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Different p-values for coefficients and LRT in univariate cox regression (coxph R)
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I have used Cox PH to test the relationship between one predictor and survival for 6 patients. Cox PH was used since the predictor is continuous. Using the "cox.zph" function, there was no significant violation of the proportional hazards assumption. However, I am confused about the interpretation in this case. The p-value for the predictor coefficient was non-significant (p=0.23), but the likelihood ratio test for the model gave p=0.03. As I understand it, the latter tests whether the coefficients used in the model significantly differ from 0, and therefore whether or not they contribute to the model. Since only one variable is used as predictor here, the results therefore seem conflicting. However, the Wald test and Score (logrank) tests shown for the model when using the "summary" function also show p=0.2 and p=0.1, respectively, which are more in line with the coefficient p-value. By reading up on this topic I have come to understand that the LRT, Wald and log-rank tests are asymptotically equivalent, but that the latter two have issues with a small sample size and that the LRT is preferrable in such cases. Given the few samples used in this case, this can explain the differences for the LTR, Wald, and log-rank tests for the model. But the p-value calculated for the coefficient itself is, as far as I understand it, also based on a Wald test. In the case of a univariate model, I suppose that the model Wald test and the coefficient Wald test may even be the same test. By this logic, I would suppose that the coefficient Wald test would also be underpowered with respect to the sample size and that the LRT test may be a more appropriate substitute for the coefficient Wald test, since the test is univariate. Is this the case?
The results of the test are shown below:
Call:
coxph(formula = SurvObj ~ n, data = surv_data)
n= 6, number of events= 3
coef exp(coef) se(coef) z Pr(>|z|)
n -0.09064 0.91335 0.07580 -1.196 0.232
exp(coef) exp(-coef) lower .95 upper .95
n 0.9133 1.095 0.7873 1.06
Concordance= 0.9 (se = 0.227 )
Rsquare= 0.542 (max possible= 0.745 )
Likelihood ratio test= 4.69 on 1 df, p=0.03
Wald test = 1.43 on 1 df, p=0.2
Score (logrank) test = 2.24 on 1 df, p=0.1
survival cox-model likelihood-ratio univariate
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up vote
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I have used Cox PH to test the relationship between one predictor and survival for 6 patients. Cox PH was used since the predictor is continuous. Using the "cox.zph" function, there was no significant violation of the proportional hazards assumption. However, I am confused about the interpretation in this case. The p-value for the predictor coefficient was non-significant (p=0.23), but the likelihood ratio test for the model gave p=0.03. As I understand it, the latter tests whether the coefficients used in the model significantly differ from 0, and therefore whether or not they contribute to the model. Since only one variable is used as predictor here, the results therefore seem conflicting. However, the Wald test and Score (logrank) tests shown for the model when using the "summary" function also show p=0.2 and p=0.1, respectively, which are more in line with the coefficient p-value. By reading up on this topic I have come to understand that the LRT, Wald and log-rank tests are asymptotically equivalent, but that the latter two have issues with a small sample size and that the LRT is preferrable in such cases. Given the few samples used in this case, this can explain the differences for the LTR, Wald, and log-rank tests for the model. But the p-value calculated for the coefficient itself is, as far as I understand it, also based on a Wald test. In the case of a univariate model, I suppose that the model Wald test and the coefficient Wald test may even be the same test. By this logic, I would suppose that the coefficient Wald test would also be underpowered with respect to the sample size and that the LRT test may be a more appropriate substitute for the coefficient Wald test, since the test is univariate. Is this the case?
