Respuesta :
Using the z-distribution, it is found that the 95% confidence interval for the true proportion of EKHS students who have seen The Office is (0.5354, 0.7246).
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:
- 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 [tex]z = 1.96[/tex].
- Out of 100 students, 63 report seeing The Office, hence [tex]n = 100, \pi = \frac{63}{100} = 0.63[/tex].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.63 - 1.96\sqrt{\frac{0.63(0.37)}{100}} = 0.5354[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.63 + 1.96\sqrt{\frac{0.63(0.37)}{100}} = 0.7246[/tex]
The 95% confidence interval for the true proportion of EKHS students who have seen The Office is (0.5354, 0.7246).
You can learn more about the use of the z-distribution to build a confidence interval at https://brainly.com/question/25730047
