Question

In analyzing the city and highway fuel efficiencies of many cars and​ trucks, the mean difference in fuel efficiencies for the 642 vehicles was 7.35 mpg with a standard deviation of 2.48 mpg. Find a​ 95% confidence interval for this difference and interpret it in context. The​ 95% confidence interval for the difference in fuel efficiencies is left parenthesis nothing comma nothing right parenthesis mpg.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 7.35

Population standard deviation =    = 2.48

Sample size = n =642

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96   ( Using z table )

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

= 1.96 * ( 2.48 /  642 )

E= 0.1918
At 95% confidence interval estimate of the population mean
is,

- E < < + E

7.35 - 0.1918 <   < 7.35+0.1918

7.1582 <   < 7.5418

( 7.1582 , 7.5418 )