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STAT292- Assignment 3 ANOVA Solved
There are five questions, worth a total of 100 marks. Question 1 starts on page 2.
Assignment Guidelines (one more time)
You are encouraged to discuss assignments with other students, but your submitted work must be your own.
The following Assignment Guidelines are helpful for all the assignments in Parts 2 and 3 of the course.
When you carry out a statistical test of hypothesis, you should state the following, when relevant:
• Model equation.
• Assumptions about the data, and comments about whether diagnostic graphs support those assumptions.
• Null and alternative hypotheses.
• ANOVA Table (if relevant), p-value.
• Statistical conclusions. For example, “We reject H0 and conclude HA, that µ1 and µ2 differ at the 5% significance level”.
• Interpretation of the statistical conclusions back to the original problem, using the original meaning of the response variable and any factors or covariates. For example, if comparing heights of two groups, “Female and male adults have different mean heights, with males being taller on average”.
Assignment Guidelines
1. Skink Temperatures
Skinks are tested for their preferred daytime temperature. Each one is placed in a long tank which is warmer at one end, cooler at the other. The temperature at the position where it settles is recorded. There are four different species of skink, and we wish to test (at the 5% level of significance) whether the species differ in their preferred temperature.
The following table gives the data.
Species
Preferred temperatures (oC)
Total
Mean
A
18
21 22
18
20
19
19
23
17
22
199
19.9
B
24
18 19
21
20
17
23
22
22
19
205
20.5
C
22
21 24
19
25
18
23
21
24
22
219
21.9
D
21
19 26
24
25
21
20
20
27
25
228
22.8
SAS output is given on pages 3 and 4.
(a) When running the experiment, other possible factors such as time of day, light,amount of food recently eaten, are kept as near constant as possible. Why?
(b) The skinks are not put in the tank together. Why?
(c) Give values of n and p (the number of treatments) for this experiment. How many degrees of freedom are in the Treatments row, the Error row and the Total row of the ANOVA table? (Give the algebraic expressions and the actual values for this experiment.)
(d) Use the output to write up the ANOVA in the style suggested in the AssignmentGuidelines on page 1. You should include a statement of the (complete) model equation, and also comments on whether the assumptions are satisfied. Use a 5% significance level for the ANOVA test.
One-Way Analysis of Variance
Dependent Variable: Temperature
Source DF Sum of Squares Mean Square F Value Pr F
Model 3 52.0750000 17.3583333 3.06 0.0402
Error 36 203.9000000 5.6638889
Corrected Total 39 255.9750000
One-Way Analysis of Variance
Levene's Test for Homogeneity of Temperature Variance ANOVA of Squared Deviations from Group Means
Source DF Sum of Squares Mean Square F Value Pr F
Species 3 86.1428 28.7143 1.36 0.2691
Error 36 757.4 21.0391
Means and Descriptive Statistics
Mean of Std. Dev. of Minimum of Maximum of
Species
Temperature Temperature Temperature Temperature
21.275 2.5619253577 17 27
A 19.9 2.0248456731 17 23
B 20.5 2.2730302828 17 24
C 21.9 2.2335820757 18 25
D 22.8 2.8982753492 19 27
2. Nasal Sprays
Improvement in breathing airflow is measured for twenty-five people suffering from nasal congestion. They were treated with either a saline spray (A) or one of four nasal sprays (B, C, D, E) available over the counter in pharmacies.
Spray
Airflow improvement
Total
Mean
A
15 10
16
14
8
63
12.6
B
25 41
37
44
26
173
34.6
C
21 6
9
15
14
65
13.0
D
16 7
24
22
15
84
16.8
E
24 15
39
34
30
142
28.4
527
21.08
Relevant SAS output follows, on pages 5 to 7.
Write a report that compares the five treatments using the guidelines on page 1. Ensure that you comment on all the included SAS output. Use a 5% significance level for all statistical tests. Make a recommendation for either a single best nasal spray, or a group of best choices which are similar in their effects; refer to the Tukey test to justify your decision.
