Rewrite this in its simplest form (please, doing an algebra 2 test)
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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%.
<h3>What is a confidence interval of proportions?</h3>A confidence interval of proportions is given by:
In which:
is the sample proportion.
- z is the critical value.
- n is the sample size.
The margin of error is given by:
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:
Then, the <em>margin of error</em> is of:
More can be learned about the z-distribution at brainly.com/question/25890103