At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Join our Q&A platform and connect with professionals ready to provide precise answers to your questions in various areas. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
Answer:
The minimum sample size we should anticipate using is 601.
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]
We have no history with this characteristic, so we have no idea as to what the proportion might be.
This means that we use [tex]\pi = 0.5[/tex], which is when the largest sample size will be needed.
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].
What is the minimum sample size we should anticipate using?
This is n for which M = 0.04. So
[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*0.5[/tex]
[tex]\sqrt{n} = \frac{1.96*0.5}{0.04}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96*0.5}{0.04})^2[/tex]
[tex]n = 600.25[/tex]
Rounding up, 601
The minimum sample size we should anticipate using is 601.
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.
