0
$\begingroup$

I have a question about the 95% confidence interval of KM and Cox. Excuse me that I am not familiar with their deep statistic theory but I thought the p-value, HR and 95%CI should be close to each other in terms of a two-group analysis.

Here is my example.

group   OS  OS.time
C1  0   0.77260274
C2  1   2.61369863
C1  1   2.136986301
C1  0   1.095890411
C2  0   0.821917808
C2  1   1.967123288
C1  0   7.131506849
C1  0   2.794520548
C2  1   4.093150685
C2  1   4.997260274
C1  0   1.120547945
C2  0   1.6
C1  0   0.775342466
C2  0   1.493150685
C1  0   7.02739726
C2  1   3.342465753
C2  0   1.506849315
C1  0   2.073972603
C1  0   8.224657534
C2  0   4.438356164
C2  0   8.180821918
C2  1   3.698630137
C2  0   1.997260274
C1  0   1.490410959
C2  0   4.243835616
C2  1   1.726027397
C2  1   3.695890411
C1  0   3.405479452
C2  1   1.361643836
C2  0   3.876712329
C1  0   2.117808219
C2  0   7.408219178
C1  0   3.994520548
C2  0   0.164383562
C1  0   5.178082192

First, I used KM analysis and calculate 95%CI manually,

fitd=survdiff(Surv(OS.time, OS)~ group, data=tmp, na.action=na.exclude)
p.val <- 1-pchisq(fitd$chisq, length(fitd$n)-1) #0.127
HR = (fitd$obs[2]/fitd$exp[2])/(fitd$obs[1]/fitd$exp[1]) #0.232
up95 = exp(log(HR) + qnorm(0.975)*sqrt(1/fitd$exp[2]+1/fitd$exp[1])) #0.874
low95 = exp(log(HR) - qnorm(0.975)*sqrt(1/fitd$exp[2]+1/fitd$exp[1])) #0.061

Second, I used univariate cox analysis and it calculate 95%CI automatically,

cox <- coxph(Surv(OS.time, OS) ~ group, data = tmp)
summary(cox)
> summary(cox)
Call:
coxph(formula = Surv(OS.time, OS) ~ group, data = tmp)

  n= 35, number of events= 10 

           coef exp(coef) se(coef)      z Pr(>|z|)
groupC2 -1.4768    0.2284   1.0573 -1.397    0.162

        exp(coef) exp(-coef) lower .95 upper .95
groupC2    0.2284      4.379   0.02875     1.814

Concordance= 0.589  (se = 0.09 )
Rsquare= 0.078   (max possible= 0.795 )
Likelihood ratio test= 2.83  on 1 df,   p=0.09
Wald test            = 1.95  on 1 df,   p=0.2
Score (logrank) test = 2.33  on 1 df,   p=0.1

You could see the pvalue for both is not significant, and HR is quite close to 0.23, but 95%CI is quite different in terms of the upper.95 is more than 1, which confused me a lot.

Could anyone explain these two CI to me and which one should I use?

$\endgroup$
0

1 Answer 1

0
$\begingroup$

If I understood correctly, you wanted to know why two calculation methods (manual and automatic) give a different results and which method is correct.

(1) The results of automatic method should be correct. I´ve used that package and manually tested the results before and they were ok.

(2) The results of manual method are controversial and not correct. With manual method HR is 0.232 and CI95% is 0.061-0.874. If this would be true, your p-value should be smaller than 0.05 and significant because both CI95-limits and HR are below 0. Thus there is an error somewhere.

Conclusion: Different results are most likely due to a mistake in manual method calculations. Automatic method should be preferred.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.