One-Way Analysis of Variance
Results: Nasal Spray Example
The ANOVA Procedure
Class Level Information Class Levels Values Spray 5 A B C D E
Number of Observations Read 25 Number of Observations Used 25
Dependent Variable: Improvement
Source DF Sum of Squares Mean Square F Value Pr F
Model 4 1959.440000 489.860000 9.73 0.0002
Error 20 1006.400000 50.320000
Corrected Total 24 2965.840000
The ANOVA Procedure
Nasal Spray Example
Levene's Test for Homogeneity of Improvement Variance
ANOVA of Squared Deviations from Group Means
Source DF Sum of Squares Mean Square F Value Pr F
Type 4 11893.9 2973.5 1.59 0.2156
Error 20 37393.0 1869.6
Level of Improvement
Type N Mean Std Dev
A 5 12.6000000 3.43511281
B 5 34.6000000 8.67755726
C 5 13.0000000 5.78791845
D 5 16.8000000 6.68580586
E 5 28.4000000 9.28977933
Tukey's Studentized Range (HSD) Test for Improvement
Note: This test controls the Type I experimentwise error rate, but it generally has a higher Type II error rate than REGWQ.
Alpha 0.05
Error Degrees of Freedom 20
Error Mean Square 50.32
Critical Value of Studentized Range 4.23186
Minimum Significant Difference 13.425
Means with the same letter are not significantly different.
Tukey Grouping Mean N Type
A 34.600 5 B
A
B A 28.400 5 E
B
B C 16.800 5 D
C
C 13.000 5 C
C
C 12.600 5 A
Nonparametric One-Way ANOVA
The NPAR1WAY Procedure: Nasal Spray Example
Wilcoxon Scores (Rank Sums) for Variable Improvement
Classified by Variable Type
Sum of Expected Std Dev Mean
Type N Scores Under H0 Under H0 Score
A 5 36.50 65.0 14.682756 7.30
B 5 108.00 65.0 14.682756 21.60
C 5 35.00 65.0 14.682756 7.00
D 5 55.50 65.0 14.682756 11.10
E 5 90.00 65.0 14.682756 18.00
Average scores were used for ties.
Kruskal-Wallis Test
Chi-Square 15.8695
DF 4
Pr Chi-Square 0.0032
3. Forensic dental X-rays
The extent to which X-rays can penetrate tooth enamel has been suggested as a suitable mechanism for differentiating between females and males in forensic medicine (e.g., think about shows like ‘CSI’ and parts of ‘NCIS’). The table below gives spectropenetration gradients for one tooth from each of eight females and eight males.
Gender
Y = spectropenetration gradient
Mean
Std. dev.
Female
4.8 5.3 3.7 4.1 5.6 4.0 3.6 5.0
4.5125
0.7605
Male
4.9 5.4 5.0 5.5 5.4 6.6 6.3 4.3
5.4250
0.7440
Note that a high reading reflects a fast drop-off in X-ray penetration, with less penetration by X-rays.
(a) Explain why the teeth have been sampled from eight different people of eachsex, and not eight teeth from one female and eight from one male.
(b) Given that the researcher could afford to test n = 16 subjects, explain the advantages of choosing eight from each group.
(c) SAS output from an ANOVA is on pages 9 and 10. Write a report, followingthe guidelines on page 1.
(d) Explain why there is no point doing a Tukey test with this data.
One-Way Analysis of Variance
Results: X-ray Penetration Gradient
The ANOVA Procedure
Class Level Information Class Levels Values
Gender 2 Female Male
Number of Observations Read 16
Number of Observations Used 16
Dependent Variable: Xray_grad
Source DF Sum of Squares Mean Square F Value Pr F
Model 1 3.33062500 3.33062500 5.88 0.0294
Error 14 7.92375000 0.56598214
Corrected Total 15 11.25437500
R-Square Coeff Var Root MSE Xray_grad Mean
0.295940 15.14099 0.752318 4.968750
Source DF Anova SS Mean Square F Value Pr F
Gender 1 3.33062500 3.33062500 5.88 0.0294
The ANOVA Procedure
X-ray Penetration Gradient
Levene's Test for Homogeneity of Xray_grad Variance
ANOVA of Squared Deviations from Group Means
Source DF Sum of Squares Mean Square F Value Pr F
Gender 1 0.00189 0.00189 0.01 0.9305
Error 14 3.3504 0.2393
Level of Xray_grad
Gender N Mean Std Dev
Female 8 4.51250000 0.76052144
Male 8 5.42500000 0.74402381
4. Personality types
In psychology, there are tests to classify people into one of many personality types. An experiment is run to find the extent of the influence of personality type on the subject’s score in a certain test. A random sample of four personality types is taken, and within each type a random sample of ten subjects is taken. Each subject is given the test, and the score Y is recorded, with data as follows:
Type
Test Score, Y
T1
50
52
44
49 60
51
40
41
54
39
T2
63
45
48
49 65
55
47
58
57
56
T3
50
52
47
48 44
56
55
39
51
53
T4
39
38
51
50 53
53
59
41
45
48
(a) Explain why this is a random effects design, rather than a fixed effects design.
