Answer :
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 [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
For this problem, we have that:
[tex]n = 1900, \pi = 0.7[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]M = 1.96\sqrt{\frac{0.7*0.3}{1900}} = 0.0206[/tex]
The margin of error is 0.0206 = 2.06 percentage points.
The lower limit of this interval is:
[tex]\pi - M = 0.7 - 0.0206 = 0.6794[/tex]
The upper limit of this interval is:
[tex]\pi + M = 0.7 + 0.0206 = 0.7206[/tex]
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).