NAME PLS 205 Final [Total Points in Exam = 100] …...PLS 205 Final [Total Points in Exam = 100]...

NAME _______________________

PLS 205 Final [Total Points in Exam = 100]

March 12, 2014

Due Date: Tuesday, March 17, by 5:00 pm; 122 Robbins Hall

Include your SAS programs, include only the critical parts of the SAS output, and discuss each result.

NO POINTS WILL BE AWARDED TO OUTPUTS WITHOUT A SENTENCE EXPLAINING

THE CONCLUSION.

Clarification questions should be directed to Miguel and Dr. Dubcovsky by e-mail only. No consultation

with other students is allowed during the exam period (including SAS programming questions). Exams

with more than one unlikely identical mistake will receive zeroes, and the incident will be referred

to the Office of Student Judicial Affairs.

Problem 1 (Split-Split Plot) [25 points]

A grower would like to test if fruit size in blueberry is affected by fungicide application, insecticide

application, and insect pollination. The treatments the grower experimented with are:

Fungicide – regular fungicide application or no fungicide application

Insecticide – regular insecticide application or no insecticide application

Pollination – 25% of the flowers pollinated or 100% of the flowers pollinated

To conduct the experiment the grower picked 4 field sections with 1.8 m wide x 12 m long dimensions.

Each field section was divided into two 1.8 m x 6 m subsections and the fungicide treatments were

randomly assigned to each subsection. Each subsection was divided into 1.8 m x 3 m sub-sub section and

the insecticide treatment was randomly assigned to these sections. Finally the sub-sub sections were

divided into sections of 1.8 m x 1.5 m where the pollination treatments were randomly assigned. To apply

the 25% pollination treatment, the grower covered the plants with a bee proof net when the flowering for

the plants reached 25%. To insure that the flowers were being pollinated the grower installed bumble bee

colonies within each cage. When the plants were ready to harvest the grower weighed all of the ripe fruit

in bulk from each 1.8 m x 1.5 m section and divided the number by the total number of ripe fruit weighed

to get an average fruit weight. While harvesting one of the workers accidentally fell on a box of berries

corresponding to the Full fungicide, None insecticide, 25% pollination, in block 2. This data was removed

and is indicated by the word MISSING. The data is summarized below:

data berries;

input Block Fungicide $ Insecticide $ Pollination Response;

Cards;

1 Full None 25 0.687

1 Full None 100 0.361

1 Full Full 25 0.87

1 Full Full 100 0.558

1 None None 25 0.442

1 None None 100 0.382

1 None Full 25 0.39

1 None Full 100 0.462

2 Full None 25 .

2 Full None 100 0.401

2 Full Full 25 0.592

2 Full Full 100 0.566


2 None None 100 0.381

2 None Full 25 0.395

2 None Full 100 0.626


3 Full None 100 0.377


3 Full Full 100 0.655


3 None None 100 0.382


3 None Full 100 0.472




4 Full Full 100 0.557


4 None None 100 0.382


4 None Full 100 0.475

;

*Proc Print;

Proc GLM data = berries;

Class Block Fungicide Insecticide Pollination;

Model Response = Block Fungicide

Block*Fungicide

Insecticide Fungicide*Insecticide

Block*Fungicide*Insecticide

Pollination Pollination*Insecticide Pollination*Fungicide

Pollination*Fungicide*Insecticide;

Random Block Block*Fungicide Block*Fungicide*Insecticide / test;

Output out = PR p = pred r = res;

Proc Sort data = berries;

by Fungicide Insecticide;


Class Block Insecticide Pollination;

Model Response = Block Pollination;

Random Block / test;

by Fungicide Insecticide;

Proc Sort data = berries;

by Pollination;


Class Block Fungicide Insecticide;

Model Response = Block Fungicide

Block*Fungicide

Insecticide Fungicide*Insecticide;

Random Block Block*Fungicide / test;

LSMeans Fungicide*Insecticide;

by Pollination;

run;

quit;

[Don’t worry about testing the assumptions]

1. [5 points] Use the table in the appendix to describe the experimental design.

Design: RCBD with a Spit-Split plot

Response Variable: Average blueberry weight

Experimental Unit: Main-plot: 1.8m x 6m section Subplot: 1.8m x 3m Sub-subplot: 1.8m x 1.5m

Class

Variable

Block or

Treatment

Number of

Levels

Fixed or

Random Description

1 Block 4 Random 1.8m x 12m section

2 Treatment 2 Fixed Fungicide

3 Treatment 2 Fixed Insecticide

4 Treatment 2 Fixed Pollination

Subsamples? NO

Covariable? NO

2. [5 points] Perform the appropriate ANOVA and report the significance of the different treatments and

interactions. THERE IS NO NEED TO TEST ASSUMPTIONS. Dependent Variable: Response

Source DF Type III SS Mean Square F Value Pr > F

Block 3 0.005373 0.001791 0.52 0.6966

Error 3 0.010294 0.003431

Error: MS(Block*Fungicide)


* Fungicide 1 0.160389 0.160389 46.56 0.0059

Error 3.0907 0.010646 0.003445

Error: 0.989*MS(Block*Fungicide) + 0.011*MS(Error)

* This test assumes one or more other fixed effects are zero.


Block*Fungicide 3 0.010294 0.003431 1.22 0.3790

Error 6.2382 0.017581 0.002818

Error: 0.9882*MS(Block*Fungic*Insecti) + 0.0118*MS(Error)


* Insecticide 1 0.022746 0.022746 8.01 0.0276

* Fungicide*Insecticid 1 0.005235 0.005235 1.84 0.2199

Error 6.4649 0.018350 0.002838

Error: 0.9774*MS(Block*Fungic*Insecti) + 0.0226*MS(Error)



Block*Fungic*Insecti 6 0.016779 0.002797 0.60 0.7240

* Pollination 1 0.067526 0.067526 14.54 0.0029

* Insectici*Pollinatio 1 0.093669 0.093669 20.17 0.0009

* Fungicide*Pollinatio 1 0.094712 0.094712 20.40 0.0009

Fungic*Insect*Pollin 1 0.001553 0.001553 0.33 0.5748

Error: MS(Error) 11 0.051081 0.004644


The ANOVA results suggest that the interaction between fungicide and insecticide are not

significant (P = 0.2199), the interaction between insecticide and pollination are significant (P =

0.0009), and the interaction between fungicide and pollination are significant (P = 0.0009). These

results suggest that we analyze the simple effects for fungicide and insecticide treatments at the

different pollination levels. The results also suggest that we should draw our conclusions about

pollination at each fungicide by insecticide treatment combinations. The three way interaction is

not significant (P = 0.5748).

3. [5 points] Based on your results for question 2; please perform the appropriate tests to analyze the

simple effects?

Analysis of pollination within each level of fungicide and insecticide combination: Fungicide=Full Insecticide=Full


Block 3 0.023022 0.007674 0.58 0.6691

Pollination 1 0.013366 0.013366 1.00 0.3903

Error: MS(Error) 3 0.039964 0.013321

When the full fungicide and insecticide treatments are applied there are no significant differences

between pollinations (P = 0.3903).

