A simple random sample of $3,500$ people age 18 or over is taken in a large
town to estimate the percentage of people (age 18 and over in that town) who
read newspers. It turns out that $2,487$ people in the sample are newspaper
readers. 12 The population percentage is estimated as
$$
\frac{2,487}{3,500} \times 100 \% \approx 71 \%
$$
The standard error is estimated as 0.8 of $1 \%,$ because
$$
\sqrt{3,500} \times \sqrt{0.71 \times 0.29} \approx 27, \quad \frac{27}{3,500} \times 100 \% \approx 0.8 \text { of } 1 \%
$$
$$
\begin{array}{l}{\text { (a) Is } 0.8 \text { of } 1 \% \text { the right SE? Answer yes or no, and explain. }} \\ {\text { (b) } 71 \% \pm 1.6 \% \text { is a }}\end{array}
$$