91Ó°ÊÓ

A survey of music tastes was recently done by public policy polling.

When asked, $x=80$ $n=571$people acknowledged downloading music illegally.

As a result, the sample's estimated proportion is

$$\begin{gathered} p^{\text{'}}=\frac{x}{n} \\ =\frac{80}{571} \\ =0.14 \end{gathered}$$

It$p^{\text{'}}$is the proportion of observations in a random sample of size n that belong to a class of interest, the confidence interval on the proportion $p$of the community that belongs to this class is about $100(1-\mathrm{Î\pm })\%$.

$p^{\text{'}}-z_{\frac{\mathrm{Î\pm }}{2}}\sqrt{\frac{p^{\text{'}}\left(1-p^{\text{'}}\right)}{n}}≤p≤p^{\text{'}}+z_{\frac{\mathrm{Î\pm }}{2}}\sqrt{\frac{p^{\text{'}}\left(1-p^{\text{'}}\right)}{n}}$ $\left(1\right)$

where z/2 is the upper and a/2 percentage point of the standard normal distribution. for a 99% two-sided confidence interval,

$$\begin{gathered} \frac{\mathrm{Î\pm }}{2}=\frac{1-0.99}{2} \\ =0.005 \end{gathered}$$

and

$z_{\frac{\mathrm{Î\pm }}{2}}=2.58$

A two-sided confidence interval for the population proportion can be calculated using the formulae.

$0.14-2.58\sqrt{\frac{0.14(1-0.14)}{571}}≤p≤0.14+2.58\sqrt{\frac{0.14(1-0.14)}{571}}$,

$0.14-0.0374≤p≤0.14+0.0374$.

As a result, the $99\%$confidence interval $p$is

$0.103≤p≤0.177$

Many people refuse to participate in telephonic surveys.

You have no way of knowing who is responding to the surveys if they do respond.

Decreasing the confidence interval, causing the error bound to fall, resulting in a narrower confidence interval