Question

Measuring Treadwear. Refer to Exercise 10.126 and find a $95\%$ confidence interval for the mean difference in measurement by the weight and groove methods.

Short answer

$(1.59,5.92)$ is the data's $95\%$ confidence interval.

Step-by-step solution

Given Information

The given data is,

$\begin{array}{|cc|}\hline \text{Method}& \text{Method}\\ 30.5& 28.7\\ 30.9& 25.9\\ 31.9& 23.3\\ 30.4& 23.1\\ 27.3& 23.7\\ 20.4& 20.9\\ 24.5& 16.1\\ 20.9& 19.9\\ 18.9& 15.2\\ 13.7& 11.5\\ 11.4& 11.2\\ \hline\end{array}$

Explanation

Below is the given data,

$\begin{array}{|cccc|}\hline \text{Method}& \text{Method}& \text{Difference}=\mathrm{d}& d^2\\ 30.5& 28.7& 1.8& 3.24\\ 30.9& 25.9& 5& 25\\ 31.9& 23.3& 8.6& 73.96\\ 30.4& 23.1& 7.3& 53.29\\ 27.3& 23.7& 3.6& 12.96\\ 20.4& 20.9& -0.5& 0.25\\ 24.5& 16.1& 8.4& 70.56\\ 20.9& 19.9& 1& 1\\ 18.9& 15.2& 3.7& 13.69\\ 13.7& 11.5& 2.2& 4.84\\ 11.4& 11.2& 0.2& 0.04\\ & & 41.3& 258.83\\ \hline\end{array}$

Explanation

The sample mean is,

$\overline{d}=\frac{\sum d}{n}$

$=\frac{41.3}{11}$

localid="1651285684383" $=3.7545$

The standard deviation is,

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

$=\sqrt{\frac{258.83-\frac{(44.3{)}^2}{11}}{11-1}}$

$=3.2213$

Explanation

The degree of freedom is,

$\text{dof}=n-1$

$=11-1$

$=10$

The level of significance's critical value is $\pm 2.228$.

The confidence level is,

$CI=\overline{d}\pm t_{\frac{a}{2}}\frac{s_d}{\sqrt{n}}$

$=3.7545\pm (2.228)\frac{3.2213}{\sqrt{11}}$

$=(1.59,5.92)$