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 with equal variances.

Question

contestada

Respuesta :

akiran007 akiran007 · Answer 1 · 2021-04-11T14:43:37+02:00

Answer:

CI= [4.324 , 5.6758]

Step-by-step explanation:

The 90 % confidence interval for the normal distribution is given by

(x`1- x2` ) ± t ∝/2 (υ) .Sp √1/n1 + 1/n2

where Sp = √[n1-1 (s1)² + n2-1 (s2)²]/n1+n2-2

and n1= 12 n2=10

x1`= 16 km x`2 = 11km

s1= 1km s2= 0.8km

and degrees of freedom υ= n1+n2-2= 20

Here ∝/2 = 0.05

From the table we get the value of t∝/2 (υ) to be 1.725

Putting the values

Sp = √[n1-1 (s1)² + n2-1 (s2)²]/n1+n2-2

Sp= √11(1) + 9(0.8)²/20

Sp= √0.838=0.915

CI = (x`1- x2` ) ± t ∝/2 (υ) .Sp √1/n1 + 1/n2

CI = (16-11)± 1.725 * 0.915√1/12+ 1/10

CI = (5) ± 0.675819

CI=[ 4.324 , 5.6758]