Discover new knowledge and insights with IDNLearn.com's extensive Q&A platform. Join our community to receive prompt, thorough responses from knowledgeable experts.
Sagot :
Answer:
A sample size of 1068 is required.
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:
[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].
CBS wishes to have 95% confidence and a margin of error in its estimate of ±0.03. What sample size is required?
We need a sample size of n, and is found when M = 0.03.
We dont have an estimate for the true proportion, so we use [tex]\pi = 0.5[/tex], which is when the largest sample size will be needed.
[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 = (\frac{1.96*0.5}{0.03})^2[/tex]
[tex]n = 1067.1[/tex]
Rounding up,
A sample size of 1068 is required.
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Find clear answers at IDNLearn.com. Thanks for stopping by, and come back for more reliable solutions.