Check if $REPLY is in a range of numbers

The [ command/shell builtin has comparison tests, so you can just do

if [ "$REPLY" -ge 1 ] && [ "$REPLY" -le 32 ]; then REPLY=-2;
elif [ "$REPLY" -ge 33 ] && [ "$REPLY" -le 48 ]; then REPLY=-1; fi

where -ge means greater-or-equal-to (and so on). The [ command is just a command, not special syntax (it's actually the same as test: check out man test), so it NEEDS the space after it. If you write [$REPLY it will try to find a command named [$REPLY and execute it, which won't work. The same goes for closing ].

Here, we're using the && shell operator to run the second command only if the first is successful. [ also supports -a to and two tests, but it's deprecated and its usage should be discouraged as it causes arguments not to be parseable reliably.

Edit: to test if the number is integer (if that can happen in your code), first do the test

if [[ "$REPLY" =~ ^[0-9]+$ ]]; then
   existing code
else echo "$REPLY is not an integer" >&2 && exit 1; fi

Of course all these bracket expressions return 0 (true) or 1 (false) and can be combined. Not only you can put everything in the same bracket, you can also do

if [[ "$REPLY" =~ ^[0-9]+$ ]] && [ "$REPLY" -ge 1 ] && [ "$REPLY" -le 32 ]; then ...

or something similar.

You could simply say:

((REPLY>=1 && REPLY<=32)) && REPLY=-2
((REPLY>=33 && REPLY<=48)) && REPLY=-1

Quoting from the manual:

((...))

(( expression ))

The arithmetic expression is evaluated according to the rules described below (see Shell Arithmetic). If the value of the expression is non-zero, the return status is 0; otherwise the return status is 1. This is exactly equivalent to

let "expression"

You could do something like this:

#!/usr/bin/env bash
read -p "- Audio Quality [scale from -2 to 10] ? "
if [ -n "$REPLY" ] ; then
    ABITRATE="-aq $REPLY"
fi

echo "You chose : $ABITRATE : $REPLY"
## If 0 < $REPLY < 33 and $REPLY is a number
if [[ "$REPLY" =~ ^[0-9]+$ && "$REPLY" -gt 0 && "$REPLY" -lt 33 ]]
then
    echo "GOOD"
else
    echo "BAD"
fi