
Patrick B. answered 07/26/20
Math and computer tutor/teacher
The problem can be restated as:
" What is the 99% confidence interval for the proportion of blue marbles to be 49% or less."
Let p be the desired proportion and N be the # of marbles in the bag
the confidence constant is z=
the error margin is E = z*sqrt( p * (1-p)/ n)
E^2 = z^2 * (p * (1-p)/ n) <--- the error margin as a function of the proportion and N
N*E^2/z^2 = p(1-p)
p^2 - p - N*E^2/z^2 = 0
Quadratic formula says:
p = [ 1 +or- sqrt ( 1 + 4* N*E^2/z^2)/2 <--- the proportion as a function of N and error margin
N = p(1-p)*z^2/E^2 <--- the sample size as a function of the proportion and error margin