A model $M$ of countable similarity type and cardinality $\kappa$ is expandable if every consistent extension $T_{1}$ of its complete theory with $|T_{1}|\leq \kappa$ is satisfiable in $M$ and it is compactly expandable if every such extension which additionally is finitely satisfiable in $M$ is satisfiable in $M$ . In the countable case and in the case of a model of cardinality $\geq 2^{\omega}$ of a superstable theory without the finite cover property the notions of saturation, expandability and compactness for (...) expandability agree. The question of the existence of compactly expandable models which are not expandable is open. Here we present a test which serves to prove that a compactly expandable model of cardinality $\geq 2^{\omega}$ of a superstable theory is expandable. It is stated in terms of the existence of a certain elementary submodel whose corresponding theory of pairs of models satisfies a weak elimination of Ramsey quantifiers. (shrink)