Answer :
Using the z-distribution, we have that:
a) A sample of 601 is needed.
b) A sample of 93 is needed.
c) A. ​Yes, using the additional survey information from part​ (b) dramatically reduces the sample size.
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]\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 z-score that has a p-value of [tex]\frac{1+\alpha}{2}[/tex].
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
95% confidence level, hence[tex]\alpha = 0.95[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.95}{2} = 0.975[/tex], so [tex]z = 1.96[/tex].
For this problem, we consider that we want it to be within 4%.
Item a:
- The sample size is n for which M = 0.04.
- There is no estimate, hence [tex]\pi = 0.5[/tex]
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.04 = 1.96\sqrt{\frac{0.5(0.5)}{n}}[/tex]
[tex]0.04\sqrt{n} = 1.96\sqrt{0.5(0.5)}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.5(0.5)}}{0.04}[/tex]
[tex](\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.5(0.5)}}{0.04}\right)^2[/tex]
[tex]n = 600.25[/tex]
Rounding up:
A sample of 601 is needed.
Item b:
The estimate is [tex]\pi = 0.96[/tex], hence:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.04 = 1.96\sqrt{\frac{0.96(0.04)}{n}}[/tex]
[tex]0.04\sqrt{n} = 1.96\sqrt{0.96(0.04)}[/tex]
[tex]\sqrt{n} = \frac{1.96\sqrt{0.96(0.04)}}{0.04}[/tex]
[tex](\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.96(0.04)}}{0.04}\right)^2[/tex]
[tex]n = 92.2[/tex]
Rounding up:
A sample of 93 is needed.
Item c:
The closer the estimate is to [tex]\pi = 0.5[/tex], the larger the sample size needed, hence, the correct option is A.
For more on the z-distribution, you can check brainly.com/question/25404151