Question

Researchers were interested in comparing two methods for estimating tire wear. The first method used the amount of weight lost by a tire. The second method used the amount of wear in the grooves of the tire. A random sample of $16$tires was obtained. Both methods were used to estimate the total distance traveled by each tire. The table provides the two estimates (in thousands of miles) for each tire.

a. Make a dot-plot of the difference (Weight – Groove) in the estimate of wear for each tire using the two methods.

b. Describe what the graph reveals about whether the two methods give similar estimates of tire wear, on average.

c. Calculate the mean difference and the standard deviation of the differences. Interpret the mean difference.

Short answer

Part a. The dot plot is as follows:

Part b. We note that 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.

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

Step-by-step solution

Part a). Step 1. Explanation

The dot plot is as follows:

Part b). Step 1. Explanation

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

We note that$15ofthe16$ 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.

Part c). Step 1. Explanation

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 :

$$\begin{gathered} \overline{x}=\frac{{\displaystyle \sum _{i-1}^n}{x}_i}{n} \\ =\frac{10.2+2.7+6.4+5.3+7.0+1.8+5+8.6+7.3+3.6+-0.3+8.4+1+3.7+2.2+0.2}{16} \\ =\frac{72.9}{16} \\ =4.5563 \end{gathered}$$

The sample variance is then as:

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

The sample standard deviation is then,

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

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