Respuesta :
Answer:
We must survey at least 48 people from this younger age group.
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]
90% confidence level
So [tex]\alpha = 0.1[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].
How many of this younger age group must we survey in order to estimate the proportion of non-grads to within .10 with 90% confidence? Use the value of p from the over-50 age group.
This is n when [tex]M = 0.1, \pi = 0.23[/tex]
So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.1 = 1.645\sqrt{\frac{0.23*0.77}{n}}[/tex]
[tex]0.1\sqrt{n} = 1.645\sqrt{0.23*0.77}[/tex]
[tex]\sqrt{n} = \frac{1.645\sqrt{0.23*0.77}}{0.1}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.645\sqrt{0.23*0.77}}{0.1})^{2}[/tex]
[tex]n = 47.92[/tex]
Rouding up,
We must survey at least 48 people from this younger age group.
