Question

Homes in nearby college town have a mean value of 158,950 dollars. It is assumed that the homes in the vicinity of the college have a higher value. To test this claim, a random sample of 16 homes is chosen from around the college. Their mean valuation is 162,460 dollars and the sample standard deviation is 5400 dollars. compute a 90% confident interval for the true mean value of homes

Answer 1

n= 16, = 162460, s= 5400, c= 90%

formula for confidence interval is

calculate t critical value for c= 90% with df = n-1 = 16 -1 = 15

using t table we get

critical value =  1.753

162460 − 2366.618 < μ < 162460 + 2366.618  

160093.382  < μ <  164826.618

confidence interval is = ( 160093.382 , 164826.618 )