The margin of error for a sample proportion is given by
[tex] z_{\alpha/2}\sqrt{ \frac{p(1-p)}{n} }[/tex]
where: [tex]z_{\alpha/2}[/tex] is the z score associated with the confidence level, p is the sample prortion and n is the sample size.
We assume a confidence level of 95%, then [tex]z_{\alpha/2}=1.96[/tex]
p = 52% = 0.52 and n = 2251
Therefore, margin of error =
[tex]z_{\alpha/2}\sqrt{ \frac{p(1-p)}{n} }= 1.96\times \sqrt{ \frac{0.52(1-0.52)}{2251} } \\ = 1.96\times \sqrt{ \frac{0.52(0.48)}{2251} } = 1.96\times \sqrt{ \frac{0.2496}{2251} } \\ = 1.96\times \sqrt{0.000110884} =1.96\times 0.0105 \\ =0.0206=2.1\%[/tex]
The the interval that is likely to contain the true population proportion is between 49.9% and 54.1%.
