Question

Drinking Habits. In a nationwide survey, conducted by Pulse Opinion Research, LLC for Rasmussen Reports, $1000$ American adults were asked, among other things, whether they drink alcoholic beverages at least once a week; $38\%$ said "yes." Determine and interpret a $95\%$ confidence interval for the proportion, $p$, of all American adults who drink alcoholic beverages at least once a week.

Short answer

The confidence interval for the proportion of all-American adults who drink alcoholic beverages at least once a week is $0.350$ to $0.410$, according to the $95\%$ confidence interval.

Step-by-step solution

Given Information

When $x$and $n-x$are both $5$the one-proportion z-interval technique is appropriate if the number of proportions is larger than one.

It has a $95\%$confidence interval, which means that $\alpha =0.05$.

Explanation

To find that

$z_{\alpha /2}=z_{0.05/2}$

$=1.96$

The confidence interval for $p$is form

$p^{\text{'}}-z_{\alpha /2}\sqrt{\frac{p^{\text{'}}\left(1-p^{\text{'}}\right)}{n}}\text{to}{p}^{\text{'}}+z_{\alpha /2}\sqrt{\frac{p^{\text{'}}\left(1-p^{\text{'}}\right)}{n}}$$\text{i.e.}0.38-1.960\sqrt{\frac{0.38(1-0.38)}{1000}}\text{to}0.38-1.960\sqrt{\frac{0.38(1-0.38)}{1000}}$$\text{i.e.}0.38\pm 1.960(0.0153)$

$0.38\pm 0.0299$

$(0.350,0.410)$

The $95\%$confidence interval for the percentage of all adult Americans who drink alcoholic beverages at least once a week is $(0.350,0.410)$.