Question

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

a. What is the standard error for this estimate of the percentage of all young Americans who earned a high school diploma?
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.
c. Report the 95% confidence interval for the percentage of all young Americans who earned a high school diploma.
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.

Answer 1

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.

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{(\pi(1-\pi))/(n)}`

In which

z is the z-score that has a p-value of
`1 - (\alpha)/(2)`.

Standard error:

The standard error is:

`s = \sqrt{(\pi(1-\pi))/(n)}`

Margin of error:

The margin of error is:

`M = z\sqrt{(\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 = (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{(\pi(1-\pi))/(n)} = \sqrt{(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 - (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.