Using the z-distribution, as we are working with proportions, it is found that:
1. 1068 people have to be sampled.
2. 2430 people have to be sampled.
3. A level of confidence of 89% is used.
What is a confidence interval of proportions?
A confidence interval of proportions is given by:
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which:
- [tex]\pi[/tex] is the sample proportion.
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
Item a:
There is no estimate for the proportion, hence [tex]\pi = 0.5[/tex] is used. Considering a margin of error of 3% with 95% confidence, we have that [tex]M = 0.03, z = 1.96[/tex], hence:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.03 = 1.96\sqrt{\frac{0.5(0.5)}{n}}[/tex]
[tex]0.03\sqrt{n} = 1.96(0.5)[/tex]
[tex]\sqrt{n} = \frac{1.96(0.5)}{0.03}[/tex]
[tex](\sqrt{n})^2 = \left(\frac{1.96(0.5)}{0.03}\right)^2[/tex]
[tex]n = 1067.1[/tex]
Rounding up, 1068 people have to be sampled.
Item 2:
The margin of error is inverse proportional to the square root of the sample size, hence to reduce the margin of error in third, the sample size has to be multiplied by 9, that is:
9 x 270 = 2430 people have to be sampled.
Item 3:
We have that:
[tex]n = 47000, M = 0.002, \pi = 0.08[/tex].
Hence:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.002 = z\sqrt{\frac{0.08(0.92)}{47000}}[/tex]
[tex]0.00125z = 0.002[/tex]
[tex]z = 1.6[/tex]
Which is the critical value corresponding to a confidence level of 89%.
More can be learned about the z-distribution at https://brainly.com/question/25890103
