Question

According to a "how to stop bullying" Web site, 15% of students report experiencing bullying one to three times within the most recent month. Let's assume the standard deviation is 4.5% of students. Joseph collects data from 186 students at a medium-sized school in Iowa and finds that only 11% reported this rate of bullying. What is his 95% confidence interval?

Answer 1

Answer:

His 95% confidence interval is (0.065, 0.155).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

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

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

$n = 186, \pi = 0.11$

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.11 - 1.96\sqrt{\frac{0.11*0.89}{186}} = 0.065$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.11 + 1.96\sqrt{\frac{0.11*0.89}{186}} = 0.155$

His 95% confidence interval is (0.065, 0.155).