The results of the test are shown below:
Call:
coxph(formula = SurvObj ~ n, data = surv_data)
n= 6, number of events= 3
coef exp(coef) se(coef) z Pr(>|z|)
n -0.09064 0.91335 0.07580 -1.196 0.232
exp(coef) exp(-coef) lower .95 upper .95
n 0.9133 1.095 0.7873 1.06
Concordance= 0.9 (se = 0.227 )
Rsquare= 0.542 (max possible= 0.745 )
Likelihood ratio test= 4.69 on 1 df, p=0.03
Wald test = 1.43 on 1 df, p=0.2
Score (logrank) test = 2.24 on 1 df, p=0.1
survival cox-model likelihood-ratio univariate
asked 3 hours ago
zbox
61
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zbox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have used Cox PH to test the relationship between one predictor and survival for 6 patients. Cox PH was used since the predictor is continuous. Using the "cox.zph" function, there was no significant violation of the proportional hazards assumption. However, I am confused about the interpretation in this case. The p-value for the predictor coefficient was non-significant (p=0.23), but the likelihood ratio test for the model gave p=0.03. As I understand it, the latter tests whether the coefficients used in the model significantly differ from 0, and therefore whether or not they contribute to the model. Since only one variable is used as predictor here, the results therefore seem conflicting. However, the Wald test and Score (logrank) tests shown for the model when using the "summary" function also show p=0.2 and p=0.1, respectively, which are more in line with the coefficient p-value. By reading up on this topic I have come to understand that the LRT, Wald and log-rank tests are asymptotically equivalent, but that the latter two have issues with a small sample size and that the LRT is preferrable in such cases. Given the few samples used in this case, this can explain the differences for the LTR, Wald, and log-rank tests for the model. But the p-value calculated for the coefficient itself is, as far as I understand it, also based on a Wald test. In the case of a univariate model, I suppose that the model Wald test and the coefficient Wald test may even be the same test. By this logic, I would suppose that the coefficient Wald test would also be underpowered with respect to the sample size and that the LRT test may be a more appropriate substitute for the coefficient Wald test, since the test is univariate. Is this the case?
The results of the test are shown below:
Call:
coxph(formula = SurvObj ~ n, data = surv_data)
n= 6, number of events= 3
coef exp(coef) se(coef) z Pr(>|z|)
n -0.09064 0.91335 0.07580 -1.196 0.232
exp(coef) exp(-coef) lower .95 upper .95
n 0.9133 1.095 0.7873 1.06
Concordance= 0.9 (se = 0.227 )
Rsquare= 0.542 (max possible= 0.745 )
Likelihood ratio test= 4.69 on 1 df, p=0.03
Wald test = 1.43 on 1 df, p=0.2
Score (logrank) test = 2.24 on 1 df, p=0.1
survival cox-model likelihood-ratio univariate
asked 3 hours ago
zbox
61
New contributor
zbox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have used Cox PH to test the relationship between one predictor and survival for 6 patients. Cox PH was used since the predictor is continuous. Using the "cox.zph" function, there was no significant violation of the proportional hazards assumption. However, I am confused about the interpretation in this case. The p-value for the predictor coefficient was non-significant (p=0.23), but the likelihood ratio test for the model gave p=0.03. As I understand it, the latter tests whether the coefficients used in the model significantly differ from 0, and therefore whether or not they contribute to the model. Since only one variable is used as predictor here, the results therefore seem conflicting. However, the Wald test and Score (logrank) tests shown for the model when using the "summary" function also show p=0.2 and p=0.1, respectively, which are more in line with the coefficient p-value. By reading up on this topic I have come to understand that the LRT, Wald and log-rank tests are asymptotically equivalent, but that the latter two have issues with a small sample size and that the LRT is preferrable in such cases. Given the few samples used in this case, this can explain the differences for the LTR, Wald, and log-rank tests for the model. But the p-value calculated for the coefficient itself is, as far as I understand it, also based on a Wald test. In the case of a univariate model, I suppose that the model Wald test and the coefficient Wald test may even be the same test. By this logic, I would suppose that the coefficient Wald test would also be underpowered with respect to the sample size and that the LRT test may be a more appropriate substitute for the coefficient Wald test, since the test is univariate. Is this the case?
The results of the test are shown below:
Call:
coxph(formula = SurvObj ~ n, data = surv_data)
n= 6, number of events= 3
coef exp(coef) se(coef) z Pr(>|z|)
n -0.09064 0.91335 0.07580 -1.196 0.232
exp(coef) exp(-coef) lower .95 upper .95
n 0.9133 1.095 0.7873 1.06
Concordance= 0.9 (se = 0.227 )
Rsquare= 0.542 (max possible= 0.745 )
Likelihood ratio test= 4.69 on 1 df, p=0.03
Wald test = 1.43 on 1 df, p=0.2
Score (logrank) test = 2.24 on 1 df, p=0.1
survival cox-model likelihood-ratio univariate
survival cox-model likelihood-ratio univariate
asked 3 hours ago
zbox
61
New contributor
zbox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
zbox
61
New contributor
zbox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
zbox
61
New contributor
zbox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
zbox
61
asked 3 hours ago
zbox
61
61
New contributor
zbox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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zbox is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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1 Answer
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"I suppose that the model Wald test and the coefficient Wald test may even be the same test" You are right on it. In fact you can check that $z^2=$ Chi square, $(-1.196)^2 = 1.430$.
