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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?

Sagot :

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:

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which:

  • [tex]\pi[/tex] is the sample proportion.
  • z is the critical value.
  • n is the sample size.

The margin of error is given by:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

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:

[tex]n = 80, \pi = \frac{26}{80} = 0.325[/tex]

Then, the margin of error is of:

[tex]M = 1.645\sqrt{\frac{0.325(0.675)}{80}} = 0.0524[/tex]

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

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