|Well in any case, you may wish to consider arranging your data vertically
|and adding 3 more bits of resolution

|Now in your case, if you wish to extend the data's dynamic range by three
|more bits you'll need (50 samples) * (11bits/sample) = 550 bits. The first
|48 samples can be grouped together in sets of 8 each and would require 528
|bits or 66 bytes. The last 2 samples would be grouped together and would
|require an additional 11 bytes if they were vertical. However, it would
|make sense to keep these horizontal and use only 4 bytes.

I think a slightly easier approach would be to keep the bottom 8 bits
of each 11-bit total in "horizontal" format and just put the top 3 bits
in vertical format, at least for the first 48 of them.  The two oddballs
you could store horizontally, optionally packing them both into a byte.
This would yield a total of 48 bytes for storing LSB's and 18 bytes for
storing 48 sets of MSB's, plus one for the oddballs; total 67.  Note
that your "vertical" math now only needs to be an increment-if-carry,
rather than an add.

Another method, somewhat alluded to above, would be to store MSB's
horizontally, but packed two per byte.  This would need a total of 75
bytes; more than the above, but less than 100.