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13% of a sample of 200 students do not like ice cream. what is the 95% confidence interval to describe the total percentage of students who do not like ice cream?

Respuesta :

Using the z-distribution, the 95% confidence interval to describe the total percentage of students who do not like ice cream is:

(8.34%, 17.66%).

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.

In this problem, we have a 95% confidence level, hence[tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so the critical value is z = 1.96.

For this problem, the estimate and the sample size are given, respectively, by:

[tex]\pi = 0.13, n = 200[/tex]

Hence the bounds of the interval are:

  • [tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.13 - 1.96\sqrt{\frac{0.13(0.87)}{200}} = 0.0834[/tex]
  • [tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.13 + 1.96\sqrt{\frac{0.13(0.87)}{200}} = 0.1766[/tex]

As a percentage, the interval is:

(8.34%, 17.66%).

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

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