Respuesta :
Answer:
We would need a sample of size at least 74.
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].
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
99% confidence level
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].
How large of a sample size is needed to estimate it to within 0.15 of the true value with 99% confidence?
We need a sample of size at least n
n is found when [tex]M = 0.15[/tex]
We don't know the proportion, so we estimate [tex]\pi = 0.5[/tex], which is the case for which we are going to need the largest sample size.
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.15 = 2.575\sqrt{\frac{0.5*0.5}{n}}[/tex]
[tex]0.15\sqrt{n} = 2.575*0.5[/tex]
[tex]\sqrt{n} = \frac{2.575*0.5}{0.15}[/tex]
[tex](\sqrt{n})^{2} = (\frac{2.575*0.5}{0.15})^{2}[/tex]
[tex]n = 73.67[/tex]
Rounding up
We would need a sample of size at least 74.
