L7 GLM Binomial

8/3/2019 L7 GLM Binomial

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Lecture 7

GLMs IIBinomial Family

Olivier MISSA, [email protected]

Advanced Research Skills
mailto:[email protected]:[email protected]


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Outline

Continue our Introduction to Generalized Linear Models.

In this lecture:

Illustrate the use of GLMs for

proportion andbinary data.


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Binary & Proportion data tend to follow

the Binomial distribution

The Canonical link of this glm family

is the logitfunction:

The variance reaches a maximum for intermediate valuesof pand a minimum at either 0% or 100%.

Reminder

)1(log

pp

ppnVar 1pnMean


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In R, binary/proportion data can be entered

into a model as a response in three different ways:

as a numeric vector

(holding the number or proportion of successes)

as a logical vector or a factor

(TRUE or the first factor level will be considered successes).

as a two-column matrix(the first column holding the number of successes and

the second column the number of failures).

Three ways to work with binary data


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Toxicity to tobacco budworm (moth) of different doses

of trans-cypermethrin.Batches of 20 moths (of each sex) were put in contact

for three days with increasing doses of the pyrethroid.

1st Example

Dose (micrograms)

Sex 1 2 4 8 16 32

Male 1 4 9 13 18 20

Female 0 2 6 10 12 16

Number of dead mothsout of 20 tested

> (dose numdead (sex SF


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1st Example

> modb summary(modb)

Coefficients:

Estimate Std. Error z value Pr(>|z|)

(Intercept) -1.71578 0.32233 -5.323 1.02e-07 ***

sexM -0.21194 0.51523 -0.411 0.68082

dose 0.11568 0.02379 4.863 1.16e-06 ***sexM:dose 0.18156 0.06692 2.713 0.00666 **

---

(Dispersion parameter for binomial family taken to be 1)

Null deviance: 124.876 on 11 degrees of freedom

Residual deviance: 18.164 on 8 degrees of freedom

AIC: 56.275

What is modelled is the proportion of successes

n

ii

i

iiiiyn

ynynyyyD

1

ln)()/ln(2


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1st Example

> ldose modb2 summary(modb2)

Coefficients:


(Intercept) -2.9935 0.5527 -5.416 6.09e-08 ***

sexM 0.1750 0.7783 0.225 0.822

ldose 0.9060 0.1671 5.422 5.89e-08 ***sexM:ldose 0.3529 0.2700 1.307 0.191

---




AIC: 43.104


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1st Example> drop1(modb2, test="Chisq")

Single term deletions

Model:SF ~ sex * ldose

Df Deviance AIC LRT Pr(Chi)

4.994 43.104

sex:ldose 1 6.757 42.867 1.763 0.1842

> modb3 summary(modb3)

Coefficients:


(Intercept) -3.4732 0.4685 -7.413 1.23e-13 ***

sexM 1.1007 0.3558 3.093 0.00198 **

ldose 1.0642 0.1311 8.119 4.70e-16 ***---




AIC: 42.867


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1st Example> drop1(modb3, test="Chisq")

Single term deletions

Model:

SF ~ sex + ldose

Df Deviance AIC LRT Pr(Chi)

6.757 42.867

sex 1 16.984 51.094 10.227 0.001384 **

ldose 1 118.799 152.909 112.042 < 2.2e-16 ***

> shapiro.test(residuals(modb3), type="deviance")

Shapiro-Wilk normality test

data: residuals(modb3, type = "deviance")

W = 0.9666, p-value = 0.8725


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1st Example

> par(mfrow=c(2,2))

> plot(modb3)


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1st Example> plot( c(0,1) ~ c(1,32), type="n", log="x",

xlab="dose", ylab="Probability")

> text(dose, numdead/20, labels=as.character(sex) )> ld lines (ld, predict(modb3, data.frame(ldose=log2(ld),

sex=factor(rep("M", length(ld)), levels=levels(sex))),

type="response") )

> lines (ld, predict(modb3, data.frame(ldose=log2(ld),

sex=factor(rep("F", length(ld)), levels=levels(sex))),type="response"), lty=2, col="red" )


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1st Example> modbp AIC(modbp)[1] 41.87836

> modbc AIC(modbc)

[1] 43.8663

> AIC(modb3)

[1] 42.86747


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> summary(modb3)

Coefficients:


(Intercept) -3.4732 0.4685 -7.413 1.23e-13 ***

sexM 1.1007 0.3558 3.093 0.00198 **

ldose 1.0642 0.1311 8.119 4.70e-16 ***

---

> exp(modb3$coeff) ## careful it may be misleading

(Intercept) sexM ldose

0.031019 3.006400 2.898560 ## odds ration: p / (1-p)

> exp(modb3$coeff[1]+modb3$coeff[2]) ## odds for males

(Intercept)

0.09325553

1st Example

logit scale

)1(log

p

p

Every doubling of the dose will leadto an increase in the odds of dying

over surviving by a factor of 2.899


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Erythrocyte Sedimentation Rate in a group of patients.

Two groups : 20 (ill) mm/hourQ: Is it related to globulin & fibrinogen level in the blood ?

2nd Example

> data("plasma", package="HSAUR")

> str(plasma)

'data.frame': 32 obs. of 3 variables:

$ fibrinogen: num 2.52 2.56 2.19 2.18 3.41 2.46 3.22 2.21 ...

$ globulin : int 38 31 33 31 37 36 38 37 39 41 ...

$ ESR : Factor w/ 2 levels "ESR < 20","ESR > 20": 1 1 ...

> summary(plasma)

fibrinogen globulin ESR

Min. :2.090 Min. :28.00 ESR < 20:26

1st Qu.:2.290 1st Qu.:31.75 ESR > 20: 6

Median :2.600 Median :36.00

Mean :2.789 Mean :35.66

3rd Qu.:3.167 3rd Qu.:38.00

Max. :5.060 Max. :46.00


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2nd Example> stripchart(globulin ~ ESR, vertical=T, data=plasma,

xlab="Erythrocyte Sedimentation Rate (mm/hr)",

ylab="Globulin blood level", method="jitter" )

> stripchart(fibrinogen ~ ESR, vertical=T, data=plasma,

xlab="Erythrocyte Sedimentation Rate (mm/hr)",

ylab="Fibrinogen blood level", method="jitter" )


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2nd Example> mod1 summary(mod1)

Coefficients:Estimate Std. Error z value Pr(>|z|)

(Intercept) -6.8451 2.7703 -2.471 0.0135 *

fibrinogen 1.8271 0.9009 2.028 0.0425 *

---




AIC: 28.840

> mod2 AIC(mod2)

[1] 28.97111

factor


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2nd Example> anova(mod1, mod2, test="Chisq")

Analysis of Deviance Table

Model 1: ESR ~ fibrinogen

Model 2: ESR ~ fibrinogen + globulin

Resid. Df Resid. Dev Df Deviance P(>|Chi|)

1 30 24.8404

2 29 22.9711 1 1.8692 0.1716

> summary(mod2)

Coefficients:


(Intercept) -12.7921 5.7963 -2.207 0.0273 *

fibrinogen 1.9104 0.9710 1.967 0.0491 *

globulin 0.1558 0.1195 1.303 0.1925---




AIC: 28.971

The difference in terms ofDeviance between thesemodels is not significant,which leads us to select

the least complex model


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2nd Example> shapiro.test(residuals(mod1, type="deviance"))

Shapiro-Wilk normality test

data: residuals(mod1, type = "deviance")

W = 0.6863, p-value = 5.465e-07

> par(mfrow=c(2,2))

> plot(mod1)

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