Schoenfled Residua test shows proportionality hazard assumptions holds but Kaplan-Meier plots...
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Schoenfled Residua test shows proportionality hazard assumptions holds but Kaplan-Meier plots intersect
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$begingroup$
"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
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
asked 15 hours ago
Omar RafiqueOmar Rafique
456
$endgroup$
add a comment |
$begingroup$
"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
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
asked 15 hours ago
Omar RafiqueOmar Rafique
456
$endgroup$
2
$begingroup$
If you cannot reject the null hypothesis, it does not mean that it is true.
$endgroup$
– Michael M
14 hours ago
$begingroup$
This reasoning accounts for the p-value. What about the Schoenfled residual plots being flat....
$endgroup$
– Omar Rafique
12 hours ago
add a comment |
$begingroup$
"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
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
asked 15 hours ago
Omar RafiqueOmar Rafique
456
$endgroup$
"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
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
asked 15 hours ago
Omar RafiqueOmar Rafique
456
asked 15 hours ago
Omar RafiqueOmar Rafique
456
edited 15 hours ago
Omar Rafique
asked 15 hours ago
Omar RafiqueOmar Rafique
456
asked 15 hours ago
Omar RafiqueOmar Rafique
456
asked 15 hours ago
Omar RafiqueOmar Rafique
456
456
2
$begingroup$
If you cannot reject the null hypothesis, it does not mean that it is true.
$endgroup$
– Michael M
14 hours ago
$begingroup$
This reasoning accounts for the p-value. What about the Schoenfled residual plots being flat....
$endgroup$
– Omar Rafique
12 hours ago
add a comment |
2
$begingroup$
If you cannot reject the null hypothesis, it does not mean that it is true.
$endgroup$
– Michael M
14 hours ago
$begingroup$
This reasoning accounts for the p-value. What about the Schoenfled residual plots being flat....
$endgroup$
– Omar Rafique
12 hours ago
2
2
$begingroup$
If you cannot reject the null hypothesis, it does not mean that it is true.
$endgroup$
– Michael M
14 hours ago
$begingroup$
If you cannot reject the null hypothesis, it does not mean that it is true.
$endgroup$
– Michael M
14 hours ago
$begingroup$
This reasoning accounts for the p-value. What about the Schoenfled residual plots being flat....
$endgroup$
– Omar Rafique
12 hours ago
$begingroup$
This reasoning accounts for the p-value. What about the Schoenfled residual plots being flat....
$endgroup$
– Omar Rafique
12 hours ago
add a comment |
2 Answers
2
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oldest
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$begingroup$
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.
answered 13 hours ago
EdMEdM
22.2k23496
$endgroup$
add a comment |
$begingroup$
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.
answered 13 hours ago
igoR87igoR87
1368
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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.
answered 13 hours ago
EdMEdM
22.2k23496
$endgroup$
add a comment |
$begingroup$
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.
answered 13 hours ago
EdMEdM
22.2k23496
$endgroup$
add a comment |
$begingroup$
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.
answered 13 hours ago
EdMEdM
22.2k23496
$endgroup$
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.
answered 13 hours ago
EdMEdM
22.2k23496
answered 13 hours ago
EdMEdM
22.2k23496
answered 13 hours ago
EdMEdM
22.2k23496
answered 13 hours ago
EdMEdM
22.2k23496
22.2k23496
add a comment |
add a comment |
$begingroup$
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.
answered 13 hours ago
igoR87igoR87
1368
$endgroup$
add a comment |
$begingroup$
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.
answered 13 hours ago
igoR87igoR87
1368
$endgroup$
add a comment |
$begingroup$
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.
answered 13 hours ago
igoR87igoR87
1368
$endgroup$
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.
answered 13 hours ago
igoR87igoR87
1368
edited 13 hours ago
answered 13 hours ago
igoR87igoR87
1368
answered 13 hours ago
igoR87igoR87
1368
answered 13 hours ago
igoR87igoR87
1368
1368
add a comment |
add a comment |
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2
$begingroup$
If you cannot reject the null hypothesis, it does not mean that it is true.
$endgroup$
– Michael M
14 hours ago
$begingroup$
This reasoning accounts for the p-value. What about the Schoenfled residual plots being flat....
$endgroup$
– Omar Rafique
12 hours ago