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Les test statistiques ´ el´ ementaires avec R version de travail ... pleine de coquilles Lo¨ ıc PONGER [email protected] USM 503 - R´ egulation et Dynamique des G´ enomes Mus´ eum National d’Histoire Naturelle 43 rue Cuvier 75005 Paris 19 f´ evrier 2010 Lo¨ ıc PONGER ([email protected]) Les test statistiques ´ el´ ementaires avec R 19 f´ evrier 2010 1 / 76

Les test statistiques el ementaires avec R - edu.mnhn.fredu.mnhn.fr/pluginfile.php/5617/mod_resource/content/0/J3-Tests... · normalit e ou taille echantillon normalit e homosc edasticit

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Page 1: Les test statistiques el ementaires avec R - edu.mnhn.fredu.mnhn.fr/pluginfile.php/5617/mod_resource/content/0/J3-Tests... · normalit e ou taille echantillon normalit e homosc edasticit

Les test statistiques elementaires avec Rversion de travail ... pleine de coquilles

Loıc PONGER

[email protected] 503 - Regulation et Dynamique des Genomes

Museum National d’Histoire Naturelle43 rue Cuvier75005 Paris

19 fevrier 2010

Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 1 / 76

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variable(s) quantitative(s)comparaison de moyennes/medianes

2 echantillons

1 echantillonet une valeur

theorique

3 echantil-lons ou plus

test deStudent/Welch

t.test()

test deWilcoxon

wilcox.test()

test deStudent

t.test()

test deWilcoxon

wilcox.test()

ANOVAaov()

TukeyHSD()

pairwise.t.test()

test deKruskal-Walliskruskal.test()

pairwise.wilcox.test()

normalite ou taille echantillon

normalite

homoscedasticite

Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 2 / 76

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Plan

1 Comparaison de moyennes et de medianesTest de Student

Test de Student pour un echantillonTest de Student pour deux echantillons appariesTest de Student pour deux echantillons independants

Test de Wilcoxon (Mann-Whitney)Test de Wilcoxon pour un echantillonTest de Wilcoxon pour deux echantillons appariesTest de Wilcoxon pour deux echantillons independants

Analyse de variance (ANOVA)Test de Kruskal-Wallis

2 Correlation et independance

3 Comparaison de distributions (normalite)

4 Comparaison de variances

5 A la decouverte des tests sous RLoıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 3 / 76

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Plan

1 Comparaison de moyennes et de medianesTest de Student

Test de Student pour un echantillonTest de Student pour deux echantillons appariesTest de Student pour deux echantillons independants

Test de Wilcoxon (Mann-Whitney)Test de Wilcoxon pour un echantillonTest de Wilcoxon pour deux echantillons appariesTest de Wilcoxon pour deux echantillons independants

Analyse de variance (ANOVA)Test de Kruskal-Wallis

2 Correlation et independance

3 Comparaison de distributions (normalite)

4 Comparaison de variances

5 A la decouverte des tests sous RLoıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 4 / 76

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Test de Student pour un echantillon

Rappels de stats

1 Comparaison d’une moyenne observee a une valeur theorique

2 Conditions d’application : X doit etre distribuee selon une loi normale(theoreme central limite).

X suit une loi normale oun est grand

3 Hypotheses :

HO : µ == µ0

H1 : µ 6= µ0 (”two.sided”), µ < µ0 (”less”) ou µ > µ0 (”greater”)

4 Statistique : sous HO , ...

... t = x−µ0√s

n−1

suit une loi de Student a n − 1 ddl

Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 5 / 76

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Test de Student pour un echantillon

Exemple

On a mesure la glycemie (en g/L) chez 21 patients (fichier gly.dat). Est-ceque le taux de glucose de ces patients differe de la valeur de reference, 1g/L ?

> insulin = read.table("gly.dat")> hist(insulin[, 1])

Histogram of insulin[, 1]

insulin[, 1]

Fre

quen

cy

0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4

01

23

45

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Test de Student pour un echantillon

Utilisation de la fonction t.test()

x les donnees, un vecteur ou une liste

mu une valeur theorique

alternative test unilateral ou bilateral

> t.test(insulin, mu = 1, alternative = "two.sided")

One Sample t-test

data: insulint = 3.2957, df = 20, p-value = 0.003612alternative hypothesis: true mean is not equal to 195 percent confidence interval:1.045847 1.203962sample estimates:mean of x1.124905

Note : n etant faible, il serait necessaire de tester la normalite de la distribution.

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Test de Student pour deux echantillons apparies

Rappels de stats

1 Comparaison des moyennes de deux echantillons apparies.

2 Preambule : calcul des differences de toutes les paires (Y = X1 − X2),calcul de la moyenne des differences (Y ), calcul de la variance desdifferences(s2

Y )

3 Condition d’application : Y doit etre distribuee selon une loi normale(theoreme central limite).

Y suit une loi normale oun est grand

4 Hypotheses :

HO : µ1 == µ2

H1 : µ1 6= µ2 (”two.sided”), µ1 < µ2 (”less”) ou µ1 > µ2 (”greater”)

5 Statistique : sous HO , ...

... t = y−0qsy

n−1

suit une loi de Student a n − 1 ddlLoıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 8 / 76

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Test de Student pour deux echantillons apparies

Exemple

On a mesure le poid de 46 jeunes filles anorexiques avant et apres untraitement (fichier anorexic.dat, donnees issues de Larry Winner’s website). La moyenne passe de 82,89 lb a 87,47 lb (1 lb = 0,45 kg). Est-ce

que le traitement a un effet significatif sur le poids des jeunes filles ?

> anx = read.table("anorexic.dat", head = T, sep = ",")> boxplot(anx[, c(3, 4)])

before after

7075

8085

9095

100

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Test de Student pour deux echantillons apparies

Utilisation de la fonction t.test()

x les valeurs de X1, un vecteur ou une liste

y les valeurs de X2, un vecteur ou une liste

paired TRUE

alternative test unilateral ou bilateral

> t.test(anx$before, anx$after, paired = T, alt = "less")

Paired t-test

data: anx$before and anx$aftert = -4.16, df = 45, p-value = 7.063e-05alternative hypothesis: true difference in means is less than 095 percent confidence interval:

-Inf -2.731276sample estimates:mean of the differences

-4.580435

Note : n etant eleve, le test de normalite n’est pas necessaire

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Test de Student pour deux echantillons independants

Rappels de stats

1 Conditions d’utilisation : X1 et X2 doivent etre distribuees selon uneloi normale (theoreme central limite).

X1 et X2 suivent une loi normale oun1 et n2 sont grands

2 Hypotheses :

HO : µ1 == µ2

H1 : µ1 6= µ2 (”two.sided”), µ1 < µ2 (”less”) ou µ1 > µ2 (”greater”)

3 Statistique : Sous H0, ...

Si les variances sont egales (test de Student sensus stricto) ,

t = x1−x2

sq

( 1n1

+ 1n2

)avec s =

√(n1−1)s2

1 +(n2−1)s22

(n1+n2−2) suit une loi de Student a

n1 + n2 − 2ddlSi les variances sont differentes (test de Welch), t = x1−x2

s avec

s =√

s21

n1+

s22

(n2suit une loi de Student a

(s21 /n1+s2

2 /n2)2

(s21 /n1)2/(n1−1)+(s2

2 /n2)2/(n2−1)ddl

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Test de Student pour deux echantillons independants

Exemple

Des chercheurs etudient la phylogenie des canides en comparant desdonnees morphometriques (largeur de la mandibule, en cm) de chiensmodernes de Thaılande a celles des loups indiens (fichier loup.dat). Cesdonnees suggerent-elles une difference de la largeur de la mandibule entreles chiens thaıs et les loups indiens ?

> loups = read.table("loups.dat", head = T)> boxplot(loups)

Thai Loup

910

1112

13

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Test de Student pour deux echantillons independants

Utilisation de la fonction t.test()

x valeurs de X1

y valeurs de X2

paired FALSE (car independants)

var.equal TRUE (test de Student) ou FALSE (test de Welch)

alternative test unilateral ou bilateral

> t.test(loups$Thai, loups$Loup, var.equal = F, alternative = "two.sided")

Welch Two Sample t-test

data: loups$Thai and loups$Loupt = -3.7733, df = 54.639, p-value = 0.0003984alternative hypothesis: true difference in means is not equal to 095 percent confidence interval:-1.5503324 -0.4746676sample estimates:mean of x mean of y10.61562 11.62813

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Plan

1 Comparaison de moyennes et de medianesTest de Student

Test de Student pour un echantillonTest de Student pour deux echantillons appariesTest de Student pour deux echantillons independants

Test de Wilcoxon (Mann-Whitney)Test de Wilcoxon pour un echantillonTest de Wilcoxon pour deux echantillons appariesTest de Wilcoxon pour deux echantillons independants

Analyse de variance (ANOVA)Test de Kruskal-Wallis

2 Correlation et independance

3 Comparaison de distributions (normalite)

4 Comparaison de variances

5 A la decouverte des tests sous RLoıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 14 / 76

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Test de Wilcoxon pour un echantillon

Rappels de stats

1 Comparaison d’une mediane observee et d’une valeur theorique2 Hypotheses :

HO : med == med0

H1 : med 6= med0 (”two.sided”), med < med0 (”less”) ou med > med0

(”greater”)

3 Statistique :

A chaque Xi , on associe sa valeur absolue Zi = |Xi |On classe les Zi et a chaque Zi , on associe son rang Ri

On calcule V =∑

Ri pour tous les i tel que Xi > 0Sous H0 et n petit, V suit une loi de distribution connue (dependant den)

Sous H0 et n grand, Z = V−E(V )sqrt(V (V )) suit une loi normale centree reduite

avec n = n1 + n2, E(V ) =n(n+1)

4(somme totale des rangs :

n(n+1)2

) et V (V ) =n(n+1)(2n+1)

24

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Test de Wicoxon pour un echantillon

Exemple

Hollander et Wolfe (1973) ont mesure un indice de depression (Hamiltonindex) chez 9 patients victimes de depression et d’anxiete avant et apres ledebut d’une therapie (administration de tranquillisants). Pour chaquepatient, les chercheurs ont calcule la difference ”(avant − apres)”. Leschercheurs attendent donc une baisse de l’indice ((avant − apres) > 0) encas de reussite. En revanche, en cas d’absence d’effet, la difference doitetre nulle ((avant − apres) == 0).

> avant.moins.apres = c(0.952, -0.147, 1.022, 0.43, 0.62, 0.59,+ 0.49, -0.08, 0.01)

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Test de Wilcoxon pour un echantillon

Utilisation de la fonction wilcox.test()

x les donnees, un vecteur ou une liste

mu une valeur theorique

alternative test unilateral ou bilateral

exact pour n petit, calcule la p-value selon la table, sinon approx.normale

> wilcox.test(avant.moins.apres, mu = 0, alternative = "greater",+ conf.int = T)

Wilcoxon signed rank test

data: avant.moins.apresV = 40, p-value = 0.01953alternative hypothesis: true location is greater than 095 percent confidence interval:0.175 Infsample estimates:(pseudo)median

0.46

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Test de Wilcoxon pour deux echantillons apparies

Rappels de stats

1 Comparaison des medianes de deux echantillons apparies

2 Hypotheses :

HO : med1 == med2

H1 : med1 6= med2 (”two.sided”), med1 < med2 (”less”) oumed1 > med2 (”greater”)

3 Statistique :

On calcule la difference entre les elements de chaque paireXi = Ai − Bi puis on compare les differences a 0 (test de Wilcoxonpour un echantillon).On calcule V =

∑Ri pour tous les i tel que Xi > 0

Sous H0 et n petit, V suit une loi de distribution connue (dependant den)

Sous H0 et n grand, Z = V−E(V )sqrt(V (V )) , avec E (V ) = n(n+1)

4 (somme totale

des rangs : n(n+1)2 )et V (V ) = n(n+1)(2n+1)

24 , suit une loi normale centreereduite

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Test de Wilcoxon pour deux echantillons apparies

Exemple

Hollander et Wolfe (1973) ont mesure un indice de depression (Hamiltonindex) chez 9 patients victimes de depression et d’anxiete avant et apres ledebut d’une therapie (administration de tranquilisants). Les chercheursattendent une baisse de l’indice en cas de reussite.

> avant <- c(1.83, 0.5, 1.62, 2.48, 1.68, 1.88, 1.55, 3.06, 1.3)> apres <- c(0.878, 0.647, 0.598, 2.05, 1.06, 1.29, 1.06, 3.14,+ 1.29)

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Test de Wilcoxon pour deux echantillons apparies

Utilisation de wilcox.test()

x les valeurs du premier echantillon, un vecteur ou une liste

y les valeurs du second echantillon, un vecteur ou une liste

paired echantillons apparies

alternative test unilateral ou bilateral

exact pour n petit, calcule la p-value selon la table, sinon approx.normale

> wilcox.test(avant, apres, paired = T, alternative = "greater")

Wilcoxon signed rank test

data: avant and apresV = 40, p-value = 0.01953alternative hypothesis: true location shift is greater than 0

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Test de Wilcoxon pour deux echantillons independants

Rappels de stats

1 Comparaison des medianes de deux echantillons independants2 Hypotheses :

HO : med1 == med2

H1 : med1 6= med2 (”two.sided”), med1 < med2 (”less”) oumed1 > med2 (”greater”)

3 Statistique :

On reunit et on ordonne les valeurs de X1 et de X2. A chaque valeur,on associe son rang.On calcule W la somme des rangs des valeurs de X1

Sous H0 et n petit, W suit une loi de distribution connue (dependantde n)

Sous H0 et n grand, Z = W−E(W )sqrt(V (X )) suit une loi normale centree reduite

avec n = n1 + n2, E(W ) =n(n+1)

4(somme totale des rangs :

n(n+1)2

) et V (W ) =n(n+1)(2n+1)

24

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Test de Wicoxon pour deux echantillons independants

Exemple

Des chercheurs ont compte le nombre moyen d’echange de chromatidessoeurs chez des individus issus de deux population differentes (Margolin,1988).

> native.American = c(8.5, 9.48, 8.65, 8.16, 8.83, 7.76, 8.63)> caucasian = c(8.27, 8.2, 8.25, 8.14, 9, 8.1, 7.2, 8.32, 7.7)

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Test de Wilcoxon pour deux echantillons independants

Utilisation de wilcox.test()

x les valeurs du premier echantillon, un vecteur ou une liste

y les valeurs du second echantillon, un vecteur ou une liste

alternative test unilateral ou bilateral

exact pour n petit, calcule la p-value selon la table, sinon approx.normale

> wilcox.test(native.American, caucasian, alternative = "two.sided")

Wilcoxon rank sum test

data: native.American and caucasianW = 47, p-value = 0.1142alternative hypothesis: true location shift is not equal to 0

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Plan

1 Comparaison de moyennes et de medianesTest de Student

Test de Student pour un echantillonTest de Student pour deux echantillons appariesTest de Student pour deux echantillons independants

Test de Wilcoxon (Mann-Whitney)Test de Wilcoxon pour un echantillonTest de Wilcoxon pour deux echantillons appariesTest de Wilcoxon pour deux echantillons independants

Analyse de variance (ANOVA)Test de Kruskal-Wallis

2 Correlation et independance

3 Comparaison de distributions (normalite)

4 Comparaison de variances

5 A la decouverte des tests sous RLoıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 24 / 76

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L’analyse de variance

Rappels de stats

1 Comparaison de k moyennes issues de k echantillons independants

2 Hypotheses :

HO : toutes les moyennes sont egalesH1 : au moins deux moyennes sont differentes

3 Statistique :

Calcul des variances intergroupe et intragroupe (ou residuelle)Calcul du rapport F = inter/intraSous H0, F suit une loi de Fischer a k − 1 et n − k ddl

4 Validation du modele :

normalite des residushomoscedasticite des residus

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L’analyse de variance

Exemple

On a mesure la longueur des œufs de coucous presents dans les nids de 6especes d’oiseaux. Y a t-il une difference de longueur ? Si oui, pourquelle(s) espece(s) ? Les donnees sont dans le fichier cuckoo.dat.

> cuckoo = read.table("cuckoo.dat", head = T)

HDGE_SPRW PIED_WTAIL TREE_PIPIT

2021

2223

2425

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L’analyse de variance

Utilisation de la fonction aov()

formula la formule a modeliser

data le tableau de donnees (si formula contient les noms descolonnes)

> model = aov(cuckoo$length ~ cuckoo$species)> model = aov(length ~ species, cuckoo)> summary(model)

Df Sum Sq Mean Sq F value Pr(>F)species 5 42.940 8.588 10.388 3.152e-08 ***Residuals 114 94.248 0.827---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

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L’analyse de variance

Validation du modele

verifier la normalite des residus

verifier l’homoscedasticite des residus

> qqnorm(model$res)> shapiro.test(model$res)

Shapiro-Wilk normality test

data: model$resW = 0.9804, p-value = 0.07762

> bartlett.test(model$res, cuckoo$species)

Bartlett test of homogeneity of variances

data: model$res and cuckoo$speciesBartlett's K-squared = 4.4794, df = 5, p-value = 0.4826

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−2 −1 0 1 2

−2

−1

01

2

Normal Q−Q Plot

Theoretical Quantiles

Sam

ple

Qua

ntile

s

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Test de Tukey HSD

Rappels de stats

1 Comparaison multiple de moyennes, correction pour les comparaisonsmultiples (α)

2 Conditions d’application : normalite et homoscedasticite des variables

3 Hypothese : H0 : les moyennes sont egales

4 Statistique : sous H0, Qa,b = max(Xa,Xb)−min(Xa,Xb)SE suit une loi des

etendues studentisees avec SE, l’ecart type des variables etudiees (ecart type residuel)

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Test de Tukey HSD

Utilisation de TukeyHSD()

x le model de l’ANOVA

> TukeyHSD(model)

Tukey multiple comparisons of means95% family-wise confidence level

Fit: aov(formula = length ~ species, data = cuckoo)

$speciesdiff lwr upr p adj

MDW_PIPIT-HDGE_SPRW -0.82253968 -1.629133605 -0.01594576 0.0428621PIED_WTAIL-HDGE_SPRW -0.21809524 -1.197559436 0.76136896 0.9872190ROBIN-HDGE_SPRW -0.54642857 -1.511003196 0.41814605 0.5726153TREE_PIPIT-HDGE_SPRW -0.03142857 -1.010892769 0.94803563 0.9999990WREN-HDGE_SPRW -1.99142857 -2.970892769 -1.01196437 0.0000006PIED_WTAIL-MDW_PIPIT 0.60444444 -0.181375330 1.39026422 0.2324603ROBIN-MDW_PIPIT 0.27611111 -0.491069969 1.04329219 0.9021876TREE_PIPIT-MDW_PIPIT 0.79111111 0.005291337 1.57693089 0.0474619WREN-MDW_PIPIT -1.16888889 -1.954708663 -0.38306911 0.0004861ROBIN-PIED_WTAIL -0.32833333 -1.275604766 0.61893810 0.9155004TREE_PIPIT-PIED_WTAIL 0.18666667 -0.775762072 1.14909541 0.9932186WREN-PIED_WTAIL -1.77333333 -2.735762072 -0.81090459 0.0000070TREE_PIPIT-ROBIN 0.51500000 -0.432271433 1.46227143 0.6159630WREN-ROBIN -1.44500000 -2.392271433 -0.49772857 0.0003183WREN-TREE_PIPIT -1.96000000 -2.922428738 -0.99757126 0.0000006

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Test de Tukey HSD

> plot(TukeyHSD(model))

−3 −2 −1 0 1

WR

EN

−T

RE

E_P

IPIT

TR

EE

_PIP

IT−

MD

W_P

IPIT

95% family−wise confidence level

Differences in mean levels of species

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Plan

1 Comparaison de moyennes et de medianesTest de Student

Test de Student pour un echantillonTest de Student pour deux echantillons appariesTest de Student pour deux echantillons independants

Test de Wilcoxon (Mann-Whitney)Test de Wilcoxon pour un echantillonTest de Wilcoxon pour deux echantillons appariesTest de Wilcoxon pour deux echantillons independants

Analyse de variance (ANOVA)Test de Kruskal-Wallis

2 Correlation et independance

3 Comparaison de distributions (normalite)

4 Comparaison de variances

5 A la decouverte des tests sous RLoıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 32 / 76

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Test de Kruskal-Wallis

Rappels de stats

1 Comparaison de k medianes

2 Hypothese :

HO : toutes les medianes sont egalesH1 : aux moins deux medianes sont differentes

3 Statistique : sous H0, H = 12N∗(N+1) ∗

∑ R2i

ni− 3 ∗ (n + 1) suit

approximativement une loi de χ2 a k-1 ddl

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Test de Kruskal-Wallis

Exemple

On etudie la production de bouchons de bouteilles issus de trois machinesdifferentes (donnees artificielles, Kruskal et Wallis, 1952).

> standard = c(340, 345, 330, 342, 338)> modif1 = c(339, 333, 344)> modif2 = c(347, 343, 349, 355)> bouchons = c(340, 345, 330, 342, 338, 339, 333,+ 344, 347, 343, 349, 355)> machines = c(rep(1, 5), rep(2, 3), rep(3, 4))

1 2 3

330

335

340

345

350

355

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Test de Kruskal-Wallis

Utilisation de la fonction kruskal.test()

x une liste de vecteurs, un vecteur avec toutes les valeurs

y les groupes (si x est un vecteur avec toutes les valeurs)

> kruskal.test(list(standard, modif1, modif2))

Kruskal-Wallis rank sum test

data: list(standard, modif1, modif2)Kruskal-Wallis chi-squared = 5.6564, df = 2, p-value = 0.05912

> kruskal.test(bouchons, machines)

Kruskal-Wallis rank sum test

data: bouchons and machinesKruskal-Wallis chi-squared = 5.6564, df = 2, p-value = 0.05912

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Test de Wilcoxon (comparaison multiple)

Analyse post-hoc

Utilisation de Wilcoxon pour la comparaison de chaques paires de variablesavec une correction pour les comparaisons multiples (α)

g les valeurs numeriques

x le facteur (avec plus de deux modalites)

alternative test unilateral ou bilateral

p.adjust correction pour les tests multiples

> pairwise.wilcox.test(bouchons, machines, p.adjust = "bonferroni")Pairwise comparisons using Wilcoxon rank sum test

data: bouchons and machines

1 22 1.000 -3 0.095 0.343

P value adjustment method: bonferroni

Note : etape inutile dans cet exemple car le test de Kruskal-Wallis n’est pas

significatif (a 5 %) ! !Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 36 / 76

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Plan

1 Comparaison de moyennes et de medianes

2 Correlation et independanceTest de correlation de Pearson et regression lineaireTest de SpearmanTest du χ2

3 Comparaison de distributions (normalite)

4 Comparaison de variances

5 A la decouverte des tests sous R

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Etude de correlation

variablesquantitatives

variablesqualitatives

test du χ2

chisq.test()

test dePearson

cor.test()