Fungicide=Full Insecticide=None


Block 3 0.001363 0.000454 9.08 0.1008

Pollination 1 0.169344 0.169344 3386.88 0.0003

Error: MS(Error) 2 0.000100 0.000050000

When the full fungicide treatment and no insecticide treatments are applied there are highly

significant differences between pollinations (P = 0.0003).

Fungicide=None Insecticide=Full


Block 3 0.009556 0.003185 1.08 0.4767

Pollination 1 0.026796 0.026796 9.05 0.0573

Error: MS(Error) 3 0.008880 0.002960

When no fungicide is applied but the full insecticide treatment is applied there are no significant

differences between pollination but are very close to being significant (P = 0.0573).

Fungicide=None Insecticide=None


Block 3 0.002025 0.000675 0.95 0.5171

Pollination 1 0.012961 0.012961 18.20 0.0236

Error: MS(Error) 3 0.002137 0.000712

When no fungicide or insecticide is applied there are significant differences between pollinations (P

= 0.0236).

Analysis of fungicide and insecticide at different levels of pollination:

Pollination=25


Block 3 0.011583 0.003861 0.58 0.6689

Error 3 0.020090 0.006697



* Fungicide 1 0.204054 0.204054 30.56 0.0098

Error 3.2054 0.021403 0.006677

Error: 0.9643*MS(Block*Fungicide) + 0.0357*MS(Error)



Block*Fungicide 3 0.020090 0.006697 1.09 0.4342

* Insecticide 1 0.005612 0.005612 0.91 0.3834

Fungicide*Insecticid 1 0.002842 0.002842 0.46 0.5269

Error: MS(Error) 5 0.030757 0.006151


When 25% of the flowers are pollinated there are no significant differences between insecticide

applications and there are no significant interactions (P = 0.834 and P = 0.5269, respectively).

There are significant differences between fungicide (P = 0.0098).

Pollination=100


Block 3 0.007504 0.002501 1.26 0.4276

* Fungicide 1 0.004658 0.004658 2.34 0.2235

Error 3 0.005969 0.001990




Block*Fungicide 3 0.005969 0.001990 0.93 0.4814

* Insecticide 1 0.113064 0.113064 52.91 0.0003

Fungicide*Insecticid 1 0.006765 0.006765 3.17 0.1255

Error: MS(Error) 6 0.012821 0.002137


When 100% of the flowers are pollinated there are no significant effects between fungicides and

there are no significant interactions (P = 0.2235 and P = 0.1255, respectively). There are significant

differences in average berry weight between insecticide treatments (P = 0.0003).

4. [5 points] Report the appropriate means for Fungicide and Insecticide at the different levels of

pollination.

Remember that there is missing data; therefore; it is necessary to use the LSMeans statement. Pollination=25

Fungicide Insecticide Response LSMEAN

Full Full 0.66575000


Full None 0.67741667

None Full 0.39300000

None None 0.46225000

Pollination=100


Full Full 0.58400000

Full None 0.37475000

None Full 0.50875000

None None 0.38175000

5. [5 points] Report the recommendations you will give on the application of fungicide and insecticide if

the grower desires to have larger fruit.

Based on significant interactions with pollination, separate recommendations should be provided at the

different levels of pollination.

At 25% pollination the grower gets significantly larger fruit by applying fungicide but there is no

significant effect of insecticide.

At the 100% pollination there was not enough evidence to suggest that there are significant differences

between fungicide treatments but the application of insecticide has a positive effect on fruit size.

In conclusion, if the grower desires larger fruit size, at 25% pollination they should apply fungicide (the

insecticide application is not beneficial) and at 100% pollination they should apply insecticide (the

fungicide application is not very beneficial).

Problem 2 (Mixed Model) [?? points]

An agronomist wants to test the effect of four treatments (two tillage practices and to cover crop

methods) on processing tomato yields and to extend his conclusions to the Central Valley. The two tillage

practices include Standard tillage (ST) vs. Conservation tillage (CT) and the two cover crop treatments

include No Cover crop (NO) vs. Cover crop (CC). The researcher is also wants to test if there are

significant interactions between the cover crop and tillage.

To conduct the experiment the researcher randomly picks six random locations in the Central Valley in

California. In each location he selects one field, divides it in four homogeneous sections with 4 large plots

in each section. In each location he randomly assigns each of the four possible cover by tillage treatment

combinations to one of the 4 large plots available within each section. The researcher measures tomato

yield in each plot and records the yield in tons per acre. The data is reported below:

data mixedfinal;

input Location $ Block Tillage $ Cover $ Response;

cards;

loc1 1 ST NO 54

loc2 1 ST NO 65

loc3 1 ST NO 42

loc4 1 ST NO 49

loc5 1 ST NO 55

loc6 1 ST NO 33

loc1 1 ST CC 58

loc2 1 ST CC 68

loc3 1 ST CC 54

loc4 1 ST CC 47

loc5 1 ST CC 48

loc6 1 ST CC 46

loc1 1 CT NO 52

loc2 1 CT NO 62

loc3 1 CT NO 51

loc4 1 CT NO 51

loc5 1 CT NO 59

loc6 1 CT NO 34

loc1 1 CT CC 69

loc2 1 CT CC 64

loc3 1 CT CC 55

loc4 1 CT CC 61

loc5 1 CT CC 52

loc6 1 CT CC 59

loc1 2 ST NO 54

loc2 2 ST NO 59

loc3 2 ST NO 41

loc4 2 ST NO 48

loc5 2 ST NO 50

loc6 2 ST NO 32

loc1 2 ST CC 53

loc2 2 ST CC 71

loc3 2 ST CC 64

loc4 2 ST CC 47

loc5 2 ST CC 56

loc6 2 ST CC 41

loc1 2 CT NO 50

loc2 2 CT NO 55

loc3 2 CT NO 52

loc4 2 CT NO 63

loc5 2 CT NO 57

loc6 2 CT NO 42

loc1 2 CT CC 58

loc2 2 CT CC 69

loc3 2 CT CC 66

loc4 2 CT CC 55

loc5 2 CT CC 60

loc6 2 CT CC 48

loc1 3 ST NO 55

loc2 3 ST NO 54

loc3 3 ST NO 36

loc4 3 ST NO 52

loc5 3 ST NO 51

loc6 3 ST NO 38

loc1 3 ST CC 52

loc2 3 ST CC 70

loc3 3 ST CC 51

loc4 3 ST CC 51

loc5 3 ST CC 58

loc6 3 ST CC 40

loc1 3 CT NO 63

loc2 3 CT NO 55

loc3 3 CT NO 51

loc4 3 CT NO 53

loc5 3 CT NO 62

loc6 3 CT NO 35

loc1 3 CT CC 56

loc2 3 CT CC 73

loc3 3 CT CC 59

loc4 3 CT CC 58

loc5 3 CT CC 63

loc6 3 CT CC 58

loc1 4 ST NO 51

loc2 4 ST NO 50

loc3 4 ST NO 51

loc4 4 ST NO 49

loc5 4 ST NO 57

loc6 4 ST NO 28

loc1 4 ST CC 52

loc2 4 ST CC 61

loc3 4 ST CC 56

loc4 4 ST CC 47

loc5 4 ST CC 48

loc6 4 ST CC 41

loc1 4 CT NO 52

loc2 4 CT NO 58

loc3 4 CT NO 46

loc4 4 CT NO 53

loc5 4 CT NO 53

loc6 4 CT NO 41

loc1 4 CT CC 59

loc2 4 CT CC 65

loc3 4 CT CC 57

loc4 4 CT CC 56

loc5 4 CT CC 52

loc6 4 CT CC 57

;

