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