Respuesta :
Answer:
The margin of error for a 99% confidence interval is of 0.0639, that is, approximately 0.06.
The margin of error for a 95% confidence interval is of 0.0486, that is, approximately 0.05.
The margin of error for a 90% confidence interval is of 0.0408, that is, approximately 0.04.
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]
350 citizens, 240 responded favorably:
This means that [tex]n = 350, \pi = \frac{240}{350} = 0.6857[/tex]
99% confidence level
So [tex]\alpha = 0.01[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.01}{2} = 0.995[/tex], so [tex]Z = 2.575[/tex].
[tex]M = 2.575\sqrt{\frac{0.6857*0.3143}{350}} = 0.0639[/tex]
The margin of error for a 99% confidence interval is of 0.0639, that is, approximately 0.06.
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].
[tex]M = 1.96\sqrt{\frac{0.6857*0.3143}{350}} = 0.0486[/tex]
The margin of error for a 95% confidence interval is of 0.0486, that is, approximately 0.05.
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].
[tex]M = 1.645\sqrt{\frac{0.6857*0.3143}{350}} = 0.0408[/tex]
The margin of error for a 90% confidence interval is of 0.0408, that is, approximately 0.04.
