Question

a random sample of 30 gas stations in a region gives the statistics below. Complete parts a) through c)

                

y=$3.29, s= $0.25

                 

                    a) Find a 95% confidence interval for the mean price of regular gasoline in the region.

                 

                    b) Find the 90% confidence interval for the mean.

                 

                    c) If we had the same statistics from a sample of 80 stations, what would the 95% confidence interval be now?

Answer 1

Part a)

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 30- 1 ) = 2.045
3.29 ± t(0.05/2, 30 -1) * 0.25/√(30)
Lower Limit = 3.29 - t(0.05/2, 30 -1) 0.25/√(30)
Lower Limit = 3.1967
Upper Limit = 3.29 + t(0.05/2, 30 -1) 0.25/√(30)
Upper Limit = 3.3833
95% Confidence interval is ( 3.1967 , 3.3833 )

Part b)
Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.1 /2, 30- 1 ) = 1.699
3.29 ± t(0.1/2, 30 -1) * 0.25/√(30)
Lower Limit = 3.29 - t(0.1/2, 30 -1) 0.25/√(30)
Lower Limit = 3.2125
Upper Limit = 3.29 + t(0.1/2, 30 -1) 0.25/√(30)
Upper Limit = 3.3675
90% Confidence interval is ( 3.2125 , 3.3675 )

Part c)
Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.05 /2, 80- 1 ) = 1.99
3.29 ± t(0.05/2, 80 -1) * 0.25/√(80)
Lower Limit = 3.29 - t(0.05/2, 80 -1) 0.25/√(80)
Lower Limit = 3.2344
Upper Limit = 3.29 + t(0.05/2, 80 -1) 0.25/√(80)
Upper Limit = 3.3456
95% Confidence interval is ( 3.2344 , 3.3456 )