Question

A study done by researchers at a university concluded that 70 70​% of all student athletes in this country have been subjected to some form of hazing. The study is based on responses from 1 comma 900 1,900 athletes. What are the margin of error and​ 95% confidence interval for the​ study? The margin of error is nothing .

Answer 1

Answer:

The margin of error is 0.0206 = 2.06 percentage points.

The 95% confidence interval for the proportion of all student athletes in this country have been subjected to some form of hazing is (0.6794, 0.7206).

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:

$n = 1900, \pi = 0.7$

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 margin of error is:

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

$M = 1.96\sqrt{\frac{0.7*0.3}{1900}} = 0.0206$

The margin of error is 0.0206 = 2.06 percentage points.

The lower limit of this interval is:

$\pi - M = 0.7 - 0.0206 = 0.6794$

The upper limit of this interval is:

$\pi + M = 0.7 + 0.0206 = 0.7206$

The 95% confidence interval for the proportion of all student athletes in this country have been subjected to some form of hazing is (0.6794, 0.7206).