Question

The yearbook committee polled 80 randomly selected students from a class of 320 ninth graders to see if they would be willing to pay more for a yearbook if their names were printed on the front. of the students who were surveyed, 26 of them said they would be willing to pay extra.
with a desired confidence interval of 90%, which has a z*-score of 1.645, what is the margin of error of this survey?

Answer 1

Using the z-distribution, as we are working with a proportion, it is found that the margin of error for the 90% confidence interval is of 0.0524 = 5.24%.

What is a confidence interval of proportions?

A confidence interval of proportions is given by:

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which:

• $\pi$ is the sample proportion.

• z is the critical value.

• n is the sample size.

The margin of error is given by:

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

In this problem, the critical value is given as z = 1.645, and since 26 out of 80 students said they would be willing to pay extra:

$n = 80, \pi = \frac{26}{80} = 0.325$

Then, the margin of error is of:

$M = 1.645\sqrt{\frac{0.325(0.675)}{80}} = 0.0524$

More can be learned about the z-distribution at brainly.com/question/25890103