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Schoenfled Residua test shows proportionality hazard assumptions holds but Kaplan-Meier plots intersect


Violation of Cox Proportional Hazards by a continuous variableCheck hazard proportional assuption in a large coxphWhat does the “z” in cox.zph mean in RLate Cross of Kaplan-Meier Curves - Does it matter?time varying coefficients in cox proportional hazard modelHow does time factor into Cox regression or a Cox proportional hazards model?Why are Kaplan-Meier curves crossing when Cox PH assumption is not violated (Global Shoenfeld non-significant)?Cox time-dependent coefficient continues to violate the PH assumptionViolation of proportional hazard assumption with big sample size - how to correct for it?Hazard ratio for more than two groups






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








1












$begingroup$


"If Kaplan-Meier plots cross each other then proportional hazard assumption does not hold". The issue I am facing is that I got the Kaplam-Meier plot(bleow). We can clearly see that it is overlapping.
But when I plot the Schoenfled residual plots, it suggests otherwise because the black solid line is flat(image below). Also the p-values(below) for Schoenfled residual plots are not significant, suggesting that proportional hazard assumption holds
enter image description hereenter image description here




ftest <- cox.zph(fitcox)
ftest
p
as.factor(C)2 0.945
as.factor(C)3 0.922
as.factor(C)4 0.717
GLOBAL 0.915




One may argue that the three hazard ratios are calculated w.r.t. the red plot. Red plot does not intersect the blue and black plots. So it is understandable that proportional hazard assumption holds.
But red plot does intersect the green one, although only a little...Is that not enough to violate the proportional hazard assumption?






cox-model kaplan-meier proportional-hazards schoenfeld-residuals







share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    If you cannot reject the null hypothesis, it does not mean that it is true.
    $endgroup$
    – Michael M
    7 hours ago











  • $begingroup$
    This reasoning accounts for the p-value. What about the Schoenfled residual plots being flat....
    $endgroup$
    – Omar Rafique
    5 hours ago

















1












$begingroup$


"If Kaplan-Meier plots cross each other then proportional hazard assumption does not hold". The issue I am facing is that I got the Kaplam-Meier plot(bleow). We can clearly see that it is overlapping.
But when I plot the Schoenfled residual plots, it suggests otherwise because the black solid line is flat(image below). Also the p-values(below) for Schoenfled residual plots are not significant, suggesting that proportional hazard assumption holds
enter image description hereenter image description here




ftest <- cox.zph(fitcox)
ftest
p
as.factor(C)2 0.945
as.factor(C)3 0.922
as.factor(C)4 0.717
GLOBAL 0.915




One may argue that the three hazard ratios are calculated w.r.t. the red plot. Red plot does not intersect the blue and black plots. So it is understandable that proportional hazard assumption holds.
But red plot does intersect the green one, although only a little...Is that not enough to violate the proportional hazard assumption?






cox-model kaplan-meier proportional-hazards schoenfeld-residuals







share|cite|improve this question











$endgroup$







  • 2




    $begingroup$
    If you cannot reject the null hypothesis, it does not mean that it is true.
    $endgroup$
    – Michael M
    7 hours ago











  • $begingroup$
    This reasoning accounts for the p-value. What about the Schoenfled residual plots being flat....
    $endgroup$
    – Omar Rafique
    5 hours ago













1












1








1


1



$begingroup$


"If Kaplan-Meier plots cross each other then proportional hazard assumption does not hold". The issue I am facing is that I got the Kaplam-Meier plot(bleow). We can clearly see that it is overlapping.
But when I plot the Schoenfled residual plots, it suggests otherwise because the black solid line is flat(image below). Also the p-values(below) for Schoenfled residual plots are not significant, suggesting that proportional hazard assumption holds
enter image description hereenter image description here




ftest <- cox.zph(fitcox)
ftest
p
as.factor(C)2 0.945
as.factor(C)3 0.922
as.factor(C)4 0.717
GLOBAL 0.915




One may argue that the three hazard ratios are calculated w.r.t. the red plot. Red plot does not intersect the blue and black plots. So it is understandable that proportional hazard assumption holds.
But red plot does intersect the green one, although only a little...Is that not enough to violate the proportional hazard assumption?






cox-model kaplan-meier proportional-hazards schoenfeld-residuals







share|cite|improve this question











$endgroup$




"If Kaplan-Meier plots cross each other then proportional hazard assumption does not hold". The issue I am facing is that I got the Kaplam-Meier plot(bleow). We can clearly see that it is overlapping.
But when I plot the Schoenfled residual plots, it suggests otherwise because the black solid line is flat(image below). Also the p-values(below) for Schoenfled residual plots are not significant, suggesting that proportional hazard assumption holds
enter image description hereenter image description here




ftest <- cox.zph(fitcox)
ftest
p
as.factor(C)2 0.945
as.factor(C)3 0.922
as.factor(C)4 0.717
GLOBAL 0.915




One may argue that the three hazard ratios are calculated w.r.t. the red plot. Red plot does not intersect the blue and black plots. So it is understandable that proportional hazard assumption holds.
But red plot does intersect the green one, although only a little...Is that not enough to violate the proportional hazard assumption?






cox-model kaplan-meier proportional-hazards schoenfeld-residuals




cox-model kaplan-meier proportional-hazards schoenfeld-residuals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 8 hours ago







Omar Rafique

















asked 8 hours ago









Omar RafiqueOmar Rafique

406




406







  • 2




    $begingroup$
    If you cannot reject the null hypothesis, it does not mean that it is true.
    $endgroup$
    – Michael M
    7 hours ago











  • $begingroup$
    This reasoning accounts for the p-value. What about the Schoenfled residual plots being flat....
    $endgroup$
    – Omar Rafique
    5 hours ago












  • 2




    $begingroup$
    If you cannot reject the null hypothesis, it does not mean that it is true.
    $endgroup$
    – Michael M
    7 hours ago











  • $begingroup$
    This reasoning accounts for the p-value. What about the Schoenfled residual plots being flat....
    $endgroup$
    – Omar Rafique
    5 hours ago







2




2




$begingroup$
If you cannot reject the null hypothesis, it does not mean that it is true.
$endgroup$
– Michael M
7 hours ago





$begingroup$
If you cannot reject the null hypothesis, it does not mean that it is true.
$endgroup$
– Michael M
7 hours ago













$begingroup$
This reasoning accounts for the p-value. What about the Schoenfled residual plots being flat....
$endgroup$
– Omar Rafique
5 hours ago




$begingroup$
This reasoning accounts for the p-value. What about the Schoenfled residual plots being flat....
$endgroup$
– Omar Rafique
5 hours ago










2 Answers
2






active

oldest

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2












$begingroup$

It’s not clear that the overlaps among the K-M curves are so bad. There might be some crossing at very early times and curves come close to each other at some later times but that type of variability might not be inconsistent with proportional hazards.



You will have to use your judgement about the underlying subject matter to decide whether this is close enough to proportional hazards for your purposes. You can’t strictly prove that proportional hazards hold so the judgement is whether there is enough evidence against them to matter for your application.






share|cite|improve this answer









$endgroup$




















    2












    $begingroup$

    You are comparing descriptive data (kaplan meier lines are crossing) with inference test (schoenfeld test) which in case of a not significant test usually seem to contradict because there is usually some descriptive difference. Imagine someone checking for normal distribution: a not significant Kolmogorov-Smirnov test (= inference test) doesn't mean that the QQ plot (= descriptive data) follows perfectly a normal distribution. Same is true for a not significant t-test where means are not exactly the same. And so on. And as always with tests of significance: they depend on sample size.



    In this example I would say that the hazards are not perfectly proportional which can be seen in the kaplan meier plots. But this is not a significant violation of the assumption judged by the schoenfeld test. The problem may arise if one strictly follows the scentence you quoted "If Kaplan-Meier plots cross each other then proportional hazard assumption does not hold" which I would question because sometimes there may be "a little" scrossing like here what not means that proportional assumption must be wrong. If this were true there would be no need for a significance test like the schoenfeld test.






    share|cite|improve this answer











    $endgroup$













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      2 Answers
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      2 Answers
      2






      active

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      active

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      2












      $begingroup$

      It’s not clear that the overlaps among the K-M curves are so bad. There might be some crossing at very early times and curves come close to each other at some later times but that type of variability might not be inconsistent with proportional hazards.



      You will have to use your judgement about the underlying subject matter to decide whether this is close enough to proportional hazards for your purposes. You can’t strictly prove that proportional hazards hold so the judgement is whether there is enough evidence against them to matter for your application.






      share|cite|improve this answer









      $endgroup$

















        2












        $begingroup$

        It’s not clear that the overlaps among the K-M curves are so bad. There might be some crossing at very early times and curves come close to each other at some later times but that type of variability might not be inconsistent with proportional hazards.



        You will have to use your judgement about the underlying subject matter to decide whether this is close enough to proportional hazards for your purposes. You can’t strictly prove that proportional hazards hold so the judgement is whether there is enough evidence against them to matter for your application.






        share|cite|improve this answer









        $endgroup$















          2












          2








          2





          $begingroup$

          It’s not clear that the overlaps among the K-M curves are so bad. There might be some crossing at very early times and curves come close to each other at some later times but that type of variability might not be inconsistent with proportional hazards.



