Answer:
A sample of 16577 is required.
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 z-score that has a p-value of
.
The margin of error is of:
![M = z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
99% confidence level
So
, z is the value of Z that has a p-value of
, so
.
She wants to estimate the proportion using a 99% confidence interval with a margin of error of at most 0.01. How large a sample size would be required?
We have no estimation for the proportion, and thus we use
, which is when the largest sample size will be needed.
The sample size is n for which M = 0.01. So
![M = z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
![0.01 = 2.575\sqrt{\frac{0.5*0.5}{n}}](https://tex.z-dn.net/?f=0.01%20%3D%202.575%5Csqrt%7B%5Cfrac%7B0.5%2A0.5%7D%7Bn%7D%7D)
![0.01\sqrt{n} = 2.575*0.5](https://tex.z-dn.net/?f=0.01%5Csqrt%7Bn%7D%20%3D%202.575%2A0.5)
![\sqrt{n} = \frac{2.575*0.5}{0.01}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B2.575%2A0.5%7D%7B0.01%7D)
![(\sqrt{n})^2 = (\frac{2.575*0.5}{0.01})^2](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E2%20%3D%20%28%5Cfrac%7B2.575%2A0.5%7D%7B0.01%7D%29%5E2)
![n = 16576.6](https://tex.z-dn.net/?f=n%20%3D%2016576.6)
Rounding up:
A sample of 16577 is required.