(b) Some SAS output is given on pages 12 and 13. Note that the boxplots do notinclude estimates of group means, since any differences in population means are not the focus of this investigation.
Present a report and your conclusions. Include in your report comments on whether the relevant assumptions seem satisfied. Give your estimated components of variance, plus the percentage of the total variance of Y that is due to personality, along with the percentage unexplained,
Do you think personality type is important in determining the score on this particular test?
SAS Output for Personality Type Example
Box Plot
One-Way Analysis of Variance
Results
The ANOVA Procedure
Class Level Information
Class
Levels
Values
PersType
4
T1 T2 T3 T4
Number of Observations Read
40
Number of Observations Used
40
Dependent Variable: Score
Source
DF
Sum of Squares
Mean Square
F Value
Pr F
Model
3
279.675000
93.225000
2.23
0.1017
Error
36
1506.700000
41.852778
Corrected Total
39
1786.375000
R-Square
Coeff Var
Root MSE
Score Mean
0.156560
12.97117
6.469372
49.87500
Source
DF
Anova SS
Mean Square
F Value
Pr F
PersType
3
279.6750000
93.2250000
2.23
0.1017
SAS Output for Personality Type Example
Levene's Test for Homogeneity of Score Variance ANOVA of Squared Deviations from Group Means
Source
DF
Sum of Squares
Mean Square
F Value
Pr F
PersType
3
2400.7
800.2
0.49
0.6926
Error
36
59004.7
1639.0
5. Phytoremediation
Phytoremediation (New Scientist, 20 Dec 1997, p.26) is a process by which plants are used to remove toxic metals from the soil. For example, sunflowers were used around Chernobyl, where there was radioactive contamination from a nuclear power station accident.
Certain plants take up toxic metals (e.g. zinc, cadmium, uranium) and accumulate them in their vacuoles as protection against chewing insects and infection.
Suppose that four species of plant were tested, at lower and higher soil pH, for their uptake of zinc, Y , measured in parts per million (ppm) of dry plant weight at the end of the trial.
Uptake of zinc, Y, (ppm):
Soil pH
Plant Name
5.5 (acid)
7 (neutral)
Lettuce
250 470 330
400 310 430
Martin red fescue
2850 2380 3130
1070 960 1300
Alpine pennycress
6340 4280 5170
2880 4330 3050
Bladder campion
3690 4750 5100
2360 1990 2140
(a) What kind of design is this? Give the model equation, including an interactionterm.
(b) SAS analysis of the data using the model from part (a) was tried on both rawdata Y and transformed data log Y . Diagnostic graphs from both analyses are given on pages 15 and 16. Explain, with reasons, whether it is better to analyse Y or log Y .
(c) Further SAS output is given on pages 17 to 19. Present a report and yourconclusions, following the usual guidelines. Use a 5% significance level.
SAS Output for Phytoremediation Example
SAS Output for Phytoremediation Example
SAS Output for Phytoremediation Example
Linear Models
The GLM Procedure
Class Level Information
Class
Levels
Values
pH
2
acid neutral
Plant
4
AlpineP BladderC Lettuce MartinRF
Number of Observations Read
24
Number of Observations Used
24
Dependent Variable: logZinc
Source
DF
Sum of Squares
Mean Square
F Value
Pr F
Model
7
24.03027190
3.43289599
92.71
<.0001
Error
16
0.59245521
0.03702845
Corrected Total
23
24.62272711
R-Square
Coeff Var
Root MSE
logZinc Mean
0.975939
2.589699
0.192428
7.430507
Source
DF
Type I SS
Mean Square
F Value
Pr F
pH
1
1.46932364
1.46932364
39.68
<.0001
Plant
3
21.65807393
7.21935798
194.97
<.0001
pH*Plant
3
0.90287433
0.30095811
8.13
0.0016
Source
DF
Type III SS
Mean Square
F Value
Pr F
pH
1
1.46932364
1.46932364
39.68
<.0001
Plant
3
21.65807393
7.21935798
194.97
<.0001
pH*Plant
3
0.90287433
0.30095811
8.13
0.0016
SAS Output for Phytoremediation Example
Alternative interaction graph for phytoremediation example
Assignment Guidelines (one more time)
You are encouraged to discuss assignments with other students, but your submitted work must be your own.
