Question

a countrys education department reported that in 2015, 68.9% of students enrolled in college or a trade school within 12 months of graduating high school. in 2017, a random sample of 100 individuals who graduated from high school 12 months prior was selected. from this sample, 105 students were found to be enrolled in college or a trade school. complete parts a through c.

a. construct a 95% confidence interval to estimate the actual proportion of students enrolled in college or a trade school within 12 months of graduating from high school in 2017.

the confidence interval has a lower limit of 0.510 and an upper limit of 0.650.
(round to three decimal places as needed.)

b. what is the margin of error for this sample?

the margin of error is
(round to three decimal places as needed.)

Explanation:

Step1: Recall the formula for margin of error

The formula for the margin of error \(E\) for a proportion confidence interval is \(E=\text{Upper limit}-\text{Sample proportion}\) or \(E = \text{Sample proportion}-\text{Lower limit}\). The sample proportion \(\hat{p}=\frac{105}{200}=0.525\)

Step2: Calculate the margin of error

Using \(E=\text{Upper limit}-\hat{p}\), with upper limit \(U = 0.656\) and \(\hat{p}=0.525\)
\(E=0.656 - 0.525=0.131\)

Using \(E=\hat{p}-\text{Lower limit}\), with lower limit \(L = 0.510\) and \(\hat{p}=0.525\)
\(E=0.525 - 0.510 = 0.015\) (This is wrong, we should use the correct formula \(E=\frac{U - L}{2}\))

The correct formula for margin of error for a confidence interval of proportion \((L,U)\) is \(E=\frac{U - L}{2}\)

Substitute \(L = 0.510\) and \(U=0.656\) into the formula:
\(E=\frac{0.656- 0.510}{2}=\frac{0.146}{2}=0.073\)

Answer:

The margin of error is \(0.073\)