Some people believe that higher-octane fuels result in better gas mileage for th
ID: 3264511 • Letter: S
Question
Some people believe that higher-octane fuels result in better gas mileage for their car. To test this claim, a researcher randomly selected
1111
individuals (and their cars) to participate in the study. Each participant received 10 gallons of gas and drove his car on a closed course. The number of miles driven until the car ran out of gas was recorded. A coin flip was used to determine whether the car was filled up with 87-octane or92-octane first, and the driver did not know which fuel was in the tank.
87-octane
234234
257257
244244
216216
113113
288288
316316
229229
193193
205205
546546
92-octane
238238
239239
230230
224224
118118
297297
351351
241241
185185
209209
562562
(a) Why is it important that the matching be done by driver and car?
A.
So that all the trials can be done at the same time
B.
How someone drives and the car they drive result in different fuel consumption
C.
Cuts the cost of doing the research
D.
So that each driver can determine which fuel is best for their car
(b) Why is it important to conduct the study on a closed track?
A.
So that all drivers and cars have similar driving conditions
B.
So that the researcher can watch the drivers
C.
So that each car travels the same distance
(c) The normal probability plots for miles on87-octane and miles on 92-octane are shown.
-2007000 10087 OctanePercent
-2007000 10092 OctanePercent
Are either of these variables normally distributed?
A.
No, neither variable is normally distributed.
B.
Yes, 92-octane is normally distributed.
C.
Yes, both variables are normally distributed.
D.
Yes, 87-octane is normally distributed.
(d) The differences are computed as 92-octane minus 87-octane. The normal probability plot of the differences is shown. Is there reason to believe that the differences are normally distributed?
Yes
No
(e) The researchers used a statistical software package to determine whether the mileage from 92-octane is greater than the mileage from 87-octane. The results are given below.
Paired T-Test and CI: 92 Octane, 87 Octane
Paired T for 92 Octane87Octane
N
Mean
StDev
SE Mean
92 Octane
11
258.273
108.924
32.842
87 Octane
11
263.091
115.161115.161
34.722
Difference
11
4.818
14.682
4.427
T-Test of mean difference
equals=0
(vs
greater than>0):
T-Valueequals=1.09
P-Value=0.151
What do you conclude at
=0.05?
Why?
A.
Reject the null hypothesis because the P-value is greater than
0.05.
B.
Fail to reject the null hypothesis because the P-value is greater than
0.050.05.
C.
Reject the null hypothesis because the P-value is less than
0.05.
D.
Fail to reject the null hypothesis because the P-value is less than
0.05.
87-octane
234234
257257
244244
216216
113113
288288
316316
229229
193193
205205
546546
92-octane
238238
239239
230230
224224
118118
297297
351351
241241
185185
209209
562562
Explanation / Answer
Part a )
How someone drives and the car they drive result in different fuel consumption
Part b )
So that all drivers and cars have similar driving conditions
Part C )
Histogram for 87 octance
Histogram for 92 octance
No, neither variable is normally distributed.
Part d ) Yes,
Part e ) Here , P-value > 0.05 so we fail to reject H0
Fail to reject the null hypothesis because the P-value is greater than 0.05.
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