The sample size, n required to produce a margin of error, M, in a probability proportion distribution is given by:
[tex]n= \frac{p(1-p)z_{\alpha/2}^2}{M^2} [/tex]
The above formular shows that the sample size is inversely proportional to the Margin of error.
Then we have:
[tex]n= \frac{k}{M^2} [/tex]
where k is some constant.
Now suppose we decrease M by a factor of 3,
[tex]\left(i.e.\ \frac{M_1}{M_2} =3\right)[/tex]
[tex]n_1= \frac{k}{M_1^2} \\ \\ n_2= \frac{k}{M_2^2} \\ \\ \frac{n_2}{n_1} = \frac{k}{M_2^2} / \frac{k}{M_1^2} = \frac{k}{M_2^2} \times \frac{M_1^2}{k} = \frac{M_1^2}{M^2_2} =\left( \frac{M_1}{M_2} \right)^2=3^2=9 \\ \\ n_2=9n_1[/tex]
Therefore, if we decrease the Margin of error by a factor of 3, we need to increase the sample size by a factor of 9.
