My poorly written lecture notes say that any Hermitian operator does have a complete set of orthogonal eigenstates with real corresponding eigenvalues and is therefore an observable.
In the article Observables, it is said that in order for a Hermitian operator to be observable its eigenvectors must form a complete set.
Best Answer
According to the postulates of quantum mechanics, each observable $p$ quantity is associated with an operator $\hat{p}$ that acts on the wavefunction $\psi$.
The relationship is given by the eigenvalue equation: $$ \hat{p}\psi = p\psi. $$
$\hat{p}$ is an operator, which means nothing on its own. $p$ is the eigenvalues, the observable which is a number.
For instance, if $p$ is the momentum:
$\hat{p} = \frac{\hbar}{i}\nabla $, i.e. a functional operator so quite useless on its own;
Acting of a plane wave $\psi = e^{ikx}$, $\hat{p}\psi = \hbar k\,\psi $. I.e. the observable momentum is $p=\hbar k$.