Respuesta :
Answer:
The sample size required is of 2305.
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]
A preliminary survey of adult Americans has estimated this proportion to be somewhere around 0.40.
This means that [tex]\pi = 0.4[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
The sample size required to estimate this proportion with 95% confidence and a margin of error of 0.02.
The sample size required is of n.
n is found when M = 0.02. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.02 = 1.96\sqrt{0.4*0.6}{n}}[/tex]
[tex]0.02\sqrt{n} = 1.96\sqrt{0.4*0.6}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.4*0.6}}{0.02}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96\sqrt{0.4*0.6}}{0.02})^2[/tex]
[tex]n = 2304.96[/tex]
Rounding up
The sample size required is of 2305.
