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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
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    $\begingroup$ I think that arguing about p-values with a sample size of $n=6$ is a bit of a tall order... Yes, there are not absolute cut-offs for a minimum sample size for a test to be applicable but that does mean the test might not be underpowered. $\endgroup$ – usεr11852 Oct 26 '18 at 23:37
  • $\begingroup$ I am voting to leave this open. It is not about R, really. The same sort of question could come up with any statistical package. $\endgroup$ – Peter Flom Oct 27 '18 at 13:06
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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.

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    $\begingroup$ 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. $\endgroup$ – gung Oct 26 '18 at 20:06
  • $\begingroup$ I agree that n=6 is way too little to be able to draw any firm conclusion, even if it would have turned out significant. And even if it could be motivated to choose the LRT p-value here, I still would not present it as a significant finding in a paper, if for no other reason than that it will be interpreted as suspicious by the reader. At this point, I am mainly interested in it from a theoretical point view. But as I understand it from the cited answer, whether the LRT is appropriate here would depend on whether the variable is normally distributed. $\endgroup$ – zbox Oct 28 '18 at 0:16

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