Respuesta :
Answer:
a) We need a sample size of at least 705.
b) We need a sample size of at least 1692.
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]
90% confidence level
So [tex]\alpha = 0.1[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].
a) If she uses 1999 estimate of 11.8% obtained from the Current Population Survey.
We need a sample of size at least n
n is found when [tex]M = 0.02, \pi = 0.118[/tex].
So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.02 = 1.645\sqrt{\frac{0.118*0.882}{n}}[/tex]
[tex]0.02\sqrt{n} = 1.645\sqrt{0.118*0.882}[/tex]
[tex]\sqrt{n} = \frac{1.645\sqrt{0.118*0.882}}{0.02}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.645\sqrt{0.118*0.882}}{0.02})^{2}[/tex]
[tex]n = 704.08[/tex]
Rounding up
We need a sample size of at least 705.
b) She does not use any estimate.
Same thing as above, we just use [tex]\pi = 0.5[/tex] when do not use any estimate.
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.02 = 1.645\sqrt{\frac{0.5*0.5}{n}}[/tex]
[tex]0.02\sqrt{n} = 1.645\sqrt{0.5*0.5}[/tex]
[tex]\sqrt{n} = \frac{1.645\sqrt{0.5*0.5}}{0.02}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.645\sqrt{0.5*0.5}}{0.02})^{2}[/tex]
[tex]n = 1691.2[/tex]
Rounding up
We need a sample size of at least 1692.
