Question

An experiment reported in Popular Science compared fuel economies for two types of similarly equipped diesel mini-trucks. Let us suppose that 12 Volkswagen and 10 Toyota trucks were tested in 90- kilometer-per-hour steady-paced trials. If the 12 Volkswagen trucks averaged 16 kilometers per liter with a standard deviation of 1.0 kilometer per liter and the 10 Toyota trucks averaged 11 kilometers per liter with a standard deviation of 0.8 kilometer per liter, construct a 90% confidence interval for the difference between the average kilometers per liter for these two mini-trucks. Assume that the distances per liter for the truck models are approximately normally distributed.

Answer 1

Solution

n1 =  12

n2 =  10

x1-bar =  16

x2-bar =  11

s1 =  1

s2 =  0.8

% =  90

Degrees of freedom = n1 + n2 - 2 =  12 + 10 -2 =  20

Pooled s = √(((n1 - 1) * s1^2 + (n2 - 1) * s2^2)/DOF) =  √(((12 - 1) * 1^2 + ( 10 - 1) * 0.8^2)/(12 + 10 -2)) =  0.915423399

SE = Pooled s * √((1/n1) + (1/n2)) =  0.915423399307665 * √((1/12) + (1/10)) =  0.391960882

t- score =  1.724718218

Width of the confidence interval = t * SE =  1.7247182182138 * 0.391960882401973 =  0.676022075

Lower Limit of the confidence interval = (x1-bar - x2-bar) - width =  5 - 0.676022074705839 =  4.323977925

Upper Limit of the confidence interval = (x1-bar - x2-bar) + width =  5 + 0.676022074705839 =  5.676022075

The 90% confidence interval is [4.324, 5.676]