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. Step 2 of 2 : Suppose a sample of 292 tenth graders is drawn. Of the students sampled, 240 read above the eighth grade level. Using the data, construct the 80% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level. Round your answers to three decimal places.

Answer 1

Answer:

The 80% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.149, 0.207).

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

Suppose a sample of 292 tenth graders is drawn. Of the students sampled, 240 read above the eighth grade level.

So 292 - 240 = 52 read below or at eight grade level, and that $n = 292, \pi = \frac{52}{292} = 0.178$

80% confidence level

So $\alpha = 0.2$, z is the value of Z that has a pvalue of $1 - \frac{0.2}{2} = 0.9$, so $Z = 1.28$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.178 - 1.28\sqrt{\frac{0.178*0.822}{292}} = 0.149$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.178 + 1.28\sqrt{\frac{0.178*0.822}{292}} = 0.207$

The 80% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.149, 0.207).