365    0.1897
4.75    89    1    0.1603    0.0136    0.1348    0.1878
5    79    3    0.1548    0.0135    0.1295    0.1822

ltable _t _d if _st, by(anychemo) interval(0(0.25)5) noadjust
Beg.                                 Std.
Interval     Total   Deaths   Lost    Survival    Error     [95% Conf. Int.]
anychemo = 0
0     0      1097      159     10     0.8551    0.0106     0.8328    0.8746
0     1       928      148      9     0.7187    0.0136     0.6910    0.7444
1     1       771      124      4     0.6031    0.0149     0.5733    0.6315
1     1       643       83      3     0.5253    0.0152     0.4950    0.5546
1     2       557       80      3     0.4498    0.0152     0.4198    0.4793
2     2       474       66      3     0.3872    0.0149     0.3580    0.4163
2     2       405       51      2     0.3384    0.0145     0.3101    0.3669
2     2       352       45      5     0.2952    0.0140     0.2680    0.3228
2     2       302       29      6     0.2668    0.0136     0.2405    0.2938
2     3       267       33     12     0.2338    0.0131     0.2087    0.2599
3     3       222       20     10     0.2128    0.0127     0.1884    0.2382
3     3       192       13      7     0.1984    0.0125     0.1745    0.2234
3     4       172       11      8     0.1857    0.0123     0.1623    0.2103
4     4       153        9      9     0.1748    0.0121     0.1518    0.1990
4     4       135       10      7     0.1618    0.0118     0.1394    0.1857
4     4       118        8      6     0.1508    0.0117     0.1288    0.1745
4     4       104        6     10     0.1421    0.0115     0.1205    0.1656
4     5        88       10      4     0.1260    0.0113     0.1049    0.1491
5     5        74        3      6     0.1209    0.0112     0.1000    0.1439
5     .        65        0     65     0.1209    0.0112     0.1000    0.1439
anychemo = 1
0     0       871       51      1     0.9414    0.0080     0.9237    0.9552
0     1       819       88      4     0.8403    0.0124     0.8142    0.8630
1     1       727      102      5     0.7224    0.0152     0.6913    0.7509
1     1       620       77      3     0.6327    0.0164     0.5996    0.6638
1     2       540       68      3     0.5530    0.0169     0.5191    0.5855
2     2       469       71      4     0.4693    0.0170     0.4355    0.5023
2     2       394       44      5     0.4169    0.0169     0.3837    0.4497
2     2       345       38      6     0.3710    0.0166     0.3385    0.4034
2     2       301       30      5     0.3340    0.0162     0.3023    0.3659
2     3       266       28     10     0.2988    0.0158     0.2681    0.3301
3     3       228       29      4     0.2608    0.0153     0.2313    0.2912
3     3       195       15      8     0.2408    0.0150     0.2120    0.2706
3     4       172       14      7     0.2212    0.0147     0.1931    0.2505
4     4       151       12      6     0.2036    0.0143     0.1762    0.2324
4     4       133        9      6     0.1898    0.0141     0.1630    0.2182
4     4       118        9      7     0.1753    0.0138     0.1492    0.2033
4     4       102        6      5     0.1650    0.0136     0.1393    0.1926
4     5        91        1      2     0.1632    0.0136     0.1376    0.1908
5     5        88        3      6     0.1576    0.0135     0.1322    0.1851
5     .        79        0     79     0.1576    0.0135     0.1322    0.1851

[Winta Mehtsun]

* Correction with appropriate delimiters now

sts list, at(0(0.25)5) by(anychemo) enter

         failure _d:  laststatus == 1
   analysis time _t:  postopfuyr
  enter on or after:  time .25
  exit on or before:  time 5

              Beg.                      Survivor      Std.
    Time     Total     Fail             Function     Error     [95% Conf. Int.]
-------------------------------------------------------------------------------
anychemo=0 
       0         0        0              1.0000         .          .         .
     .25         0        0              1.0000         .          .         .
      .5       930      159              0.8544    0.0107     0.8321    0.8740
     .75       771      149              0.7166    0.0137     0.6888    0.7424
       1       644      123              0.6019    0.0149     0.5720    0.6304
    1.25       559       83              0.5240    0.0152     0.4937    0.5533
     1.5       474       83              0.4457    0.0152     0.4157    0.4752
    1.75       407       63              0.3858    0.0149     0.3566    0.4149
       2       353       51              0.3371    0.0145     0.3088    0.3656
    2.25       304       45              0.2938    0.0140     0.2666    0.3214
     2.5       268       29              0.2655    0.0136     0.2392    0.2925
    2.75       223       33              0.2322    0.0131     0.2070    0.2582
       3       193       20              0.2110    0.0127     0.1866    0.2364
    3.25       173       13              0.1964    0.0125     0.1726    0.2214
     3.5       154       11              0.1836    0.0122     0.1603    0.2082
    3.75       136        9              0.1724    0.0120     0.1496    0.1967
       4       118       11              0.1581    0.0118     0.1358    0.1820
    4.25       105        7              0.1485    0.0116     0.1265    0.1720
     4.5        89        6              0.1395    0.0115     0.1180    0.1629
    4.75        75       10              0.1234    0.0112     0.1025    0.1464
       5        65        3              0.1180    0.0112     0.0972    0.1410
anychemo=1 
       0         0        0              1.0000         .          .         .
     .25         0        0              1.0000         .          .         .
      .5       821       51              0.9414    0.0080     0.9236    0.9552
     .75       727       90              0.8378    0.0125     0.8116    0.8607
       1       620      101              0.7206    0.0153     0.6895    0.7493
    1.25       541       76              0.6321    0.0164     0.5990    0.6633
     1.5       469       69              0.5512    0.0170     0.5173    0.5837
    1.75       395       70              0.4685    0.0171     0.4347    0.5015
       2       345       45              0.4147    0.0169     0.3815    0.4476
    2.25       302       37              0.3698    0.0166     0.3373    0.4023
     2.5       268       30              0.3326    0.0162     0.3010    0.3646
    2.75       229       28              0.2970    0.0158     0.2663    0.3283
       3       196       29              0.2588    0.0153     0.2292    0.2892
    3.25       173       15              0.2386    0.0150     0.2098    0.2685
     3.5       152       14              0.2187    0.0146     0.1907    0.2480
    3.75       134       12              0.2011    0.0143     0.1738    0.2299
       4       119        9              0.1870    0.0141     0.1603    0.2154
    4.25       103        9              0.1724    0.0138     0.1463    0.2003
     4.5        93        6              0.1621    0.0136     0.1365    0.1897
    4.75        89        1              0.1603    0.0136     0.1348    0.1878
       5        79        3              0.1548    0.0135     0.1295    0.1822
-------------------------------------------------------------------------------
Note: Survivor function is calculated over full data and evaluated at indicated times; it is not calculated from aggregates
      shown at left.

[Paul Dickman]

See #8. If you use -ltable- without grouping time into intervals then you should get estimates that are identical to the Kaplan-Meier estimates. If you do want to group time into intervals then I suggest using the actuarial adjustment (i.e., without the noadjust option); estimates should then be similar to the K-M.