Respuesta :
Answer:
A sample of 307 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 of:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In this question, we have that:
We dont know the true proportion of women police officers, so we use [tex]\pi = 0.5[/tex], which is when the largest sample size would be needed.
92% confidence level
So [tex]\alpha = 0.08[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.08}{2} = 0.96[/tex], so [tex]Z = 1.75[/tex].
How large should a sample be if the margin of error is .05 for a 92% confidence interval
We need a sample of n, and n is found when [tex]M = 0.05[/tex]. So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.05 = 1.75\sqrt{\frac{0.5*0.5}{n}}[/tex]
[tex]0.05\sqrt{n} = 1.75*0.5[/tex]
[tex]\sqrt{n} = 17.5[/tex]
[tex](\sqrt{n})^2 = (17.5)^2[/tex]
[tex]n = 306.25[/tex]
Rounding up
A sample of 307 is needed.
