Respuesta :
Answer:
A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07
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 of the interval is given by:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In this problem, we have that:
[tex]p = 0.69[/tex]
99.5% confidence level
So [tex]\alpha = 0.005[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.005}{2} = 0.9975[/tex], so [tex]Z = 2.81[/tex].
Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.07?
This is n when M = 0.07. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.07 = 2.81\sqrt{\frac{0.69*0.31}{n}}[/tex]
[tex]0.07\sqrt{n} = 1.2996[/tex]
[tex]\sqrt{n} = \frac{1.2996}{0.07}[/tex]
[tex]\sqrt{n} = 18.5658[/tex]
[tex](\sqrt{n})^{2} = (18.5658)^{2}[/tex]
[tex]n = 345[/tex]
A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07