The following Assignment Guidelines are helpful for all the assignments in Parts 2 and 3 of the course.
When you carry out a statistical test of hypothesis, you should state the following, when relevant:
• Model equation.
• Assumptions about the data, and comments about whether diagnostic graphs support those assumptions.
• Null and alternative hypotheses.
• ANOVA Table (if relevant), p-value.
• Statistical conclusions. For example, “We reject H0 and conclude HA, that µ1 and µ2 differ at the 5% significance level”.
• Interpretation of the statistical conclusions back to the original problem, using the original meaning of the response variable and any factors or covariates. For example, if comparing heights of two groups, “Female and male adults have different mean heights, with males being taller on average”.
Assignment Guidelines
1. Skink Temperatures
Skinks are tested for their preferred daytime temperature. Each one is placed in a long tank which is warmer at one end, cooler at the other. The temperature at the position where it settles is recorded. There are four different species of skink, and we wish to test (at the 5% level of significance) whether the species differ in their preferred temperature.
The following table gives the data.
Species
Preferred temperatures (oC)
Total
Mean
A
18
21 22
18
20
19
19
23
17
22
199
19.9
B
24
18 19
21
20
17
23
22
22
19
205
20.5
C
22
21 24
19
25
18
23
21
24
22
219
21.9
D
21
19 26
24
25
21
20
20
27
25
228
22.8
SAS output is given on pages 3 and 4.
(a) When running the experiment, other possible factors such as time of day, light,amount of food recently eaten, are kept as near constant as possible. Why?
(b) The skinks are not put in the tank together. Why?
(c) Give values of n and p (the number of treatments) for this experiment. How many degrees of freedom are in the Treatments row, the Error row and the Total row of the ANOVA table? (Give the algebraic expressions and the actual values for this experiment.)
(d) Use the output to write up the ANOVA in the style suggested in the AssignmentGuidelines on page 1. You should include a statement of the (complete) model equation, and also comments on whether the assumptions are satisfied. Use a 5% significance level for the ANOVA test.
One-Way Analysis of Variance
Dependent Variable: Temperature
Source DF Sum of Squares Mean Square F Value Pr F
Model 3 52.0750000 17.3583333 3.06 0.0402
Error 36 203.9000000 5.6638889
Corrected Total 39 255.9750000
One-Way Analysis of Variance
Levene's Test for Homogeneity of Temperature Variance ANOVA of Squared Deviations from Group Means
Source DF Sum of Squares Mean Square F Value Pr F
Species 3 86.1428 28.7143 1.36 0.2691
Error 36 757.4 21.0391
Means and Descriptive Statistics
Mean of Std. Dev. of Minimum of Maximum of
Species
Temperature Temperature Temperature Temperature
21.275 2.5619253577 17 27
A 19.9 2.0248456731 17 23
B 20.5 2.2730302828 17 24
C 21.9 2.2335820757 18 25
D 22.8 2.8982753492 19 27
2. Nasal Sprays
Improvement in breathing airflow is measured for twenty-five people suffering from nasal congestion. They were treated with either a saline spray (A) or one of four nasal sprays (B, C, D, E) available over the counter in pharmacies.
Spray
Airflow improvement
Total
Mean
A
15 10
16
14
8
63
12.6
B
25 41
37
44
26
173
34.6
C
21 6
9
15
14
65
13.0
D
16 7
24
22
15
84
16.8
E
24 15
39
34
30
142
28.4
527
21.08
Relevant SAS output follows, on pages 5 to 7.
Write a report that compares the five treatments using the guidelines on page 1. Ensure that you comment on all the included SAS output. Use a 5% significance level for all statistical tests. Make a recommendation for either a single best nasal spray, or a group of best choices which are similar in their effects; refer to the Tukey test to justify your decision.
