Computers in some vehicles calculate various quantities related to performance.

Q: Computers in some vehicles calculate various quantities related to performance.

a) H0: = 0 Ha: 0 difference mean = 2.685 s = 2.745 hence t = (2.685 -0)/(2.745/sqrt(20)) = 4.37438 df =n-1 = 19 p-value = P(T > 4.37438 ) = 0.000 since p-value < 0.05 we reject the null We have strong evidence that the results of the two computations are different. mean sd computer 43.125 4.257733 driver 40.44 4.597528

ID: 3064045 • Letter: C

Question

Computers in some vehicles calculate various quantities related to performance. One of these is the fuel efficiency, or gas mileage, usually expressed as miles per gallon (mpg). For one vehicle equipped in this way, the miles per gallon were recorded each time the gas tank was filled, and the computer was then reset. In addition to the computer calculating miles per gallon, the driver also recorded the miles per gallon by dividing the miles driven by the number of gallons at fill-up. The driver wants to determine if these calculations are different.

(a) State the appropriate H0.

H0: 0

H0: > 0

H0: = 0

H0: 0

State the appropriate Ha.

Ha: = 0

Ha: < 0

Ha: 0

(b) Carry out the test. (Round your answer for t to four decimal places.)
t =

Give the degrees of freedom.
df=

Give the P-value. (Round your answer to four decimal places.)
p-value=

Interpret the result.

We have strong evidence that the results of the two computations are different.

We do not have sufficient evidence that the results of the two computations are different.

I CANNOT USE EXCEL TO CALCULATE THE T-TEST, SO PLEASE SHOW YOUR WORK SO I CAN UNDERSTAND HOW TO CALCULATE IT BY HAND.

Fill-up 1 2 3 4 5 6 7 8 9 10 Computer 41.6 50.6 36.8 37.5 34.5 44.7 47.7 43.2 47.4 42.3 Driver 36.5 44.2 37.2 35.6 30.5 40.5 40.0 41.0 42.8 39.2 Fill-up 11 12 13 14 15 16 17 18 19 20 Computer 43.1 44.7 48.1 46.1 46.6 39.2 37.4 43.4 44.3 43.3 Driver 38.8 44.5 45.4 45.3 45.7 34.2 35.2 39.8 44.9 47.5

Explanation / Answer

a)
H0: = 0
Ha: 0

difference

mean = 2.685

s = 2.745

hence

t = (2.685 -0)/(2.745/sqrt(20))

= 4.37438

df =n-1 = 19

p-value = P(T > 4.37438 ) = 0.000

since p-value < 0.05

we reject the null

We have strong evidence that the results of the two computations are different.

mean sd computer 43.125 4.257733 driver 40.44 4.597528

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Computers in some vehicles calculate various quantities related to performance.

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Computers in some vehicles calculate various quantities related to performance.

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