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 3957 students classified as Juniors and find that 3226 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?

Step-by-step explanation:

Using the formula

p +/- z* √(p(1-p) / n)

Where p is sample proportion = 3226/3957 = 0.8153, n = 3957 and z* is 2.576

0.8153 + (2.576 √(0.8153(1-0.8153) / 3957))

0.8153 + (2.576 √ (0.8153(0.1847) /3957)

0.8513 + (2.576 √(0.1572/3957)

0.8513 + (2.576 √0.00003974)

0.8513 + 2.576(0.0063)

0.8513 + 0.016

0.8670 ~ 87%

For the lower interview

0.8513 - 0.016

= 0.8353 ~ 84%

Thus, at 99% confidence interval is 84% and 87%

answer: a) (0,0), (1,1), (2,8), and (3,27).

step-by-step explanation:

0^3=0

1^3=1

2^3=8

3^3=27

in effect, the exponential equation for this would me y=3^x. each x-coordinate cubed is giving the y coordinate, so a is your answer : )

answer: .75 km

step-by-step explanation: