Review Session
Warm-up
Pair the term in the first column with the most related term in the second column and explain the relationship of each pair of terms:
accuracy linear contrast
precision randomization
regression pseudoreplication
nesting independence
orthogonality replication
Deciding between contrasts and means separation (8)
1. How many questions in total are you asking?
More than (t – 1) à Means separation
(t – 1) or less à Continue…
2. Are the questions motivated by the data or by the treatment structure?
By the data à Means separation
By the treatment structure à Continue…
3. Do the questions form an orthogonal set?
No à Means separation
Yes à Orthogonal contrasts
V2.6 (t = 7) Maintaining the EER at 5%, use the appropriate mean comparisons to declare if:
a. 1-in-2 management significantly increases regeneration compared to shelterwood.
b. Selective logging promotes significantly more regeneration than: i) no management at all, and ii) clear felling.
c. The 5% increase in cut shrubs under seed-tree management compared to shelterwood has a significant impact on chaparral regeneration.
Deciding among methods of means separation (6)
Fixed-range Multiple-range
LSD REGWQ
Dunnetts
Tukey
Scheffe
Management Practice / % of shrubs cutControl / 0
Selective Logging / 10
Gap Logging / 25
1-in-2 / 50
Shelterwood / 75
Seed-Tree / 80
Clear Felling / 95
Answer this question assuming no other analysis will be performed on the data.
Using the most sensitive test which controls MEER, which management practice(s) significantly increase(s) regeneration compared to the control?
Answer this question assuming no other analysis will be performed on the data.
Using the most sensitive test which controls MEER, which management practice(s) result(s) in the maximum regeneration rates?
Answer this question assuming no other analysis will be performed on the data.
Using the most sensitive fixed-range test which controls MEER, which practice(s) would you recommend?
Answer these questions assuming no other analysis will be performed on the data.
Using the most sensitive test which controls MEER:
1. Which practice(s) would you recommend?
2. What is the character of the response between regeneration rate and % clearing? (e.g. linear, quadratic, cubic, etc.)
Interpreting means separation tables (5)
Tukey's Studentized Range (HSD) Test for Growth
Alpha 0.05
Error Degrees of Freedom 18
Error Mean Square 18.31657
Critical Value of Studentized Range 4.67313
Minimum Significant Difference 10.00
Tukey Grouping Mean Practice
A 54 1n2
A B 46 ShW
B 42 STr
C B 38 GLo
C D 29 ClF
D 26 Sel
D 20 Con
Question: In presenting data, should the order be most desirable à least desirable?
Comparison-wise error rate (CER) vs. Experiment-wise error rate (EER) (5)
Spin. What's the chance of landing on "A"?
Spin again. What's the chance of landing on "A"?
What's the chance that both spins landed on "A"?
Spin 2A / B / C / D
Spin 1 / A / AA / AB / AC / AD
B / BA / BB / BC / BD
C / CA / CB / CC / CD
D / DA / DB / DC / DD
How lucky are you? What's the chance that ten spins will all be "A"?
What's the chance that at least one spin will not be "A"?
Recognizing Experimental Designs
A professor wants his graduate student to carry out an experiment to see the effects of four different light conditions (100, 600, 1100, 1600 mmol/m2s) on photosynthetic rates in soybean leaves. He has 16 plants and four growth chambers, each of which has four isolated compartments where light intensity is regulated independently. The four growth chambers are a various ages with different levels of precision in temperature control.
The professor's suggested design:
Experimental design 1: Assign one light level to each growth chamber (e.g. light levels 100, 600, 1100, 1600 assigned randomly to growth chambers A, B, C, and D, respectively) and randomly assign one plant to each of the four compartments in each chamber. After one week, measure the photosynthetic rate of two randomly selected leaves from each plant.
The student suggests an alternative design:
Experimental design 2: Randomize the four light levels within each chamber (one to each of the four isolated compartments) and randomly assign one plant to each of the four compartments in each chamber. After one week, measure the photosynthetic rate of two randomly selected leaves from each plant.
Questions:
1. What are the designs of these two experiments?
2. Which is the better design?
3. What is the number of replications in each design?
Variance Components (3)
A population geneticist wants to test the significance of the differences in anther length among three races of Berberis spp. adapted to open grasslands (G), forests (F), and riparian areas (R). He randomly selects four populations from each environment and then six individuals from each population. Finally, he randomly selects two anthers from each individual and measures them.
Source DF SS MS F p
Total __ 5109.9
Races __ 845.2 ______
Populations __ 620.9 ______
Individuals __ 1801.2 ______
Error __ 1842.6 ______
1. Calculate the degrees of freedom and mean squares.
2. Are there significant differences in anther length among races?
3. Is there significant variation among populations of the same race?
4. Is there significant variation among individuals of the same population?
5. Calculate the proportion of variation contributed at each level.
Non-additivity in RCBD (6)
Observed / TreatmentBlock / A / B / Mean / βj
1 / 12 / 28 / 20 / -5
2 / 18 / 42 / 30 / 5
Mean / 15 / 35 / 25
τi / -10 / 10
Predicted / Treatment / Error / Treatment
Block / A / B / Block / A / B
1 / 10 / 30 / 1 / 2 / -2
2 / 20 / 40 / 2 / -2 / 2
One rep per cell: Yij = m + ti + bj + eij
where eij = (tibj) + Experimental error
>1 rep per cell: Yijk = m + ti + bj + (tibj) + eijk
where eijk = Experimental error
A Tukey 1-df test for Nonadditivity (1 rep per cell)
asks the same question as
an F-test for the Block*Treatment interaction (>1 reps per cell)
Determination of contrast coefficients (for orthogonal contrasts and Scheffe group comparisons) (4)
