Respuesta :
Answer:
The 95% confidence interval for the proportion of all students who would report drinking any alcohol in the past year is (0.818, 0.856).
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 = 1474, \pi = \frac{1234}{1474} = 0.8372[/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 lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.8372 - 1.96\sqrt{\frac{0.8372*0.1628}{1474}} = 0.818[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.8372 + 1.96\sqrt{\frac{0.8372*0.1628}{1474}} = 0.856[/tex]
The 95% confidence interval for the proportion of all students who would report drinking any alcohol in the past year is (0.818, 0.856).
