Does the gravitational force of 9.8m/s^2 change depending on your elevation?
Yes. It is inversely proportional to the square of the distance from the center of the Earth.
At the top of Mt. Everest (h = 8.84 km), weight and "g" would be reduced by a factor of approximately
[R/(R+h)]^2 = [6371/6379.84]^2 = 0.9972
which is 0.28% less than at sea level.
(R = 6371 km is the radius of the Earth)
There are additional corrections due to the presence of the mountain itself. Stricly speaking, the formula I used applies only if one is located a distance h above a perfect sphere with mass distributed as a function of r only.