assumptions of the anova

Assumptions of the ANOVA The error terms are randomly, independently, and

normally distributed, with a mean of zero and a common variance.– There should be no systematic patterns among the residuals– The distribution of residuals should be symmetric (not skewed)– there should be no relationship between the size of the error

variance and the mean for different treatments or blocks– The error variances for different treatment levels or different

blocks should be homogeneous (of similar magnitude) The main effects are additive

– the magnitude of differences among treatments in one block should be similar in all other blocks

– i.e., there is no interaction between treatments and blocks

If the ANOVA assumptions are violated:

Affects sensitivity of the F testSignificance level of mean comparisons may

be much different than they appear to beCan lead to invalid conclusions

Diagnostics Use descriptive statistics to test assumptions

before you analyze the data– Means, medians and quartiles for each group

(histograms, box plots)– Tests for normality, additivity– Compare variances for each group

Examine residuals after fitting the model in your analysis– Descriptive statistics of residuals– Normal plot of residuals– Plots of residuals in order of observation– Relationship between residuals and predicted values

(fitted values)

SAS Box Plots

Look For

Outliers Skewness Common

Variance

Caution

Not many observations per group

Mean

Median

Outlier (>1.5*IQR)

Quartile (25%)Min

IQR

IQR = interquartile range (25% - 75%)

Additivity

Yij = + i + ij CRD

Yij = + i + j + ij RBD

Linear additive model for each experimental design

Implies that a treatment effect is the same for all blocks and that the block effect is the same for all treatments

When the assumption would not be correct...

Water Table

When there is an interaction between blocks and treatments - the model is no longer additive– may be multiplicative; for example, when one treatment

always exceeds another by a certain percentage

Two nitrogen treatments applied to 3 blocks

1 2 3

Differences between treatments might be greater in block 3

SAS interaction plot

Test is applicable to any two-way classification such as RBD classified by blocks and treatments

Testing Additivity --- Tukey’s test

Compute SS for nonadditivity = (Q2*N)/(SST*SSB) with 1 df

The error term is partitioned into nonadditivity and residual and can be tested with F

Compute a table with raw data, treatment means, treatment effects ( ), block means and block effects ( )

..j. YY

...i YY

ij i. .. . j ..Q Y Y Y Y Y N = t*r

Test can also be done with SAS

Residuals Residuals are the error terms – what is left over after

accounting for all of the effects in the model

ij ij ie Y Y.. T

ij ij i je Y Y.. B T

Yij = + i + ij CRDYij = + i + ij CRD

Yij = + i + j + ij RBD

Independence Independence implies that the error (residual) for

one observation is unrelated to the error for another– Adjacent plots are more similar than randomly

scattered plots– So the best insurance is randomization– In some cases it may be better to throw out a

randomization that could lead to biased estimates of treatment effects

– Observations in a time series may be correlated (and randomization may not be possible)

Normality Look at stem leaf plots, boxplots of residuals Normal probability plots Minor deviations from normality are not

generally a problem for the ANOVA

Normality

-3 -2 -1 0 1 2 3

-150

-100

-50

0

50

100

150

200

250

300

350

Normal Probability Plot from Original Data

Quantiles of standard normal

Res

idua

ls

NormalityNormal Probability Plot from the same Data

After Transformation

-4

-3

-2

-1

0

1

2

3

-3 -2 -1 0 1 2 3

Quantiles of standard normal

Res

idua

ls

Homogeneity of Variances

Logic would tell us that differences required for significance would be greater for the two highly variable treatments

ReplicatesTreatment 1 2 3 4 5 Total Mean s2

A 3 1 5 4 2 15 3 2.5 B 6 8 7 4 5 30 6 2.5 C 12 6 9 3 15 45 9 22.5 D 20 14 11 17 8 70 14 22.5

If we analyzed together:

Source df SS MS FTreatments 3 330 110 8.8**Error 16 200 12.5

LSD=4.74

Source df SS MS FTreatments 1 22.5 22.5 9*Error 8 20.0 2.5

Source df SS MS FTreatments 1 62.5 62.5 2.78 Error 8 180 22.5

Analysis for A and B

Analysis for C and D

Conclusions would be different if we analyzed the two groups separately:

Test the effect of a new vitamin on the weights of animals.

What you see What the ANOVA assumes

Relationships of Means and Variances Most common cause of heterogeneity of variance

Take each observation and remove the general mean, the treatment effects and the block effects; what is left will be the error term for that observation

The model = Block effect =

Treatment effect =

so ... then ... Finally ...

