Answer:
a. The margin of error for the survey is of 0.0308 = 3.08%.
b. The 95% confidence interval that is likely to contain the exact percent of all people who favor the home team winning is (65.96%, 78.04%).
Step-by-step explanation:
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 z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
The margin of error of the survey is:
[tex]M = \sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
The confidence interval can be written as:
[tex]\pi \pm zM[/tex]
In a survey of 212 people at the local track and field championship, 72% favored the home team winning.
This means that [tex]n = 212, \pi = 0.72[/tex]
a. Find the margin of error for the survey.
[tex]M = \sqrt{\frac{0.72*0.28}{212}} = 0.0308[/tex]
The margin of error for the survey is of 0.0308 = 3.08%.
b. Give the 95% confidence interval that is likely to contain the exact percent of all people who favor the home team winning.
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
Lower bound:
[tex]\pi - zM = 0.72 - 1.96*0.0308 = 0.6596[/tex]
Upper bound:
[tex]\pi + zM = 0.72 + 1.96*0.0308 = 0.7804[/tex]
As percent:
0.6596*100% = 65.96%
0.7804*100% = 78.04%.
The 95% confidence interval that is likely to contain the exact percent of all people who favor the home team winning is (65.96%, 78.04%).