Regressionlineaire

lm()

test deSpearman

cor.test()

normalite

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Plan

1 Comparaison de moyennes et de medianes

2 Correlation et independanceTest de correlation de Pearson et regression lineaireTest de SpearmanTest du χ2

3 Comparaison de distributions (normalite)

4 Comparaison de variances

5 A la decouverte des tests sous R

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Test de correlation de Pearson

Rappels de stats

1 test la presence d’une correlation lineaire entre deux variables

2 Conditions d’application : normalite des variables, lien lineaire entreles variables

3 Hypotheses :

HO : r == 0H1 : r 6= 0

4 Statistique : sous H0, t = rq1−r2

n−2

suit une loi de Student a n-2 ddl

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Test de correlation de Pearson

Exemple

On a mesure la consomation moyenne en cigarettes et la mortalite dans 25differentes categories professionnelles (Moore et al., 1989). Y a t-il unecorrelation entre la consommation en cigarette et la mortalite ?

> smoke = read.table("smoke.dat", head = T, sep = "\t")●

70 80 90 100 110 120 130

6080

100

120

140

smoke$Smoking

smok

e$M

orta

lity

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Test de correlation de Pearson

Utilisation de la fonction cor.test()

x,y les valeurs des deux variables

method ”pearson”

> cor.test(smoke$Mortality, smoke$Smoking, method = "pearson")

Pearson's product-moment correlation

data: smoke$Mortality and smoke$Smokingt = 4.9222, df = 23, p-value = 5.658e-05alternative hypothesis: true correlation is not equal to 095 percent confidence interval:0.4478559 0.8662224sample estimates:

cor0.7162398

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Regression lineaire

Rappels de stats

1 Construire un modele lineaire de la forme Y = aX + b et test ducoefficient de determination

2 Conditions d’application : normalite des variables, lien lineaire entreles variables

3 Hypotheses :

HO : R2 == 0H1 : R2 6= 0

4 Statistique : sous H0, F = R2(n−2)1−r2 suit une loi de Fischer-Snedecors a

1 et n-2 ddl

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Regression lineaire

Utilisation de la fonction lm()

formula la formule du modele lineaire

data le tableau de donnee (si les noms de colonnes sont utilisesdans formula

> model = lm(smoke$Mortality ~ smoke$Smoking)> summary(model)

Call:lm(formula = smoke$Mortality ~ smoke$Smoking)

Residuals:Min 1Q Median 3Q Max

-30.107 -17.892 3.145 14.132 31.732

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

(Intercept) -2.8853 23.0337 -0.125 0.901smoke$Smoking 1.0875 0.2209 4.922 5.66e-05 ***---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 18.62 on 23 degrees of freedomMultiple R-squared: 0.513, Adjusted R-squared: 0.4918F-statistic: 24.23 on 1 and 23 DF, p-value: 5.658e-05

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Regression lineaire

Utilisation des fonctions lm() et anova()

formula la formule du modele lineaire

data le tableau de donnee (si les noms de colonnes sont utilisesdans formula

> model = lm(smoke$Mortality ~ smoke$Smoking)> anova(model)

Analysis of Variance Table

Response: smoke$MortalityDf Sum Sq Mean Sq F value Pr(>F)

smoke$Smoking 1 8395.7 8395.7 24.228 5.658e-05 ***Residuals 23 7970.3 346.5---Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

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Plan

1 Comparaison de moyennes et de medianes

2 Correlation et independanceTest de correlation de Pearson et regression lineaireTest de SpearmanTest du χ2

3 Comparaison de distributions (normalite)

4 Comparaison de variances

5 A la decouverte des tests sous R

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Test de correlation de Spearman

Rappels de stats

1 Tester la presence d’une correlation entre deux variables2 Hypotheses :

HO : r == 0H1 : r 6= 0, r < 0 ou r > 0

3 Statistique : sous H0, ...

..., si n est petit, r = 1− 6 ∗P

d2i

n(n2−1) suit un loi determinee.

..., si n est grand, Z = r−E(r)√V (r)

suit une loi normale centree-reduite.

avec di etant la difference de rang entre les xi et les yi , E(r) = 0 et V (r) = 1/(n − 1)

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Test de correlation de Spearman

Exemple

On a recense les positions sur la ligne de depart et sur la ligne d’arrivee dechaque concurrents durant les courses de la Winston Cup (Nascar) entre1975 et 2003. Y a t-il un lien entre les deux variables ?

