Answer :
Answer:
We need a sample size of at least 1161.
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 for the interval is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
For this problem, we have that:
[tex]\pi = 0.22[/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].
How large a sample should she take?
We need a sample size of at least n.
n is found when [tex]M = 0.02[/tex]
So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.02 = 1.645\sqrt{\frac{0.22*0.78}{n}}[/tex]
[tex]0.02\sqrt{n} = 1.645\sqrt{0.22*0.78}[/tex]
[tex]\sqrt{n} = \frac{1.645\sqrt{0.22*0.78}}{0.02}[/tex]
[tex](\sqrt{n})^{2} = (\frac{1.645\sqrt{0.22*0.78}}{0.02})^{2}[/tex]
[tex]n = 1160.88[/tex]
Rounding up
We need a sample size of at least 1161.