          You will have to use your judgement about the underlying subject matter to decide whether this is close enough to proportional hazards for your purposes. You can’t strictly prove that proportional hazards hold so the judgement is whether there is enough evidence against them to matter for your application.






          share|cite|improve this answer









          $endgroup$



          It’s not clear that the overlaps among the K-M curves are so bad. There might be some crossing at very early times and curves come close to each other at some later times but that type of variability might not be inconsistent with proportional hazards.



          You will have to use your judgement about the underlying subject matter to decide whether this is close enough to proportional hazards for your purposes. You can’t strictly prove that proportional hazards hold so the judgement is whether there is enough evidence against them to matter for your application.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 6 hours ago









          EdMEdM

          22.2k23496




          22.2k23496























              2












              $begingroup$

              You are comparing descriptive data (kaplan meier lines are crossing) with inference test (schoenfeld test) which in case of a not significant test usually seem to contradict because there is usually some descriptive difference. Imagine someone checking for normal distribution: a not significant Kolmogorov-Smirnov test (= inference test) doesn't mean that the QQ plot (= descriptive data) follows perfectly a normal distribution. Same is true for a not significant t-test where means are not exactly the same. And so on. And as always with tests of significance: they depend on sample size.



              In this example I would say that the hazards are not perfectly proportional which can be seen in the kaplan meier plots. But this is not a significant violation of the assumption judged by the schoenfeld test. The problem may arise if one strictly follows the scentence you quoted "If Kaplan-Meier plots cross each other then proportional hazard assumption does not hold" which I would question because sometimes there may be "a little" scrossing like here what not means that proportional assumption must be wrong. If this were true there would be no need for a significance test like the schoenfeld test.






              share|cite|improve this answer











              $endgroup$

















                2












                $begingroup$

                You are comparing descriptive data (kaplan meier lines are crossing) with inference test (schoenfeld test) which in case of a not significant test usually seem to contradict because there is usually some descriptive difference. Imagine someone checking for normal distribution: a not significant Kolmogorov-Smirnov test (= inference test) doesn't mean that the QQ plot (= descriptive data) follows perfectly a normal distribution. Same is true for a not significant t-test where means are not exactly the same. And so on. And as always with tests of significance: they depend on sample size.



                In this example I would say that the hazards are not perfectly proportional which can be seen in the kaplan meier plots. But this is not a significant violation of the assumption judged by the schoenfeld test. The problem may arise if one strictly follows the scentence you quoted "If Kaplan-Meier plots cross each other then proportional hazard assumption does not hold" which I would question because sometimes there may be "a little" scrossing like here what not means that proportional assumption must be wrong. If this were true there would be no need for a significance test like the schoenfeld test.






                share|cite|improve this answer











                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  You are comparing descriptive data (kaplan meier lines are crossing) with inference test (schoenfeld test) which in case of a not significant test usually seem to contradict because there is usually some descriptive difference. Imagine someone checking for normal distribution: a not significant Kolmogorov-Smirnov test (= inference test) doesn't mean that the QQ plot (= descriptive data) follows perfectly a normal distribution. Same is true for a not significant t-test where means are not exactly the same. And so on. And as always with tests of significance: they depend on sample size.



                  In this example I would say that the hazards are not perfectly proportional which can be seen in the kaplan meier plots. But this is not a significant violation of the assumption judged by the schoenfeld test. The problem may arise if one strictly follows the scentence you quoted "If Kaplan-Meier plots cross each other then proportional hazard assumption does not hold" which I would question because sometimes there may be "a little" scrossing like here what not means that proportional assumption must be wrong. If this were true there would be no need for a significance test like the schoenfeld test.






                  share|cite|improve this answer











                  $endgroup$



                  You are comparing descriptive data (kaplan meier lines are crossing) with inference test (schoenfeld test) which in case of a not significant test usually seem to contradict because there is usually some descriptive difference. Imagine someone checking for normal distribution: a not significant Kolmogorov-Smirnov test (= inference test) doesn't mean that the QQ plot (= descriptive data) follows perfectly a normal distribution. Same is true for a not significant t-test where means are not exactly the same. And so on. And as always with tests of significance: they depend on sample size.



                  In this example I would say that the hazards are not perfectly proportional which can be seen in the kaplan meier plots. But this is not a significant violation of the assumption judged by the schoenfeld test. The problem may arise if one strictly follows the scentence you quoted "If Kaplan-Meier plots cross each other then proportional hazard assumption does not hold" which I would question because sometimes there may be "a little" scrossing like here what not means that proportional assumption must be wrong. If this were true there would be no need for a significance test like the schoenfeld test.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 6 hours ago

























                  answered 6 hours ago









                  igoR87igoR87

                  1266




                  1266



























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