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Analysis of variance April 16, 2009 Contents Comparison of several - - PowerPoint PPT Presentation

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Analysis of variance April 16, 2009 Contents Comparison of several groups One-way ANOVA Two-way ANOVA Interaction Model checking Acknowledgement for use of presentation Julie Lyng Forman, Dept. of Biostatistics (2008),

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Analysis of variance April 16, 2009

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Contents

  • Comparison of several groups
  • One-way ANOVA
  • Two-way ANOVA

– Interaction

  • Model checking

Acknowledgement for use of presentation

  • Julie Lyng Forman, Dept. of Biostatistics (2008),
  • Lene Theil Skovgaard, Dept. of Biostatistics (2007,

2006)

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Marc Andersen StatGroup ApS e-mail: mja@statgroup.dk http://staff.pubhealth.ku.dk/~pd/V+R/html/

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ANOVA, April 2009 1

Comparison of 2 or more groups number different same

  • f groups

individuals individual 2 unpaired paired t-test t-test ≥2

  • neway

twoway analysis of variance analysis of variance One-way analysis of variance:

  • Do the distributions differ between the groups?
  • Do the levels differ between the groups?
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ANOVA, April 2009 2

Example: 22 bypass-patients, 3 different kinds of ventilation during anaesthesia, randomized

Group I 50% N2O, 50% O2 for 24 hours Group II 50% N2O, 50% O2 during operation Group III 30–50% O2 (no N2O) for 24 hours Gr.I Gr.II Gr.III n 8 9 5 Mean 316.6 256.4 278.0 SD 58.7 37.1 33.8

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ANOVA, April 2009 3

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ANOVA, April 2009 4

One-way ANOVA

  • one-way:

because we only have one critera for classification of the

  • bservations, here ventilation method
  • ANalysis Of VAriance:

because we compare the variance between groups with the variance within groups

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ANOVA, April 2009 5

Model: Yij = µi + εij

j’th observation individual in group no. i deviation mean of group no. i

Observations are assumed be independent and to follow a normal distribution (within each group) with the same variance. εij ∼ N(0, σ2) or equivalently Yij ∼ N(µi, σ2) Model assumptions must be checked!

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ANOVA, April 2009 6

Hypothesis testing Usual approach

  • Null hypothesis: group means are equal, H0 : µi = µ
  • Alternative hypothesis: group means are not equal
  • We show the means are not equal by rejecting the null hypothesis
  • f equality (ref DGA, 8.5 Hypothesis Testing)
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ANOVA, April 2009 7

ANOVA math: Sums of squares

Decomposition of ’deviation from grand mean’: yij − ¯ y· = (yij − ¯ yi) + (¯ yi − ¯ y·) yij j’th observation in i’th group ¯ yi average in i’th group ¯ y. total average Decomposition of variation (sums of squares):

  • i,j

(yij − ¯ y·)2

  • total variation

=

  • i,j

(yij − ¯ yi)2

  • within groups

+

  • i,j

(¯ yi − ¯ y·)2

  • between groups
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ANOVA, April 2009 8

Decomposition of variation: total = between + within SStotal = SSbetween + SSwithin (n − 1) = (k − 1) + (n − k) F-test statistic: F = MSbetween MSwithin = SSbetween/(k − 1) SSwithin/(N − k) Reject the null hypothesis if F is large, i.e. if the variation between groups is too large compared to the variation within groups.

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ANOVA, April 2009 9

Usually the analysis is summarized in an Analysis of variance table Variation df SS MS F P Between k − 1 SSb SSb/dfb MSb/MSw P(F(dfb, dfw) > Fobs) Within n − k SSw SSw/dfw Total n − 1 SStot

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ANOVA, April 2009 10

Analysis of variance table - Anaestesia example df SS MS F P Between 2 15515.88 7757.9 3.71 0.04 Within 19 39716.09 2090.3 Total 21 55231.97 F = 3.71 ∼ F(2, 19) ⇒ P = 0.04 Weak evidence of non-equality of the three means

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ANOVA, April 2009 11

Analysis of variance in SAS

To define the anaestesia data in SAS, we write

data ex_redcell; input grp redcell; cards; 1 243 1 251 1 275 . . . . . . 3 293 3 328 ; run;

The variable redcell contains all the measurements of the outcome and grp contains the method of ventilation for each individual.

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ANOVA, April 2009 12

Analysis of variance program:

proc glm data=ex_redcell; class grp; model redcell=grp / solution; run;

General Linear Models Procedure Dependent Variable: REDCELL Sum of Mean Source DF Squares Square F Value Pr > F Model 2 15515.7664 7757.8832 3.71 0.0436 Error 19 39716.0972 2090.3209 Corrected Total 21 55231.8636 R-Square C.V. Root MSE REDCELL Mean 0.280921 16.14252 45.7200 283.227 Source DF Type I SS Mean Square F Value Pr > F GRP 2 15515.7664 7757.8832 3.71 0.0436 Source DF Type III SS Mean Square F Value Pr > F GRP 2 15515.7664 7757.8832 3.71 0.0436

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ANOVA, April 2009 13

The option solution outputs parameter estimates:

T for H0: Pr > |T| Std Error of Parameter Estimate Parameter=0 Estimate INTERCEPT 278.0000000 B 13.60 0.0001 20.44661784 GRP 1 38.6250000 B 1.48 0.1548 26.06442584 2

  • 21.5555556 B
  • 0.85

0.4085 25.50141290 3 0.0000000 B . . . NOTE: The X’X matrix has been found to be singular and a generalized inverse was used to solve the normal equations. Estimates followed by the letter ’B’ are biased, and are not unique estimators of the parameters.

  • Group 3 (the last group) is the reference group
  • The estimates for the other groups refer to differences to this

reference group

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ANOVA, April 2009 14

Interpreting the estimates

Some issues:

  • Clinical significance
  • Statistical significance
  • Provide confidence interval
  • Does it make sense?
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ANOVA, April 2009 15

Multiple comparisons

The F-test show, that there is a difference — but where? Pairwise t-tests are not suitable due to risk of mass significance Recall a significance level of α = 0.05 means 5% chance of wrongfully rejecting a true hypothesis (type I error) The chance of at least one type I error goes up with the number of tests (for k groups, we have m = k(k − 1)/2 possible tests,

the actual significance level can be as bad as: 1 − (1 − α)m, e.g. for k=5: 0.40)

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ANOVA, April 2009 16

There is no completely satisfactory solution. Approximative solutions:

  • 1. Select a (small) number of relevant comparisons in the planning

stage.

  • 2. Make a graph of the average ±2 × SEM and judge visually (!),

perhaps supplemented with F-tests on subsets of groups.

  • 3. Modify the t-tests by multiplying the P-values with the number
  • f tests, the socalled Bonferroni correction (conservative)
  • 4. Use a correction for multiple testing (Dunnett, Tukey) or a

(prespecified) multiple testing procedure

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ANOVA, April 2009 17

Tukey multiple comparisons in SAS: proc glm data=ex_redcell; class grp; model redcell=grp / solution; lsmeans grp / adjust=tukey pdiff cl; run;

The GLM Procedure Least Squares Means Adjustment for Multiple Comparisons: Tukey-Kramer Least Squares Means for effect grp Pr > |t| for H0: LSMean(i)=LSMean(j) Dependent Variable: redcell i/j 1 2 3 1 0.0355 0.3215 2 0.0355 0.6802 3 0.3215 0.6802 Least Squares Means for Effect grp Difference Simultaneous 95% Between Confidence Limits for i j Means LSMean(i)-LSMean(j) 1 2 60.180556 3.742064 116.619047 1 3 38.625000

  • 27.590379

104.840379 2 3

  • 21.555556
  • 86.340628

43.229517

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ANOVA, April 2009 18

Visual assessment: the bars represent confidence intervals for the means.

proc gplot data=ex_redcell; plot redcell*grp / haxis=axis1 vaxis=axis2 frame; axis1 order=(1 to 3 by 1)

  • ffset=(8,8)

label=(H=3 ’gruppe nr.’) value=(H=2) minor=NONE; axis2

  • ffset=(1,1) value=(H=2) minor=NONE

label=(A=90 R=0 H=3 ’red cell foliate’); symbol1 v=circle i=std2mjt l=1 h=2 w=2; run;

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ANOVA, April 2009 19

Model checking

Check if the assumptions are reasonable: (If not the analysis is unreliable!)

  • Variance homogeneity may be checked by performing

Levenes test (or Bartletts test).

