Question

PLEASE HELP ME

Of 357 randomly selected medical students, 30 said that they planned to work in a rural 1) community. Find a 95% confidence interval for the true proportion of all medical students who plan to work in a rural community.

Answer 1

The 95% confidence interval for the true proportion is found from the following formula:
\[p-1.96\sqrt{\frac{p(1-p)}{n}} < \pi < p+1.96\sqrt{\frac{p(1-p)}{n}}\]
where p =357 and n = 357

Answer 2

so i just fill in the numbers and that will be my answer??

Answer 3

Basically yes. You will need to do some calculations to get the two values of probability that are the upper and lower limit.

Answer 4

and both p and n are 357?

Answer 5

Sorry, my bad.
p = 30/357
n = 357

Answer 6

30 divided 357?

Answer 7

\[p=\frac{30}{357}\]
\[n=357\]