Respuesta :
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.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of .
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
98% confidence level
So [tex]\alpha = 0.02[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.02}{2} = 0.99[/tex], so [tex]Z = 2.327[/tex].
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 [tex]\pi = 0.5[/tex]
So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.001 = 2.327\sqrt{\frac{0.5*0.5}{n}}[/tex]
[tex]0.001\sqrt{n} = 2.327*0.5[/tex]
[tex]\sqrt{n} = \frac{2.327*0.5}{0.001}[/tex]
[tex](\sqrt{n})^{2} = (\frac{2.327*0.5}{0.001})^{2}[/tex]
[tex]n = 1353732.25[/tex]
Rounding up
A sample size of at least 1,353,733 is required.
