Question

A poll by Gallup asked, "If you won $10$ million dollars in the lottery, would you continue to work or stop working?' Of the $1039$ American adults surveyed, $707$ said that they would continue working. Obtain a $95\%$ confidence interval for the proportion of all American adults who would continue working if they won 10 million dollars in the lottery.

Short answer

The confidence range for the proportion of all American adults who would continue working if they won $10$million in the lottery is $0.651$and $0.708$at a$95\%$ confidence level.

Step-by-step solution

Given Information

"Would you continue to work or stop working if you won $10$millions in the lottery?" Gallup questioned in a poll. $707$out of$1039$ individuals in the United States stated they would keep working.

Explanation

The sample percentage $p^{\text{'}}=\frac{x+2}{n+4}$was calculated from the data.

$\frac{707+2}{1039+4}=0.6798$

The one-proportion $z$-interval technique is adequate because $x$and $n-x$are both $5$or more.

The confidence interval is $95\%$, which means $\alpha =0.05$.

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

The $p$confidence interval is of the form

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

i.e. $0.6798-1.960\sqrt{\frac{0.6798(1-0.6798)}{50}}\text{to}0.6798-1.960\sqrt{\frac{0.6798(1-0.6798)}{50}}$

i.e. $(0.6798-0.0282)\text{to}(0.6798+0.0282)$

i.e.$(0.651,0.708)$