Introduction to the Practice of Statistics
Introduction to the Practice of Statistics
9th Edition
ISBN: 9781319013387
Author: David S. Moore, George P. McCabe, Bruce A. Craig
Publisher: W. H. Freeman
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Chapter 13, Problem 35E

(a)

To determine

To graph: The plot for means.

(a)

Expert Solution
Check Mark

Explanation of Solution

Graph: To obtain the interaction plot, use Minitab. Follow the steps below:

Step 1: Insert the data into Minitab worksheet.

Step 2: Go to Stat > ANOVA > Interaction plot.

Step 3: Select “Means” in the column for Responses and select “Location and Car” in the column for Factors.

Step 4: Click OK.

The obtained interaction plot is obtained as shown below:

Introduction to the Practice of Statistics, Chapter 13, Problem 35E

Interpretation: The interaction plot represents that there is a little interaction between two Chromium means. There may be an interaction for the larger values.

(b)

To determine

To find: The pooled standard deviation.

(b)

Expert Solution
Check Mark

Answer to Problem 35E

Solution: The required pooled standard deviation is 0.12777.

Explanation of Solution

Calculation: The pooled standard deviation is calculated by using the formula as shown below:

sp(n11)s12+(n21)s22+(n31)s32+(n41)s42n1+n2+n3+n44

The pooled standard deviation is obtained as follows:

 sp=(n1)(s12+s22+s32+s42)4(n1)=(s12+s22+s32+s42)4

=0.122+0.122+0.142+0.1324=0.12777

Interpretation: The pooling of standard deviation can be applied here because the data values satisfies the condition that the value of maximum standard deviation is less than the twice of the minimum standard deviation value, i.e., σmax<2×σmin.

(c)

Section 1:

To determine

The contrast to compare the mean for new with old car.

(c)

Section 1:

Expert Solution
Check Mark

Answer to Problem 35E

Solution: The contrast of compared groups is as follows:

c1=1/2×(μNewcity+μNewhighway)1/2×(μOldcity+μOldhighway)

Explanation of Solution

A contrast (c) is defined as a combination of population means, c=i=1nai×μi where ai is chosen as i=1nai=0 and μi is the sample means. The contrast to compare the groups of interest can be obtained as:

c1=1/2×(μNewcity+μNewhighway)1/2×(μOldcity+μOldhighway)

Section 2:

To determine

The contrast to compare the means for new car while driving in city.

Section 2:

Expert Solution
Check Mark

Answer to Problem 35E

Solution: The contrast of compared groups is as follows:

c2=1×(μNewcity)1×(μNewHighway)

Explanation of Solution

A contrast (c) is defined as a combination of population means, c=i=1nai×μi where ai is chosen as i=1nai=0 and μi is the sample means. The contrast to compare the groups of interest can be obtained as:

c2=1×(μNewcity)1×(μNewHighway)

Section 3:

To determine

The contrast to compare the mean for highway with city while driving an old car.

Section 3:

Expert Solution
Check Mark

Answer to Problem 35E

Solution: The contrast of compared groups is as follows:

c3=1×(μOldhighway)1×(μOldcity)

Explanation of Solution

A contrast (c) is defined as a combination of population means, c=i=1nai×μi where ai is chosen as i=1nai=0 and μi is the sample means. The contrast to compare the groups of interest can be obtained as:

c3=1×(μOldhighway)1×(μOldcity)

(d)

To determine

The benefit of repeated measurements.

(d)

Expert Solution
Check Mark

Answer to Problem 35E

Solution: The repeated measure of study reduces the within group variation.

Explanation of Solution

The benefit of observing the group of participants in repeated manner rather than observing the random participants in each treatment reduces the within-group variation or within group errors. It is observed that by subjecting the same individuals to four treatments, rather than four individuals to one treatment each, the group within group’s variability is reduced. The repeated measure of study reduces the within-group variability which improves the F-statistic to define the significance of test.

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