Question

A sample of 1000 college students at NC State University were randomly selected for a survey. Among the survey participants, 108 students suggested that classes begin at 8 AM instead of 8:30 AM. The sample proportion is 0.108.

What is the lower endpoint for the 90% confidence interval? Give your answer to three decimal places.

(Note that due to the randomization of the questions, the numbers in this question might be different from the previous question.)

Answer:

Answer 1

Solution:

n = 1000

x = 108

= x/n = 0.108

Our aim is to construct 90% confidence interval.

c = 0.90

= 1- c = 1- 0.90 = 0.10

  /2 = 0.10 2 = 0.05 and 1- /2 = 0.950

= 1.645

Now , the margin of error is given by

E = /2 *  

= 1.645 * [(0.108*(1 - 0.108)/1000]

=  0.016

Confidence interval for proportion is

     E

Lower endpoint of the interval is - E i.e. (0.108 -  0.016) = 0.092

The lower endpoint for the 90% confidence interval is 0.092