Answer:
The margin of error is 0.025 = 2.5 percentage points
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].
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
99% confidence level
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].
The sample proportion is 0.102.
[tex]p = 0.102[/tex]
1000 college students at NC State University were randomly selected for a survey.
So [tex]n = 1000[/tex]
What is the margin of error for a 99% confidence interval for this sample?
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]M = 2.575\sqrt{\frac{0.102*0.898}{1000}} = 0.025[/tex]
