Get the information you need with the help of IDNLearn.com's extensive Q&A platform. Ask your questions and get detailed, reliable answers from our community of knowledgeable experts.
Sagot :
Using the z-distribution, it is found that the needed sample sizes are:
a) 242
b) 1842
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]
99% confidence level, hence[tex]\alpha = 0.99[/tex], z is the value of Z that has a p-value of [tex]\frac{1+0.99}{2} = 0.995[/tex], so [tex]z = 2.575[/tex].
Item a:
The estimate is:
[tex]\pi = 0.223 - 0.189 = 0.034[/tex]
The sample size is n for which M = 0.03, then:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.03 = 2.575\sqrt{\frac{0.034(0.966)}{n}}[/tex]
[tex]0.03\sqrt{n} = 2.575\sqrt{0.034(0.966)}[/tex]
[tex]\sqrt{n} = \frac{2.575\sqrt{0.034(0.966)}}{0.03}[/tex]
[tex](\sqrt{n})^2 = \left(\frac{2.575\sqrt{0.034(0.966)}}{0.03}\right)^2[/tex]
[tex]n = 241.97[/tex]
Rounding up, a sample of 242 is needed.
Item b:
No prior estimates, hence [tex]\pi = 0.5[/tex] is used.
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.03 = 2.575\sqrt{\frac{0.5(0.5)}{n}}[/tex]
[tex]0.03\sqrt{n} = 2.575\sqrt{0.5(0.5)}[/tex]
[tex]\sqrt{n} = \frac{2.575\sqrt{0.5(0.5)}}{0.03}[/tex]
[tex](\sqrt{n})^2 = \left(\frac{2.575\sqrt{0.5(0.5)}}{0.03}\right)^2[/tex]
[tex]n = 1841.8[/tex]
Rounding up, a sample of 1842 is needed.
For more on the z-distribution, you can check https://brainly.com/question/25404151
We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Discover the answers you need at IDNLearn.com. Thank you for visiting, and we hope to see you again for more solutions.
