Chapter 12: Q. 17 (page 762)

Paired tires Exercise 69 in Chapter 8 (page 519 ) 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 scatterplot below displays the two estimates (in thousands of miles) for each tire.
Computer output from a least-squares regression analysis of these data is shown below. 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 (0.5785,1.001).
(b) Researchers want to test whether there is a difference between the two methods of estimating tire wear.
Explain why an appropriate pair of hypotheses for this test is $H_{0} : β = 1 versus H_{a} : β \neq 1$
(c) What conclusion would you draw for this significance test based on your interval in part (a)? Justify your answer.

Short Answer

Expert verified

a) $(0.5787, 1.0017)$

b) $H_{0} : β = 1$

$H_{a} : β \neq 1$

c) There is not sufficient evidence to support the claim of a difference.

Step by step solution

Given information(Part a)

Given:

$n = 16 b = 0.79021 S E_{b} = 0.07104$

Explanation(Part a)

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

$d f = n - 2 = 16 - 2 = 14$

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" $b - t^{*} \times S E_{b} = 0.79021 - 2.977 \times 0.07104 = 0.5787$

localid="1650543092212" $b + t^{*} \times S E_{b} = 0.79021 + 2.977 \times 0.07104 = 1.0017$

The slight deviation is due to rounding errors.

Given Information(Part b)

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

Explanation(Part b)

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

$H_{0} : β = 1$

The alternative hypothesis states the opposite of the null hypothesis:

$H_{a} : β \neq 1$

Given Information(Part c)

Given:

$H_{0} : β = 1$

$H_{a} : β \neq 1$

Explanation(Part c)

$H_{0} : β = 1$

$H_{a} : β \neq 1$

Confidence interval is given in exercise part (a):

$(0.5785, 1.001)$

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