One-Way Analysis of Variance
Results: Nasal Spray Example
The ANOVA Procedure
Class Level Information Class Levels Values Spray 5 A B C D E
Number of Observations Read 25 Number of Observations Used 25
Dependent Variable: Improvement
Source DF Sum of Squares Mean Square F Value Pr F
Model 4 1959.440000 489.860000 9.73 0.0002
Error 20 1006.400000 50.320000
Corrected Total 24 2965.840000
The ANOVA Procedure
Nasal Spray Example
Levene's Test for Homogeneity of Improvement Variance
ANOVA of Squared Deviations from Group Means
Source DF Sum of Squares Mean Square F Value Pr F
Type 4 11893.9 2973.5 1.59 0.2156
Error 20 37393.0 1869.6
Level of Improvement
Type N Mean Std Dev
A 5 12.6000000 3.43511281
B 5 34.6000000 8.67755726
C 5 13.0000000 5.78791845
D 5 16.8000000 6.68580586
E 5 28.4000000 9.28977933
Tukey's Studentized Range (HSD) Test for Improvement
Note: This test controls the Type I experimentwise error rate, but it generally has a higher Type II error rate than REGWQ.
Alpha 0.05
Error Degrees of Freedom 20
Error Mean Square 50.32
Critical Value of Studentized Range 4.23186
Minimum Significant Difference 13.425
Means with the same letter are not significantly different.
Tukey Grouping Mean N Type
A 34.600 5 B
A
B A 28.400 5 E
B
B C 16.800 5 D
C
C 13.000 5 C
C
C 12.600 5 A
Nonparametric One-Way ANOVA
The NPAR1WAY Procedure: Nasal Spray Example
Wilcoxon Scores (Rank Sums) for Variable Improvement
Classified by Variable Type
Sum of Expected Std Dev Mean
Type N Scores Under H0 Under H0 Score
A 5 36.50 65.0 14.682756 7.30
B 5 108.00 65.0 14.682756 21.60
C 5 35.00 65.0 14.682756 7.00
D 5 55.50 65.0 14.682756 11.10
E 5 90.00 65.0 14.682756 18.00
Average scores were used for ties.
Kruskal-Wallis Test
Chi-Square 15.8695
DF 4
Pr Chi-Square 0.0032
3. Forensic dental X-rays
The extent to which X-rays can penetrate tooth enamel has been suggested as a suitable mechanism for differentiating between females and males in forensic medicine (e.g., think about shows like ‘CSI’ and parts of ‘NCIS’). The table below gives spectropenetration gradients for one tooth from each of eight females and eight males.
Gender
Y = spectropenetration gradient
Mean
Std. dev.
Female
4.8 5.3 3.7 4.1 5.6 4.0 3.6 5.0
4.5125
0.7605
Male
4.9 5.4 5.0 5.5 5.4 6.6 6.3 4.3
5.4250
0.7440
Note that a high reading reflects a fast drop-off in X-ray penetration, with less penetration by X-rays.
(a) Explain why the teeth have been sampled from eight different people of eachsex, and not eight teeth from one female and eight from one male.
(b) Given that the researcher could afford to test n = 16 subjects, explain the advantages of choosing eight from each group.
(c) SAS output from an ANOVA is on pages 9 and 10. Write a report, followingthe guidelines on page 1.
(d) Explain why there is no point doing a Tukey test with this data.
One-Way Analysis of Variance
Results: X-ray Penetration Gradient
The ANOVA Procedure
Class Level Information Class Levels Values
Gender 2 Female Male
Number of Observations Read 16
Number of Observations Used 16
Dependent Variable: Xray_grad
Source DF Sum of Squares Mean Square F Value Pr F
Model 1 3.33062500 3.33062500 5.88 0.0294
Error 14 7.92375000 0.56598214
Corrected Total 15 11.25437500
R-Square Coeff Var Root MSE Xray_grad Mean
0.295940 15.14099 0.752318 4.968750
Source DF Anova SS Mean Square F Value Pr F
Gender 1 3.33062500 3.33062500 5.88 0.0294
The ANOVA Procedure
X-ray Penetration Gradient
Levene's Test for Homogeneity of Xray_grad Variance
ANOVA of Squared Deviations from Group Means
Source DF Sum of Squares Mean Square F Value Pr F
Gender 1 0.00189 0.00189 0.01 0.9305
Error 14 3.3504 0.2393
Level of Xray_grad
Gender N Mean Std Dev
Female 8 4.51250000 0.76052144
Male 8 5.42500000 0.74402381
4. Personality types
In psychology, there are tests to classify people into one of many personality types. An experiment is run to find the extent of the influence of personality type on the subject’s score in a certain test. A random sample of four personality types is taken, and within each type a random sample of ten subjects is taken. Each subject is given the test, and the score Y is recorded, with data as follows:
Type
Test Score, Y
T1
50
52
44
49 60
51
40
41
54
39
T2
63
45
48
49 65
55
47
58
57
56
T3
50
52
47
48 44
56
55
39
51
53
T4
39
38
51
50 53
53
59
41
45
48
(a) Explain why this is a random effects design, rather than a fixed effects design.
