Respuesta :
Answer:
The sample size n = 4225
Step-by-step explanation:
We will use maximum error formula = [tex]\frac{z_{a} S.D}{\sqrt{n} }[/tex]
but we will find sample size "n"
[tex]\sqrt{n} = \frac{z_{a} S.D}{maximum error}[/tex]
Squaring on both sides , we get
[tex]n = (\frac{z_{a} S.D}{maximum error})^2[/tex]
Given 99% confidence interval (z value) = 2.56
given maximum error = 0.02
[tex]n = (\frac{z_{a} p(1-p)}{maximum error})^2[/tex]
n≤ [tex](\frac{2.56X\frac{1}{2} }{0.02}) ^{2}[/tex] ( here S.D = p(1-p) ≤ 1/2
on simplification , we get n = 4225
Conclusion:
The sample size of two samples is n = 4225
verification:-
We will use maximum error formula = [tex]\frac{z_{a} S.D}{\sqrt{n} }[/tex] = [tex]\frac{2.56 1/2}{\ \sqrt{4225} }[/tex] = 0.0196
substitute all values and simplify we get maximum error is 0.02
Using the z-distribution, it is found that samples of size 4155 are needed.
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 of:
[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} = 0.99[/tex], so [tex]z = 2.575[/tex].
The "worst case" scenario is an estimate of [tex]\pi = 0.5[/tex].
The sample size is n for which M = 0.02, hence:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.02 = 2.575\sqrt{\frac{0.5(0.5)}{n}}[/tex]
[tex]0.02\sqrt{n} = 2.575(0.5)[/tex]
[tex]\sqrt{n} = \frac{2.575(0.5)}{0.02}[/tex]
[tex](\sqrt{n})^2 = \left(\frac{2.575(0.5)}{0.02}\right)^2[/tex]
[tex]n = 4144[/tex]
Rounding up, samples of size 4155 are needed.
A similar problem is given at https://brainly.com/question/23905122