Question 919273
ME = K/n^2  (K = zSD)... same for both experiments
ME1 = K/100^2 vs ME2 = K/500^2
ME1/ME2 = {{{(1/100^2) / (1/500^2 )}}}=  500^2/100^2 = 25
probable error is reduced by 1/25 with the larger sample