Question

10. In a survey of 212 people at the local track and field championship, 72% favored the home team

Answer 1

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 $\pi$, 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 z-score that has a p-value of $1 - \frac{\alpha}{2}$.

The margin of error of the survey is:

$M = \sqrt{\frac{\pi(1-\pi)}{n}}$

The confidence interval can be written as:

$\pi \pm zM$

In a survey of 212 people at the local track and field championship, 72% favored the home team winning.

This means that $n = 212, \pi = 0.72$

a. Find the margin of error for the survey.

$M = \sqrt{\frac{0.72*0.28}{212}} = 0.0308$

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

Lower bound:

$\pi - zM = 0.72 - 1.96*0.0308 = 0.6596$

Upper bound:

$\pi + zM = 0.72 + 1.96*0.0308 = 0.7804$

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%).