Question

In a simple random sample of 1500 young Americans 1305 had earned a high school diploma.

Answer 1

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 $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

Standard error:

The standard error is:

$s = \sqrt{\frac{\pi(1-\pi)}{n}}$

Margin of error:

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}} = zs$

The confidence interval is:

Sample proportion plus/minus margin of error. So

$(\pi - M, \pi + M)$

In a simple random sample of 1500 young Americans 1305 had earned a high school diploma.

This means that $n = 1500, \pi = \frac{1305}{1500} = 0.87$

a. What is the standard error for this estimate of the percentage of all young Americans who earned a high school diploma?

$s = \sqrt{\frac{\pi(1-\pi)}{n}} = \sqrt{\frac{0.87*0.13}{1500}} = 0.0087$

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 $\alpha = 0.05$, z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

Then

$M = zs = 1.96*0.0087 = 0.0171$

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.