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
, and a confidence level of
, we have the following confidence interval of proportions.
![\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=%5Cpi%20%5Cpm%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
In which
z is the z-score that has a p-value of
.
The margin of error of the survey is:
![M = \sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=M%20%3D%20%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
The confidence interval can be written as:
![\pi \pm zM](https://tex.z-dn.net/?f=%5Cpi%20%5Cpm%20zM)
In a survey of 212 people at the local track and field championship, 72% favored the home team winning.
This means that ![n = 212, \pi = 0.72](https://tex.z-dn.net/?f=n%20%3D%20212%2C%20%5Cpi%20%3D%200.72)
a. Find the margin of error for the survey.
![M = \sqrt{\frac{0.72*0.28}{212}} = 0.0308](https://tex.z-dn.net/?f=M%20%3D%20%5Csqrt%7B%5Cfrac%7B0.72%2A0.28%7D%7B212%7D%7D%20%3D%200.0308)
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
, z is the value of Z that has a p-value of
, so
.
Lower bound:
![\pi - zM = 0.72 - 1.96*0.0308 = 0.6596](https://tex.z-dn.net/?f=%5Cpi%20-%20zM%20%3D%200.72%20-%201.96%2A0.0308%20%3D%200.6596)
Upper bound:
![\pi + zM = 0.72 + 1.96*0.0308 = 0.7804](https://tex.z-dn.net/?f=%5Cpi%20%2B%20zM%20%3D%200.72%20%2B%201.96%2A0.0308%20%3D%200.7804)
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%).