Respuesta :
Answer:
a) The standard error for this estimate of the percentage of all young Americans who earned a high school diploma is 0.87%.
b) The margin of error, using a 95% confidence level, for estimating the percentage of all young Americans who earned a high school diploma is of 1.71%.
c) The 95% confidence interval for the percentage of all young Americans who earned a high school diploma is (85.29%, 88.71%).
d) The lower bound of the confidence interval is above 80%, which means that the confidence interval supports the claim that the percentage of young Americans who cam high school diplomas has increased.
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].
Standard error:
The standard error is:
[tex]s = \sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
Margin of error:
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}} = zs[/tex]
The confidence interval is:
Sample proportion plus/minus margin of error. So
[tex](\pi - M, \pi + M)[/tex]
In a simple random sample of 1500 young Americans 1305 had earned a high school diploma.
This means that [tex]n = 1500, \pi = \frac{1305}{1500} = 0.87[/tex]
a. What is the standard error for this estimate of the percentage of all young Americans who earned a high school diploma?
[tex]s = \sqrt{\frac{\pi(1-\pi)}{n}} = \sqrt{\frac{0.87*0.13}{1500}} = 0.0087[/tex]
0.0087*100% = 0.87%.
The standard error for this estimate of the percentage of all young Americans who earned a high school diploma is 0.87%.
b. Find the margin of error, using a 95% confidence level, for estimating the percentage of all young Americans who earned a high school diploma.
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
[tex]M = zs = 1.96*0.0087 = 0.0171[/tex]
0.0171*100% = 1.71%
The margin of error, using a 95% confidence level, for estimating the percentage of all young Americans who earned a high school diploma is of 1.71%.
c. Report the 95% confidence interval for the percentage of all young Americans who earned a high school diploma.
87% - 1.71% = 85.29%
87% + 1.71% = 88.71%.
The 95% confidence interval for the percentage of all young Americans who earned a high school diploma is (85.29%, 88.71%).
d. Suppose that in the past, 80% of all young Americans earned high school diplomas. Does the confidence interval you found in part c support or refute the claim that the percentage of young Americans who cam high school diplomas has increased? Explain.
The lower bound of the confidence interval is above 80%, which means that the confidence interval supports the claim that the percentage of young Americans who cam high school diplomas has increased.
