Respuesta :
Answer:
You will need to sample at least 3108 students.
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].
In this problem, we have that:
[tex]\pi = 0.75[/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 margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
How many students will I need to survey if I want to estimate, with 99% confidence, the true proportion to within 2%?
You need a sample size of at least n.
n is found when M = 0.02. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.02 = 2.575\sqrt{\frac{0.75*0.25}{n}}[/tex]
[tex]0.02\sqrt{n} = 1.115[/tex]
[tex]\sqrt{n} = \frac{1.115}{0.02}[/tex]
[tex]\sqrt{n} = 55.75[/tex]
[tex]\sqrt{n}^{2} = (55.75)^{2}[/tex]
[tex]n = 3108[/tex]
You will need to sample at least 3108 students.
