Question

In a recent Zogby International Poll, nine of 48 respondents rated the likelihood of a terrorist attack in their community as “likely” or “very likely.” Use the “plus-four” method to create a 97% confidence interval for the proportion of American adults who believe that a terrorist attack in their community is likely or very likely. Explain what this confidence interval means in the context of the problem.

Short answer

The confidence interval in the context of the problem means that with a$97\%$confidence interval for the proportion of American adults who believe that a terrorist attack in their community is likely or very likely is between $8.9\%$to$33.5\%$.

Step-by-step solution

introduction

Given $x=9$and $n=48$

$\stackrel{}{p\text{'}}=\frac{x}{n}=\frac{9}{48}=0.1875$

Confidence interval =$97\%$

$\alpha =1-0.97=0.03$

explanation

Now,$z_{\frac{a}{2}}=2.17$

we have,

Error bound= $Z\frac{a}{2}\sqrt{\frac{p^{\mathrm{\prime }}{q}^{\mathrm{\prime }}}{n}}$

$=2.17\sqrt{\frac{0.212\times 0.788}{52}}=0.123$

Class interval = $p\text{'}$$\pm$error bound

$\Rightarrow (0.212\pm 0.123)$

role="math" localid="1650888106268" $=(0.089,0.335)$

Hence, the proportion of American adults who believe that a terrorist attack in their community is likely or very likely is between$8.9\mathrm{\%}\text{to}33.5\mathrm{\%}$.