Q. 63 - The Practice of Statistics for AP

Question

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 $20$ of these records:

$15.8 13.6 15.6 19.1 22.4 15.6 22.5 17.2 19.4 22.6$
$19.4 18.0 14.6 18.7 21.0 14.8 22.6 21.5 14.3 20.9$
Construct and interpret a $95\%$ confidence interval for the mean fuel efficiency M for this vehicle.

Step-by-Step Solution

Answer

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

Step 1: Given information

The mpg values for a random sample of $20$of 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

Step 2: Explanation

Determine the mean by:

 $$\begin{gathered} \overline{x}=\frac{\text{ sum of the samples }}{\text{ total number of sumple }} \\ =\frac{15.8+13.6+\cdots +14.3+20.9}{20} \\ =18.48 \end{gathered}$$

Then determine the standard deviation by:

$$\begin{gathered} s_x=\sqrt{\frac{\sum {\left(x_i-\overline{x}\right)}^2}{n-1}} \\ =\sqrt{\frac{(15.8-18.48{)}^2+\cdots +(20.9-18.48{)}^2}{20-1}} \\ =3.1158 \end{gathered}$$

Step 3: Explanation

Determine the  degree of freedom by:

$df=n-1$

$df=20-1$

$df=1$

Convert the confidence level is $95\%$ into decimal:
$\frac{95}{100}=0.95$
Determine the column by:
$\frac{1-c}{2}=\frac{1-0.95}{2}=0.025$
Calculate the critical value $t^{*}$, from the table $B$.

Utilize the row $df$ and column.
Thus, critical value $t^{*}=2.093$.

Step 4: Explanation

Determine the margin of error:

$E=t^{*}\frac{s_x}{\sqrt{n}}$

$=2.093\times \frac{3.1158}{\sqrt{20}}$

$=1.4582$
Finally, determine the population mean as:
$\overline{x}-E<\mu <\overline{x}+E$

$18.48-1.4582<\mu <18.48+1.4582$

$17.0218<\mu <19.9382$
Therefore, the population mean is from $17.0218 mpg$ to $19.9382 mpg$.