91Ó°ÊÓ

T is given:

$$\begin{gathered} n=16 \\ c=0.95 \end{gathered}$$

The mean is:

$$\begin{gathered} \stackrel{¯}{x}=\frac{{\displaystyle \underset{i-1}{\overset{n}{∑}}}{x}_i}{n} \\ =\frac{10.2+2.7+6.4+5.3+7+1.8+5+8.6+7.3+3.6+-0.3+8.4+1+3.7+2.2+0.2}{16} \\ =\frac{72.9}{16} \\ =4.5563 \end{gathered}$$

The sample variance is then as:

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

The sample standard deviation is then,

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

Now, the degree of freedom will be:

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

Now, the value of t will be:

$t_{\mathrm{Î\pm }/2}=2.131$

Now, the confidence interval will be:

$$\begin{gathered} \stackrel{¯}{x}-t_{\mathrm{Î\pm }/2}×\frac{s}{\sqrt{n}} \\ =4.5563-2.131×\frac{3.2255}{\sqrt{16}} \\ =2.8379 \\ \stackrel{¯}{x}+t_{\mathrm{Î\pm }/2}×\frac{s}{\sqrt{n}} \\ =4.5563+2.131+\frac{3.2255}{\sqrt{16}} \\ =6.2747 \end{gathered}$$

Thus we conclude that we are$95\%$ confident that the true mean of the tires undergoing the weight method is between$2.8379and6.2747$ higher than the true mean wear of the tires undergoing the groove method.