Use a 15 K ohm, or slightly smaller resistor.Sounds good - I like the idea of testing it with a resistor.
It looks like it should trip if I connect a 47KΩ between the ground screw and the load side of the Levitron switch in the box ?
Will turning the switch on immediately trip the GFI ?
A 47K ohm resistor may or may not work. at 120V, 47K ohms will only give you a 2.5 mA "leakage" current. A GFCI is not required to trip below 5 mA (I'm still trying to determine if that is the nominal, or upper spec limit), and is required not to trip below 1.9 - 2.0 mA. You also have to worry about the tolerance of the resistor. Most common resistors are +/- 10%, so your 47K resistor might be 51.7K, with a current of only 2.3 mA. Also, your voltage could be below nominal. If your "120V" was only 105V (during brown outs voltages can drop below 100V, IIRC), your test current might only be 105 / 51700 = 2.0 mA.
I'd recommend designing the test circuit to provide a minimum fault current of 6 mA under worst case conditions. 100V / 0.006A = 16.7K ohm. 15 K ohm is a standard resistor value that will be less than 16.7 K ohm even if it is at it's upper spec limit on tolerance.
What you really want to test is that the trip mechanism actually works, as it is the most likely failure point of the GFCI. Shift in trip limit is less likely, and also a less dangerous "fail" condition for the GFCI (as long as it stays under ~10 mA.) Also, testing the actual trip limit will require some precision measuring equipment ($$$.)
Brew on