10. In a survey of 212 people at the local track and field championship, 72% favored the home team winning
a. Find the margin of error for the survey.
b. Give the 95% confidence interval that is likely to contain the exact percent of all people who
favor the home team winning.

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.

In which

z is the z-score that has a p-value of .

The margin of error of the survey is:

The confidence interval can be written as:

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

This means that

a. Find the margin of error for the survey.

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:

Upper bound:

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%).

the discriminant is 64 + 4*1*12 = 112

this is prime.

step-by-step explanation: