Question

Offshore Drilling. In the February $2013$article "Offshore Drilling Support High as Deepwater Horizon Oil Spill Trial Opens," E. Swanson reported on a HuffPost and YouGov poll that asked Americans what they think about increased offshore drilling for oil and natural gas. Of the $1000$U.S, adults surveyed, $280$ said that they were opposed. Find a $99\%$ confidence interval for the proportion of all U.S. adults who, at the time, opposed increased offshore drilling for oil and natural gas.

Short answer

As a result, the $99\%$confidence interval for the proportion of all U.S. adults who opposed increasing offshore drilling for oil and natural gas at the time $(0.244,0.318)$.

The confidence interval for the proportion of all U.S. adults who opposed additional offshore drilling for oil and natural gas between $0.244$and $0.318$is $95\%$sure.

Step-by-step solution

Given information

Total adults surveyed is $1000$

Number of Adults opposed is$280$

Explanation

From the given data

The sample proportion is,

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

$\frac{280+2}{1000+4}=0.2809$

Here $x$and $n-x$are both $5$or greater, so the one-proportion $z-$interval procedure is appropriate. It has a $99\%$confidence interval, which means that $\alpha =0.05$

To find

$z_{\frac{a}{2}}=Z_{\frac{0.05}{2}}$

$=2.575$

The confidence interval is

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

$=0.2809\pm 2.575\sqrt{\frac{0.2809(1-0.2809)}{1000+4}}$

$=0.2809\pm 2.575(0.0142)$

role="math" localid="1651297944531" $=0.2809\pm 0.03660$

$=(0.244,0.318)$

As a result, the $99\%$confidence interval for the proportion of all U.S. adults who opposed increasing offshore drilling for oil and natural gas at the time $(0.244,0.318)$.

The confidence interval for the proportion of all U.S. adults who opposed additional offshore drilling for oil and natural gas between $0.244$and $0.318$is $95\%$sure.