Respuesta :
Using the z-distribution, it is found that the 99.7% confidence interval for the percentage of the students in the population who hope to have children someday is (68.5%, 81.5%).
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 99.7% confidence level, hence by the Empirical Rule the critical value is z = 3.
The sample size and the estimate are given, respectively, by:
[tex]n = 400, \pi = \frac{300}{400} = 0.75[/tex].
As a proportion, the bounds of the interval are given by:
- [tex]0.75 - 3\sqrt{\frac{0.75(0.25)}{400}} = 0.685[/tex]
- [tex]0.75 + 3\sqrt{\frac{0.75(0.25)}{400}} = 0.815[/tex]
As a percentage, the interval is (68.5%, 81.5%).
More can be learned about the z-distribution at https://brainly.com/question/25890103
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