Respuesta :
Answer:
A sample of 2084 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]
In an earlier study, the population proportion was estimated to be 0.19.
This means that [tex]\pi = 0.19[/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].
How large a sample would be required in order to estimate the fraction of tenth graders reading at or below the eighth grade level at the 98% confidence level with an error of at most 0.02?
This is n for which M = 0.02. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.02 = 2.327\sqrt{\frac{0.19*0.81}{n}}[/tex]
[tex]0.02\sqrt{n} = 2.327\sqrt{0.19*0.81}[/tex]
[tex]\sqrt{n} = \frac{2.327\sqrt{0.19*0.81}}{0.02}[/tex]
[tex](\sqrt{n})^2 = (\frac{2.327\sqrt{0.19*0.81}}{0.02})^2[/tex]
[tex]n = 2083.4[/tex]
Rounding up:
A sample of 2084 is needed.
