Answer:
6546 students would need to be sampled.
Step-by-step explanation:
In a sample with a number n of people surveyed with a probability of a success of , and a confidence level of
, we have the following confidence interval of proportions.
In which
z is the z-score that has a p-value of .
The margin of error is:
The dean randomly selects 200 students and finds that 118 of them are receiving financial aid.
This means that
90% confidence level
So , z is the value of Z that has a p-value of
, so
.
If the dean wanted to estimate the proportion of all students receiving financial aid to within 1% with 90% reliability, how many students would need to be sampled?
n students would need to be sampled, and n is found when M = 0.01. So
Rounding up:
6546 students would need to be sampled.