39The base being bulky cannot abstract a proton as easily from a hindered carbon in the substrate. Hence it chooses the proton of that carbon which is least hindered, beta to the leaving group. Steric hindrance affects almost everything.
C2H5O- is a little bulkier compared to CH3O- but not as much as to facilitate a Hoffman elimination I think...strange. When only the beta carbon is hindered, it ruins chances of SN, not of Zaitsev elimination.
Note the difference where Zaitsev elimination normally prefers that beta carbon which has more methyl groups or has more hyperconjugative support in the transition state.