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
, and a confidence level of
, we have the following confidence interval of proportions.
![\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=%5Cpi%20%5Cpm%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
In which
z is the zscore that has a pvalue of
.
For this problem, we have that:
![n = 1474, \pi = \frac{1234}{1474} = 0.8372](https://tex.z-dn.net/?f=n%20%3D%201474%2C%20%5Cpi%20%3D%20%5Cfrac%7B1234%7D%7B1474%7D%20%3D%200.8372)
95% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
The lower limit of this interval is:
![\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.8372 - 1.96\sqrt{\frac{0.8372*0.1628}{1474}} = 0.818](https://tex.z-dn.net/?f=%5Cpi%20-%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.8372%20-%201.96%5Csqrt%7B%5Cfrac%7B0.8372%2A0.1628%7D%7B1474%7D%7D%20%3D%200.818)
The upper limit of this interval is:
![\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.8372 + 1.96\sqrt{\frac{0.8372*0.1628}{1474}} = 0.856](https://tex.z-dn.net/?f=%5Cpi%20%2B%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D%20%3D%200.8372%20%2B%201.96%5Csqrt%7B%5Cfrac%7B0.8372%2A0.1628%7D%7B1474%7D%7D%20%3D%200.856)
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).