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
, and a confidence level of
, we have the following confidence interval of proportions.
In which z is the z-score that has a p-value of
.
The margin of error is:
![M = z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
95% confidence level, hence
, z is the value of Z that has a p-value of
, so
.
For this problem, we consider that we want it to be within 4%.
Item a:
- The sample size is <u>n for which M = 0.04.</u>
- There is no estimate, hence
![\pi = 0.5](https://tex.z-dn.net/?f=%5Cpi%20%3D%200.5)
![M = z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
![0.04 = 1.96\sqrt{\frac{0.5(0.5)}{n}}](https://tex.z-dn.net/?f=0.04%20%3D%201.96%5Csqrt%7B%5Cfrac%7B0.5%280.5%29%7D%7Bn%7D%7D)
![0.04\sqrt{n} = 1.96\sqrt{0.5(0.5)}](https://tex.z-dn.net/?f=0.04%5Csqrt%7Bn%7D%20%3D%201.96%5Csqrt%7B0.5%280.5%29%7D)
![\sqrt{n} = \frac{1.96\sqrt{0.5(0.5)}}{0.04}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B1.96%5Csqrt%7B0.5%280.5%29%7D%7D%7B0.04%7D)
![(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.5(0.5)}}{0.04}\right)^2](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E2%20%3D%20%5Cleft%28%5Cfrac%7B1.96%5Csqrt%7B0.5%280.5%29%7D%7D%7B0.04%7D%5Cright%29%5E2)
![n = 600.25](https://tex.z-dn.net/?f=n%20%3D%20600.25)
Rounding up:
A sample of 601 is needed.
Item b:
The estimate is
, hence:
![M = z\sqrt{\frac{\pi(1-\pi)}{n}}](https://tex.z-dn.net/?f=M%20%3D%20z%5Csqrt%7B%5Cfrac%7B%5Cpi%281-%5Cpi%29%7D%7Bn%7D%7D)
![0.04 = 1.96\sqrt{\frac{0.96(0.04)}{n}}](https://tex.z-dn.net/?f=0.04%20%3D%201.96%5Csqrt%7B%5Cfrac%7B0.96%280.04%29%7D%7Bn%7D%7D)
![0.04\sqrt{n} = 1.96\sqrt{0.96(0.04)}](https://tex.z-dn.net/?f=0.04%5Csqrt%7Bn%7D%20%3D%201.96%5Csqrt%7B0.96%280.04%29%7D)
![\sqrt{n} = \frac{1.96\sqrt{0.96(0.04)}}{0.04}](https://tex.z-dn.net/?f=%5Csqrt%7Bn%7D%20%3D%20%5Cfrac%7B1.96%5Csqrt%7B0.96%280.04%29%7D%7D%7B0.04%7D)
![(\sqrt{n})^2 = \left(\frac{1.96\sqrt{0.96(0.04)}}{0.04}\right)^2](https://tex.z-dn.net/?f=%28%5Csqrt%7Bn%7D%29%5E2%20%3D%20%5Cleft%28%5Cfrac%7B1.96%5Csqrt%7B0.96%280.04%29%7D%7D%7B0.04%7D%5Cright%29%5E2)
![n = 92.2](https://tex.z-dn.net/?f=n%20%3D%2092.2)
Rounding up:
A sample of 93 is needed.
Item c:
The closer the estimate is to
, 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