(b) Some SAS output is given on pages 12 and 13. Note that the boxplots do notinclude estimates of group means, since any differences in population means are not the focus of this investigation.
Present a report and your conclusions. Include in your report comments on whether the relevant assumptions seem satisfied. Give your estimated components of variance, plus the percentage of the total variance of Y that is due to personality, along with the percentage unexplained,
Do you think personality type is important in determining the score on this particular test?
SAS Output for Personality Type Example
Box Plot
One-Way Analysis of Variance
Results
The ANOVA Procedure
Class Level Information
Class
Levels
Values
PersType
4
T1 T2 T3 T4
Number of Observations Read
40
Number of Observations Used
40
Dependent Variable: Score
Source
DF
Sum of Squares
Mean Square
F Value
Pr F
Model
3
279.675000
93.225000
2.23
0.1017
Error
36
1506.700000
41.852778
Corrected Total
39
1786.375000
R-Square
Coeff Var
Root MSE
Score Mean
0.156560
12.97117
6.469372
49.87500
Source
DF
Anova SS
Mean Square
F Value
Pr F
PersType
3
279.6750000
93.2250000
2.23
0.1017
SAS Output for Personality Type Example
Levene's Test for Homogeneity of Score Variance ANOVA of Squared Deviations from Group Means
Source
DF
Sum of Squares
Mean Square
F Value
Pr F
PersType
3
2400.7
800.2
0.49
0.6926
Error
36
59004.7
1639.0
5. Phytoremediation
Phytoremediation (New Scientist, 20 Dec 1997, p.26) is a process by which plants are used to remove toxic metals from the soil. For example, sunflowers were used around Chernobyl, where there was radioactive contamination from a nuclear power station accident.
Certain plants take up toxic metals (e.g. zinc, cadmium, uranium) and accumulate them in their vacuoles as protection against chewing insects and infection.
Suppose that four species of plant were tested, at lower and higher soil pH, for their uptake of zinc, Y , measured in parts per million (ppm) of dry plant weight at the end of the trial.
Uptake of zinc, Y, (ppm):
Soil pH
Plant Name
5.5 (acid)
7 (neutral)
Lettuce
250 470 330
400 310 430
Martin red fescue
2850 2380 3130
1070 960 1300
Alpine pennycress
6340 4280 5170
2880 4330 3050
Bladder campion
3690 4750 5100
2360 1990 2140
(a) What kind of design is this? Give the model equation, including an interactionterm.
(b) SAS analysis of the data using the model from part (a) was tried on both rawdata Y and transformed data log Y . Diagnostic graphs from both analyses are given on pages 15 and 16. Explain, with reasons, whether it is better to analyse Y or log Y .
(c) Further SAS output is given on pages 17 to 19. Present a report and yourconclusions, following the usual guidelines. Use a 5% significance level.
SAS Output for Phytoremediation Example
SAS Output for Phytoremediation Example
SAS Output for Phytoremediation Example
Linear Models
The GLM Procedure
Class Level Information
Class
Levels
Values
pH
2
acid neutral
Plant
4
AlpineP BladderC Lettuce MartinRF
Number of Observations Read
24
Number of Observations Used
24
Dependent Variable: logZinc
Source
DF
Sum of Squares
Mean Square
F Value
Pr F
Model
7
24.03027190
3.43289599
92.71
<.0001
Error
16
0.59245521
0.03702845
Corrected Total
23
24.62272711
R-Square
Coeff Var
Root MSE
logZinc Mean
0.975939
2.589699
0.192428
7.430507
Source
DF
Type I SS
Mean Square
F Value
Pr F
pH
1
1.46932364
1.46932364
39.68
<.0001
Plant
3
21.65807393
7.21935798
194.97
<.0001
pH*Plant
3
0.90287433
0.30095811
8.13
0.0016
Source
DF
Type III SS
Mean Square
F Value
Pr F
pH
1
1.46932364
1.46932364
39.68
<.0001
Plant
3
21.65807393
7.21935798
194.97
<.0001
pH*Plant
3
0.90287433
0.30095811
8.13
0.0016
SAS Output for Phytoremediation Example
Alternative interaction graph for phytoremediation example
1 file (3.0MB)
