91Ó°ÊÓ

The size of a simple random sample from a population, as well as the number of successes.

$x=8,n=40,95\%\text{level}$

$$\begin{gathered} ∴n-x=40-8 \\ =32 \end{gathered}$$, here $x$and $n-x$are both $5$or greater.

The sample proportion $p^{\text{'}}=\frac{x}{n}$is calculated from the data.

$\frac{8}{40}=0.2$

The size of a simple random sample from a population, as well as the number of successes.

$x=8,n=40,95\%\text{level}$

$$\begin{gathered} ∴n-x=40-8 \\ =32 \end{gathered}$$, here $x$and $n-x$ are both $5$ or greater.

The confidence interval is $95\%$, which means $\mathrm{Î\pm }=0.05$.

It is discovered thatlocalid="1651326317744" $$\begin{gathered} z_{a/2}=z_{0.05/2} \\ =1.96 \end{gathered}$$

The $p$confidence interval is of the form

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

i.e. localid="1651326715537" $0.2-1.96\sqrt{\frac{0.2(1-0.2)}{40}}\text{to}0.2+1.96\sqrt{\frac{0.2(1-0.2)}{40}}$

i.e. localid="1651326727514" $0.2-0.124\text{to}0.2+0.124$

i.e.localid="1651326739386" $0.076\text{to}0.324$