Question

Step 2. Suppose a sample of 1134 tenth graders is drawn. Of the students sampled, 930 read above the eighth grade level. Using the data, construct the 99% 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: Lower endpoint: Upper endpoint:

Answer 1

We would be looking at the first question here as:

The sample proportion here is computed as:
p = x/n = 930/1134 = 0.8201

From standard normal tables, we have here:
P(-2.576 < Z < 2.576) = 0.99

The confidence interval thus is obtained here as:

$p \pm z^*\sqrt{\frac{p(1-p)}{n}}$

$0.8201 \pm 2.576*\sqrt{\frac{0.8201(1-0.8201)}{1134}}$

$0.8201 \pm 0.0294$

This is the required confidence interval here. (rounded to 3 decimal places )