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Chapter 8: Q. 63 (page 518)

63. Give it some gas! Computers in some vehicles calculate various quantities related to performance. One of these is 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. Here are the mpg values for a random sample of 20of these records:

15.813.615.619.122.415.622.517.219.422.6
19.418.014.618.721.014.822.621.514.320.9
Construct and interpret a 95% confidence interval for the mean fuel efficiency M for this vehicle.

Short Answer

Expert verified

The mean fuel efficiency for the vehicle is between 17.0218mpg and 19.9382mpg, according to 95% confidence.

Step by step solution

01

Given information

The mpg values for a random sample of 20of these records:

15.8
13.6
15.6
19.1
22.4
15.6
22.5
17.2
19.4
22.6
19.4
18.0
14.6
18.7
21.0
14.8
22.6
21.5
14.3
20.9
02

Explanation

Determine the mean by:

x¯=sumofthesamplestotalnumberof sumple=15.8+13.6+⋯+14.3+20.920=18.48

Then determine the standard deviation by:

sx=∑xi-x¯2n-1=(15.8−18.48)2+⋯+(20.9−18.48)220−1=3.1158

03

Explanation

Determine the degree of freedom by:

df=n-1

df=20-1

df=1

Convert the confidence level is 95% into decimal:
95100=0.95
Determine the column by:
1-c2=1-0.952=0.025
Calculate the critical value t*, from the table B.

Utilize the row df and column.
Thus, critical value t*=2.093.

04

Explanation

Determine the margin of error:

E=t*sxn

=2.093×3.115820

=1.4582
Finally, determine the population mean as:
x¯-E<μ<x¯+E

18.48-1.4582<μ<18.48+1.4582

17.0218<μ<19.9382
Therefore, the population mean is from 17.0218mpgto 19.9382mpg.

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