Respuesta :
Answer:
a) A sample size of at least 251 students is needed.
b) A sample of at least 97 students is needed.
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
We need a sample size of at least n.
n is found when M = 0.08.
We wont predict a proportion value for the school, so we use [tex]\pi = 0.5[/tex]
So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.08 = 1.96\sqrt{\frac{0.5*0.5}{n}}[/tex]
[tex]0.08\sqrt{n} = 1.96\sqrt{0.5*0.5}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.5*0.5}}{0.08}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.96\sqrt{0.5*0.5}}{0.08})^{2}[/tex]
[tex]n = 150.1[/tex]
Rounding up
A sample size of at least 251 is needed.
(b) you use the results from the surveyed school as a guideline.
Now we have that [tex]\pi = 0.2[/tex]
So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.08 = 1.96\sqrt{\frac{0.2*0.8}{n}}[/tex]
[tex]0.08\sqrt{n} = 1.96\sqrt{0.2*0.8}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.2*0.8}}{0.08}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.96\sqrt{0.2*0.8}}{0.08})^{2}[/tex]
[tex]n = 96.04[/tex]
Rounding up
A sample of at least 97 students is needed.
