Question

In analyzing the city and highway fuel efficiencies of many cars and​ trucks, the mean difference in fuel efficiencies for the 611vehicles was 7.38 mpg with a standard deviation of 2.51

mpg. Find a​ 95% confidence interval for this difference and interpret it in context. Round to 2 decimals.

Answer 1

Solution :

given that

= 7.38

= 2.51

n = 611

at 95%confidence level z is

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05/ 2 = 0.025

Z_/2 = Z_0.025 =1.96

Margin of error = E = Z_/2* ( /n)

= 1.96* (2.51 / 611)

= 0.20

At 99% confidence interval estimate of the population mean is,

- E < < + E

7.38-0.20 < < 7.38+0.20

7.18< <7.58