If I am transmitting at 14 bps for example, at the end of 10 seconds I will still have exactly the same amount of data, regardless of whether I transmit 4 bit characters or 8 bit characters. I could transmit 32bit floats over the same transport and I would

*still* be transmitting at 14 bps.

A DTMF tone is a "packet" that holds 4 bits of data, I'm curious how you think I'm misunderstanding it.

If you can fit even the 26 letters in common English usage (not counting numbers or punctuation) into the 16 DTMF symbols, recommend contacting a patent attorney immediately to apply for protection of your intellectual miracle.

If you can fit the "number" 242 into a single DTMF tone then perhaps you should do the same.

Bits are bits, and it doesn't matter what those bits represent. Your transfer rate is entirely dependent on how many DTMF tone bursts you can send per unit time, not if they are to be interpreted as numbers or letters. If you are only transmitting 4 bit "numbers" then it's true you will get twice as many

*symbols* across per unit time, but those N 8 bit values, or N*2 4 bit values represent exactly the same amount of data (N*8 bits).

--edit--

I just realized that instead of "numbers" you probably actually mean "digits" as in the 0-9 that we typically use the DTMF to represent, but the data/bits argument still stands.