workshop in r & glms: #2 diane srivastava university of british columbia srivast@zoology.ubc.ca
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Workshop in R & GLMs: #2
Diane Srivastava
University of British Columbia
srivast@zoology.ubc.ca
Start by loading your Lakedata_06 dataset:
diane<-read.table(file.choose(),sep=";",header=TRUE)
Dataframes
Two ways to identify a column (called "treatment") in
your dataframe (called "diane"):
diane$treatment
OR
attach(diane); treatment
At end of session, remember to: detach(diane)
Summary statistics
length (x)
mean (x)
var (x)
cor (x,y)
sum (x)
summary (x) minimum, maximum, mean, median, quartiles
What is the correlation between two variables in your dataset?
Factors
• A factor has several discrete levels (e.g. control,
herbicide)
• If a vector contains text, R automatically assumes it
is a factor.
• To manually convert numeric vector to a factor:
x <- as.factor(x)
• To check if your vector is a factor, and what the
levels are:
is.factor(x) ; levels(x)
ExerciseMake lake area into a factor called AreaFactor:
Area 0 to 5 ha: small
Area 5.1 to 10: medium
Area > 10 ha: large
You will need to:
1. Tell R how long AreaFactor will be.
AreaFactor<-Area; AreaFactor[1:length(Area)]<-"medium"
2. Assign cells in AreaFactor to each of the 3 levels
3. Make AreaFactor into a factor, then check that it is a factor
ExerciseMake lake area into a factor called AreaFactor:
Area 0 to 5 ha: small
Area 5.1 to 10: medium
Area > 10 ha: large
You will need to:
1. Tell R how long AreaFactor will be.
AreaFactor<-Area; AreaFactor[1:length(Area)]<-"medium"
2. Assign cells in AreaFactor to each of the 3 levels
AreaFactor[Area<5.1]<-“small"; AreaFactor[Area>10]<-“large"
3. Make AreaFactor into a factor, then check that it is a factorAreaFactor<-as.factor(AreaFactor); is.factor(AreaFactor)
Linear regression
model <- lm (y ~ x, data = diane)
invent a name for your model
linear model
insert youry vector name here
insert yourx vector name here
insert yourdataframename here
ALT+126
Linear regression
model <- lm (Species ~ Elevation, data = diane)
summary (model)
Call:lm(formula = Species ~ Elevation, data = diane)
Residuals: Min 1Q Median 3Q Max -7.29467 -2.75041 -0.04947 1.83054 15.00270
Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 9.421568 2.426983 3.882 0.000551 ***Elevation -0.002609 0.003663 -0.712 0.482070 ---Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 4.811 on 29 degrees of freedomMultiple R-Squared: 0.01719, Adjusted R-squared: -0.0167 F-statistic: 0.5071 on 1 and 29 DF, p-value: 0.4821
Linear regression
model2 <- lm (Species ~ AreaFactor, data = diane)
summary (model2)
Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 11.833 1.441 8.210 6.17e-09 ***AreaFactormedium -2.405 1.723 -1.396 0.174 AreaFactorsmall -8.288 1.792 -4.626 7.72e-05 ***
Large has mean species richness of 11.8
Medium has mean species richness of 11.8 - 2.4 = 9.4
Small has a mean species richness of 11.8 - 8.3 = 3.5
mean(Species[AreaFactor=="medio"])
ANOVA
model2 <- lm (Species ~ AreaFactor, data = diane)
anova (model2)
Analysis of Variance Table
Response: Species Df Sum Sq Mean Sq F value Pr(>F) AreaFactor 2 333.85 166.92 13.393 8.297e-05 ***Residuals 28 348.99 12.46
F tests in regression
model3 <- lm (Species ~ Elevation + pH, data = diane)
anova (model, model3)
Model 1: Species ~ ElevationModel 2: Species ~ Elevation + pH
Res.Df RSS Df Sum of Sq F Pr(>F) 1 29 671.10 2 28 502.06 1 169.05 9.4279 0.004715 **
F 1, 28 = 9.43
• Fit the model: Species~pH
• Fit the model: Species~pH+pH2
("pH2" is just pH2)
• Use the ANOVA command to decide whether species richness is a linear or quadratic function of pH
Exercise
Distributions: not so normal!
