Question

The National Assessment of Educational Progress (NAEP) includes a "long-term trend" study that tracks reading and mathematics skills over time, and obtains demographic information. In the 2012 study, a random sample of 9000 17-year-old students was selected.24 The NAEP sample used a multistage design, but the overall effect is quite similar to an SRS of 17-year-olds who are still in school. In the sample, 51% of students had at least one parent who was a college graduate. Estimate, with 99% confidence, the proportion of all 17-year-old students in 2012 who had at least one parent graduate from college.

Answer 1

Answer:

The 99% confidence interval estimate for the proportion of all 17-year-old students in 2012 who had at least one parent graduate from college is (0.4964, 0.5236).

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 zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

Sample of 9000, 51% of students had at least one parent who was a college graduate.

This means that $n = 9000, \pi = 0.51$

99% confidence level

So $\alpha = 0.01$, z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$, so $Z = 2.575$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.51 - 2.575\sqrt{\frac{0.51*0.49}{9000}} = 0.4964$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.51 + 2.575\sqrt{\frac{0.51*0.49}{9000}} = 0.5236$

The 99% confidence interval estimate for the proportion of all 17-year-old students in 2012 who had at least one parent graduate from college is (0.4964, 0.5236).