BIO503: Lecture 4Statistical models in R

Stefan Bentink

bentink@jimmy.harvard.edu

Statistical tests in R

Just some examples:

> t.test()

> pairwise.t.test()

> chisq.test()

> fisher.test()

> ks.test()

> …

One sample t-test

> data(ChickWeight)> t.test(ChickWeight[, 1], mu = 100)

One Sample t-test

data: ChickWeight[, 1]

t = 7.3805, df = 577, p-value = 5.529e-13

alternative hypothesis: true mean is not equal to 100

95 percent confidence interval:

116.0121 127.6246

sample estimates:

mean of x

121.8183

Two sample t-test

> t.test(ChickWeight$weight[ChickWeight$Diet == "1"],

+ ChickWeight$weight[ChickWeight$Diet =="2"])

Welch Two Sample t-test

data: ChickWeight$weight[ChickWeight$Diet == "1"] and ChickWeight$weight[ChickWeight$Diet == "2"]

t = -2.6378, df = 201.384, p-value = 0.008995

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval:

-34.899942 -5.042482

sample estimates:

mean of x mean of y

102.6455 122.6167

Linear Regression Models

residual error

regression coefficient

dependent variable

intercept independent variable

Using the methods of least squares, we can derive the following estimators:

Our goal is to test the hypothesis: 0^

We can do this with a T test:

)(

0^

^

SEt

under the null hypothesis, this follows a T distribution with (n-1) df.

Installing ISwR Package

Please install the ISwR package on your computer.

This package contains all the data sets used in Peter Dalgaard's book Introductory Statistics with R.

To load the package into your current R session:

> library(ISwR)

To find out more information, including what objects are contained in a package:

> library(help=ISwR)

Example Dataset tlc

We'll be using the dataset tlc that exists in ISwR package.

To load this dataset:

> data(tlc) # tlc = total lung capacity

What kind of object is tlc?

> class(tlc)

To learn about this dataset:

> help(tlc)

By using the attach command, we release the columns of the data.frame into the workspace.

> attach(tlc)

> age

Linear Regression with lm

Is there a linear relationship between height and Total Lung Capacity (TLC)?

> lmObject <- lm(tlc ~ height, data=tlc)

What kind of object is lmObject?

> class(lmObject)

The model object represents an object that encapsulates the results of a model fit.

The desired quantities can be obtained using extractor functions.

A basic extractor function is summary:

> summary(lmObject)

Interpreting the Output from lm

Call:... Stores the call to the lm function. Handy for keeping track of what model you fit.

Residuals:...

Coefficients:...

Summary stats of the residuals from the model. Recall: residuals should follow approximately a standard Normal distribution if the model is a good fit.

Estimates from the model. Standard error. T statistics. P-value.

Residual standard error:...

22

22

n

xy

n

SS iiRES

Residual variance

22

^^

n

SS

n

RES

Plug in estimates, to get RSE:

Interpreting the Output from lm

Multiple R-squared:...

Pearson correlation coefficient2 of y, x (e.g. tlc, height).

Do cor(tlc[,4], height))^2 to check.

F statistic: This is the F test that the regression coefficient is zero.

Adjust R-squared:...

Visualizing Our Fitted Model

So what does it really mean, to have fit a linear model of TLC and height?

Plot the data:

> TLC <- tlc[,4]

> plot(height, TLC)

Add the regression line we fit with our model:

> abline(lmObject)

Values Fitted by the Linear Model

Retrieve the fitted values using the extractor function fitted:

> fitted(lmObject)

Convince yourself that these are the points that fall on the line we just made in the previous plot.

> plot(height, TLC)

> abline(lmObject)

> points(height, fitted(lmObject), pch=20, col="red")

To grab the residual values:

> resid(lmObject)

Eyeballing Good Fit

We can use the information from the residuals and fitted values to create plots to see how good our model fit is.

> plot(height, TLC)

> abline(lmObject)

Use the segments function to draw vertical lines from the fitted values to the real data values.

> segments(height, fitted(lmObject), height, TLC)

We can also take a look at the residuals.

> plot(height, resid(lmObject))

And use a QQ plot to see if the residuals are normally distributed.

> qqnorm(resid(lmObject))

> qqline(resid(lmObject))

Confidence Interval BandsConfidence bands are added to regression lines to reflect the

uncertainty about a true parameter that cannot be observed i.e. the true regression coefficient .