• Review assumptions for parametric stats (ANOVA, regressions)
• Why don’t transformations always work?
• Introduce non-normal distributions
Tests before ANOVA, t-tests
• Normality
• Constant variances
Tests for normality: exercise
data<-c(rep(0:6,c(42,30,10,5,5,4,4)));data
How many datapoints are there?
Tests for normality: exercise
• Shapiro-Wilks (if sig, not normal)
shapiro.test (data)
If error message, make sure the stats package is loaded, then try again:
library(stats); shapiro.test (data)
Tests for normality: exercise
• Kolmogorov-Smirnov (if sig, not normal)
ks.test(data,”pnorm”,mean(data),sd=sqrt(var(data)))
Tests for normality: exercise
• Quantile-quantile plot (if wavers substantially off 1:1 line, not normal)
par(pty="s")
qqnorm(data); qqline(data)opens up a single plot window
Tests for normality: exercise
-2 -1 0 1 2
01
23
45
6
Normal Q-Q Plot
Theoretical Quantiles
Sa
mp
le Q
ua
ntil
es
Tests for normality: exercise
If the distribution isn´t normal, what is it?
freq.data<-table(data); freq.data
barplot(freq.data)
0 1 2 3 4 5 6
01
02
03
04
0
Non-normal distributions• Poisson (count data, left-skewed, variance =
mean)
• Negative binomial (count data, left-skewed, variance >> mean)
• Binomial (binary or proportion data, left-skewed, variance constrained by 0,1)
• Gamma (variance increases as square of mean, often used for survival data)
Exercise
model2 <- lm (Species ~ AreaFactor, data = diane)
anova (model2)
1. Test for normality of residuals
resid2<- resid (model2)
...you do the rest!
2. Test for homogeneity of variances
summary (lm (abs (resid2) ~ AreaFactor))
Regression diagnostics
1. Residuals are normally distributed
2. Absolute value of residuals do not change with predicted value (homoscedastcity)
3. Residuals show no pattern with predicted values (i.e. the function “fits”)
4. No datapoint has undue leverage on the model.
Regression diagnostics
model3 <- lm (Species ~ Elevation + pH, data = diane)
par(mfrow=c(2,2)); plot(model3)
1. Residuals are normally distributed
• Straight “Normal Q-Q plot”
Theoretical Quantiles
Std
. d
evia
nce
res
id.
2. Absolute residuals do not change with predicted values
• No fan shape in Residuals vs fitted plot
• No upward (or downward) trend in Scale-location plot
Fitted values
Sq
rt (
abs
(SD
res
id.)
)
Fitted values
Res
idu
als MALO
BUENO
Examples of neutral and fan-shapes
3. Residuals show no pattern
Curved residual plots result from fitting a straight line to non-linear data (e.g. quadratic)
4. No unusual leverage
Cook’s distance > 1 indicates a point with undue leverage (large change in model fit when removed)
Transformations
Try transforming your y-variable to improve the regression diagnostic plot
• replace Species with log(Species)
• replace Species with sqrt(Species)
Poisson distribution
• Frequency data
• Lots of zero (or minimum value) data
• Variance increases with the mean
1. Correct for correlation between mean and variance by log-transforming y (but log (0) is undefined!!)
2. Use non-parametric statistics (but low power)
3. Use a “generalized linear model” specifying a Poisson distribution
What do you do with Poisson data?
The problem: Hard to transform data to satisfy all requirements!
Tarea: Janka example
Janka dataset: Asks if Janka hardness values are a good estimate of timber density? N=36
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