How to test that an individual differ from the group? I have an odd question concerning the way one could assess that a value collected on a single individual differs from values collected on a group of people N (clinical research). This is a lot like assessing that an individual is considered as an outlier I guess.
I was just thinking of two ways of doing it:


*

*firstly one could just calculate the mean and standard deviation of the group (N values) and then assess the number of standard deviation the score of the individual is located from the group;

*secondly I was just thinking of performing a one sample t test where the mean of values collected on the group differ from the single value collected on the individual (instead of testing against 0, testing against the value of the individual).
I would like to know if this would be a correct approach or if there might be other tests I do not know about.
Thank you!
 A: You're basically interested in knowing how probable your test value $V$ is given the distribution that generated the $N$ other values. Also, from your selected tests, it sounds like we are testing for extremeness of a value relative to others (as opposed to, for example, inliers, or "splitting two modes").
I'd suggest against using either approach you gave. The former only makes sense if your population is well-approximated by a normal. However, the distribution of the $N$ values can really be almost anything (Cauchy, Exponential, etc), so the z-score may be meaningless. (Although Chebyshev Inequality gives rough bounds if you can estimate the mean and standard deviation well enough).
The t-test is not useful here because you cannot (or maybe don't want to) assume the null distribution of either the $N$ values nor your single value $V$ come from an approximately normal distribution.
Why not just compute the percentile of your test value $V$ relative to the $N$ values? 
If we assume that your $N+1$ values are sampled iid from the same underlying distribution $P$ with CDF $F$, then $P(F(V)>0.99) < 0.01$. We can use the $N$ values to develop the empirical distribution function for $F$, from which we get the estimate of the percentile of $V$ as $\hat{F}(V)$. 
If $\hat{F}(V)>0.99$ we can reject with approximately $99%$ confidence.
Update based on comments
As pointed out by @whuber, a parametric approach will probably be a better use of your data (provided you can justify some parametric family). If your data are reasonably normal, then you can fit a $t$-distribution to your $N$ points. The cool thing about the $t$-distribution is that it is the exact predictive distribution for a new data point, given the observed data come from a normal distribution.
So, let's say you have a sample of $N$ points from some Normal distribution, but you don't know the mean $\mu$ or standard deviation ($\sigma$). However, you have estimates of the mean and standard deviation using your $N$ points (i.e., $\hat{\mu} = \bar{x},\;\hat{\sigma}=s$), and you want to use these estimates to construct an interval that will contain the next value with some probability (say, 95% chance). 
You can use the $t$-distribution to construct your (central) prediction intervals for the new observation $y$ from the N existing data points $\mathbf{x}$like so (central means its symmetric about the sample mean):
$$ \mathrm{Let}\;\;I_{\alpha,N} := \bar{x} \pm t_{\alpha/2}s_N\sqrt{1+\frac{1}{N}},\; \mathrm{then}$$
$$P(y \in I_{\alpha}) = 1-\alpha$$
So, in your case, you have your $N$ points and your $y$ is the "outlier" to be tested. You can set $\alpha$ to, say, .05 or .01. Then, if your point is outside this interval, it means that a point at least this extreme only happens 0.05 or .01 of the time, and so seems "unusual" give the other data.
That is one way forward.
