Question

The ability to find a job after graduation is very important to GSU students as it is to the students at most colleges and universities.

Suppose we take a poll (random sample) of 3613 students classified as Juniors and find that 2956 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.

Answer 1

Solution :

Given that,

n = 3613

x = 2956

Point estimate = sample proportion = = x / n = 2956/3613=0.818

1 -   = 1- 0.818 =0.182

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* (((0.818*0.182) /3613 )

E = 0.017

A 99% confidence interval for proportion p is ,

- E < p < + E

0.818-0.017 < p < 0.818+0.017

0.801< p < 0.835