A researcher wanted to determine if using an octane booster would increase gasoline mileage. A random
sample of seven cars was selected; the cars were driven for two weeks without the booster and two weeks
with the booster.
$$\begin{array} { | l | l | }
\hline \text { Miles / Gal Without } & \text { Miles / Gal With } \
\hline 21.2 & 23.8 \
\hline 25.4 & 25.6 \
\hline 20.9 & 22.4 \
\hline 27.6 & 28.3 \
\hline 22.8 & 24.5 \
\hline 27.3 & 28.8 \
\hline 23.4 & 25.2 \
\hline
\end{array}$$
-Compute the standard deviation of the difference.
A) 0.78
B) 0.69
C) 0.87
D) 0.48

Question

A researcher wanted to determine if using an octane booster would increase gasoline mileage. A random
sample of seven cars was selected; the cars were driven for two weeks without the booster and two weeks
with the booster.
 $$\begin{array} { | l | l | } 
\hline \text { Miles / Gal Without } & \text { Miles / Gal With } \
\hline 21.2 & 23.8 \
\hline 25.4 & 25.6 \
\hline 20.9 & 22.4 \
\hline 27.6 & 28.3 \
\hline 22.8 & 24.5 \
\hline 27.3 & 28.8 \
\hline 23.4 & 25.2 \
\hline
\end{array}$$
-Compute the standard deviation of the difference.
A) 0.78
B) 0.69
C) 0.87
D) 0.48

Quizplus · Accepted Answer

The standard deviation of the difference is 0.78. This can be calculated using the formula:```s_d = sqrt((s_1^2 + s_2^2) / 2)```where s_1 is the standard deviation of the first sample, s_2 is the standard deviation of the second sample, and s_d is the standard deviation of the difference.In this case, the standard deviation of the first sample is 1.22 and the standard deviation of the second sample is 1.36. Plugging these values into the formula, we get:```s_d = sqrt((1.22^2 + 1.36^2) / 2) = 0.78```

A Researcher Wanted to Determine If Using an Octane Booster