Solution: According to the problem, if $A\cap B=B$, then $B\subseteq A$. We will discuss this in three cases:

1. When $p+1 > 2p-1$, or equivalently $p 2$, we have $B=\{x|p+1\leqslant x\leqslant 2p-1\}$. If $B\subseteq A$, then we have the system of inequalities $\begin{cases} 2p-1\leqslant 5 \\\\ p+1\geqslant -2\end{cases}$. Solving this system, we get $(-3\leqslant p\leqslant 3)$. However, since $p > 2$, the condition is satisfied when $2 < p\leqslant 3$.

In summary, the condition $A\cap B=B$ holds when $p\leqslant 3$. Therefore, the range of values for $p$ is $\boxed{p\leqslant 3}$.