Respuesta :
Answer:
a. 0.87
b. The margin of error is of 0.0124.
c. The 90% confidence interval for the population proportion of adults who think the future health of Social Security is a major economic concern is (0.8576, 0.8824).
d. The 95% confidence interval for this population proportion is (0.8553, 0.8847).
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 z-score that has a p-value of [tex]1 - \frac{\alpha}{2}[/tex].
The margin of error is of:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
The confidence interval is:
[tex]\pi \pm M[/tex]
a. What is the point estimate of the population proportion of adults who think the future health of Social Security is a major economic concern.
The sample proportion, that is, 1740 out of n = 2000. So
[tex]\pi = \frac{1740}{2000} = 0.87[/tex]
0.87 is the point estimate.
b. At 90% confidence, what is the margin of error?
90% confidence level
So [tex]\alpha = 0.1[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.1}{2} = 0.95[/tex], so [tex]Z = 1.645[/tex].
Then the margin of error is of:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]M = 1.645\sqrt{\frac{0.87*0.13}{2000}}[/tex]
[tex]M = 0.0124[/tex]
The margin of error is of 0.0124.
c. Develop a 90% confidence interval for the population proportion of adults who think the future health of Social Security is a major economic concern
[tex]\pi - M = 0.87 - 0.0124 = 0.8576[/tex]
[tex]\pi + M = 0.87 + 0.0124 = 0.8824[/tex]
The 90% confidence interval for the population proportion of adults who think the future health of Social Security is a major economic concern is (0.8576, 0.8824).
d.
95% confidence level
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a p-value of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].
Then the margin of error is of:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]M = 1.96\sqrt{\frac{0.87*0.13}{2000}}[/tex]
[tex]M = 0.0147[/tex]
The confidence interval will be of:
[tex]\pi - M = 0.87 - 0.0147 = 0.8553[/tex]
[tex]\pi + M = 0.87 + 0.0147 = 0.8847[/tex]
The 95% confidence interval for this population proportion is (0.8553, 0.8847).
