Paired tires Exercise 75 in Chapter 10 (page 686) compared two methods for estimating tire wear. The first method used the amount of weight lost by a tire. The second method used the amount of wear in the grooves of the tire. A random sample of 16 tires was obtained. Both methods were used to estimate the total distance traveled by each tire. The following scatterplot displays the two estimates (in thousands of miles) for each tire. 13

Here is some computer output from a least squares regression analysis of these data. assume that the conditions for regression inference are met.

a. Verify that the $99\%$confidence interval for the slope of the population regression line is $\left(0.5787,1.0017\right)$

b. Researchers want to test whether there is a difference in the two methods of estimating tire wear. Explain why the researchers might want to test the hypotheses $H_0:{\beta }_1=1$versus $H_{\alpha }:{\beta }_1\ne 1$.

c. Compute the standardized test statistic and P-value for the test in part (b). What conclusion would you draw at the $\alpha =0.01$significance level?

d. Does the confidence interval in part (a) lead to the same conclusion as the test in part (c)? Explain your answer.

Given information are

we have to verify that the$99\%$confidence interval for the slope of the population regression line is $\left(0.5787,1.0017\right)$

Formulas used for the boundaries of the confidence interval is

$b-t^{*}\times S{E}_b$

$b+t^{*}\times S{E}_b$

The degree of freedom is

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

$t*=2.977$

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

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

Given information are

we have to Explain why the researchers might want to test the hypotheses $H_0:\beta =1\&H_{\alpha }:\beta \ne 1$

Tire wear is calculated using the same measurement units in both variables. If you want to know if there's a difference, assume there isn't one and that both require the same amount of increase, which gives you the null hypothesis.

$H_0:\beta =1$

The null hypothesis statement is the opposite of the alternative hypothesis statement:

$H_{\alpha }:\beta \ne 1$

Given information are

we have to compute the standardized test statistic and P-value for the test in part (b).

Test statistic is calculated as

$$\begin{gathered} t=\frac{b-{\beta }_0}{S{E}_b} \\ =\frac{0.79021-1}{0.07104} \\ =-2.953 \end{gathered}$$

The degree of freedom is computed as

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

$\left|-2.953\right|=2.953$is between $2.624$and $2.977$, the t values for $p=0.01\&p=0.005$respectively, according to Table B in the row for$14$degrees of freedom. As a result,

$0.005<p<0.01$

Finding the P-value using the tail probability

This is due to the fact that this is a two-tailed test (i.e., it is a test for $\beta \ne 1$). To compensate, the p-interval must be doubled.therefore, the p-value ranges from localid="1654219899195" $0.01\&0.02$

The reason for this is that p- values greater than our $\alpha =0.01$significance level fail to reject $H_0$. There is insufficient evidence to infer that there is a difference between the two tire wear calculation methodologies.

Given information are

we have to determine does the confidence interval in part (a) lead to the same conclusion as the test in part (c)

$H_0:\beta =1$

$H_{\alpha }:\beta \ne 1$

confidence interval in part (a) is $0.5785,1.001$

The confidence interval is $1,$ therefore $\beta =1$ is likely, implying that there is insufficient evidence to support the assertion of a difference