Question

A simple random sample of 35 colleges and universities in the United States has a mean tuition of $18,900 with a standard deviation of $10,400. Construct an 80% 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,

Point estimate = sample mean =     = 18,900

Population standard deviation =    =10,400

Sample size n =35

At 80% confidence level the z is ,

Z/2 = 1.282 ( Using z table )

Margin of error = E =   Z/2    * ( /n)
= 1.282 * ( 10,400 / 35)

= 2254
At 95% confidence interval n mean
is,

- E < < + E

18,900 - 2254 <   < 18,900 +   2254

(16646 ,21154)