~Finding sample size with the finite population correction~

~For those of you who are mathematically inclined, this is a very nice exercise in your algebra skills.

~Recall that to find the minimum sample size (without the finite population correction), all you do is to interchange E and √(n) in the formula for the margin of error.  Then square the result to get n and round up, if necessary. Here is the formula for the margin of error for review purposes: (for means)

                E = Z_a/2s/√(n)

~When the finite population correction is used, it multiplies the right side of this formula and becomes:

              E = Z_a/2[s/√(n)]√[(N-n)/(N-1)]

~This is a lot more complicated, as you can see.

~The technique of interchanging E with √(n) is not going to work, since n is also involved in the last factor.

~In algebra, when solving for a quantity, it cannot occur anywhere else in the equation. (it must be isolated on one side of the equation).

~To do this, we must "free"  those terms than contain n.

~Your first step is to "break" the square root in the last factor over the top (numerator) and bottom (denominator). We will get this:

               E = Z_a/2[s/√(n)][√(N-n)] / [√(N-1)]

~Now, multiply both sides by √(N-1).  We get:

              E √(N-1) = Z_a/2(s)[√(N-n)] / [√(n)]

~Divide both sides by Z_a/2s, to get:

             E √(N-1) / Z_a/2s) = √(N-n) / √(n)

~Now, we have all terms containing n on one side without appearing anywhere else.  Since the right side is a square root divided by a another square root, we can eliminate these square roots by squaring both sides. We then will have the following:

            E² (N-1) / (Z_a/2)²s² = (N-n)/n

~Now, to isolate n.  Since this is a mathematical proportion (two fractions equal to each other), we can use the property of cross multiplication (product of the means = product of the extremes).
We then get: [after multiplying (Z_a/2)²s²across N-n on the right side]

      n E² (N-1) = N (Z_a/2)²s² - n(Z_a/2)²s²

~This is not looking so nice!  But, stick with it...

~Since the term on the left side has n and the 2nd term on the right side has n, let's get those on the same side by transposing the one on the right side to the left side (i.e., add it to both sides of the equation).

~Once both of these terms are on the left side, we can factor out a common factor of n.  That gives us this:

     n[ E²(N-1) + (Z_a/2)²s²] = N (Z_a/2)²)s²

~Finally, divide both sides by the factor (in front of n), to get :

       n =  [N (Z_a/2)²s²] / [E²(N-1)+(Z_a/2)²s²]

~That's it!...what a relief!...hope you could follow all the steps..

~NOTE:  For proportions, a similar derivation is used. The end result is as follows:   

  n = [N (Z_a/2)²(P CAP)(Q CAP)] /[E²(N-1)+(Z_a/2)²(P CAP)(Q CAP)]

~It's very unlikely that your teacher will hold you responsible for these derivations...however, you never know...