In: Statistics and Probability
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.
Solution :
Given that,
Point estimate = sample mean =
= 18700
Population standard deviation =
= 10800
Sample size = n =35
At 99% confidence level the z is ,
= 1 - 99% = 1 - 0.99 = 0.01
/ 2 = 0.01 / 2 = 0.005
Z/2 = Z0.005 = 2.576 ( Using z table )
Margin of error = E = Z/2* ( /n)
=2.576 * (10800 / 35)
= 4703
At 99% confidence interval is,
- E < < + E
18700 - 4703 < < 18700+4703
13997< < 23403
(18700, 23403)