Introduction to the Practice of Statistics: w/CrunchIt/EESEE Access Card
Introduction to the Practice of Statistics: w/CrunchIt/EESEE Access Card
8th Edition
ISBN: 9781464158933
Author: David S. Moore, George P. McCabe, Bruce A. Craig
Publisher: W. H. Freeman
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Chapter 13, Problem 47E

(a)

To determine

To find: The means for each species by water combination and plot these means.

(a)

Expert Solution
Check Mark

Answer to Problem 47E

Solution: For Fresh biomass, the first three means for Species-1 are 109.1, 165.1, and 168.82. The first three means for Species-2 are 116.4, 156.8, and 254.88. The first three means for Species-3 are 55.60, 78.9, and 90.3. The first three means for Species-4 are 35.13, 58.33, and 94.54.

For Dry biomass, the first three means for Species-1 are 40.56, 63.86, and 71.00. The first three means for Species-2 are 34.50, 57.36, and 79.60. The first three means for Species-3 are 26.25, 31.87, and 36.24. The first three means for Species-4 are 15.53, 23.29, and 37.05.

Explanation of Solution

Calculation: To obtain the means for the response variable Fresh biomass, Minitab is used. The steps to be followed are:

Step 1: Go to the Minitab worksheet.

Step 2: Go to Stat > Basic Statistics > Descriptive Statistics.

Step 3: Enter the variable ‘Fresh biomass’ in the ‘Variable’ option.

Step 4: Enter the variables ‘Species and Water’ in the ‘By Variable’ option.

Step 5: Go to statistics and select ‘Mean.

Step 6: Go to graph and select Individual value plot’.

Step 7: Click Ok.

The obtained result shows the means for each species by water level. Here, few of them are given: The first three means for Species-1 are 109.1, 165.1, and 168.82. The first three means for Species-2 are 116.4, 156.8, and 254.88. The first three means for species-3 are 55.60, 78.9, and 90.3. The first three means for Species-4 are 35.13, 58.33, and 94.54. The obtained graph is given below:

Introduction to the Practice of Statistics: w/CrunchIt/EESEE Access Card, Chapter 13, Problem 47E , additional homework tip  1

To obtain the means for the response variable Dry biomass, Minitab is used. The steps to be followed are:

Step 1: Go to the Minitab worksheet.

Step 2: Go to Stat > Basic Statistics > Descriptive Statistics.

Step 3: Enter the variable ‘Dry biomass’ in the ‘Variable’ option.

Step 4: Enter the variables ‘Species and Water’ in the ‘By Variable’ option.

Step 5: Go to statistics and select ‘Mean.

Step 6: Go to graph and select Individual value plot’.

Step 7: Click Ok.

The obtained result shows the means for each species by water level. Here, few of them are given: The first three means for Species-1 are 40.56, 63.86, and 71.00. The first three means for Species-2 are 34.50, 57.36, and 79.60. The first three means for species-3 are 26.25, 31.87, and 36.24. The first three means for Species-4 are 15.53, 23.29, and 37.05. The obtained graph is given below:

Introduction to the Practice of Statistics: w/CrunchIt/EESEE Access Card, Chapter 13, Problem 47E , additional homework tip  2

(b)

To determine

To find: The standard deviation for each species by water combination and plot these means.

(b)

Expert Solution
Check Mark

Answer to Problem 47E

Solution: For Fresh biomass, the first three standard deviations for Species-1 are 20.9, 29.1, and 18.87. The first three standard deviations for Species-2 are 29.3, 46.9, and 13.94. The first three standard deviations for Species-3 are 13.20, 29.5, and 28.3. The first three standard deviations for Species-4 are 11.63, 6.79, and 13.93.

For Dry biomass, the first three standard deviations for Species-1 are 5.58, 7.51, and 6.03. The first three standard deviations for Species-2 are 11.61, 6.15, and 13.09. The first three standard deviations for Species-3 are 6.43, 11.32, and 11.27. The first three standard deviations for Species-4 are 4.89, 3.33, and 5.19.

Explanation of Solution

Calculation: To obtain the means for Fresh biomass, Minitab is used. The steps to be followed are:

Step 1: Go to the Minitab worksheet.

Step 2: Go to Stat > Basic Statistics > Descriptive Statistics.

Step 3: Enter the variable ‘Fresh biomass’ in the ‘Variable’ option.

Step 4: Enter the variables ‘Species and Water’ in the ‘By Variable’ option.

