Answer:
With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is of at least 216.
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:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
For this problem, we have that:
[tex]p = 0.1[/tex]
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is
We need a sample size of at least n, in which n is found M = 0.04.
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.04 = 1.96\sqrt{\frac{0.1*0.9}{n}}[/tex]
[tex]0.04\sqrt{n} = 0.588[/tex]
[tex]\sqrt{n} = \frac{0.588}{0.04}[/tex]
[tex]\sqrt{n} = 14.7[/tex]
[tex](\sqrt{n})^{2} = (14.7)^{2}[/tex]
[tex]n = 216[/tex]
With a .95 probability, the sample size that needs to be taken if the desired margin of error is .04 or less is of at least 216.
