Why base current is affected most on changing beta in voltage divider bias configuration of BJT?
What if beta were infinity? The base voltage would be exactly defined by the potential divider formed by R1 and R2 and the 16 volt supply: -

i.e. Vb would be 16V \$\times \dfrac{9.1}{9.1+62}\$ = 2.048 volts.

If beta dropped from infinity, the current taken by the base (although small) would lower that 2.048 volts a little bit. With a base current of 20 uA, Vb drops to about 1.89 V (a drop of about 7.8%).

I converted the 16 volt source to a current source and paralleled R1 and R2 then subtracted 20 uA to get the new base voltage by the way

I'm just trying to demonstrate that a fundamentally massive swing in beta from around 90 to infinity results in a change in Vb of about 8%. This change in bias voltage has a knock-on effect on Ic and Ib of approximately the same amount.

If R1 and R2 were lowered in value by 10 the influence of the 20 uA base bias current would take Vb from 2.048 volts to 2.03 volts i.e. a drop of about 1%. 

This is why "strong" bias resistor values are preferred because (in conjunction with an emitter resistor) the change in beta from one transistor to another (or from one temperature to another) results in a far smaller change in quiescent operating conditions.