Examining the error terms

..ij i j ijY Y e

i. ..i Y Y

.j ..j Y Y

.. i. .. . j ..ij ijY Y Y Y Y Y e

i. . j ..ij ijY Y Y Y e

i. . j ..ij ije Y Y Y Y

Looking at the error components

Trt. I II III IV Mean A 47 52 62 51 53 B 50 54 67 57 57 C 57 53 69 57 59 D 54 65 74 59 63 Mean 52 56 68 56 58

Trt. I II III IV Mean A 0 1 -1 0 0 B -1 -1 0 2 0 C 4 -4 0 0 0 D -3 4 1 -2 0Mean 0 0 0 0 0

e11 = 47 – 52 – 53 + 58 = 0

Looking at the error components

Trt. I II III IV Mean A .18 .30 .28 .44 0.3 B .32 .4 .42 .46 0.4 C 2.0 3.0 1.8 2.8 2.4 D 2.5 3.3 2.5 3.3 2.9 E 108 140 135 165 137 F 127 153 148 176 151 Mean 40 50 48 58 49

Trt. I II III IV A 8.88 -1.00 0.98 -8.86 B 8.92 -1.00 1.02 -8.94 C 8.60 -0.40 0.40 -8.60 D 8.60 -0.60 0.60 -8.60 E -20.00 2.00 -1.00 19.00 F -15.00 1.00 2.00 16.00

e11 = 0.18 – 40 - 0.3 + 49 = 8.88

Predicted values

Remember ..ij i j ijY Y e

i. . j .. ijY Y Y e

Predicted value ..ij i jY Y

i. . j ..Y Y Y

Plots of ije vs ijY should be random

Plots of ije vs ijY will be autocorrelated

Residual Plots A valuable tool for examining the validity of assumptions

for ANOVA – should see a random scattering of points on the plot

For simple models, there may be a limited number of groups on the Predicted axis

Look for random dispersion of residuals above and below zero

Residual Plots – Outlier Detection Recheck data input and correct obvious errors If an outlier is suspected, could look at studentized

residuals (ij)

Treat as a missing plot if too extreme (e.g. ij > 3 or 4)

ij

ijij

e

es

ij

ie

i

r 1s MSEr

For a CRD

Predicted Values

Residuals (eij)

Outliers

Visual Scores Values are discrete

– Do not follow a normal distribution– Range of possible values is limited

Alternatives?

Residual Plot of Stand Ratings

Predicted Values3 4 5 6 7 80

1020

Plant Stand Ratings

Score

Freq

uenc

y

Are the errors randomly distributed? Residuals are not randomly distributed around zero

– they follow a pattern Model may not be adequate

– e.g., fitting a straight regression line when response is curvilinear

Model not adequate

-1.5

-1

-0.5

0

0.5

1

1.5

0 5 10 15 20 25

Predicted values

Resi

dual

s

Are variances homogeneous?

-160

-120

-80

-40

0

40

80

120

160

0 20 40 60 80 100 120 140 160

Predicted values

Resi

dual

s

In this example the variance of the errors increases with the mean (note fan shape)

Cannot assume a common variance for all treatments

Homogeneity Quick Test (F Max Test) By examining the ratio of the largest variance to

the smallest and comparing with a probability table of ratios, you can get a quick test.

The null hypothesis is that variances are equal, so if your computed ratio is greater than the table value (Kuehl, Table VIII), you reject the null hypothesis.

(min)(max)

2

2

ss Where t = number of independent

variances (mean squares) that you are comparingv = degrees of freedom associated with each mean square

An Example

An RBD experiment with four blocks to determine the effect of salinity on the application of N and P on sorghum

5437.99/9.03 = 602.21 Table value (t=7, v=r-1=3) = 72.9602.21>72.9Reject null hypothesis and conclude that variances are NOT homogeneous (equal)

Treatment Variance N 19.54 P 1492.27

N+P 98.21 S 5437.99

N+S 9.03 P+S 496.58

N+P+S 22.94

Other HOV tests are more sensitive If the quick test indicates that variances are not equal

(homogeneous), no need to test further But if quick test indicates that variances ARE

homogeneous, you may want to go further with a Levene (Med) test or Bartlett’s test which are more sensitive.

This is especially true for values of t and v that are relatively small.

F max, Levene (Med), and Bartlett’s tests can be adapted to evaluate homogeneity of error variances from different sites in multilocational trials.

Homogeneity of Variances - Tests

Johnson (1981) compared 56 tests for homogeneity of variance and found the Levene (Med) test to be one of the best. – Based on deviations of observations from the median

for each treatment group. Test statistic is compared to a critical F,t-1,N-t value.

