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A principal of a large high school wants to estimate the true proportion of high school students who use the community’s public library. To do so, he selects a random sample of 50 students and asks them if they use the community’s public library. The 95% confidence interval for the true proportion of all students who use the community’s public library is 0. 25 to 0. 34. If the principal had used a sample size of 200 students rather than 50 students, how would the length of the second interval compare to the original interval?

Sagot :

Considering the margin of error of the z-distribution, as we are working with a proportion, it is found that the second interval would be twice as long as the first.

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]

From it, we have that the margin of error is inverse proportional to the square root of the sample size. Hence, multiplying the sample size by 4, for example, from 50 to 200, will generate an interval that is twice as long.

More can be learned about the margin of error of a confidence interval at https://brainly.com/question/25890103

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