Question

A simple random sample of 35 colleges and universities in the United States has a mean tuition of 18700 with a standard deviation of 10800. Construct a 99% confidence interval for the mean tuition for all colleges and universities in the United States. Round the answers to the nearest whole number.

Answer 1

Solution :

Given that,

t /2,df = 2.728

Margin of error = E = t/2,df * (s /n)

= 2.728 * (10800 / 35)

Margin of error = E = 4980

The 90% confidence interval estimate of the population mean is,

- E < < + E

18700 - 4980 < < 18700 + 4980

13720 < < 23680

(13720 , 23680)