Chapter 10.3, Problem 79E

To determine

To construct and interpret a $95\%$ confidence interval for the true mean difference (Weight minus Groove) in the estimates of tire wear these two methods in the population of tires.

Answer to Problem 79E

We are $95\%$ confident that the true mean of the tires undergoing the weight method is between $2.8379\text{ and }6.2747$ higher than the true mean wear of the tires undergoing the groove method.

Explanation of Solution

T is given:

  $\begin{array}{l}n=16\\ c=0.95\end{array}$

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,

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

Now, the degree of freedom will be:

  $\begin{array}{l}df=n-1\\ =16-1\\ =15\end{array}$

Now, the value of t will be:

  $t_{\alpha /2}=2.131$

Now, the confidence interval will be:

  $\begin{array}{l}\overline{x}-t_{\alpha /2}\times \frac{s}{\sqrt{n}}\\ =4.5563-2.131\times \frac{3.2255}{\sqrt{16}}\\ =2.8379\\ \overline{x}+t_{\alpha /2}\times \frac{s}{\sqrt{n}}\\ =4.5563+2.131\times \frac{3.2255}{\sqrt{16}}\\ =6.2747\end{array}$

Thus we conclude that we are $95\%$ confident that the true mean of the tires undergoing the weight method is between $2.8379\text{ and }6.2747$ higher than the true mean wear of the tires undergoing the groove method.