The more narrow the confidence band is, suggests a well-determined line.

Using the predict function without any other input arguments yields the fitted values predicted by the model.

> predict(lmObject)

To compute the confidence interval bands for the fitted values, you need to specify the interval argument:

> predict(lmObject, interval="confidence")

Visualizing Confidence Interval Bands

Go ahead and plot these values:

> pp <- predict(lmObject, interval="confidence")

> plot(height, pp[,1], type="l", ylim=range(pp))

> lines(height, pp[,2], col="red", lwd=3)

> lines(height, pp[,3], col="blue", lwd=3)

What's the problem?

Predicting on New Data

Our problem will be solved if height was ordered sequentially.

One solution: predict fitted values on new (ordered) height values.

Create a sequence of numbers that go from min(height) to max(height) approximately.

> range(height)

> new <- data.frame(height = seq(from=120, to=200, by=2))

Compute new predictions:

> pp.new <- predict(lmObject, new, interval="confidence")

Plotting Confidence Interval Bands

Now we can make the plot. First the fitted values: > plot(new$height, pp.new[,1], type="l")

Then the upper interval band:> lines(new$height, pp.new[,2], col="red",

lwd=2)

Finally the lower interval band:> lines(new$height, pp.new[,3], col="blue",

lwd=2)

Prediction Interval Bands

Prediction interval bands reflect the uncertainty about future observations.

Prediction bands are generally wider than the confidence interval bands.

> predict(lmObject, interval="prediction")

Plot the fitted data with the prediction interval bands and confidence interval bands superimposed.

> pp.pred <- predict(lmObject, new, interval="prediction")

> pp.conf <- predict(lmObject, new, interval="confidence")

Visualizing Fitted Values, Prediction and Confidence BandsNote: instead of using lines individually, we can also use the

function matlines which plots columns of a matrix. > help(matlines)

> plot(new$height, pp.pred[,1], ylim=range(pp.pred, pp.conf))

Add the prediction bands:> matlines(new$height, pp.pred[,-1], type=c("l",

"l"), lwd=c(3,3), col=c("red", "blue"), lty=c(1,1))

Add the confidence bands:> matlines(new$height, pp.conf[,-1], type=c("l",

"l"), lwd=c(3,3),col=c("red", "blue"), lty=c(2,2))

Tutorial 1

ANOVA Models

The (one-way) ANOVA model generalizes linear regression. – One factor that has G levels. – For each level we have J replicates.

ijiijY j = 1

j = JJ re

pli

cate

s

i = 1 i = Gi = 2 …

…

G levels

ANOVA Model:

Analysis of Variance Models

j = 1

j = JJ re

pli

cate

s

i = 1 i = Gi = 2 …

…

G levels

1

_

x Gx_

ix_

_

x

ijx

11x

i j

iB xxSSD2__

i j

iijW xxSSD2_

i j

ijTOTALWB xxSSDSSDSSD2_

Total variation:

Variation between groups:

Variation within groups:

ANOVA is all about splitting variation up.

ANOVA ModelsQuestion: is there a significant relationship between our factor A

and the response variable Y? If yes, then ideally the variation within groups should be small. The variation between groups should be large.

Variation between groups:

Variation within groups:

1B

SSDMS B

B

1W

SSDMS W

W

Our test statistic:W

B

MS

MSF

General idea:Under the null hypothesis that the factor A has no effect: F = 1.Large values of F indicate the factor is significant.

B = # levels

W = n - B

ANOVA ExampleLet's use a different dataset:

> library(MASS)

> data(ChickWeight)

> attach(ChickWeight)

The factor Diet has 4 levels.

> levels(Diet)

> anova(lm(weight ~ Diet, data=ChickWeight))

Analysis of Variance Table

Response: weight

Df Sum Sq Mean Sq F value Pr(>F)

Diet 3 155863 51954 10.81 6.433e-07

Residuals 574 2758693 4806

Two-way ANOVAWe can fit a two-way ANOVA:

> anova(lm(weight ~ Diet + Chick, data=ChickWeight))

Analysis of Variance Table

Response: weight

Df Sum Sq Mean Sq F value Pr(>F)

Diet 3 155863 51954 11.5045 2.571e-07

Chick 46 374243 8136 1.8015 0.001359

Residuals 528 2384450 4516

The interpretation of the model output is sequential, from the bottom to the top.

This line tests the model: weight ~ Diet + Chick

This line tests the model: weight ~ Diet vs weight ~ Diet + Chick.

Tutorial 2

Multiple Linear Regression

Multiple explanatory variables to explain a response variable.

Can I explain the values of the response variable by the levels of the explanatory variables?

Do I need all explanatory variables to explain the response variable?

Specifying ModelsIn R we use model formula to specify the model we want to fit to

our data. y ~ x Simple Linear Regressiony ~ x – 1 Simple Linear Regression

without the intercept (line goes through origin)

y ~ x1 + x2 + x3 Multiple Regressiony ~ x + I(x^2) Quadratic Regressionlog(y) ~ x1 + x2 Multiple Regression of

Transformed VariableFor factors A, B:y ~ A 1-way ANOVA y ~ A + B 2-way ANOVAy ~ A*B 2-way ANOVA + interaction term

Fit multiple regression model to a data.frame

> ##Get data.frame

> cfseal <- read.table("cfseal.txt", header=T,

+ sep="\t")

> heart.log <- log(cfseal$heart)

> cfseal.log <- cfseal

> cfseal.log[,1] <- heart.log

> colnames(cfseal.log)[1] <- "heart.log"

> ##Fit model

> seal.lm <- lm(heart.log ~ ., data=cfseal.log)

Update models and model selection Some handy functions to know about:

new.model <- update(old.model, new.formula)

Model Selection functions available in the MASS package

drop1, dropterm

add1, addterm

step, stepAIC

Similarly,

anova(modObj, test="Chisq")

Generalized Linear Models

Linear regression models hinge on the assumption that the response variable follows a Normal distribution.

Generalized linear models are able to handle non-Normal response variables and transformations to linearity.

Logistic Regression

When faced with a binary response Y = (0,1), we use logistic regression.

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Logistic regression

Problem 2 – Logistic Regression

Read in the anaesthetic data set, data file: anaesthetic.txt.

Covariates:

move binary numeric vector for patient movement

(1 = movement, 0 = no movement)

conc anaethestic concentration

Goal: estimate how the concentration of movement varies with increasing concentration of the anesthetic agent.

Fit the Logistic Regression Model

> anes.logit <- glm(move ~ conc, family=binomial(link=logit), data=anesthetic)

The output summary looks like this: > summary(anes.logit)

Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) 6.469 2.418 2.675 0.00748 **conc -5.567 2.044 -2.724 0.00645 **

Estimates of P(Y=1) are given by: > fitted.values(anes.logit)

Estimating Log Odds Ratio

To get back the log odds ratio

> anes.logit$linear.predictors

> plot(anesthetic$conc, anes.logit$linear.predictors)

> abline(coefficients(anes.logit))

Looks like the odds of not moving increase significantly when you increase the concentration of the anesthetic agent beyond 0.8.

Problem 3 – Multiple Logistic RegressionRead in data set birthwt.txt.

low indicator of birth weight less than 2.5kg age mother's age in years lwt mother's weight in pounds at last menstrual period race mother's race (1 = white, 2 = black, 3 = other) smoke smoking status during pregnancy ptl number of previous premature labours ht history of hypertension ui presence of uterine irritability ftv number of physician visits during the first trimester bwt birth weight in grams

We fit a logistic regression using the glm function and using the binomial family.

Problem 4 - Poisson Regression

Poisson regression is often used for the analysis of count data or the calculation of rates associated with a rare event or disease.

Example: schooldata.csv.

We can fit the Poisson regression model using the glm function and the poisson family.

Survival Analysis

library(survival)

Example: aml leukemia data

Kaplan-Meier curve

fit1 <- survfit(Surv(aml$time[1:11],aml$status[1:11])~1)

summary(fit1)

plot(fit1)

Log-rank test

survdiff(Surv(time, status)~x, data=aml)

Survival analysis

> cp <- coxph(Surv(aml$time,

+ aml$status)~x,data=aml)

>

> summary(cp)

>

> plot(survfit(Surv(aml$time,aml$status)~x,

+ data=aml),col=c("red","green"),lwd=2)