*Proc Print data = mixedfinal;

Proc GLM data = mixedfinal;

Class Location Block Tillage Cover;

Model Response = Location Block(Location)

Tillage Cover Tillage*Cover

Location*Tillage Location*Cover

Location*Tillage*Cover;

Random Location Block(Location) Location*Tillage Location*Cover

Location*Tillage*Cover / test;

Means Cover Tillage;

Proc Varcomp data = mixedfinal Method = Type1;

class Location Block Tillage Cover;

Model Response = Tillage Cover Tillage*Cover

Location Block(Location)

Location*Tillage Location*Cover

Location*Tillage*Cover / fixed = 3;

Proc Gplot data = mixedfinal;

** Two-way Plots **;

symbol1 i=std1mtj v=none color=BLUE mode = INCLUDE;

plot Response * Cover / vaxis = 47 to 58 by 1;




plot Response * Tillage / vaxis = 47 to 58 by 1;




symbol2 i=std1mtj v=none color=BLACK mode = INCLUDE;

symbol3 i=std1mtj v=none color=GREEN mode = INCLUDE;

symbol4 i=std1mtj v=none color=ORANGE mode = INCLUDE;

symbol5 i=std1mtj v=none color=RED mode = INCLUDE;

symbol6 i=std1mtj v=none color=PURPLE mode = INCLUDE;

plot Response * Location = Cover;

run;

quit;

1. [6 points] Perform appropriate analysis of variance on the data. Report the results of the analysis and

answer the following questions:

a) Are there significant differences in yield between tillage methods or cover crop methods?

b) Is the interaction between tillage and cover crop significant?

c) Are the effects of tillage, cover crop, and their interaction consistent across locations? If yes,

produce a plot showing the interaction.

Dependent Variable: Response


Location 5 3520.677083 704.135417 4.19 0.0594

Error 5.6932 956.799535 168.059491

Error: MS(Block(Location)) + MS(Location*Tillage) + MS(Location*Cover) -

MS(Locati*Tillage*Cover) - MS(Error)



Block(Location) 18 359.562500 19.975694 1.04 0.4312

Locati*Tillage*Cover 5 111.177083 22.235417 1.16 0.3403

Error: MS(Error) 54 1034.187500 19.151620


* Tillage 1 625.260417 625.260417 16.04 0.0103

Error 5 194.927083 38.985417

Error: MS(Location*Tillage)



* Cover 1 969.010417 969.010417 6.44 0.0520

Error 5 752.427083 150.485417

Error: MS(Location*Cover)



Tillage*Cover 1 29.260417 29.260417 1.32 0.3032

Location*Tillage 5 194.927083 38.985417 1.75 0.2764

Location*Cover 5 752.427083 150.485417 6.77 0.0280

Error 5 111.177083 22.235417

Error: MS(Locati*Tillage*Cover)

a) The ANOVA results suggest that there are significant differences between the tillage treatments (P =

0.0103) and there is no significant differences between the cover crop treatments (P = 0.0520) although it

is very close to being significant.

b) The interaction between tillage and cover crop is not significant

c) The effects of tillage is consistent across locations but the effect of Cover varies among locations.

However, since location is a random factor we still want to describe the effect of cover across all locations

If yes show an interaction plot for the significant interaction.

Above is a plot of the interaction between the cover and location. It shows that the cover crop

treatment is better in most locations except for locations 4 and 5 where it doesn’t seem to make any

difference whether a cover crop is used or not.

2. [5 points] Create two plots using SAS, R, or Excel where the first plot shows the main effects of tillage

and the second plot shows the main effects of the crop treatments. Based on this result and the previous

ANOVA provide your conclusion and recommendations for the complete Central Valley.

Main Effects of Cover Crop:

Main Effects of Tillage:

Based on the mean and ANOVA results we can recommend the application of Conservative tillage across

the Central Valley. Based on the current results the use of cover crops cannot be recommended uniformly

across the Central valley. However, given the marginal NS result for cover crop and the significant

interaction between location and cover crop, it would be worth doing a follow up study to identify

locations within the Central Valley where the use of cover crop is significantly better.

3. [5 points] Please report the means of the main effects of tillage and cover crop. Is the difference

between standard tillage and conventional tillage larger than the difference between no cover crop

and cover crop? Does the largest difference correspond to the most significant result? If not explain

why.

The means of cover crop across all locations, blocks, and tillage methods:

Level of N Response

Cover Mean Std Dev

CC 48 56.4375 8.02365187

NO 48 50.0833 8.75392516

difference 6.3541667

The difference in tomato yield between having a cover crop and not having a cover crop is 6.35.

The means of tillage across all locations, blocks, and tillage methods:

Level of N Response

Tillage Mean Std Dev

CT 48 55.8125 7.87848201

ST 48 50.7083 9.28727872

difference 5.1041667

The difference in tomato yield between having standard tillage and having conservation tillage is 5.10.

The difference between the two cover crop treatments is larger than the difference between the two

tillage treatments. This larger difference is in the less significant factor. This is due to the fact that

we are using loc*cover as an error term, and since the cover crop response varies across locations,

this error is large and the differences are not significant. By contrast the interaction between tillage

and location is not significant, resulting in a smaller error for tillage.

The expected means squares (EMS) for tillage is:

Source Type III Expected Mean Square

Tillage Var(Error) + 4 Var(Locati*Tillage*Cover) + 8 Var(Location*Tillage) +

Q(Tillage,Tillage*Cover)

The correct error term for tillage is the interaction between location and tillage:

Location*Tillage Var(Error) + 4 Var(Locati*Tillage*Cover) + 8 Var(Location*Tillage)

The EMS for cover crop is:


Cover Var(Error) + 4 Var(Locati*Tillage*Cover) + 8 Var(Location*Cover) + Q(Cover,Tillage*Cover)

The correct error term for cover crop is the interaction between location and cover crop:

Location*Cover Var(Error) + 4 Var(Locati*Tillage*Cover) + 8 Var(Location*Cover)

Notice that the mean squares for Location*Tillage is smaller than the means squares for

Location*Cover.


Tillage*Cover 1 29.260417 29.260417 1.32 0.3032

Location*Tillage 5 194.927083 38.985417 1.75 0.2764

Location*Cover 5 752.427083 150.485417 6.77 0.0280

Error 5 111.177083 22.235417

Error: MS(Locati*Tillage*Cover)

The variance due to the Location*Cover interaction is approximately 4 times larger than the

variance due to Location*Tillage.

In conclusion, the variance due to Location*Cover is much larger than the variance due to the

Location*Tillage interaction, which explains the lack of significant differences for cover crops

across Central Valley as a whole.

4. [4 points] Using the same model as in question 1, estimate the variance components for all of the

random effects in your model. Which of these random effects had the largest variance?

