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 mpg were recorded each time the gas tank was filled, and the computer was then reset. In addition to the computer computing mpg, the driver also recorded the mpg by dividing the miles driven by the number of gallons at each fill-up. The following data are the differences between the computer's and the driver's calculations for a random sample of 20 records. The driver wants to determine if these calculations are different. Assume the standard deviation of a difference to be ? = 3.0.

4.0 7.5 ?0.6 1.7 3.7 4.5 6.0 2.2 4.6 3.0
4.4 0.2 3.0 1.1 1.1 6.0 2.1 3.7 ?0.6 ?4.2
(a) State the appropriate H0 and Ha to test this suspicion.

H0: ? < 0 mpg; Ha: ? > 0 mpg

H0: ? > 0 mpg; Ha: ? < 0 mpg

H0: ? = 3 mpg; Ha: ? ? 3 mpg

H0: ? = 0 mpg; Ha: ? ? 0 mpg

H0: ? > 3 mpg; Ha: ? < 3 mpg

(b) Carry out the test. Give the P-value. (Round your answer to four decimal places.)

Interpret the result in plain language.

We conclude that ? = 3 mpg; that is, we have strong evidence that the computer's reported fuel efficiency does not differ from the driver's computed values.

We conclude that ? = 0 mpg; that is, we have strong evidence that the computer's reported fuel efficiency differs from the driver's computed values.

We conclude that ? ? 0 mpg; that is, we have strong evidence that the computer's reported fuel efficiency does not differ from the driver's computed values.

We conclude that ? ? 3 mpg; that is, we have strong evidence that the computer's reported fuel efficiency differs from the driver's computed values.

We conclude that ? ? 0 mpg; that is, we have strong evidence that the computer's reported fuel efficiency differs from the driver's computed values.

a

explanation:

part a

answer: a reasonable domain is where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

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part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

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part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).