The percentage of patients overdue for a vaccination is often of interest for a medical clinic. Some clinics examine every record to determine that percentage; in a large practice, though, taking a census of the records can be time-consuming.Cullen(1994) took a sample of the 580 children served by an Auckland family practice to estimate the proportion of interest.
a) What sample size in an SRS (without replacement) would be necessary to estimate the proportion with 95% confidence and margin of error 0.10?
This is what I have: $S^2 \approx p(1-p)$, the maximum value for the standard deviation is attained at $p=\frac{1}{2}.$
The value $Z_{\frac{\alpha}{2}}$ =1.96, then n_0:
$n_0=\frac{Z^2_{\frac{\alpha}{2}}*S^2}{ME^2}=n_0=\frac{(1.96)^2*\frac{1}{2}*(1-\frac{1}{2})}{(0.1)^2}=96.04 \approx 96$
then $n=\frac{n_o}{1+\frac{n_0}{N}}=\frac{96}{1+\frac{96}{580}}=82.3669 \approx 82$
Did i calculate it correctly?
I don´t say that your applied formulas are wrong, but on this site the formula is different:
$ \begin{alignat}{} \\ \qquad \qquad \qquad \qquad \qquad \qquad n=\frac{z^2\cdot p\cdot q+ME^2}{ME^2+z^2\cdot p\cdot \frac{q}N} \\ \\ \\ \qquad \qquad \qquad \qquad \qquad \qquad n=\frac{1.96^2\cdot 0.5\cdot 0.5+0.1^2}{0.1^2+1.96^2\cdot 0.5\cdot \frac{0.5}{580}}=83.25\approx 83 \end{alignat} $
This result corresponds with the results of the two calculators (1,2).