Reversing the algebraic sign on the velocity reverses the direction of
motion. Is the same statement true for the acceleration vector and
explain your reasoning?

Absolutely. If you designate the "right" as a positive direction, and
the "left" as the negative direction, than something traveling at
+5m/s is traveling at 5m/s to the right. Add on that that object is
accelerating at +5m/s^2, and it is accelerating to the right, or
getting faster. Now, take this same object traveling at +5m/s, and say
that it is accelerating at -5m/s^2. What is happening? The object is
slowing down. Even though it is still traveling to the right, its
acceleration is to the left. Carry this through further - one second
after this statement is made, the object has stopped. It was going at
5m/s, accelerated at -5m/s^2 in the opposite direction, and is now at
a stand-still. Carry this through yet another second further, and the
object is traveling at -5m/s (5m/s to the left). Why? It is
accelerating in the negative direction. This means that it is slowing
down while its velocity is positive, and speeding up in the negative
direction while the velocity is zero or negative.

muehlbauer-ga is pretty much right, but let me clarify.  

Reversing the algebraic sign of acceleration *DOES NOT* reverse the
direction of motion.

Reversing the algebraic sign of acceleration *DOES* reverse the
direction of acceleration.

If a body is traveling in a positive direction (positive velocity) has
a negative acceleration, that means it is slowing down.  Eventually,
it will stop and begin traveling in a negative direction.

v = at + v0;   where v is instantaneous velocity; v0 is initial
velocity, t is time and a is acceleration.  v can be positive if a is
negative when v0 > -at.  When t >= -v0/a then the direction of travel
becomes negative.