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 1382 tenth graders is drawn. Of the students sampled, 996 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.

THANK YOU

Answer 1

Solution :

Given that,

n = 1382

x = 996

Point estimate = sample proportion = = x / n = 0.721

1 - = 0.279

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.576 * (((0.721*0.279) / 1382)

= 0.031

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.721 - 0.031 < p < 0.721 + 0.031

0.690 < p < 0.752

The 99% confidence interval is : (0.690 , 0.752)