Respuesta :
Answer:
We need a sample size of at least 383.
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]
85% confidence level
So [tex]\alpha = 0.15[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.15}{2} = 0.925[/tex], so [tex]Z = 1.44[/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 85% confidence level with an error of at most 0.03
We need a sample size of at least n.
n is found with [tex]M = 0.03, \pi = 0.21[/tex]
Then
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.03 = 1.44\sqrt{\frac{0.21*0.79}{n}}[/tex]
[tex]0.03\sqrt{n} = 1.44\sqrt{0.21*0.79}[/tex]
[tex]\sqrt{n} = \frac{1.44\sqrt{0.21*0.79}}{0.03}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.44\sqrt{0.21*0.79}}{0.03})^{2}[/tex]
[tex]n = 382.23[/tex]
Rounding up
We need a sample size of at least 383.
