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 3597 students classified as Juniors and find that 3099 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

Answer: (0.8468, 0.8764)

Step-by-step explanation:

Formula to find the confidence interval for population proportion is given by :-

$\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

, where $\hat{p}$  = sample proportion.

z* = Critical value

n= Sample size.

Let p be the true proportion of GSU Juniors who believe that they will, immediately, be employed after graduation.

Given : Sample size = 3597

Number of students  believe that they will find a job immediately after graduation= 3099

Then,  $\hat{p}=\dfrac{3099}{3597}\approx0.8616$

We know that , Critical value for 99% confidence interval = z*=2.576  (By z-table)

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

$0.8616\pm(2.576)\sqrt{\dfrac{0.8616(1-0.8616)}{3597}}$

$0.8616\pm (2.576)\sqrt{0.0000331513594662}$

$\approx0.8616\pm0.0148\\\\=(0.8616-0.0148,\ 0.8616+0.0148)=(0.8468,\ 0.8764)$

Hence, the 99 % confidence interval for the proportion of GSU Juniors who believe that they will, immediately, be employed after graduation. = (0.8468, 0.8764)