Chapter 10.3, Problem 75E

(a)

To determine

To make a dot plot of the difference (Weight minus Groove) in the estimate of wear for each tire using the two methods.

Explanation of Solution

The dot plot is as follows:

(b)

To determine

To describe what the graph reveals about whether the two methods give similar estimates of tier wear on average.

Explanation of Solution

From the above graph in part (a), we have,

We note that $15\text{ of the 16}$ dots in the dot plot lie to the right to zero which implies that most of the differences (Weight minus Groove) are positive and thus mean total distance traveled with the weight loss method appears to exceed the mean total distance traveled with the groove method.

(c)

To determine

To calculate the mean difference and the standard deviation of the differences and interpret the mean difference.

Answer to Problem 75E

The mean is $4.5563$ and the standard deviation is $3.2255$ .

Explanation of Solution

It is given that:

  $10.2,2.7,6.4,5.3,7.0,1.8,5,8.6,7.3,3.6,-0.5,7.3,3.6,-0.5,8.4,1,3.7,2.2,0.2$

The mean is:

  $\begin{array}{l}\overline{x}=\frac{{\displaystyle \sum _{i=1}^n{x}_i}}{n}\\ =\frac{\begin{array}{l}10.2+2.7+6.4+5.3+7+1.8+5+8.6+7.3+3.6-0.3\\ +8.4+1+3.7+2.2+0.2\end{array}}{16}\\ =\frac{72.9}{16}\\ =4.5563\end{array}$

The sample variance is then as:

  $\begin{array}{l}{s}^2=\frac{{\displaystyle \sum {(x-\overline{x})}^2}}{n-1}\\ =\frac{\begin{array}{}$ {(10.2-4.5563)}^2+{(2.7-4.5563)}^2+{(6.4-4.5563)}^2+{(5.3-4.5563)}^2+$$ {(7-4.5563)}^2+{(1.8-4.5563)}^2+{(5-4.5563)}^2+{(8.6-4.5563)}^2+{(7.3-4.5563)}^2$$ +{(3.6-4.5563)}^2+{(-0.3-4.5563)}^2+{(8.4-4.5563)}^2+{(1-4.5563)}^2+{(3.7-4.5563)}^2$$ +{(2.2-4.5563)}^2+{(0.2-4.5563)}^2$\end{array}}{16-1}\\ =10.4040\end{array}$

The sample standard deviation is then,

  $\begin{array}{l}s=\sqrt{s^2}\\ =\sqrt{10.4040}\\ =3.2255\end{array}$

The difference is $4.5563$ thousands of miles on average which varies on average by $3.2255$ thousands of miles.