Answer:
A sample size of at least 1,353,733 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 zscore that has a pvalue of .
The margin of error is:
![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)
98% confidence level
So
, z is the value of Z that has a pvalue of
, so
.
You would like to be 98% confident that you esimate is within 0.1% of the true population proportion. How large of a sample size is required?
We need a sample size of at least n.
n is found when M = 0.001.
Since we don't have an estimate for the proportion, we use the worst case scenario, that is ![\pi = 0.5](https://tex.z-dn.net/?f=%5Cpi%20%3D%200.5)
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)
![0.001 = 2.327\sqrt{\frac{0.5*0.5}{n}}](https://tex.z-dn.net/?f=0.001%20%3D%202.327%5Csqrt%7B%5Cfrac%7B0.5%2A0.5%7D%7Bn%7D%7D)
![0.001\sqrt{n} = 2.327*0.5](https://tex.z-dn.net/?f=0.001%5Csqrt%7Bn%7D%20%3D%202.327%2A0.5)
![\sqrt{n} = \frac{2.327*0.5}{0.001}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B2.327%2A0.5%7D%7B0.001%7D)
![(\sqrt{n})^{2} = (\frac{2.327*0.5}{0.001})^{2}](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E%7B2%7D%20%3D%20%28%5Cfrac%7B2.327%2A0.5%7D%7B0.001%7D%29%5E%7B2%7D)
![n = 1353732.25](https://tex.z-dn.net/?f=n%20%3D%201353732.25)
Rounding up
A sample size of at least 1,353,733 is required.