Answer:
A sample size of 2080 is needed.
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]
98% confidence level
So [tex]\alpha = 0.02[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.02}{2} = 0.99[/tex], so [tex]Z = 2.327[/tex].
Based on previous evidence, you believe the population proportion is approximately 60%.
This means that [tex]\pi = 0.6[/tex]
How large of a sample size is required?
We need a sample of n.
n is found when [tex]M = 0.025[/tex]. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.025 = 2.327\sqrt{\frac{0.6*0.4}{n}}[/tex]
[tex]0.025\sqrt{n} = 2.327\sqrt{0.6*0.4}[/tex]
[tex]\sqrt{n} = \frac{2.327\sqrt{0.6*0.4}}{0.025}[/tex]
[tex](\sqrt{n})^{2} = (\frac{2.327\sqrt{0.6*0.4}}{0.025})^{2}[/tex]
[tex]n = 2079.3[/tex]
Rounding up
A sample size of 2080 is needed.
