Answer :
Answer:
Question 1:
(b) 4,445
Question 2:
(c) 2.1%
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 of:
[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].
Question 1:
We have no previous estimate for the population proportion, so we use [tex]\pi = 0.5[/tex].
The sample size is n for which M = 0.015. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.015 = 1.96\sqrt{\frac{0.5*0.5}{n}}[/tex]
[tex]0.015\sqrt{n} = 1.96*0.5[/tex]
[tex]\sqrt{n} = \frac{1.96*0.5}{0.015}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96*0.5}{0.015})^2[/tex]
[tex]n = 4268[/tex]
Samples above this value should be used, and the smaller sample above this value is of 4445, so the answer is given by option b.
Question 2:
Now we find M for which [tex]n = 2222[/tex].
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]M = 1.96\sqrt{\frac{0.5*0.5}{2222}}[/tex]
[tex]M = 0.021[/tex]
So 2.1%, and the correct answer is given by option c.