Question

When 2251 voters were polled, 52% said they were voting yes on an initiative measure. find the margin of error and an interval that is likely to contain the true population proportion. ±2.1%; between 49.9% and 54.1% ±21%; between 31% and 73% ±4.7%; between 47.3% and 56.7% ±47.4%; between 4.6% and 99.4%?

Answer 1

The margin of error for a sample proportion is given by
$z_{\alpha/2}\sqrt{ \frac{p(1-p)}{n} }$
where: $z_{\alpha/2}$ 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 $z_{\alpha/2}=1.96$
p = 52% = 0.52 and n = 2251

Therefore, margin of error =
 $z_{\alpha/2}\sqrt{ \frac{p(1-p)}{n} }= 1.96\times \sqrt{ \frac{0.52(1-0.52)}{2251} } \\ = 1.96\times \sqrt{ \frac{0.52(0.48)}{2251} } = 1.96\times \sqrt{ \frac{0.2496}{2251} } \\ = 1.96\times \sqrt{0.000110884} =1.96\times 0.0105 \\ =0.0206=2.1\%$

The the interval that is likely to contain the true population proportion is between 49.9% and 54.1%.