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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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