Question

Suppose we take a poll (random sample) of 3907 students classified as Juniors and find that 2904 of them believe that they will find a job immediately after graduation.

What is the 99% confidence interval for the proportion of GSU Juniors who believe that they will, immediately, be employed after graduation.

1. (0.725, 0.761)
2. (0.732, 0.755)
3. (0.73, 0.757)
4. (0.736, 0.75)

Answer 1

Solution :

Given that,

n = 3907

x = 2904

Point estimate = sample proportion = = x / n = 2904/3907=0.743

1 -   = 1-0.743 =0.257

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

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

= 2.576* (((0.743*0.257) /3907 )

E = 0.018

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.743-0.018 < p < 0.743+0.018

(0.725, 0.761)