Question

Suppose a sample of 991991 tenth graders is drawn. Of the students sampled, 823823 read above the eighth grade level. Using the data, construct the 99%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 1

Let be the population proportion.

= 823/991 = 0.83

Confidence interval is given by

                                        

Two sided critical z value at 99% alpha level = Z_c = Z_0.005 = 2.58

(From the standard normal statistical table)

CI = 0.83 ± 2.58 * sqrt( 0.83*(1 - 0.83)/991) = (0.799 , 0.861)

99% Confidence interval for the proportion of tenth graders reading at or below the eighth grade level is (0.799, 0.861).