Question

September 11 A recent study asked U.S. adults to name $10$historic events that occurred in their lifetime that have had the greatest impact on the country. The most frequently chosen answer was the September $11$$,2001$, terrorist attacks, which was included by $76\%$of the $2025$randomly selected U.S. adults. Construct and interpret a $95\%$confidence interval for the proportion of all U.S. adults who would include the $9/11$attacks on their list of $10$historic events.

Short answer

For population proportion,$95\%$confidence interval is$0.7414<p<0.7786$

Step-by-step solution

Given Information

It is given that $n=2025$

$\hat{p}=76\%=0.76$

Confidence Level$=0.95$

Concept Used

Formula to be used:

$\hat{p}=\frac{x}{n}$(Sample Proportion)

$E=Z_{\alpha /2}\times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$(Margin of Error)

$(\hat{p}-E,\hat{p}+E)$(Confidence Interval)

Conditions

• Random Selection: This is satisfied as US adults are selected randomly.
• Less than $10\%$ population: Adults are less than $10\%$of US adults. This condition is satisfied.
• $n\hat{p}\text{and}n(1-\hat{p})\text{must be}\ge 10$: Here $n\hat{p}=2025\times 0.76=1539\ge 10$and$n(1-\hat{p})=2025\times (1-0.76)=486\ge 10\text{which is met}$. This condition is also satisfied.

All conditions are met.

Calculation

Confidence level is $0.95$

Level of significance is $a=0.05$

${\mathrm{Z}}_{\mathrm{c}}={\mathrm{Z}}_{\mathrm{a}/2}\text{critical value}=1.96$

Margin of error is

$E=1.96\times \sqrt{\frac{0.76(1-0.76)}{2025}}$

Confidence interval is $(0.76-0.0186,0.76+0.0186)$

data-custom-editor="chemistry" $(0.7414,0.7786)$

$E=0.0186$