Question

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​).

Answer 1

Answer:

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

Step-by-step explanation:

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

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

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

$n = 2611, \pi = \frac{825}{2611} = 0.316$

95% confidence level

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

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.316 - 1.96\sqrt{\frac{0.316*0.684}{2611}} = 0.298$

The upper limit of this interval is:

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

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