Type 1 Estimates

Variance Component Estimate

Var(Location) 33.50475

Var(Block(Location)) 0.20602

Var(Location*Tillage) 2.09375

Var(Location*Cover) 16.03125

Var(Locati*Tillage*Cover) 0.77095

Var(Error) 19.15162

The random effect that had the largest variance was location and it is estimated to be 33.50475.

Problem 3 (Factorial) [25 points]

A table grape breeder has developed two potential varieties and would like to release the variety that

produces the largest berries upon application of gibberellic acid (GA), as well as confirm that these two

new varieties are better than the traditional variety Thompson Seedless. The breeder selects three

homogeneous sections in a vineyard and plants 18 10-plant plots in each section. Within each section he

randomly assigns two plots to one of the 9 combinations ofthe three varieties and three GA treatments (0

ppm, 10 ppm, and 20 ppm). At harvest, he randomly picks 50 berries per plant per plot and weighs them

in bulk. The average individual berry weights (g) calculated for each plot are reported below:

Data Grape;

Input Block $ GA $ Var $ Id $ Wt;

Cards;

1 0 TS 1 4.15

1 0 TS 1 4.19

1 0 Var1 2 4.84

1 0 Var1 2 4.77

1 0 Var2 3 4.92

1 0 Var2 3 4.58

1 10 TS 4 4.90

1 10 TS 4 4.81

1 10 Var1 5 5.28

1 10 Var1 5 5.37

1 10 Var2 6 5.64

1 10 Var2 6 5.29

1 20 TS 7 5.45

1 20 TS 7 5.17

1 20 Var1 8 6.39

1 20 Var1 8 6.10

1 20 Var2 9 5.66

1 20 Var2 9 6.01

2 0 TS 1 4.27

2 0 TS 1 4.83

2 0 Var1 2 5.21

2 0 Var1 2 5.24

2 0 Var2 3 5.34

2 0 Var2 3 5.11

2 10 TS 4 4.83

2 10 TS 4 4.95

2 10 Var1 5 6.02

2 10 Var1 5 5.50

2 10 Var2 6 5.79

2 10 Var2 6 5.29

2 20 TS 7 5.90

2 20 TS 7 5.76

2 20 Var1 8 6.06

2 20 Var1 8 6.56

2 20 Var2 9 5.95

2 20 Var2 9 6.29

3 0 TS 1 5.01

3 0 TS 1 4.80

3 0 Var1 2 5.61

3 0 Var1 2 5.06

3 0 Var2 3 5.58

3 0 Var2 3 5.44

3 10 TS 4 5.25

3 10 TS 4 5.35

3 10 Var1 5 6.20

3 10 Var1 5 6.13

3 10 Var2 6 5.57

3 10 Var2 6 5.91

3 20 TS 7 5.67

3 20 TS 7 5.87

3 20 Var1 8 6.78

3 20 Var1 8 6.70

3 20 Var2 9 6.33

3 20 Var2 9 6.24

;

;

Proc GLM data=Grape order=data;

title "Exploratory model";

Class Block GA Var;

Model Wt = Block|GA|Var;


title 'ANOVA';

Class Block GA Var;

Model Wt=Block GA|Var;

Output out=PR p=Pred r=Res;

Proc univariate normal plot data=PR;

title 'Normality';

Var Res;


title 'Levene GA';

Class GA;

Model Wt = GA;

Means GA/hovtest=Levene;


title 'Levene Var';

Class Var;

Model Wt=Var;

Means Var/hovtest=Levene;

Proc Gplot data=Grape;

title "GA by Variety Interactions";

symbol1 i=std1mtj v=none color=BLUE;

symbol2 i=std1mtj v=none color=BLACK;

symbol3 i=std1mtj v=none color=RED;

Plot Wt*GA=Var /

Description="GA by Variety Interactions";


title 'contrast';

Class Block Id;

Model Wt = Block Id;

Output out=PRID p=PredID r=ResID;

Contrast 'Control vs new' Id -2 1 1 -2 1 1 -2 1 1;

Contrast 'NewVar' Id 0 1 -1 0 1 -1 0 1 -1;

Contrast 'Lin' Id -1 -1 -1 0 0 0 1 1 1;

Contrast 'Quad' Id 1 1 1 -2 -2 -2 1 1 1;

Contrast 'Lin*NV' Id 0 -1 1 0 0 0 0 1 -1;

Contrast 'Lin*contvsnew' Id 2 -1 -1 0 0 0 -2 1 1;

Contrast 'Quad*NV' Id 0 1 -1 0 -2 2 0 1 -1;

Contrast 'Quad*contvsnew' Id -2 1 1 4 -2 -2 -2 1 1;

Means Id/REGWQ;

Proc univariate normal plot data=PRID;

title 'Normality on ID';

Var ResID;


title 'Levene on ID';

Class ID;

Model Wt = ID;

Means ID/hovtest=Levene;

Proc sort data=Grape;

by GA;


title "Var effect by GA level";

Class Block Var;

Model Wt = Block Var;

Contrast 'Control vs New' Var -2 1 1;

Contrast 'New' Var 0 -1 1;

by GA;

Proc sort data=Grape;

by Var;


title "GA effect by Varlevel";

Class Block GA;

Model Wt = Block GA;

Contrast 'lineal' GA 1 0 -1;

Contrast 'quadratic' GA 1 -2 1;

by Var;

Run; Quit;

1. [3 points] Describe the experimental design in detail using the table in the appendix.

Design: 3x3 factorial arranged in an RCBD with two reps per treatment combination per block

Response Variable: Mean individual berry weight (g)

Experimental Unit: 10-plant plot

Class

Variable

Block or

Treatment

Number

of

Levels

Fixed or

Random Description

1 Block 3 Random Vineyard sections

2 Treatment 3 Fixed Three table grape varieties

3 Treatment 3 Fixed GA application concentration (ppm)

Subsamples? NO

Covariable? NO

2. [2 points] Test for block*treatment interactions, normality of residuals, and homogeneity of variances.


Block 2 2.77231111 1.38615556 31.39 <.0001

GA 2 11.07067778 5.53533889 125.36 <.0001

Block*GA 4 0.13141111 0.03285278 0.74 0.5705

Var 2 4.89293333 2.44646667 55.41 <.0001

Block*Var 4 0.02845556 0.00711389 0.16 0.9562

Gib*Var 4 0.23568889 0.05892222 1.33 0.2827

Block*GA*Var 8 0.36292222 0.04536528 1.03 0.4401

Since our design includes two replications per block, we can include block*treatment interactions in an

exploratory model. Block*GA (p = 0.5705), Block*Variety (p = 0.9562), and Block*GA*Treatment (p =

0.4401) interactions were all not significant. We can assume additivity between blocks and treatments.

Tests for Normality

Test Statistic p Value

Shapiro-Wilk W 0.974229 Pr < W 0.2938

The Shapiro-Wilk test resulted in a W-statistic = 0.974229 and a p = 0.2938. We can conclude the

residuals are normally distributed.

Levene's Test for Homogeneity of Wt Variance

ANOVA of Squared Deviations from Group Means

Source DF Sum of Squares Mean Square F Value Pr > F

GA 2 0.00145 0.000727 0.02 0.9832

Error 51 2.1929 0.0430

Levene’s test for the GA treatment resulted in a p = 0.9832. We can conclude variances among GA

concentrations are homogeneous.

