Answer:
For a margin of error of ± $400 , the required sample size is n = 150.0625
For a margin of error of ± $210 , the required sample size is n = 544.44
For a margin of error of ± $110 , the required sample size is n = 1984.2975
Step-by-step explanation:
a) M. E = + 400, n= ?
n= [1.96 * 2500 /400] ^2 = 150.0625
b) m. E = 210
n= [1.96 * 2500/210]^2 = 544.44
c) M.E = 110
n= [1.96 * 2500 /110] = 1984