Chapter 12.1, Problem 22E

(a)

To determine

To Explain: the 99% confidence interval for the slope of the population regression line is (0.5787, 1.0017).

Answer to Problem 22E

(0.5787, 1.0017)

Explanation of Solution

Given:

  

  

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

Formula used:

The boundaries of the confidence interval

  $\begin{array}{l}b-t^{\ast }\times S{E}_b\\ b+t^{\ast }\times S{E}_b\end{array}$

Calculation:

The degrees of freedom is

  $df=n-2=16-2=14$

The critical value can be found in table B in the mention row of $df=14$ and in the column of c=99%:

  $t^{\ast }=2.977$

The boundaries of the confidence intervals are

  $\begin{array}{l}b-t^{\ast }\times S{E}_b=0.79021-2.977\times 0.07104=0.5787\\ b+t^{\ast }\times S{E}_b=0.79021+2.977\times 0.07104=1.0017\end{array}$

The slight deviation is there due to rounding errors.

(b)

To determine

To find: that is there any difference in the two methods of the calculating the tire wears.

Answer to Problem 22E

  $H_0:\beta =1$

  $H_a:\beta \ne 1$

Explanation of Solution

Given:

  

  

The two variables both calculate the tire wear in the same measurement units. If there is require to know if there is a difference, assume that there is no difference and there both require increasing by the same amounts, resulting in the null hypothesis

  $H_0:\beta =1$

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

  $H_a:\beta \ne 1$

(c)

To determine

To find: the standardized test statistic and P-value on the basis of part (b) and conclusion at the $\alpha =0.01$ .

Answer to Problem 22E

  $\begin{array}{l}t=-2.953\\ 0.005<p<0.01\end{array}$

Explanation of Solution

Given:

  

  

Formula used:

  $t=\frac{b-{\beta }_0}{S{E}_b}$

Calculation:

Finding the test statistic

Find $b\text{and }S{E}_b$ from the mention in the row labelled “Weight” and the columns labeled “Coef” and “SE Coef”, respectively.

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

Find tail probability P

  $df=n-2=16-2=14$

Looking at Table B in the row for 14 degrees of freedom and find that $|-2.953|=2.953$ is between 2.624 and 2.977, the t values for p=0.01 and p=0.005, respectively. Therefore,

  $0.005<p<0.01$ .

Using the tail probability to find the P-value

The reason is that this is a two-tailed test (i.e. because this is a test for $\beta \ne 1$ ). It must be double the p-interval to compensate. Therefore, the p-value is between 0.01 and 0.02.

The reason is that the p-value greater than our significance level of $\alpha =0.01$ , fail to reject $H_0$ . There is not sufficient convincing proof to conclude that there is a difference between the two methods of calculating the tire wear.

(d)

To determine

To find: that the conclusion of part (a) and part (c) is same or not.

Answer to Problem 22E

There is no enough evidence to help the claim of a difference.

Explanation of Solution

Given:

  

  

  $\begin{array}{l}{H}_0:\beta =1\\ H_a:\beta \ne 1\end{array}$

Confidence interval mention in part (a)

(0.5785, 1.001)

The confidence interval 1 and therefore it is likely to get $\beta =1$ , which means that there is no enough proof to help the claim of a difference.