Levene's Test for Homogeneity of Wt Variance

ANOVA of Squared Deviations from Group Means

Source DF Sum of Squares Mean Square F Value Pr > F

Var 2 0.2646 0.1323 1.37 0.2626

Error 51 4.9159 0.0964

Levene’s test for the variety treatment resulted in a p = 0.2626. We can conclude variances among

varieties are homogeneous.

3. [4 points] Perform ANOVA with the correct model and discuss the results.

R-Square Coeff Var Root MSE Wt Mean

0.917097 3.644315 0.199708 5.480000


Block 2 2.77231111 1.38615556 34.76 <.0001

GA 2 11.07067778 5.53533889 138.79 <.0001

Var 2 4.89293333 2.44646667 61.34 <.0001

GA*Var 4 0.23568889 0.05892222 1.48 0.2258

The reduced model ANOVA explains 91.7% of the variation in the data, as indicated by the r-squared

value. The effects of GA concentration (p < 0.0001) and variety (p < 0.0001) on berry weight were both

highly significant. Block effects were also highly significant (p < 0.0001). ANOVA did not show a

significant interaction between GA and variety (p = 0.2258)

4. [3 points] Present an interaction plot of GA concentration by variety on the effect of berry weight.

Comment on the interaction plot in light of the ANOVA results.

Although the ANOVA results showed the GA*Variety interaction was not significant, the above plot

suggests otherwise. The slopes of the lines representing Thompson Seedless and Variety 1 are more or

less parallel; however, the line representing Variety 2 intersects that of Variety 1, suggesting there may be

some hidden interaction not detected by ANOVA.

5. [5 points] Are any of the components of the GA*Variety interaction significant?

Contrast DF Contrast SS Mean Square F Value Pr > F

Control vs new 1 4.66253333 4.66253333 116.90 <.0001

NewVar 1 0.23040000 0.23040000 5.78 0.0206

Lin 1 11.04454444 11.04454444 276.92 <.0001

Quad 1 0.02613333 0.02613333 0.66 0.4227

Lin*NV 1 0.23010417 0.23010417 5.77 0.0207

Lin*contvsnew 1 0.00073472 0.00073472 0.02 0.8927

Quad*NV 1 0.00031250 0.00031250 0.01 0.9299

Quad*contvsnew 1 0.00453750 0.00453750 0.11 0.7375

According to the analysis, there is an interaction between the lineal response of GA concentration and the

two new varieties.

6. [6 points]Based on your answer to question 5, answer the following questions using the appropriate

analyses:

Since the contrast Lin x New varieties is significant, then it is better to analyze the

simple effects.

Contrasts of simple effects of variety by GA:

GA=0


Control vs New 1 1.44000000 1.44000000 39.55 <.0001

New 1 0.00480000 0.00480000 0.13 0.7224

GA=10


Control vs New 1 1.69433611 1.69433611 36.84 <.0001

New 1 0.08500833 0.08500833 1.85 0.1971

GA=20


Control vs New 1 1.53346944 1.53346944 38.91 <.0001

New 1 0.37100833 0.37100833 9.41 0.0090

a. Was there a significant difference in berry weight between Thompson Seedless and the two new

varieties?

At all levels of GA the new varieties are better than the control (p < 0.0001).

b. Was there a significant difference in berry weight between the two new varieties?

The analysis by GA indicates that the New varieties are significantly different only at the highest

level of GA (p = 0.0090).

Contrasts of simple effects of GA by variety:

Var=TS


lineal 1 3.59707500 3.59707500 99.17 <.0001

quadratic 1 0.02200278 0.02200278 0.61 0.4500

Var=Var1


lineal 1 5.14830000 5.14830000 108.85 <.0001

quadratic 1 0.00284444 0.00284444 0.06 0.8101

Var=Var2


lineal 1 2.53000833 2.53000833 54.81 <.0001

quadratic 1 0.00613611 0.00613611 0.13 0.7213

c. How would you describe the response of berry weight to GA concentration?

In all the varieties the GA response is lineal (p < 0.0001).

7. [2 points] What recommendations would you make to the breeder if the difference in cost between the

10 and 20 ppm GA applications is small?

Since the response to GA is lineal, the highest GA produces the biggest berries;at the highest GA

concentration (20 ppm) the new variety 1 produced bigger berries than variety 2. Furthermore, if variety 1

produces bigger berries at 20 ppm GA compared with 10 ppm, the best combination is variety 1 at 20

ppm. Alternatively:

Ryan-Einot-Gabriel-Welsch Multiple Range Test for Wt

Alpha 0.05

Error Degrees of Freedom 43

Error Mean Square 0.039883

Number of Means 2 3 4 5 6 7 8 9

Critical Range 0.3050541 0.3312879 0.3455767 0.3551772 0.3623306 0.367953 0.367953 0.3764065

Means with the same letter

are not significantly different.

REGWQ Grouping Mean N Id

A 6.4317 6 8

B 6.0800 6 9

C 5.7500 6 5

C 5.6367 6 7

C 5.5817 6 6

Means with the same letter

are not significantly different.

REGWQ Grouping Mean N Id

D 5.1617 6 3

D 5.1217 6 2

D 5.0150 6 4

E 4.5417 6 1

Because this experiment is balanced, we can compare means using the REGWQ test, which controls EER

but is more sensitive than Tukey’s test. According to REGWQ, the largest mean berry size was produced

by variety 1 vines receiving GA application at 20 ppm. Although, berry weights were similar in varieties

1 and 2 at 0 and 10 ppm, the differences in response to GA concentration between the two varieties

resulted in a more pronounced effect in variety 1 at 20 ppm. The breeder should release variety 1.

Problem 4 [5 points]

Two levels of Nitrogen (N0 and N1) and two levels of Potassium (K0 and K1) are tested in a 2 x 2

factorial organized in a RCBD with three blocks. A third factor, Varieties (A, B, and C) is added as a split

plot of the Nitrogen Potassium combinations. Show the lay out of the field for this experiment

N0 K0 A N1 K1 B N1 K0 C N0 K1 C

N0 K0 C N1 K1 A N1 K0 A N0 K1 A

N0 K0 B N1 K1 C N1 K0 B N0 K1 B

N1 K1 B N0 K0 A N0 K1 C N1 K0 A

N1 K1 C N0 K0 B N0 K1 A N1 K0 C

N1 K1 A N0 K0 C N0 K1 B N1 K0 B

N0 K1 C N1 K0 A N0 K0 C N1 K1 A

N0 K1 B N1 K0 B N0 K0 B N1 K1 B

N0 K1 A N1 K0 C N0 K0 A N1 K1 C

Problem 5 [25 points]

Black point is a common disease of wheat in the Sacramento region. The incidence of the disease is

estimated as the % of grains with a black discoloration at the end of the grain. A researcher wants to test

the differences in resistance among four specific varieties (designated 1 to 4) and three fungicides

(designated 1-3). He selects three homogeneous blocks of land and divides each in 4 large plots, and

randomizes the four varieties separately in each plot. He plants the field in a North-South direction,

changing variety in each block. By head emergence he cut two aisles in a West-East direction dividing

each block in three horizontal sections. He then drives his tractor in a West-East zigzag applying only one

fungicide in each pass through the block. The order in which the three fungicides will be applied is

randomized independently in each of the blocks. A design of the field is included below. Since Black

point is affected by lodging (the falling spikes are more likely to show the disease), the researcher records

the lodging of each of the 12 plots in each block after the fungicide was applied and before harvest. The

data for the percentage of grains with black point and the % lodging in each plot is included in the Table

below the graph of the field.

