The CI is a random interval. Per definition it does not fit anyway to a theoretic value. This is the main problem here. The CI always belongs to a random sample, never to a fixed, given, theoretical or hypothesized value.

Thank you for your answer. But it already exists intervals under the null hypothesis for mean or proportion which bet how much the mean/proportion is likely to bounce when you only observe a sample with N individuals. And this interval is always the same for a given null hypothesis and a given N. I was wondering if such an interval could be computed for a median survival time...

Dear Audrey,

I am not sure if I understand you question correctly. But survival rates over time can easily be analyzed with a Cox regression model (coxph function in R). This function gives you also the CIs you need. In addition, you could try to compare survival proportion values fitted to a generalized linear model (GLM) with binominal error structure and logit link function.

Thank you Boran. To explain better my question: I have a single arm population (N=60) for which I would like to know if the PFS (Progression-Free Survival) median is significantly different from 5 months. I know I could use the 95%CI of the estimated value but I'd rather use a 95% interval of the theoretical value... I think the two approaches are really different. Moreover, by simulation, the second one seems more powerful. Assuming an exponential or a weibull distribution under the null hypothesis, could I predict the observed median value? Not the opposite...

Dear Audrey,

I insert a R code that might help you. It uses the parametric model survreg with assumed Weibull distribution and the data lung from the survival package.

> library (survival)

> fit1 tapply(predict(fit1,type="response"),lung$sex, median)

1 2

359.3015 533.6498

#use kaplan meier estimate to determine median and CI

> fit3 fit3

Call: survfit.formula(formula = Surv(time, status) ~ sex, data = lung)

records n.max n.start events median 0.95LCL 0.95UCL

sex=1 138 138 138 112 270 212 310

sex=2 90 90 90 53 426 348 550

#you could use also library(rms)

#fit2

Thank you very much Boran,

Indeed, this is what I use when I want to predict the median. My question is: is it the real statistical test to compare this estimated median to a theoretical one? Because for me, a hypothesis test is made under the null hypothesis and the CI we're talking about is not computed under H0. Maybe it's note clear... The usual CI is centered around the OBSERVED value, whereas the interval under the null hypothesis is centered around the THEORETICAL value (H0).

Audrey, it is not the "usual(!) CI" that is centered around the observerd values - - the CI is a function of the observed data, there is in principle no CI thinkable that would *not* be centered araound the observed values. The CI is obtained by P(lower < observed < upper) < confidence.

Under H0, the parameters do NOT have any variability. By definition, H0 is a fixed scenario. The only thing that is vaiable is the data, and thus the sample mean and the CI (and the p-value!).

Thank you Jochen, I totally agree that under H0, the parameters do not have any variability, but I would like to predict what happened when only considering a sample of the "H0-population". I mean, how much will be the sampling fluctuations? If I KNOW that the median PFS is 5-months in the whole population, what must I observe with only 60 individuals?

About the "usual CI" I think I have some language issues, because in French we talk about "intervalle de fluctuation" when using the theoretical value in the equation and about "intervalle de confiance" when using the observed value.

Let us take an example:

If you know that the proportion P is 0.40, you can bet, with a risk of 5% that the proportion, you will observe on a sample with N individuals, will be included in the following interval: P+/-1.96*sqrt( P*(1-P) / N ). Comparing the observed proportion of a N-sample to this interval is equivalent to test the null hypothesis H0: P=0.40

But if you don't know the proportion P of a population from which a N-sample is derived you will use the observed proportion "p" and bet that the proportion P is included in this CI: p+/-1.96*sqrt( p*(1-p) / N ). If 0.40 is not in this CI, is it equivalent to test H0: P=0.40?

If so, my problem is solved, if not how can I compute the 1st interval for a median survival time?

Sorry, I meant:

Examining if 0.40 is not in this CI, is it equivalent to test H0: P=0.40?

Examining 0.40 is in the CI.95 is equivalent to testing the H0: P=0.40. If the CI.95 does not include P then the H0 can be rejected at alpha=0.95. CI and test are (usually) equivalent, CIs are (usually) constructed from inverted tests.

Thank you! You respond to my question!

You could also use liklihood ratio test - assuming exponential, compare loglikelihood of data at MLE vs null lambda (ln2/med)

Thank you Milan.You're right. One of my colleague has explained to me that a theoretical median is equivalent to a theoretical proportion under the exponential distribution so an exact binomial test can be used to test H0: p(5months)=0.5 versus H1: p(5months)=0.63median of 7.5 months.

thank you everybody for your help.

That is perfect and simple if no censoring prior to 5 months - excellent.