Yuh esketit Gucci gang ya I came in with the sauce ouuuuub

Mathematics

2 answers:

IRISSAK [1]2 years ago

6 0

The answer should be true

zavuch27 [327]2 years ago

3 0

Answer:

trueeeee

Step-by-step explanation:

You might be interested in

A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than ex

amm1812

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 $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

The dean randomly selects 200 students and finds that 118 of them are receiving financial aid.

This means that $n = 200, \pi = \frac{118}{200} = 0.59$

90% confidence level

So $\alpha = 0.1$ , z is the value of Z that has a p-value of $1 - \frac{0.1}{2} = 0.95$ , so $Z = 1.645$ .

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

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.01 = 1.645\sqrt{\frac{0.59*0.41}{n}}$