Question

Out of 6,000 people, 4,500 said they spend over 20 hours per week on their phones. Construct a 75% confidence interval for the population mean of people that spend over 20 hours per week on their phones.
CI = (73.98%, 75.54%)
CI = (74.36%, 75.64%)
CI = (73.23%, 75.01%)
CI = (74.01%, 75.92%)

Answer 1

Using the z-distribution, as we are working with a proportion, it is found that the 75% confidence interval for the population mean of people that spend over 20 hours per week on their phones is given by:

CI = (74.36%, 75.64%).

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, the parameters are:

$n = 6000, \pi = \frac{4500}{6000} = 0.75, z = 1.15$

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.75 - 1.15\sqrt{\frac{0.75(0.25)}{6000}} = 0.7436$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.75 + 1.15\sqrt{\frac{0.75(0.25)}{6000}} = 0.7564$

Using percentages, the interval is given by:

CI = (74.36%, 75.64%).

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