Answer :
Answer:
The sample size is 1875.
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]
Sampling error of 0.03.
This means that [tex]M = 0.03[/tex]
99.74% confidence level
So [tex]\alpha = 0.0026[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.0026}{2} = 0.9987[/tex], so [tex]Z = 3[/tex].
25% of all adults had used the Internet for such a purpose
This means that [tex]\pi = 0.25[/tex]
What is the sample size
The sample size is n. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.03 = 3\sqrt{\frac{0.25*0.75}{n}}[/tex]
[tex]0.03\sqrt{n} = 3\sqrt{0.25*0.75}[/tex]
Simplifying by 0.03 both sides
[tex]\sqrt{n} = 100\sqrt{0.25*0.75}[/tex]
[tex](\sqrt{n})^2 = (100\sqrt{0.25*0.75})^2[/tex]
[tex]n = 1875[/tex]
The sample size is 1875.