Respuesta :
Answer:
The sample size needed if the margin of error of the confidence interval is to be about 0.04 is 18.
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]
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].
Past studies suggest this proportion will be about 0.15
This means that [tex]p = 0.15[/tex]
Find the sample size needed if the margin of error of the confidence interval is to be about 0.04
This is n when M = 0.04. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.04 = 1.96\sqrt{\frac{0.15*0.85}{n}}[/tex]
[tex]0.04\sqrt{n} = 1.96\sqrt{0.15*0.85}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.15*0.85}}{0.04}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.96\sqrt{0.15*0.85}}{0.04})^{2}[/tex]
[tex]n = 17.5[/tex]
Rounding up
The sample size needed if the margin of error of the confidence interval is to be about 0.04 is 18.