Data Fungi;

Input Bl Var Fun Y X;

Z = Y - 3.82630208*(X-15.94444); Cards; 1 1 1 39.6 14 2 1 1 52.5 17 3 1 1 47.1 15 1 1 2 54.5 16 2 1 2 63.5 17 3 1 2 63.0 16 1 1 3 60.4 17 2 1 3 68.6 19 3 1 3 67.4 18 1 2 1 55.2 15 2 2 1 55.2 16 3 2 1 61.5 16 1 2 2 76.7 16 2 2 2 60.6 15 3 2 2 68.0 16 1 2 3 78.5 16 2 2 3 62.7 14 3 2 3 80.6 18 1 3 1 38.9 15 2 3 1 58.3 16 3 3 1 45.5 16 1 3 2 73.4 16 2 3 2 53.3 15 3 3 2 67.2 16 1 3 3 63.7 15 2 3 3 66.2 17 3 3 3 73.1 17 1 4 1 38.6 15 2 4 1 46.3 16 3 4 1 51.3 16 1 4 2 56.4 15 2 4 2 63.2 16 3 4 2 59.6 15 1 4 3 65.3 16 2 4 3 50.2 13 3 4 3 70.4 18 ;

Proc GLM Data = Fungi;

Title 'One-way ANOVAs for X and Y';

Class Bl Var Fun;

Model X Y = Bl Var Bl*Var Fun Bl*Fun Var*Fun;

Random Bl Bl*Var Bl*Fun /test;

Proc GLM;

Title 'General regression';

Model Y = X;

Proc GLM Order = Data;

Title 'The ANCOVA';

Class Bl Var Fun;

Model Y = Bl Var Bl*Var Fun Bl*Fun Var*Fun X /Solution;


Output out = FungiPR p = Pred r = Res;

LSMeans Var/ StdErr Pdiff Adjust = Tukey;

LSMeans Fun/ StdErr Pdiff Adjust = Tukey;

Proc Univariate Data = FungiPR normal;

Title 'Normality of residuals in ANCOVA';

Var Res;

Proc GLM Data = Fungi;

Title 'ANOVA on Z';

Class Bl Var Fun;

Model Z = Bl Var Bl*Var Fun Bl*Fun Var*Fun;


Output out = FungiPRz p = Predz r = Resz;

Means Var/ Tukey; * or LSMean statement also correct;

Means Fun/ Tukey;

Proc Univariate Data = FungiPRz normal;

Title 'Normality of residuals using z';

Var Resz;

Proc Plot Data = FungiPRz;

Plot Resz*Predz = Var;

Proc GLM;

Title 'Homogeneity of variances Var';

Class Var ;

Model Z = Var;

Means Var / hovtest = Levene;

Proc GLM;

Title 'Homogeneity of variances Fun';

Class Fun;

Model Z = Fun;

Means Fun / hovtest = Levene;

Proc GLM;

Title 'Homogeneity of slopes Var';

Class Bl Var;

Model Z = Bl Var X Var*X;

Proc GLM;

Title 'Homogeneity of slopes Fun';

Class Bl Fun;

Model Z = Bl Fun X Fun*X;

Proc GLM;

Title 'Homogeneity of slopes Combinations Fun*Var not necessary';

Class Bl Var Fun;

Model Z = Bl Var Bl*Var Fun Bl*Fun Var*Fun Var*Fun*X;

Run;Quit;

Class Level Information

Class Levels Values

Bl 3 1 2 3

Var 4 1 2 3 4

Fun 3 1 2 3

Number of Observations Read 36

Number of Observations Used 36

One-way ANOVAs for X and Y

Dependent Variable: X

Sum of

Source DF Squares Mean Square F Value Pr > F

Model 23 41.22222222 1.79227053 2.02 0.1039

Error 12 10.66666667 0.88888889

Corrected Total 35 51.88888889

R-Square Coeff Var Root MSE X Mean

0.794433 5.913088 0.942809 15.94444


Bl 2 5.05555556 2.52777778 2.84 0.0975

Var 3 5.00000000 1.66666667 1.88 0.1876

Bl*Var 6 9.83333333 1.63888889 1.84 0.1728

Fun 2 5.72222222 2.86111111 3.22 0.0760

Bl*Fun 4 9.11111111 2.27777778 2.56 0.0926

Var*Fun 6 6.50000000 1.08333333 1.22 0.3616

Dependent Variable: Y

Sum of


Model 23 3694.325278 160.622838 4.53 0.0046

Error 12 425.384444 35.448704


R-Square Coeff Var Root MSE Y Mean

0.896744 9.939241 5.953881 59.90278


Bl 2 160.817222 80.408611 2.27 0.1460

Var 3 613.705278 204.568426 5.77 0.0111

Bl*Var 6 347.280556 57.880093 1.63 0.2211

Fun 2 2169.557222 1084.778611 30.60 <.0001

Bl*Fun 4 368.257778 92.064444 2.60 0.0897

Var*Fun 6 34.707222 5.784537 0.16 0.9820


Bl Var(Error) + 4 Var(Bl*Fun) + 3 Var(Bl*Var) + 12 Var(Bl)

Var Var(Error) + 3 Var(Bl*Var) + Q(Var,Var*Fun)

Bl*Var Var(Error) + 3 Var(Bl*Var)

Fun Var(Error) + 4 Var(Bl*Fun) + Q(Fun,Var*Fun)

Bl*Fun Var(Error) + 4 Var(Bl*Fun)

Var*Fun Var(Error) + Q(Var*Fun)

Dependent Variable: X


Bl 2 5.055556 2.527778 0.83 0.4860

Error 5.0633 15.330504 3.027778

Error: MS(Bl*Var) + MS(Bl*Fun) - MS(Error)