> nascar = read.table("nascard.dat", head = T)

0 10 20 30 40 50 60

010

2030

4050

starting

finis

hing

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Test de Spearman (Kendall)

utilisation de cor.test()

x, y les valeurs de x et de y

method ”spearman”

alternative test unilateral ou bilateral

> cor.test(nascar$finishing, nascar$starting, method = "spearman")

Spearman's rank correlation rho

data: nascar$finishing and nascar$startingS = 4.452649e+12, p-value < 2.2e-16alternative hypothesis: true rho is not equal to 0sample estimates:

rho0.3706519

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Plan

1 Comparaison de moyennes et de medianes

2 Correlation et independanceTest de correlation de Pearson et regression lineaireTest de SpearmanTest du χ2

3 Comparaison de distributions (normalite)

4 Comparaison de variances

5 A la decouverte des tests sous R

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Test du χ2

Rappels de stats

1 Tester la correlation ou l’independance de deux variables quantitativesayant n et p modalites

2 Conditions d’application : les Efftheo doivent etre superieurs a 5, sinonfaire des simulations (ou voir le test exact de Fischer)

3 Hypotheses :

HO : les deux variables sont independantesH1 : les deux variables ne sont pas independantes

4 Statistique : χ2 =∑ (Effobs−Efftheo )2

Efftheosuit une loi de χ2 a

(n − 1) ∗ (p − 1) ddl

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Test du χ2

Exemple

On aimerait savoir si la possession d’une brouette depend du milieuprofessionnel (donnees de S. Ballesteros).

sociologues banquiers archeologues

avec brouette 37 36 12 85

sans brouette 65 43 7 115

102 79 19 200

> socio <- matrix(c(37, 65, 36, 43, 12, 7), 2, 3, dimnames = list(c("Avec brouette",+ "Sans brouette"), c("Sociologue", "Banquier", "Archeologue")))

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Test du χ2

utilisation de chisq.test()

x une table ou un vecteur de type facteur

y un vecteur de type facteur (si x et un vecteur)

simulate.p.value FALSE ou TRUE si Efftheo < 5,

B nombre de simulations

> chisq.test(socio)

Pearson's Chi-squared test

data: socioX-squared = 5.2402, df = 2, p-value = 0.0728

> chisq.test(socio)$expected

Sociologue Banquier ArcheologueAvec brouette 43.35 33.575 8.075Sans brouette 58.65 45.425 10.925

> chisq.test(socio)$residuals

Sociologue Banquier ArcheologueAvec brouette -0.9644488 0.4185080 1.381238Sans brouette 0.8291626 -0.3598026 -1.187487

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Plan

1 Comparaison de moyennes et de medianes

2 Correlation et independance

3 Comparaison de distributions (normalite)Test de Shapiro-WilkTest Kolmogorov-Smirnov

4 Comparaison de variances

5 A la decouverte des tests sous R

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Comparaison de distributions

2 distributionsquelconques

test de normalite

test deShapiro

shapiro.test()

test deKolmogorov-

Smirnov

ks.test()

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Plan

1 Comparaison de moyennes et de medianes

2 Correlation et independance

3 Comparaison de distributions (normalite)Test de Shapiro-WilkTest Kolmogorov-Smirnov

4 Comparaison de variances

5 A la decouverte des tests sous R

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Test de Shapiro-Wilk

Principe

1 Tester la normalite d’une distribution

2 Hypotheses :

HO : La distribution des X suit une loi normaleH1 : La distribution des X ne suit pas une loi normale

3 Statistique :

les valeurs sont ordonnees (xi , valeur de rang i),pour chaque xi , une valeur ai correspondant a la valeur attendue sousl’hyp. H0 est calculee.

La statistique du test est : W =(Pn

i=1 ai xi )2Pn

i=1(xi−x)2

(le rapport des etendues partielles et des carres des ecarts a lamoyenne)Cette statistique est liee au graphique quantile-quantile. Plus W estpetit, plus la distribution de la variable X s’eloigne d’une distributionnormale.

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Test de Shapiro-Wilk

Exemple

On a mesure la glycemie (en g/L) chez 21 patients (fichier gly.dat).Est-ce que le taux de glucose de ces patients suit une loi normale ? Lanormalite est ,necessaire pour faire ensuite un test de Student.

> insulin = read.table("gly.dat")> hist(insulin[, 1])

Histogram of insulin[, 1]

insulin[, 1]

Fre

quen

cy

0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4

01

23

45

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Test de Shapiro-Wilk

Utilisation de shapiro.test()

x les donnees, un vecteur ou une liste

> shapiro.test(insulin[, 1])

Shapiro-Wilk normality test

data: insulin[, 1]W = 0.9411, p-value = 0.2296

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Plan

1 Comparaison de moyennes et de medianes

2 Correlation et independance

3 Comparaison de distributions (normalite)Test de Shapiro-WilkTest Kolmogorov-Smirnov

4 Comparaison de variances

5 A la decouverte des tests sous R

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Test de Kolmogorov-Smirnov

Principes

1 Ce test consiste a calculer la difference maximale existant entre lesdistributions de frequences relatives cumulees (dfrc) de deuxechantillons

