AUC/C-index confidence interval estimate Is there a way to compute an estimate of the AUC/C-index confidence interval, simply by using a mean AUC/C-index and sample size?
 A: As stated in the comments to your question and other answer, if there is no information such as the variance available (and just the estimate, which I assume you refer to with the 'mean'), AFAIK it is not possible to deduce the necessary information to calculate a confidence interval. 
If you however have access to slightly more information (i.e. the amount of cases and total sample size), you could calculate Hanley's variance of the c-index as estimate for the variance. This is used in meta-analyses of prognostic research (in the biomedical field), when variance or confidence intervals are not given, but the information is needed to pool multiple c-indices.
Hanley variance ($H.var$): if $c$ is the c-index you've found, $N$ is total sample size and $n_1$ resembles the amount of cases, variance is calculated like so:
Take 
$Q1=c/(2-c)$; and $Q2=(2*(c^2))/(1+c)$
Then
$H.var = (c*(1-c)+(n_1-1)*(Q1-(c^2))+(N-n_1-1)*(Q2- c^2))/(n_1*(N-n_1))$
So now you have variance and need to assume a distribution for the c-index which is plausible to find a confidence interval. I've been taught not to assume a normal distribution on the c-index scale, because the c-index is bounded by $[0.5,1]$ (or $[0,1]$ for that matter, but still the bounds exist). Especially when the c-index is close to these bounds a normal distribution is far off. The alternative is to transform everything to the logit scale (which transforms a $[0,1]$ bounded value to a $<-Inf,Inf>$ value), and then calculate a $1-α$% confidence interval using a normal distribution:
$c.logit=log(c/(1-c))$; and $c.var.logit=H.var/((c*(1-c))^{2})$
Then:
$CI =e^{c.logit±z_α*c.var.logit} / (1+ e^{c.logit±z_α*c.var.logit})$
A: If you have AUC calculated for multiple "subjects" or samples and want to estimate the precision of the mean AUC, you can use a simple parametric confidence interval for the mean:
$$CI=M \pm SE_M*t_{crit}=M \pm \frac{s}{\sqrt{n}}*t_{crit}$$
where $M$ is the mean, $s$ is the standard deviation, $n$ is the number of subjects, and $t_{crit}$ is the critical $t$ value with $n-1$ degrees of freedom (asymptotically approaches 1.96).
Edit: A better solution would be to use bootstrapped confidence intervals, but that is more complicated than simply using the mean and sample size. See comments. 