  • In case of variance inhomogeneity, we may also perform a

weighted analysis (Welch’s test), just as in the T-test

  • Normality may be checked through probability plots (or

histograms) of residuals, or by a numerical test on the residuals.

  • In case of non-normality, we may use the nonparametric

Kruskal-Wallis test Transformation (often logarithms) may help to achieve variance homogeneity as well as normality

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ANOVA, April 2009 20

Check of variance homogeneity and normality in SAS

proc glm data=ex_redcell; class grp; model redcell=grp; means grp / hovtest=levene welch;

  • utput out=model p=predicted r=residual;

run; Store residuals in a dataset for further model checking proc univariate normal data=model; var residual; run;

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ANOVA, April 2009 21

Output from proc glm: Test for variance homogeneity

Levene’s Test for Homogeneity of redcell Variance ANOVA of Squared Deviations from Group Means Sum of Mean Source DF Squares Square F Value Pr > F grp 2 18765720 9382860 4.14 0.0321 Error 19 43019786 2264199

and weighted anova in case of variance heterogeneity:

Welch’s ANOVA for redcell Source DF F Value Pr > F grp 2.0000 2.97 0.0928 Error 11.0646

So we are not too sure concerning the group differences.....

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ANOVA, April 2009 22

Output from proc univariate: Test for normality:

Tests for Normality Test

  • -Statistic---
  • ----p Value----

Shapiro-Wilk W 0.965996 Pr < W 0.6188 Kolmogorov-Smirnov D 0.107925 Pr > D >0.1500 Cramer-von Mises W-Sq 0.043461 Pr > W-Sq >0.2500 Anderson-Darling A-Sq 0.263301 Pr > A-Sq >0.2500

The 4 tests focus on different aspects of non-normality.

  • For small data sets, we rarely get significance
  • For large data sets, we almost always get significance
  • Could look at a probability plot instead
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ANOVA, April 2009 23

Non-parametric ANOVA, the Kruskal-Wallis test:

proc npar1way wilcoxon; exact; class grp; var redcell; run;

Wilcoxon Scores (Rank Sums) for Variable redcell Classified by Variable grp Sum of Expected Std Dev Mean grp N Scores Under H0 Under H0 Score

  • 1

8 120.0 92.00 14.651507 15.000000 2 9 77.0 103.50 14.974979 8.555556 3 5 56.0 57.50 12.763881 11.200000 Kruskal-Wallis Test Chi-Square 4.1852 DF 2 Asymptotic Pr > Chi-Square 0.1234 Exact Pr >= Chi-Square 0.1233

Again, we have ’lost’ the significance....

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ANOVA, April 2009 24

Two-way analysis of variance Two criterias for subdividing observations, A og B Data in two-way layout: (not for analysis!!) B A 1 2 · · · c 1 · · · 2 · · · . . . . . . . . . . . . r · · ·

  • Effect of both factors
  • Perhaps even interaction

(effect modification)

One factor may be ’individuals’ or “experimental units” (e.g. different treatments tried on same person)

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ANOVA, April 2009 25

Repeated measurements Example: Short term effect of enalaprilate on heart rate

Time Subject 30 60 120 average 1 96 92 86 92 91.50 2 110 106 108 114 109.50 3 89 86 85 83 85.75 4 95 78 78 83 83.50 5 128 124 118 118 122.00 6 100 98 100 94 98.00 7 72 68 67 71 69.50 8 79 75 74 74 75.50 9 100 106 104 102 103.00 average 96.56 92.56 91.11 92.33 93.14

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ANOVA, April 2009 26

Line plot (“Spaghettiogram”) Ideally the time courses are parallel.

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ANOVA, April 2009 27

Additive model: Yst = µ + αs + βt + εst The two effects (s and t) work in an additive way. The εst’s are assumed to be independent, normally distributed with mean 0, and identical variances, εst ∼ N(0, σ2) (check this!) Variational decomposition: SStotal = SSsubject + SStime + SSresidual

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ANOVA, April 2009 28

Analysis of variance table - enalaprilate example

df SS MS F P Subjects 8 8966.6 1120.8 90.60 <0.0001 Times 3 151.0 50.3 4.07 0.0180 Residual 24 296.8 12.4 Total 35 9414.3 Highly significant difference between subjects (not very interesting) Significant time differences.

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ANOVA, April 2009 29

Two-way ANOVA in SAS:

proc glm data=ex_pulse; class subject times; model hrate=subject times / solution; run;

General Linear Models Procedure Class Level Information Class Levels Values SUBJECT 9 1 2 3 4 5 6 7 8 9 TIMES 4 0 30 60 120 Number of observations in data set = 36

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ANOVA, April 2009 30

Analysis of variance table from output:

General Linear Models Procedure Dependent Variable: HRATE Sum of Mean Source DF Squares Square F Value Pr > F Model 11 9117.52778 828.86616 67.03 0.0001 Error 24 296.77778 12.36574 Corrected Total 35 9414.30556 R-Square C.V. Root MSE HRATE Mean 0.968476 3.775539 3.51650 93.1389 Source DF Type I SS Mean Square F Value Pr > F SUBJECT 8 8966.55556 1120.81944 90.64 0.0001 TIMES 3 150.97222 50.32407 4.07 0.0180 Source DF Type III SS Mean Square F Value Pr > F SUBJECT 8 8966.55556 1120.81944 90.64 0.0001 TIMES 3 150.97222 50.32407 4.07 0.0180

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ANOVA, April 2009 31

Parameter estimates from output:

T for H0: Pr > |T| Std Error of Parameter Estimate Parameter=0 Estimate INTERCEPT 102.1944444 B 50.34 0.0001 2.03024963 SUBJECT 1

  • 11.5000000 B
  • 4.62

0.0001 2.48653783 2 6.5000000 B 2.61 0.0152 2.48653783 3

  • 17.2500000 B
  • 6.94

0.0001 2.48653783 4

  • 19.5000000 B
  • 7.84

0.0001 2.48653783 5 19.0000000 B 7.64 0.0001 2.48653783 6

  • 5.0000000 B
  • 2.01

0.0557 2.48653783 7

  • 33.5000000 B
  • 13.47

0.0001 2.48653783 8

  • 27.5000000 B
  • 11.06

0.0001 2.48653783 9 0.0000000 B . . . TIMES 4.2222222 B 2.55 0.0177 1.65769189 30 0.2222222 B 0.13 0.8945 1.65769189 60

  • 1.2222222 B
  • 0.74

0.4681 1.65769189 120 0.0000000 B . . . NOTE: The X’X matrix has been found to be singular and a generalized inverse was used to solve the normal equations. Estimates followed by the letter ’B’ are biased, and are not unique estimators of the parameters.

  • subject 9 at time 120 minutes is the reference
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ANOVA, April 2009 32

Expected values for subject=3, times=30: ˆ yst = ˆ µ + ˆ αs + ˆ βt = 102.19 − 17.25 + 0.22 = 85.16 Residuals rst =

  • bserved − expected

= yst − ˆ yst ≈ εst Residual for subject 3, time 30: r32 = 86 − 85.16 = 0.84

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ANOVA, April 2009 33

Model checking

Look for:

  • differences in variances (systematic?)
  • Non-normality.
  • Lack of additivity (interaction).

Can only be tested if there is more than one observation for each combination

  • Serial correlation?

(Neighboring observations look more alike)

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ANOVA, April 2009 34

Use the residuals for model checking:

  • Probability plot of residuals.
  • Plot residuals vs expected values.
  • Plot residuals vs group.
  • Look for outliers (a large residual means observed and expected

values deviate a lot).

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ANOVA, April 2009 35

Enalaprilate example:

No systematic patterns should be present.

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ANOVA, April 2009 36

Interaction Example of two criterias for subdividing individuals: sex and smoking habits Outcome: FEV1 Here, we see an interaction between sex and smoking.

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ANOVA, April 2009 37

Possible explanations for interaction:

  • biologically different effects of smoking on males and females
  • perhaps the women do not smoke as much as the men
  • perhaps the effect is relative

(to be expressed in %)

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ANOVA, April 2009 38

Example: The effect of smoking on birth weight

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ANOVA, April 2009 39

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ANOVA, April 2009 40

Interaction:

  • There is an effect of smoking, but only for those who have been

smoking for a long time.

  • There is an effect of duration, and this effects increases with

amount of smoking The effect of duration depends upon.... amount of smoking and the effect of amount depends upon.... duration of smoking

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ANOVA, April 2009 41

Example: Fibrinogen after spleen operation 34 rats are randomized, in 2 ways

  • 17 have their spleen removed (splenectomy=yes/no)
  • 8/17 in each group are kept in high altitude

(place=altitude/control) Outcome: Fibrinogen level in mg at day 21

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ANOVA, April 2009 42

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ANOVA, April 2009 43

The usual additive model: Yspr = µ + αs + βp + εspr, εspr ∼ N(0, σ2) splenectomy (s=yes/no) and place (p=altitude/control) have an additive effect. Model with interaction Yspr = µ + αs + βp + γsp + εspr, εspr ∼ N(0, σ2) Here, we specify an interaction between splenectomy and place, i.e. the effect of living in a high altitude may be thought to depend upon whether or not you have an intact spleen. and vice versa..

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ANOVA, April 2009 44

Two-way ANOVA with interaction in SAS: proc glm data=ex_fibrinogen; class splenectomy place; model fibrinogen=place splenectomy place*splenectomy / solution;

  • utput out=model p=predicted r=residual;

run;

The GLM Procedure Class Level Information Class Levels Values splenectomy 2 no yes place 2 altitude control Number of observations 34

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ANOVA, April 2009 45

Dependent Variable: fibrinogen Sum of Source DF Squares Mean Square F Value Pr > F Model 3 138402.2949 46134.0983 7.51 0.0007 Error 30 184321.2639 6144.0421 Corrected Total 33 322723.5588 R-Square Coeff Var Root MSE fibrinogen Mean 0.428857 22.21804 78.38394 352.7941 Source DF Type I SS Mean Square F Value Pr > F place 1 57895.84355 57895.84355 9.42 0.0045 splenectomy 1 79976.50000 79976.50000 13.02 0.0011 splenectomy*place 1 529.95139 529.95139 0.09 0.7710 Source DF Type III SS Mean Square F Value Pr > F place 1 57895.84355 57895.84355 9.42 0.0045 splenectomy 1 78937.01021 78937.01021 12.85 0.0012 splenectomy*place 1 529.95139 529.95139 0.09 0.7710

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ANOVA, April 2009 46

Standard Parameter Estimate Error t Value Intercept 261.6666667 B 26.12798017 10.01 place altitude 90.5833333 B 38.08774887 2.38 place control 0.0000000 B . . splenectomy no 104.4444444 B 36.95054391 2.83 splenectomy yes 0.0000000 B . . splenectomy*place no altitude

  • 15.8194444 B

53.86421101

  • 0.29

splenectomy*place no control 0.0000000 B . . splenectomy*place yes altitude 0.0000000 B . . splenectomy*place yes control 0.0000000 B . . Parameter Pr > |t| Intercept <.0001 place altitude 0.0240 place control . splenectomy no 0.0083 splenectomy yes . splenectomy*place no altitude 0.7710 splenectomy*place no control . splenectomy*place yes altitude . splenectomy*place yes control . NOTE: The X’X matrix has been found to be singular, and a generalized inverse was used to solve the normal equations. Terms whose estimates are followed by the letter ’B’ are not uniquely estimable.

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ANOVA, April 2009 47

The reference levels are place=control, splenectomy=yes (they come last in the alphabet) so the expected fibrinogen level for these animals is intercept=261.67 For all other groups, we have to add one or more extra estimates, as shown in the table below:

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ANOVA, April 2009 48

place splenectomy control altitude 261.67 261.67 yes + 90.58 = 352.25 261.67 261.67 + 104.44 + 104.44 no + 90.58

  • 15.82

= 366.11 = 440.87

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ANOVA, April 2009 49

Model checking

Variance homogeneity may be judged from a one-way anova:

The GLM Procedure Class Level Information Class Levels Values group 4 no_altitude no_control yes_altitude yes_control Number of observations 34 Levene’s Test for Homogeneity of fibrinogen Variance ANOVA of Squared Deviations from Group Means Sum of Mean Source DF Squares Square F Value Pr > F group 3 2.3669E8 78896856 1.55 0.2211 Error 30 1.5232E9 50773012

No reason to suspect inhomogeneity

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ANOVA, April 2009 50

Normality assumption for residuals (proc univariate normal)

Tests for Normality Test

  • -Statistic---
  • ----p Value------

Shapiro-Wilk W 0.964119 Pr < W 0.3193 Kolmogorov-Smirnov D 0.124882 Pr > D >0.1500 Cramer-von Mises W-Sq 0.094165 Pr > W-Sq 0.1325 Anderson-Darling A-Sq 0.494715 Pr > A-Sq 0.2098

No reason to suspect non-normality

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ANOVA, April 2009 51

In the two-way anova, the interaction was not significant (P=0.77), so we omit it from the model:

proc glm data=ex_fibrinogen; class splenectomy place; model fibrinogen=place splenectomy / solution clparm; run;

Dependent Variable: fibrinogen Sum of Source DF Squares Mean Square F Value Pr > F Model 2 137872.3435 68936.1718 11.56 0.0002 Error 31 184851.2153 5962.9424 Corrected Total 33 322723.5588 R-Square Coeff Var Root MSE fibrinogen Mean 0.427215 21.88815 77.22009 352.7941 Source DF Type III SS Mean Square F Value Pr > F place 1 57895.84355 57895.84355 9.71 0.0039 splenectomy 1 79976.50000 79976.50000 13.41 0.0009

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ANOVA, April 2009 52

Standard Parameter Estimate Error t Value Pr > |t| Intercept 265.3888889 B 22.50900351 11.79 <.0001 place altitude 82.6736111 B 26.53221591 3.12 0.0039 place control 0.0000000 B . . . splenectomy no 97.0000000 B 26.48627265 3.66 0.0009 splenectomy yes 0.0000000 B . . .

  • Removal of spleen leads to a decrease in fibronogen of approx

97.00 mg at day 21

  • Placing in altitude leads to an increase in fibronogen of approx

82.67 mg at day 21

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ANOVA, April 2009 53

Residual plots Normality: Variance homogeneity:

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ANOVA, April 2009 54

More complicated analyses of variances

  • Three- or more-sided analysis of variance.
  • Latin squares

1 2 3 I A B C II B C A III C A B (Cochran & Cox (1957): Experimental Designs, 2.ed., Wiley)

  • Cross-over designs
  • Variance component models
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ANOVA, April 2009 55

Example of a latin square: A rabbit experiment

  • 6 rabbits
  • Vaccination at 6 diffe-

rent spots on the back

  • 6 different orders of

vaccination

  • Swelling is area of

blister (cm2)

spot rabbit order swelling 1 1 3 7.9 1 2 5 8.7 1 3 4 7.4 1 4 1 7.4 . . 6 4 4 5.8 6 5 1 6.4 6 6 3 7.7

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ANOVA, April 2009 56

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ANOVA, April 2009 57

Some illustrations:

1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4 5 5 5 5 5 5 6 6 6 6 6 6 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 3 4 4 4 4 4 4 5 5 5 5 5 5 6 6 6 6 6 6

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ANOVA, April 2009 58

Fit 3-way analysis of variance, with additive effects

proc glm; class rabbit spot order; model swelling=rabbit spot order; run;

The GLM Procedure Class Level Information Class Levels Values rabbit 6 1 2 3 4 5 6 spot 6 a b c d e f

  • rder

6 1 2 3 4 5 6 Number of observations 36

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ANOVA, April 2009 59

Dependent Variable: swelling Sum of Source DF Squares Mean Square F Value Pr > F Model 15 17.23000000 1.14866667 1.75 0.1205 Error 20 13.13000000 0.65650000 Corrected Total 35 30.36000000 R-Square Coeff Var Root MSE swelling Mean 0.567523 10.99883 0.810247 7.366667 Source DF Type III SS Mean Square F Value Pr > F rabbit 5 12.83333333 2.56666667 3.91 0.0124 spot 5 3.83333333 0.76666667 1.17 0.3592

  • rder

5 0.56333333 0.11266667 0.17 0.9701

The design is balanced, so the test of the effect of one variable (covariate) does not depend on which of the others are still in the model.

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ANOVA, April 2009 60

How about possible interactions?

proc glm; class rabbit spot order; model swelling=rabbit spot order spot*order; run;

Dependent Variable: swelling Sum of Source DF Squares Mean Square F Value Pr > F Model 35 30.36000000 0.86742857 . . Error 0.00000000 . Corrected Total 35 30.36000000 Source DF Type I SS Mean Square F Value Pr > F rabbit 5 12.83333333 2.56666667 . . spot 5 3.83333333 0.76666667 . .

  • rder

5 0.56333333 0.11266667 . . spot*order 20 13.13000000 0.65650000 . .

There is no room for interaction, since there is only one

  • bservation for each combination of spot and order!