Step 5: Go to statistics and select ‘Standard deviation’.

Step 6: Click Ok.

The obtained result shows the means for each species by water level. Here, few of them are given: The first three standard deviations for Species-1 are 20.9, 29.1, and 18.87. The first three standard deviations for Species-2 are 29.3, 46.9, and 13.94. The first three standard deviations for species-3 are 13.20, 29.5, and 28.3. The first three standard deviations for Species-4 are 11.63, 6.79, and 13.93.

To obtain the means for Dry biomass, Minitab is used. The steps to be followed are:

Step 1: Go to the Minitab worksheet.

Step 2: Go to Stat > Basic Statistics > Descriptive Statistics.

Step 3: Enter the variable ‘Dry biomass’ in the ‘Variable’ option.

Step 4: Enter the variables ‘Species and Water’ in the ‘By Variable’ option.

Step 5: Go to statistics and select ‘Standard deviation’.

Step 6: Click Ok.

The obtained result shows the means for each species by water level. Here, few of them are given: The first three standard deviations for Species-1 are 5.58, 7.51, and 6.03. The first three standard deviations for Species-2 are 11.61, 6.15, and 13.09. The first three standard deviations for species-3 are 6.43, 11.32, and 11.27. The first three standard deviations for Species-4 are 4.89, 3.33, and 5.19

The highest SD from the Fresh biomass data is 108.01 and the lowest SD is 6.79. Thus,

σmax<2×σmin108.016.79<215.90<2

The highest SD from the Dry biomass data is 35.76 and the lowest SD is 3.12. Thus,

σmax<2×σmin35.763.12<211.46<2

Hence, it is reasonable to pool the standard deviation.

(c)

To determine

To test: A two-way ANOVA for Fresh biomass and Dry biomass

(c)

Expert Solution
Check Mark

Answer to Problem 47E

Solution: A two-way ANOVA for Fresh biomass is provided below:

Source of Variation

Degree of freedom

Sum of squares

Mean sum of squares

F- value

P- value

Species

3

458295

152765

81.45

0.000

Water

6

491948

81991

43.71

0.000

Interaction

18

60334

3352

1.79

0.040

Error

84

157551

1876

Total

111

1168129

A two-way ANOVA for Dry biomass is provided below:

Source of Variation

Degree of freedom

Sum of squares

Mean sum of squares

F- value

P- value

Species

3

50524

16841.3

79.93

0.000

Water

6

56624

9437.3

44.79

0.000

Interaction

18

8419

467.7

2.22

0.008

Error

84

17698

219.7

Total

111

133265

Explanation of Solution

Calculation: To perform a Two-way ANOVA for Fresh biomass, Minitab is used. The steps to be followed are:

Step 1: Go to the Minitab worksheet.

Step 2: Go to Stat > ANOVA > Two-Way ANOVA.

Step 3: Enter the variable ‘Fresh biomass’ in the ‘Response’ option.

Step 4: Enter the variables ‘Species’ in the ‘Row Factor and ‘Water’ in the Column Factor’.

Step 5: Click Ok.

The obtained results are provided below:

Source of Variation

Degree of freedom

Sum of squares

Mean sum of squares

F- value

P- value

Species

3

458295

152765

81.45

0.000

Water

6

491948

81991

43.71

0.000

Interaction

18

60334

3352

1.79

0.040

Error

84

157551

1876

Total

111

1168129

To perform a Two-way ANOVA for Dry biomass, Minitab is used. The steps to be followed are:

Step 1: Go to the Minitab worksheet.

Step 2: Go to Stat > ANOVA > Two-Way ANOVA.

Step 3: Enter the variable ‘Dry biomass’ in the ‘Response’ option.

Step 4: Enter the variables ‘Species’ in the ‘Row Factor and ‘Water’ in the Column Factor’.

Step 5: Click Ok.

The obtained results are provided below:

Source of Variation

Degree of freedom

Sum of squares

Mean sum of squares

F- value

P- value

Species

3

50524

16841.3

79.93

0.000

Water

6

56624

9437.3

44.79

0.000

Interaction

18

8419

467.7

2.22

0.008

Error

84

17698

219.7

Total

111

133265

Conclusion: From ANOVA table above, for both Fresh biomass and Dry biomass, main effects and interaction are significant. The interaction for Fresh biomass and Dry biomass has the P-value 0.04 and 0.008, respectively.

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