91Ó°ÊÓ

New mothers breast-feed their infants is $779$

Infants involved in-person interviews and physical examinations is$434.$

From the data of the sample proportion

$p^{\text{'}}=\frac{x+2}{n+4}$

$\frac{334+2}{434+4}=0.767$

Here $x$and $n-x$are both 5 or greater, so the one-proportion $z-$interval procedure is appropriate.

It has a $95\%$confidence interval, which means that $\mathrm{Î\pm }=0.05$

To find

$z_{a/2}=z_{0.05/2}$

$=1.96$

Then the confidence interval is

$p^{\text{'}}-7_{\frac{a}{2}}\sqrt{\frac{p^{\text{'}}\left(1-p^{\text{'}}\right)}{n}}$to $p^{\text{'}}+z_{a/2}\sqrt{\frac{p^{\text{'}}\left(1-p^{\text{'}}\right)}{n}}$

$0.767Â\pm 1.960\sqrt{\frac{0.767(1-0.767)}{434+4}}$

$=0.767Â\pm 1.960(0.0202)$

$=0.767Â\pm 0.03960$

role="math" localid="1651297137932" $=(0.7274,0.8066)$

The confidence interval for the percentage of all new mothers who breastfeed their infants at least briefly lies $(0.7274,0.8066)$, according to the $95\%$confidence.

The confidence interval for the percentage of all new moms who breastfeed their infants at least briefly lies between $0.7274$and $0.8066$, according to $95\%$confidence.