4. If $0<\theta<30^{\circ}$, and $\sin \theta=k m+\frac{1}{3}(k$ is a constant and $k<0)$, then the range of values for $m$ is
4. $\frac{1}{6 k}<m<-\frac{1}{3 k}$.

Given $0<\theta<30^{\circ}$, then $0<\sin \theta<\frac{1}{2}$, i.e., $0<k m+\frac{1}{3}<$ $\frac{1}{2}$. And $k<0$, solve accordingly.