Question

1. Suppose we are interested in finding out the proportion of the population at EKHS that has seen
The Office. We contact an SRS of 100 students in the school. Of these 100 students, 63 report
seeing The Office. Find a 95% confidence interval for the true proportion of EKHS students who
have seen The Office.

Answer 1

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:

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which:

• $\pi$ is the sample proportion.

• z is the critical value.

• n is the sample size.

In this problem:

• 95% confidence level, hence$\alpha = 0.95$, z is the value of Z that has a p-value of $\frac{1+0.95}{2} = 0.975$, so $z = 1.96$.

• Out of 100 students, 63 report seeing The Office, hence $n = 100, \pi = \frac{63}{100} = 0.63$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.63 - 1.96\sqrt{\frac{0.63(0.37)}{100}} = 0.5354$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.63 + 1.96\sqrt{\frac{0.63(0.37)}{100}} = 0.7246$

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