A survey asked whether respondents favored or opposed the death penalty for people convicted of murder. Software shows the results below, where X refers to the number of the respondents who were in favor. Construct the 95% confidence interval for the proportion of the adults who were opposed to the death penalty from the confidence interval stated below for the proportion in favor.
X N Sample p 95.0% CI
1786 2611 0.684 (0.666,0.702 )
The 95% confidence interval for those opposed is:___________ (.666 . 666, .702 . 702).

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

For this problem, we have that:

1786 of the 2611 were in favor, so 2611 - 1786 = 825 were opposed. Then

[tex]n = 2611, \pi = \frac{825}{2611} = 0.316[/tex]

95% confidence level

So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.316 - 1.96\sqrt{\frac{0.316*0.684}{2611}} = 0.298[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.316 + 1.96\sqrt{\frac{0.316*0.684}{2611}} = 0.334[/tex]

The 95% confidence interval for those opposed is: (0.298, 0.334).