That's wrong except the conclusion. If a grammar is LL(1) and not containing $\epsilon$, then it is SLR(1) and so LALR(1) and LR(1).
LL(1) generate a superset of regular and subset of DCFL. LL(k) strictly generate a subset than LL(k+1) for k >=0.
SLR(1) generate DCFL. LALR(1) and LR(1) also generate DCFL. LR(k) generate the same language as LR(k+1) for k > 0. LR(0) generate the DCFL with prefix property (DPDA with empty stack).
For parser we tell power not in terms of the language their grammar can generate, but the grammars they can parse. SLR(1) < LALR(1) < LR(1), because they can parse more and more grammars in that order though all those grammars can generate the same set- DCFL. i.e., for any LR(1) grammar there is an equivalent SLR(1) grammar which can be parsed by a SLR(1) parser.