– This is now the default homogeneity of variance test in SAS (HOVTEST).

Bartlett’s test is also common– Based on a chi-square test with t-1 df– If calculated value is greater than tabular value, then

variances are heterogeneous

What to do if assumptions are violated? Divide your experiment into subsets of blocks or

treatments that meet the assumptions and conduct separate analyses

Transform the data and repeat the analysis– residuals follow another distribution (e.g., binomial, Poisson)– there is a specific relationship between means and variances– residuals of transformed data must meet the ANOVA assumptions

Use a nonparametric test– no assumptions are made about the distribution of the residuals– most are based on ranks – some information is lost– generally less powerful than parametric tests

Use a Generalized Linear Model (PROC GLIMMIX in SAS)– make the model fit the data, rather than changing the data to fit the

model

Relationships between means and variances...

Can usually tell just by looking. Do the variances increase as the means increase?

If so, construct a table of ratios of variance to means and standard deviation to means

Determine which is more nearly proportional - the ratio that remains more constant will be the one more nearly proportional

This information is necessary to know which transformation to use – the idea is to convert a known probability distribution to a normal distribution

M-C 0.3 0.01147 0.107 0.04 0.36M-V 0.4 0.00347 0.059 0.01 0.15C-C 2.4 0.3467 0.589 0.14 0.24C-V 2.9 0.2133 0.462 0.07 0.16S-C 137.0 546.0 23.367 3.98 0.17S-V 151.0 425.3 20.624 2.82 0.14

Trt Mean Var SDev Var/M SDev/M

Comparing Ratios - Which Transformation?

SDev roughly proportional to the means

The Log Transformation When the standard deviations (not the variances) of

samples are roughly proportional to the means, the log transformation is most effective

Common for counts that vary across a wide range of values– numbers of insects– number of diseased plants/plot

Also applicable if there is evidence of multiplicative rather than additive main effects– e.g., an insecticide reduces numbers of insects by 50%– e.g., early growth of seedlings may be proportional to current size

of plants

General remarks... Data with negative values cannot be

transformed with logs Zeros present a special problem If negative values or zeros are present, add 1 to

all data points before transforming You can multiply all data points by a constant

without violating any rules Do this if any of the data points are less than 1

(to avoid negative logs)

Recheck... After transformation, rerun the ANOVA on the

transformed data Recheck the transformed data against the

assumptions for the ANOVA– Look at residual plots, normal plots– Carry out Levene’s test or Bartlett’s for homogeneity of variance– Apply Tukey’s test for additivity

Beware that a transformation that corrects one violation in assumptions may introduce another

Square Root Transformation One of a family of power transformations The variance tends to be proportional to the mean

– e.g., if leaf length is normally distributed, then leaf area may require a square root transformation

Use when you have counts of rare events in time or space– number of insects caught in a trap

May follow a Poisson distribution (for discrete variables) If there are counts under 10, it is best to use square root

of Y + 0.5 Will be easier to declare significant differences in mean

separation When reporting, “detransform” the means – present

summary mean tables on original scale

Arcsin or Angular Transformation

Counts expressed as percentages or proportions of the total sample may require transformation

Follow a binomial distribution - variances tend to be small at both ends of the range of values ( close to 0 and 100%)

Not all percentage data are binomial in nature– e.g., grain protein is a continuous, quantitative variable

that would tend to follow a normal distribution If appropriate, it usually helps in mean separation

ijYarcsin

Arcsin or Angular Transformation Data should be transformed if the range of percentages is

greater than 40 May not be necessary for percentages in the range of 30-

70% If percentages are in the range of 0-30% or 70-100%, a

square root transformation may be better Do not include treatments that are fixed at 0% or at 100% Percentages are converted to an angle expressed in

degrees or in radians – express Yij as a decimal fraction – gives results in radians– 1 radian = 57.296 degrees

ijYarcsin

Summary of TransformationsType of data Issue Transformation Quantitative variable where treatment effects are proportional

Natural scale is not normal (often right- skewed) Lognormal distribution

Log

Positive integers that cover a wide range of values

Standard deviation proportional to the mean and/or nonadditivity

Log or log(Y+1)

Counts of rare events Variance = mean Poisson distribution

Square root or sqrt(Y+0.5)

Percentages, wide range of values including extremes

Variances are smaller near zero and 100 Binomial distribution

ArcSin

Reasons for Transformation We don’t use transformation just to give us

results more to our liking We transform data so that the analysis will be

valid and the conclusions correct Remember ....

– all tests of significance and mean separation should be carried out on the transformed data

– calculate means of the transformed data before “detransforming”