* Var 3 5.000000 1.666667 1.02 0.4484

Error: MS(Bl*Var) 6 9.833333 1.638889


Bl*Var 6 9.833333 1.638889 1.84 0.1728

Bl*Fun 4 9.111111 2.277778 2.56 0.0926

Var*Fun 6 6.500000 1.083333 1.22 0.3616

Error: MS(Error) 12 10.666667 0.888889


* Fun 2 5.722222 2.861111 1.26 0.3773

Error: MS(Bl*Fun) 4 9.111111 2.277778



Bl 2 160.817222 80.408611 0.70 0.5408

Error 4.7121 539.518861 114.495833



* Var 3 613.705278 204.568426 3.53 0.0880

Error: MS(Bl*Var) 6 347.280556 57.880093


Bl*Var 6 347.280556 57.880093 1.63 0.2211

Bl*Fun 4 368.257778 92.064444 2.60 0.0897

Var*Fun 6 34.707222 5.784537 0.16 0.9820

Error: MS(Error) 12 425.384444 35.448704


* Fun 2 2169.557222 1084.778611 11.78 0.0211

Error: MS(Bl*Fun) 4 368.257778 92.064444

General regression Dependent Variable: Y

Sum of


Model 1 1199.737046 1199.737046 13.97 0.0007

Error 34 2919.972677 85.881549



0.291219 15.47045 9.267230 59.90278

Source DF Type I SS Mean Square F Value Pr > F

X 1 1199.737046 1199.737046 13.97 0.0007

Parameter Estimate Error t Value Pr > |t|

Intercept -16.76541756 20.57073476 -0.82 0.4207

X 4.80845824 1.28650877 3.74 0.0007

The ANCOVA Sum of


Model 24 3850.491546 160.437148 6.56 0.0012

Error 11 269.218176 24.474380



0.934651 8.258647 4.947159 59.90278


Bl 2 39.129919 19.564960 0.80 0.4741

Var 3 705.515872 235.171957 9.61 0.0021

Bl*Var 6 111.537826 18.589638 0.76 0.6160

Fun 2 1318.414157 659.207078 26.93 <.0001

Bl*Fun 4 93.050956 23.262739 0.95 0.4714

Var*Fun 6 71.641933 11.940322 0.49 0.8046

X 1 156.166268 156.166268 6.38 0.0282

Average SLOPE

Standard

Parameter Estimate Error t Value Pr > |t|

Intercept 3.43035301 B 26.39826327 0.13 0.8990

Bl 1 -1.42677951 B 5.67330227 -0.25 0.8061

….

X 3.82630208 1.51475183 2.53 0.0282

Average slope 3.82630208


Bl Var(Error) + 3.3569 Var(Bl*Fun) + 2.5177 Var(Bl*Var) + 10.071 Var(Bl)

Var Var(Error) + 2.6809 Var(Bl*Var) + Q(Var,Var*Fun)

Bl*Var Var(Error) + 2.7602 Var(Bl*Var)

Fun Var(Error) + 3.3017 Var(Bl*Fun) + Q(Fun,Var*Fun)

Bl*Fun Var(Error) + 3.5393 Var(Bl*Fun)


X Var(Error) + Q(X)

Tests of Hypotheses for Mixed Model Analysis of Variance



Bl 2 39.129919 19.564960 1.09 0.5075

Error 1.5359 27.581309 17.957457

Error: 0.9121*MS(Bl*Var) + 0.9485*MS(Bl*Fun) - 0.8606*MS(Error)


* Var 3 705.515872 235.171957 12.54 0.0043

Error 6.4711 121.390464 18.758732

Error: 0.9713*MS(Bl*Var) + 0.0287*MS(Error)


Bl*Var 6 111.537826 18.589638 0.76 0.6160

Bl*Fun 4 93.050956 23.262739 0.95 0.4714

Var*Fun 6 71.641933 11.940322 0.49 0.8046

X 1 156.166268 156.166268 6.38 0.0282

Error: MS(Error) 11 269.218176 24.474380


* Fun 2 1318.414157 659.207078 28.24 0.0026

Error 4.6191 107.828046 23.344089

Error: 0.9329*MS(Bl*Fun) + 0.0671*MS(Error)

Least Squares Means

Adjustment for Multiple Comparisons: Tukey-Kramer

Standard LSMEAN

Var Y LSMEAN Error Pr > |t| Number

1 55.0617043 1.8911008 <.0001 1

2 67.1932726 1.6682658 <.0001 2

3 60.1681279 1.6511987 <.0001 3

4 57.1880064 1.7511080 <.0001 4

Least Squares Means for effect Var

Pr > |t| for H0: LSMean(i)=LSMean(j)


i/j 1 2 3 4

1 0.0034 0.2422 0.8686

2 0.0034 0.0504 0.0064

3 0.2422 0.0504 0.6110

4 0.8686 0.0064 0.6110

Standard LSMEAN

Fun Y LSMEAN Error Pr > |t| Number

1 50.5483869 1.5292919 <.0001 1

2 64.0273365 1.4581778 <.0001 2

3 65.1326100 1.6576195 <.0001 3

Least Squares Means for effect Fun



i/j 1 2 3

1 0.0001 0.0003

2 0.0001 0.8834

3 0.0003 0.8834

Normality of residuals in ANCOVA

Test --Statistic--- -----p Value------


ANOVA on Z

Z = Y - 3.82630208*(X-15.94444); Sum of


Model 23 2700.808117 117.426440 5.23 0.0024

Error 12 269.218176 22.434848


R-Square Coeff Var Root MSE Z Mean

0.909355 7.907054 4.736544 59.90276


Bl 2 41.371442 20.685721 0.92 0.4241

Var 3 756.249312 252.083104 11.24 0.0008

Bl*Var 6 117.650450 19.608408 0.87 0.5413

Fun 2 1582.413105 791.206553 35.27 <.0001

Bl*Fun 4 107.136797 26.784199 1.19 0.3628

Var*Fun 6 95.987011 15.997835 0.71 0.6463


Bl Var(Error) + 4 Var(Bl*Fun) + 3 Var(Bl*Var) + 12 Var(Bl)

Var Var(Error) + 3 Var(Bl*Var) + Q(Var,Var*Fun)

Bl*Var Var(Error) + 3 Var(Bl*Var)

Fun Var(Error) + 4 Var(Bl*Fun) + Q(Fun,Var*Fun)

Bl*Fun Var(Error) + 4 Var(Bl*Fun)


Tests of Hypotheses for Mixed Model Analysis of Variance

Dependent Variable: Z


Bl 2 41.371442 20.685721 0.86 0.5362

Error 2.0113 48.186457 23.957759



* Var 3 756.249312 252.083104 12.86 0.0051

Error: MS(Bl*Var) 6 117.650450 19.608408


Bl*Var 6 117.650450 19.608408 0.87 0.5413

Bl*Fun 4 107.136797 26.784199 1.19 0.3628

Var*Fun 6 95.987011 15.997835 0.71 0.6463

Error: MS(Error) 12 269.218176 22.434848


* Fun 2 1582.413105 791.206553 29.54 0.0040

Error: MS(Bl*Fun) 4 107.136797 26.784199

SAME RESULTS AS ANCOVA!

Least Squares Means

Adjustment for Multiple Comparisons: Tukey

Standard LSMEAN

Var Z LSMEAN Error Pr > |t| Number

1 55.0616873 1.5788480 <.0001 1

2 67.1932556 1.5788480 <.0001 2

3 60.1681109 1.5788480 <.0001 3

4 57.1879894 1.5788480 <.0001 4

Least Squares Means for effect Var



i/j 1 2 3 4

1 0.0008 0.1557 0.7780

2 0.0008 0.0368 0.0036

3 0.1557 0.0368 0.5600

4 0.7780 0.0036 0.5600

Standard LSMEAN

Fun Z LSMEAN Error Pr > |t| Number

1 50.5483699 1.3673224 <.0001 1

2 64.0273195 1.3673224 <.0001 2

3 65.1325929 1.3673224 <.0001 3




i/j 1 2 3

1 <.0001 <.0001

2 <.0001 0.8375

3 <.0001 0.8375

Normality of residuals using z

Tests for Normality

Test --Statistic--- -----p Value------


Plot of Resz*Predz. Symbol is value of Var.

