Respuesta :
Answer:
A sample size of 1031 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 of:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
37% of freshmen do not visit their counselors regularly.
This means that [tex]\pi = 0.37[/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].
You would like to be 98% confident that your estimate is within 3.5% of the true population proportion. How large of a sample size is required?
A sample size of n is required.
n is found when M = 0.035. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.035 = 2.327\sqrt{\frac{0.37*0.63}{n}}[/tex]
[tex]0.035\sqrt{n} = 2.327\sqrt{0.37*0.63}[/tex]
[tex]\sqrt{n} = \frac{2.327\sqrt{0.37*0.63}}{0.035}[/tex]
[tex](\sqrt{n})^2 = (\frac{2.327\sqrt{0.37*0.63}}{0.035})^2[/tex]
[tex]n = 1030.4[/tex]
Rounding up:
A sample size of 1031 is required.
