This password strength test is attempting to estimate the entropy of given passwords under the assumption that it is a word roughly obeying the character distributions of English text, using Shannon's approximation. It does not apply to randomly generated passwords, which violate these assumptions, as described in Appendex A of the NIST plublication linked on that very site.[1] As that document describes, the entropy of a randomly (not user!) selected password is not estimated in this manner, but calculated according to the same formula I provided: H = log_2 (b^L), where b is the number of letters in the alphabet (95) and L is the length of the password. (If the password had any less entropy than that, then by definition it would not have been pseudorandomly generated!)
Additionally, the algorithm employed by this site does not take into account that an intelligent password cracker is capable of exploiting the construction of passphrases.
In other words, you've taken an algorithm designed to approximate the entropy of a user-selected password and misapplied it to both randomly-generated passwords and user-selected passphrases. The fact that the algorithm is able to give you a number for these inputs does not mean it is any valid indication of how difficult such a password or passphrase would be to crack.
[1] http://csrc.nist.gov/publications/nistpubs/800-63/SP800-63V1...
EDIT: But you're right that passphrases can be easier to use by people. I personally think a password manager like keepass / lastpass / etc. is a better choice than trying to select a memorable password, though.