Question

Measuring Treadwear. R. Stichler et al. compared two methods of measuring treadwear in their paper "Measurement of Treadwear of Commercial Tires" (Rubber Age, Vol. 73:2). Eleven tires were each measured for treadwear by two methods, one based on weight and the other on groove wear. The data, in thousands of miles, are as follows.

At the$5\%$ significance level, do the data provide sufficient evidence to conclude that, on average, the two measurement methods give different results?

Short answer

The null hypothesis is rejected, and the data are adequate to establish that the two measuring methods provide different findings on average.

Step-by-step solution

Given Information

Given data is shown below

We have to explain Whether the data provide sufficient evidence to conclude that on average the two measurement methods give different result.

Explanation

The null and alternative hypothesis are:

$$\begin{gathered} H_0:{\mu }_1={\mu }_2 \\ H\alpha :{\mu }_1\ne {\mu }_2 \end{gathered}$$

The table is given below:

Mean is:

$$\begin{gathered} \overline{d}=\frac{\sum d}{n} \\ =\frac{41.3}{11} \\ =3.7545 \end{gathered}$$

Standard deviation is:

$S_d=\sqrt{\frac{\sum d_i^2-\frac{{\left(\sum d_i\right)}^2}{n}}{n-1}}$

$$\begin{gathered} =\sqrt{\frac{258.83-\frac{(41.3{)}^2}{11}}{11-1}} \\ =3.2213 \end{gathered}$$

The formula of test statistics is : $t=\frac{\overline{d}}{\frac{s_d}{\sqrt{m}}}$

substitute the given values

$t=\frac{3.7545}{\frac{3.2213}{\sqrt{11}}}$

$=3.866$

The degree of freedom is $$\begin{gathered} dof=n-1 \\ =11-1 \\ =10 \end{gathered}$$

The critical value for level of significance is $\pm 2.228$

Since, the value of test statistic is fall in the rejection region.