Respuesta :
Answer:
a) We need a sample size of at least 2401.
b) We need a sample size of at least 1936.
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].
(a) you are unwilling to predict the proportion value at your school and
We use [tex]\pi = 0.5[/tex], which is when we are going to need the largest sample size.
So we need to find n when [tex]\pi = 0.5, M = 0.02[/tex]. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.02 = 1.96\sqrt{\frac{0.5*0.5}{n}}[/tex]
[tex]0.02\sqrt{n} = 1.96*0.5[/tex]
[tex]\sqrt{n} = \frac{1.96*0.5}{0.02}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.96*0.5}{0.02})^{2}[/tex]
[tex]n = 2401[/tex]
We need a sample size of at least 2401.
(b) you use the results from the surveyed school as a guideline.
Now the same calculation, just with [tex]\pi = 0.28[/tex]. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.02 = 1.96\sqrt{\frac{0.28*0.72}{n}}[/tex]
[tex]0.02\sqrt{n} = 1.96*\sqrt{0.28*0.72}[/tex]
[tex]\sqrt{n} = \frac{1.96*\sqrt{0.28*0.72}}{0.02}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.96*\sqrt{0.28*0.72}}{0.02})^{2}[/tex]
[tex]n = 1936[/tex]
We need a sample size of at least 1936.