But for LRT, its validity also depends on the large sample if response variable is not normal. See Is the likelihood ratio test a large sample inference method?.
You want to select LRT because LRT produces the p value that you wanted. In fact, when sample size = 6, the reliability of statistical results is low.
answered 3 hours ago
a_statistician
2,269139
1
6 patients w/ 3 events is pretty tough. It isn't clear it anything can be done there w/o recourse to pretty strong assumptions.
– gung♦
3 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
"I suppose that the model Wald test and the coefficient Wald test may even be the same test" You are right on it. In fact you can check that $z^2=$ Chi square, $(-1.196)^2 = 1.430$.
But for LRT, its validity also depends on the large sample if response variable is not normal. See Is the likelihood ratio test a large sample inference method?.
You want to select LRT because LRT produces the p value that you wanted. In fact, when sample size = 6, the reliability of statistical results is low.
answered 3 hours ago
a_statistician
2,269139
1
6 patients w/ 3 events is pretty tough. It isn't clear it anything can be done there w/o recourse to pretty strong assumptions.
– gung♦
3 hours ago
add a comment |Â
up vote
3
down vote
"I suppose that the model Wald test and the coefficient Wald test may even be the same test" You are right on it. In fact you can check that $z^2=$ Chi square, $(-1.196)^2 = 1.430$.
But for LRT, its validity also depends on the large sample if response variable is not normal. See Is the likelihood ratio test a large sample inference method?.
You want to select LRT because LRT produces the p value that you wanted. In fact, when sample size = 6, the reliability of statistical results is low.
answered 3 hours ago
a_statistician
2,269139
1
6 patients w/ 3 events is pretty tough. It isn't clear it anything can be done there w/o recourse to pretty strong assumptions.
– gung♦
3 hours ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
"I suppose that the model Wald test and the coefficient Wald test may even be the same test" You are right on it. In fact you can check that $z^2=$ Chi square, $(-1.196)^2 = 1.430$.
But for LRT, its validity also depends on the large sample if response variable is not normal. See Is the likelihood ratio test a large sample inference method?.
You want to select LRT because LRT produces the p value that you wanted. In fact, when sample size = 6, the reliability of statistical results is low.
answered 3 hours ago
a_statistician
2,269139
"I suppose that the model Wald test and the coefficient Wald test may even be the same test" You are right on it. In fact you can check that $z^2=$ Chi square, $(-1.196)^2 = 1.430$.
But for LRT, its validity also depends on the large sample if response variable is not normal. See Is the likelihood ratio test a large sample inference method?.
You want to select LRT because LRT produces the p value that you wanted. In fact, when sample size = 6, the reliability of statistical results is low.
answered 3 hours ago
a_statistician
2,269139
edited 3 hours ago
answered 3 hours ago
a_statistician
2,269139
answered 3 hours ago
a_statistician
2,269139
answered 3 hours ago
a_statistician
2,269139
2,269139
1
6 patients w/ 3 events is pretty tough. It isn't clear it anything can be done there w/o recourse to pretty strong assumptions.
– gung♦
3 hours ago
add a comment |Â
1
6 patients w/ 3 events is pretty tough. It isn't clear it anything can be done there w/o recourse to pretty strong assumptions.
– gung♦
3 hours ago
1
1
6 patients w/ 3 events is pretty tough. It isn't clear it anything can be done there w/o recourse to pretty strong assumptions.
– gung♦
3 hours ago
6 patients w/ 3 events is pretty tough. It isn't clear it anything can be done there w/o recourse to pretty strong assumptions.
– gung♦
3 hours ago
add a comment |Â
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