Plot of Resz*Predz. Symbol is value of Var.

8 ˆ

‚ 3

‚

‚

‚

‚

6 ˆ

‚

‚

‚ 3

‚

‚

4 ˆ 1

‚

‚

Resz ‚ 4 3

‚ 4

‚ 1 2

2 ˆ 4

‚ 2

‚ 1 2

‚ 3

‚ 2

‚ 4 1

0 ˆ 1

‚ 2 1 4

‚ 4 2

‚ 1 3 2

‚

‚ 2

-2 ˆ 2 34

‚ 4

‚ 4

‚ 1

‚

‚ 3 1

-4 ˆ 3

‚

‚

‚

‚

‚

-6 ˆ 3

Šƒˆƒƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒƒƒˆƒƒƒƒƒƒƒƒƒƒˆƒƒ

40 45 50 55 60 65 70 75 80

Predz

Homogeneity of variances Var

Levene's Test for Homogeneity of Z Variance

Sum of Mean

Source DF Squares Square F Value Pr > F

Var 3 24941.5 8313.8 1.64 0.1996

Error 32 162209 5069.0

Homogeneity of variances Fun

Levene's Test for Homogeneity of Z Variance

Sum of Mean


Fun 2 47.7960 23.8980 0.01 0.9885

Error 33 68013.9 2061.0

Variances are homogeneous

Homogeneity of Slopes for Variety:


Block 2 69.662038 34.831019 0.47 0.6302

Variety 3 124.837767 41.612589 0.56 0.6452

X 1 1179.312568 1179.312568 15.91 0.0005

X*Variety 3 117.541162 39.180387 0.53 0.6665

Homogeneity of Slopes for Fungicide:


Block 2 95.1473158 47.5736579 1.09 0.3499

Fungicide 2 156.2513999 78.1257000 1.79 0.1854

X 1 483.9957565 483.9957565 11.09 0.0024

X*Fungicide 2 114.1416994 57.0708497 1.31 0.2864

Slopes are homogeneous

ANSWERS TO QUESTIONS

1. [5 points] Are there significant differences among varieties and among fungicides in the percentage of

black point, once the values were adjusted for lodging?

Yes both Var and Fun showed significant differences once the values were adjusted for lodging


* Var 3 705.515872 235.171957 12.54 0.0043

Error 6.4711 121.390464 18.758732

Error: 0.9713*MS(Bl*Var) + 0.0287*MS(Error)


* Fun 2 1318.414157 659.207078 28.24 0.0026

Error 4.6191 107.828046 23.344089

Error: 0.9329*MS(Bl*Fun) + 0.0671*MS(Error)

2. [5 points] Are the differences the same if the values are not adjusted for lodging? If different provide

a potential explanation of the source of the discrepancies.

The differences for Varieties were not significant in the ANOVA for the unadjusted values and the

differences among fungicides were less significant. Since there is no effect of treatments on lodging, the

simplest explanation is that the differences in lodging were generating additional variability in the % of

black point, and that once that variability was eliminated the ANCOVA had more power to see the

differences among varieties and among fungicides.

3. [3 points] Were the responses to fungicides different at the different varieties when the values were

corrected for lodging?

No. There is no significant interaction between varieties and fungicide


Bl*Var 6 111.537826 18.589638 0.76 0.6160

Bl*Fun 4 93.050956 23.262739 0.95 0.4714

Var*Fun 6 71.641933 11.940322 0.49 0.8046

X 1 156.166268 156.166268 6.38 0.0282

Error: MS(Error) 11 269.218176 24.474380

4. [5 points] Based on the previous answer (and using the adjusted values) use the most sensitive fixed

range mean comparison test controlling the experiment wise error rate to compare all four varieties

and all three fungicides separately. What would be your recommendation?

The correct analysis is for the main effects Variety and Fungicide separately. Using the Tukey test we find

that variety 2 performs significantly worse than the other three varieties (higher infection) and that the

other 3 varieties were not significantly different among each other

Var Z LSMEAN Error Pr > |t| Number

1 55.0616873 1.5788480 <.0001 1

2 67.1932556 1.5788480 <.0001 2

3 60.1681109 1.5788480 <.0001 3

4 57.1879894 1.5788480 <.0001 4

i/j 1 2 3 4

1 0.0008 0.1557 0.7780

2 0.0008 0.0368 0.0036

3 0.1557 0.0368 0.5600

4 0.7780 0.0036 0.5600

Fungicide 1 showed significantly lower values than fungicides 2 and 3 that did not differed from each other

significantly

Standard LSMEAN

Fun Z LSMEAN Error Pr > |t| Number

1 50.5483699 1.3673224 <.0001 1

2 64.0273195 1.3673224 <.0001 2

3 65.1325929 1.3673224 <.0001 3




i/j 1 2 3

1 <.0001 <.0001

2 <.0001 0.8375

3 <.0001 0.8375

The recommendation is to use fungicide 1 with any variety except variety 2

5. [4 points] Test all the required assumptions required to perform the test on the adjusted values

Residuals of the ANCOVA are normal Test --Statistic--- -----p Value------


Both Variety and Fungicide showed homogeneity of variances by Levene’s test Levene's Test for Homogeneity of Z Variance

Sum of Mean


Var 3 24941.5 8313.8 1.64 0.1996

Error 32 162209 5069.0

Sum of Mean


Fun 2 47.7960 23.8980 0.01 0.9885

Error 33 68013.9 2061.0

The slopes for the regression are homogeneous among Varieties and among fungicides

Homogeneity of Slopes for Variety:


Block 2 69.662038 34.831019 0.47 0.6302

Variety 3 124.837767 41.612589 0.56 0.6452

X 1 1179.312568 1179.312568 15.91 0.0005

X*Variety 3 117.541162 39.180387 0.53 0.6665

Homogeneity of Slopes for Fungicide:


Block 2 95.1473158 47.5736579 1.09 0.3499

Fungicide 2 156.2513999 78.1257000 1.79 0.1854

X 1 483.9957565 483.9957565 11.09 0.0024

X*Fungicide 2 114.1416994 57.0708497 1.31 0.2864

6. [3 points] Was the test with the adjusted values more efficient than the test with the unadjusted

values? Quantify the difference.

Relative efficiency of the ANCOVA to an ANOVA for %Black Point.

The effective error MS after adjustment for X is given by

An estimate of the relative precision is 35.448704 / 28.5867 = 1.24. This indicates that the covariance

analysis provides 24% more information per measurement than the ANOVA.

Appendix

5867.28666.10*23

222.41124.474380

)1(1

var

var

iableX

iableX

YAdjustedError

SSErrort

SSTRT

MS

When you are asked to "describe in detail the design of this experiment," please do so by completing the

following template:

Design:

Response Variable:

Experimental Unit:

Class

Variable

Block or

Treatment

Number of

Levels

Fixed or

Random Description

1

2

↓

n

Subsamples? YES / NO

Covariable? YES / NO

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