Answer :
Answer:
340 of the adults in the sample voted.
The 95% confidence interval estimate of the population percent of Americans that vote is (0.6391, 0.7209). This means that we are 95% sure that the true proportion of Americans that vote is between 0.6391 and 0.7209.
Step-by-step explanation:
In 2008, a random sample of 500 american adults was take and we found that 68% of them voted. How many of the 500 adults in the sample voted?
This is 68% of 500.
So 0.68*500 = 340.
340 of the adults in the sample voted.
Now construct a 95% confidence interval estimate of the population percent of Americans that vote and write a sentence to explain the confidence interval.
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence interval [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:
[tex]n = 500, p = 0.68[/tex]
95% confidence interval
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}} = 0.68 - 1.96\sqrt{\frac{0.68*0.32}{500}} = 0.6391[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.68 + 1.96\sqrt{\frac{0.68*0.32}{500}} = 0.7209[/tex]
The 95% confidence interval estimate of the population percent of Americans that vote is (0.6391, 0.7209). This means that we are 95% sure that the true proportion of Americans that vote is between 0.6391 and 0.7209.