Question

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the eighth grade level. Suppose a sample of 1537 tenth graders is drawn. Of the students sampled, 1184 read above the eighth grade level. Using the data, construct the 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level.

Answer 1

Answer:

The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.2087, 0.2507).

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 $1 - \frac{\alpha}{2}$.

For this problem, we have that:

Suppose a sample of 1537 tenth graders is drawn. Of the students sampled, 1184 read above the eighth grade level. So 1537 - 1184 = 353 read at or below this level. Then

$n = 1537, \pi = \frac{353}{1537} = 0.2297$

95% confidence level

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

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2297 - 1.96\sqrt{\frac{0.2297*0.7703}{1537}} = 0.2087$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.2297 + 1.96\sqrt{\frac{0.2297*0.7703}{1537}} = 0.2507$

The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.2087, 0.2507).