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.
a. Suppose a sample of 269 tenth graders is drawn. Of the students sampled, 224 read above the eighth grade level. Using the data, estimate the proportion of tenth graders reading at or below the eighth grade level. Enter your answer as a fraction or a decimal number rounded to three decimal places.
b. Suppose a sample of 709 tenth graders is drawn. Of the students sampled, 546 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. Round your answers to three decimal places.

Answer 1

Answer:

(a) The proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b) The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

Step-by-step explanation:

Let X = number of students who read above the eighth grade level.

(a)

A sample of n = 269 students are selected. Of these 269 students, X = 224 students who can read above the eighth grade level.

Compute the proportion of students who can read above the eighth grade level as follows:

$\hat p=\frac{X}{n}=\frac{224}{269}=0.8327$

The proportion of students who can read above the eighth grade level is 0.8327.

Compute the proportion of tenth graders reading at or below the eighth grade level as follows:

$1-\hat p=1-0.8327$

        $=0.1673$

Thus, the proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b)

the information provided is:

n = 709

X = 546

Compute the sample proportion of tenth graders reading at or below the eighth grade level as follows:

$\hat q=1-\hat p$

  $=1-\frac{X}{n}$

  $=1-\frac{546}{709}$

  $=0.2299\\\approx 0.229$

The critical value of z for 95% confidence interval is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Compute the 95% confidence interval for the population proportion as follows:

$CI=\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

     $=0.229\pm 1.96\times \sqrt{\frac{0.229(1-0.229)}{709}}\\=0.229\pm 0.03136\\=(0.19764, 0.26036)\\\approx (0.198, 0.260)$

Thus, the 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).