A simple random sample of 35 colleges and universities in the United States has a mean tuition of $19,400 with a standard deviation of $11,000. Construct a 90% confidence interval for the mean tuition for all colleges and universities in the United States. Round the answers to the nearest whole number A 90% confidence interval for the mean tuition for all colleges and universities is < μ <.

$16,341< μ <$22,459.

Step-by-step explanation:

The formula for calculating confidence interval is expressed as;

CI = xbar±(z×s/√n)

xbar is the mean tuition

z is the z score at 90% CI

s is the standard deviation

n is the sample size

Given

xbar = $19,400

s = $11,000

n = 35

z = 1.645

Substitute into the formula for calculating confidence interval

CI = xbar±(z×s/√n)

CI = 19400±(1.645×11000/√35)

CI = 19400±(1.645×11000/5.916)

CI = 19400±(1.645×1859.36)

CI = 19400±3058.65

CI = [19400-3,058.65, 19400+3,058.65]

CI = [16,341.35, 22,458.65]

CI = [16,341, 22,459] to nearest whole number.

Hence a 90% confidence interval for the mean tuition for all colleges and universities is 16,341< μ <22,459.

answer: answer is c!

ok i got to organize the

3/15 and 12/55

3*

aha

3*4 equals 12 but not

a: no

b:

8/24 12/35

no

gautie

it's c

5/18 = 25/90

5*5=25

18*5 = 90

hope that you!

one more happy customer!

< 3

step-by-step explanation:

answer: check all that apply, where the options?

step-by-step explanation: