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 12, Problem 68E

(a)

To determine

To graph: The plot ai versus i for the linear trend and the value for contrast for linear trend.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given: The value of ai is provided for the linear trend. The data table is provided for means.

Introduction to the Practice of Statistics, Chapter 12, Problem 68E , additional homework tip  1

Graph: The plot is drawn by following these steps:

Step 1: Open Excel sheet and write the data value for ai. The screenshot is shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 68E , additional homework tip  2

Step 2: INSERT > Recommended Charts > All Charts > Line Chart. The screenshot is shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 68E , additional homework tip  3

The plot is obtained as shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 68E , additional homework tip  4

The values chosen for ai are linear with i.

The contrast can be calculated by the sum of weighted means. If μi is the mean for the ith group, the contrast for means can be obtained as:

Ψ=i=1nai×μi=(2)×3.82+(1)×4.88+(0)×4.56+(1)×4.41+(2)×3.99=0.13

Interpretation: The value of ai follows linear pattern with i. The contrast for linear trend is Ψ=0.13.

(b)

To determine

To graph: The plot ai versus i for the quadratic trend and the value for contrast for quadratic trend.

(b)

Expert Solution
Check Mark

Explanation of Solution

Graph: The plot is drawn by following these steps:

Step 1: Open Excel sheet and write the data value for ai. The screenshot is shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 68E , additional homework tip  5

Step 2: INSERT > Recommended Charts > All Charts > Line Chart. The screenshot is shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 68E , additional homework tip  6

The plot is obtained as shown below:

Introduction to the Practice of Statistics, Chapter 12, Problem 68E , additional homework tip  7

The contrast can be calculated by the sum of weighted means. If μi is the mean for the group, the contrast for means can be obtained as:

Ψ=i=1nai×μi=(2)×3.82+(1)×4.88+(2)×4.56+(1)×4.41+(2)×3.99=2.79

The contrast for means for μi=5i can be obtained as:

Ψ=i=1nai×μi=(2)×5+(1)×5+(2)×5+(1)×5+(2)×5=(2121+2)×5=0

Interpretation: The value chosen, ai, follows quadratic pattern with i. If means are equal, contrast will sum to zero, which states the property of contrast such that i=1nai=0.

(c)

To determine

To test: The hypotheses for the linear, quadratic, and cubic trend.

(c)

Expert Solution
Check Mark

Answer to Problem 68E

Solution: The P-values are less than 0 .05 at 5% significance level for quadratic trend. The P-values are more than 0 .05 at 5% significance level for linear and cubic trend. So, the null hypothesis is rejected for quadratic trend and the null hypothesis fails to reject for the linear and cubic trend.

Explanation of Solution

Calculation: To test the null hypothesis, steps are as follows:

Contrasts for the linear trend can be calculated as:

Ψlinear=i=1nai×μi=(2)×3.82+(1)×4.88+(0)×4.56+(1)×4.41+(2)×3.99=0.13

Contrasts for the quadratic trend can be calculated as:

Ψquadratic=i=1nai×μi=(2)×3.82+(1)×4.88+(2)×4.56+(1)×4.41+(2)×3.99=2.79

Contrasts for the cubic trend can be calculated as:

Ψcubic=i=1nai×μi=(1)×3.82+(2)×4.88+(0)×4.56+(2)×4.41+(1)×3.99=1.11

The pooled standard deviation is obtained as:

sp=(n11)s12+(n21)s22+(n31)s32+(n41)s42+(n51)s52n1+n2+n3+n4+n55=(241)0.992+(331)0.852+(261)1.072+(301)1.432+(211)1.02224+33+26+30+215=1.094

The standard error of linear contrast can be calculated as:

SEΨlinear=sp×(2)2n1+(1)2n2+(0)2n3+(1)2n4+(2)2n5=1.094×(2)224+(1)233+(0)226+(1)230+(2)221=0.7097

The standard error of quadratic contrast can be calculated as:

SEΨquadratic=sp×(2)2n1+(1)2n2+(2)2n3+(1)2n4+(2)2n5=1.094×(2)224+(1)233+(2)226+(1)230+(2)221=0.8293

The standard error of cubic contrast can be calculated as:

SEΨcubic=sp×(1)2n1+(2)2n2+(0)2n3+(2)2n4+(1)2n5=1.094×(1)224+(2)233+(0)226+(2)230+(1)221=0.6415

The resultant t statistic values for the linear trend can be calculated as:

tlinear=ΨlinearSEΨlinear=0.130.7097=0.1832

The resultant t- statistic values for the quadratic trend can be calculated as:

tquadratic=ΨquadraticSEΨquadratic=2.790.8293=3.3643

The resultant t- statistic values for the cubic trend can be calculated as:

tcubic=ΨcubicSEΨcubic=1.110.6415=1.7303

The resultant P- values are obtained by using the standard normal distribution table as follows:

Pvaluelinear=0.57

Pvaluequadratic=0.99

Pvaluecubic=0.043

Conclusion: The P-values is less than 0 .05 at 5% significance level for cubic trend. The P-values are more than 0 .05 at 5% significance level for linear and quadratic trend. So, the null hypothesis is rejected for cubic trend and the null hypothesis fails to reject for the linear and qudratic trend.

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