Pretty much, yeah. This is really just a generalization of the principle we see in the length of a binary password. The complexity is s^l, where s is the number of symbols you can possibly draw from, and l is the number of symbols in the password.

Consider a binary string. There are two symbols in binary (1 and 0) and the number of symbols you use is the number of bits in the string. This gives us the familiar password length calculation: the complexity is 2^l.

When you start using words as the basic unit, you gain many more symbols: if you use the full Oxford English Dictionary, you have about 600,000 of them. But when you do this, your symbol is no longer a single byte: it's a word. That word can be represented as a string of bytes, yes, but it's guaranteed that this string will be one of the 600,000 byte strings that represent English words, and you can eliminate all of the others. So you can't measure your password in bytes anymore; you have to measure it in words instead. A one-word password is, therefore, of complexity (600,000^1).

Six hundred thousand possibilities sounds like a lot, but computers can crack that in seconds. Even if you restrict yourself to just printable ASCII, you can do better than that in only three characters. I don't recommend actually doing a three-character printable-ASCII password, because you'd have just under 900,000 possible passwords: still well within crackable range.

But because the complexity increases so fast as you increase the length of the password, that's how you can gain complexity back. If you use four words instead of one, you have (600,000^4) possibilities: some 130 sextillion. You'd need 70 random bits, or 11 printable ASCII characters, to do better.

And that, my friends, is why "k[0" is a better password than "antidisestablishmentarianism". But it's also why "correct horse battery staple" is better than either one (or would be, if xkcd hadn't made it so famous).