1. For trend analysis, consult the table of coefficients
2. For group comparisons, state each null hypothesis as an equation and simplify.
Consider an experiment to determine the effects of scarification (mild, strong) and acid treatments (30, 60, 90 seconds in HCl) on seed germination. The treatments:
A. No scarification or acid (control)
B. Mild scarification + 30 seconds in HCl
C. Mild scarification + 60 seconds in HCl
D. Mild scarification + 90 seconds in HCl
E. Strong scarification + 30 seconds in HCl
F. Strong scarification + 60 seconds in HCl
G. Strong scarification + 90 seconds in HCl
1. Is there an effect of seed treatment?
2. Is there a difference between mild and strong scarification?
3. Characterize the response of seed germination to HCl treatment response (i.e. linear, quadratic, cubic, etc.).
4. Is the response of seed germination to HCl treatment different at different intensities of scarification?
In SAS, use the Order = Data statement!
Example of non-orthogonal group comparisons (Scheffe):
1. Is there an effect of mild scarification on seed germination, relative to no treatment?
2. Is there an effect of strong scarification on seed germination, relative to no treatment?
Error, power, and significance questions… (2)
What determines the Type I error in an experiment?
What determines the Type II error?
What determines the power?
What do you do if p = 5.01?
One-tailed versus two-tailed tests (3)
When significant difference in one direction is either impossible or of absolutely no interest at all, a one-tailed test may be appropriate.
In general, one-tailed tests are almost never appropriate in basic research. Most often, they are mis-used to artificially inflate the power of a study.
Not answered because not asked on the exam:
Interpreting interactions in transformed data (5)
Selecting the right data transformation (2)
Understanding partial null hypotheses (1)
Where does (a = 1 – m/2) come from? (Hinz and Eagles 1976) (1)
Not answered because material from the second half of the course:
The effects of covariance on data (1)
Testing assumptions in factorial experiments (1)
Generating interactions plots in factorial experiment (1)
Not answered because covered in detail in the reading:
Why might MSE not decrease when blocks are introduced? (1)
Why partition treatment df for tests? (1)
Final tips
Average subsamples! Test all assumptions using these averages.
Be very clear about the experimental design before touching SAS.
Read all directions and descriptions extremely carefully.
To maximize your grade…
Include all SAS code.
Report p-values when stating conclusions.
When asked to choose the most appropriate method of analysis, choose one method.
Variance Components (3)
Total number of data points in the experiment:
2 anthers x 6 individuals x 4 populations x 3 races = 144
Source DF SS MS
Total 143 5109.9
Races 3-1 = 2 845.2 845.2/2 = 422.6
Populations 3(4-1) = 9 620.9 620.9/9 = 69.0
Individuals 3(4)(6-1) = 60 1801.2 1801.2/60 = 30.0
Error 143-71 = 72 1842.6 1842.6/72 = 25.6
What explains this variation observed at each level?
EXPECTED MS (EMS)
Race Var(Error) + 2 Var(Ind) + 12 Var(Pop) + 48 Var(Race) = 422.6
Population Var(Error) + 2 Var(Ind) + 12 Var(Pop) = 69.0
Individual Var(Error) + 2 Var(Ind) = 30.0
Error Var(Error) = 25.6
2. Are there significant differences in anther length among races?
H0: Var(Race) = 0 à MS(Race)/MS(Pop) = 1
3. Is there significant variation among populations of the same race?
H0: Var(Pop) = 0 à MS(Pop)/MS(Ind) = 1
4. Is there significant variation among individuals of the same population?
H0: Var(Ind) = 0 à MS(Ind)/MS(Error) = 1
Source DF SS MS F p
Total 143 5109.9
Races 2 845.2 422.6 422.6 / 69.0 = 6.12 *
Populations 9 620.9 69.0 69.0 / 30.0 = 2.30 *
Individuals 60 1801.2 30.0 30.0 / 25.6 = 1.17 NS
Error 72 1842.6 25.6
5. Calculate the proportion of variation contributed at each level.
EXPECTED MS (EMS)
Race Var(Error) + 2 Var(Ind) + 12 Var(Pop) + 48 Var(Race) = 422.6
Population Var(Error) + 2 Var(Ind) + 12 Var(Pop) = 69.0
Individual Var(Error) + 2 Var(Ind) = 30.0
Error Var(Error) = 25.6
Var(Error) = 25.6
Var(Error) + 2 Var(Ind) = 30.0
25.6 + 2 Var(Ind) = 30.0
Var(Ind) = (30.0 – 25.6)/2
Var(Ind) = 2.2
Var(Error) + 2 Var(Ind) + 12 Var(Pop) = 69.0
30.0 + 12 Var(Pop) = 69.0
Var(Pop) = (69.0 – 30.0)/12
Var(Pop) = 3.25
Var(Error) + 2 Var(Ind) + 12 Var(Pop) + 48 Var(Race) = 422.6
69.0 + 48 Var(Race) = 422.6
Var(Race) = (422.6 – 69.0)/48
Var(Race) = 7.37
Variance Components
Var(Race) 7.4 19.2%
Var(Pop) 3.3 8.6%
Var(Ind) 2.2 5.7%
Var(Error) 25.6 66.5%
Total 38.5 100.0%
In SAS:
Proc GLM;
Class Race Pop Ind;
Model Length = Race Pop(Race) Ind(Pop Race);
Random Pop(Race) Ind(Pop Race);
Test h = Race e = Pop(Race);
Test h = Pop(Race) e = Ind(Pop Race);
Proc VarComp Method = Type1;
Class Race Pop Ind;
Model Length = Race Pop(Race) Ind(Pop Race);
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