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Groovy tires 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鈥揋roove) 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.

Step by step solution

01

Dot Plot 

02

Interpretation 

The differences of distance travelled as per weight and groove are positive in most of the sample units.

So, it states - that the distance travelled as per wear and tear, according to weight base ; are mostly higher than the distance travelled as per wear and tear according to grooves basis.

03

Mean and standard Deviation of Differences 

$MeanDifference=Sumofdifferences/numberofdifferences$

$$\begin{gathered} [10.2+2.7+6.4+...(otherdifferences)+...+3.7+2.2+0.2] \\ 梅16 \\ =72.9/16 \\ =4.556 \end{gathered}$$
role="math" localid="1655229348625" $S\tan dardDeviationofdifferences=\sqrt{鈭�{(x_i-x\text{'})}^2/(n-1)}$

where $x_i=eachdifference,x\text{'}=meandifference,n=numberofunits$

• 1st tire : ${(x_{i-}x\text{'})}^2={(10.2-4.556)}^2=31.9676$
• 2nd tire : ${(x_i-x\text{'})}^2={(2.7-4.556)}^2=3.45$

Similarly finding all ${(x_i-x\text{'})}^2$and summing them, $[{(10.2-4.55625)}^2+...+{(0.2-4.55625)}^2]梅[16-1]$

= $\sqrt{56.059375/15}$

$$\begin{gathered} \sqrt{10.4039} \\ =3.225 \end{gathered}$$

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