Respuesta :
Answer:
The margin of error for the 95% confidence interval for the population proportion is of 0.0287.
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]
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].
Find the margin of error for the 95% confidence interval for the population proportion.
We have that [tex]n = 820, p = 0.227[/tex]
So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]M = 1.96\sqrt{\frac{0.227*0.773}{820}} = 0.0287[/tex]
The margin of error for the 95% confidence interval for the population proportion is of 0.0287.
