91Ó°ÊÓ

Given:

$$\begin{gathered} n=16 \\ b=0.79021 \\ S{E}_b=0.07104 \end{gathered}$$

The degrees of freedom is the sample size decreased by 2 :

$$\begin{gathered} df=n-2 \\ =16-2 \\ =14 \end{gathered}$$

The critical t-value can be found in table B in the row of d f=14 and in the column of c=99% :

$t^{*}=2.977$

The boundaries of the confidence interval then become:

localid="1650543080433" $$\begin{gathered} b-t^{*}×S{E}_b \\ =0.79021-2.977×0.07104 \\ =0.5787 \end{gathered}$$

localid="1650543092212" $$\begin{gathered} b+t^{*}×S{E}_b \\ =0.79021+2.977×0.07104 \\ =1.0017 \end{gathered}$$

The slight deviation is due to rounding errors.

Want to test whether there is a difference between the two methods of estimating tire wear.

The two variables both measure the tire wear in the same measurement units. If we want to know if there is a difference, we assume that there is no difference and thus both need to increase by the same amounts, resulting in the null hypothesis

$H_0:\mathrm{β}=1$

The alternative hypothesis states the opposite of the null hypothesis:

$H_a:\mathrm{β}≠1$

Given:

$H_0:\mathrm{β}=1$

$H_a:\mathrm{β}≠1$

$H_0:\mathrm{β}=1$

$H_a:\mathrm{β}≠1$

Confidence interval is given in exercise part (a):

$(0.5785,1.001)$

The confidence interval contains 1 and thus it is likely to obtain $\mathrm{β}=1$, which means that there is not sufficient evidence to support the claim of a difference.