A tire manufacturer is interested in testing the fuel econom

A tire manufacturer is interested in testing the fuel economy for 2 different tread patterns, A and B. Tires of each tread type are driven for 1000 miles on each of 18 different cars. The mean difference in the mileage (mileage A – mileage B) was 1.74 mpg, and the standard deviation of the difference was 1.46 mpg. Calculate a 99% confidence interval for the mean difference in fuel economy. Round intermediate and final answers to two decimal place and report in the format: (1.23, 4.56)

Solution

Note that              
Margin of Error E = t(alpha/2) * s / sqrt(n)              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.005          
X = sample mean =    1.74          
t(alpha/2) = critical t for the confidence interval =    2.89823052          
s = sample standard deviation =    1.46          
n = sample size =    18          
df = n - 1 =    17          
Thus,              
Margin of Error E =    0.997354448          
Lower bound =    0.742645552          
Upper bound =    2.737354448          
              
Thus, the confidence interval is              
              
(   0.74   ,   2.74   ) [ANSWER]