2 Hypotheses :

HO : les dfrc de X1 et de X2 sont identiquesH1 : les dfrc de X1 et de X2 sont differentes, la dfrc de X − 1 est ”plusfaible”que celle de X2, la dfrc de X − 1 est ”plus elevee”que celle de X2

3 Statistique : Sous H0, la statistique est : D = max(FX 1 − FX 2)

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Test de de Kolmogorov-Smirnov

8 9 10 11 12 13 14

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ecdf loups vs. thai

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Test de de Kolmogorov-Smirnov

Utilisation de la fonction ks.test()

Pour comparer deux distributions observees

x valeurs de X1

y valeurs de X2

alternative test unilateral ou bilateral

> ks.test(loups$Thai, loups$Loup)

Two-sample Kolmogorov-Smirnov test

data: loups$Thai and loups$LoupD = 0.4375, p-value = 0.004375alternative hypothesis: two-sided

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Test de de Kolmogorov-Smirnov

Utilisation de ks.test()

Pour comparer une distribution observee et une distribution theorique (loinormale) : Attention : les valeurs des parametres de la loi theorique ne doiventpas etre estimes a partir de la distribution observees ! ! Ceci pose un problemepour le test de normalite (voir le test de Lilliefors, library nortest).

x valeurs de X1

y loi theorique (”pnorm”, ”pt”, ...)

... parametre(s) de la loi (moyenne, variance, ...)

alternative test unilateral ou bilateral

> ks.test(loups$Thai, "pnorm", 10, 5)

One-sample Kolmogorov-Smirnov test

data: loups$ThaiD = 0.3817, p-value = 0.0001786alternative hypothesis: two-sided

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Plan

1 Comparaison de moyennes et de medianes

2 Correlation et independance

3 Comparaison de distributions (normalite)

4 Comparaison de variancesTest de Fischer-SnedecorTest de Bartlett

5 A la decouverte des tests sous R

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Plan

1 Comparaison de moyennes et de medianes

2 Correlation et independance

3 Comparaison de distributions (normalite)

4 Comparaison de variancesTest de Fischer-SnedecorTest de Bartlett

5 A la decouverte des tests sous R

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Test de Fischer-Snedecor

Rappels de stats

1 Comparer les variances de deux echantillons

2 Hypotheses :

HO : σ1 == σ2

H1 : σ1 6= σ2

3 Statistique : Sous H0, ...... F = σ1/σ2 suit une loi de Fischer a n1 − 1 et n2 − 1 ddl

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Test de Fischer-Snedecor

Exemple

Des chercheurs etudient la phylogenie des canides en comparant desdonnees morphometriques de chiens modernes de Thaılande a celles desloups indiens (fichier loup.dat). La premiere colonne correspond a lalargeur de la mandibule (en cm) sous la premiere molaire des chiens thaıs,et l’autre a des loups indiens. Est-ce que les mesures des deux especespresentent des differences de variances ?

> loups = read.table("loups.dat", head = T)> boxplot(loups)

Thai Loup

910

1112

13

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Test de Fischer-Snedecor

Utilisation de la fonction var.test()

x, y les donnees de X1 et de X2, un vecteur ou une liste

ratio 1 par defaut mais cela peut etre modifie

alternative bilateral ou unilateral

> var.test(loups$Thai, loups$Loup)

F test to compare two variances

data: loups$Thai and loups$LoupF = 2.1598, num df = 31, denom df = 31, p-value = 0.03562alternative hypothesis: true ratio of variances is not equal to 195 percent confidence interval:1.054303 4.424580sample estimates:ratio of variances

2.159826

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Plan

1 Comparaison de moyennes et de medianes

2 Correlation et independance

3 Comparaison de distributions (normalite)

4 Comparaison de variancesTest de Fischer-SnedecorTest de Bartlett

5 A la decouverte des tests sous R

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Test de Bartlett

Principe

1 Comparer les variances de k echantillons

2 Condition d’application : les variables doivent etre distribuees selon laloi normale

3 Hypothese :

HO : les k σi sont egauxH1 : au moins deux σi sont differents

4 Statistique : Sous H0, ...

... χ2 = ... suit une loi du chi2 a k − 1 ddl

Note : il existe d’autres tests (Levene, Log-anova,Cochran, ...)

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Test de Bartlett

Exemple

On a mesure la longueur des œufs de coucous presents dans les nids de 6especes d’oiseaux. Y a t-il une difference de variance entre les groupes ?

> cuckoo = read.table("cuckoo.dat", head = T)

HDGE_SPRW PIED_WTAIL TREE_PIPIT

2021

2223

2425

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Test de Bartlett

Utilisation de la fonction bartlett.test()

x,g les valeurs des Xi et les groupes correspondant

formula xg

data si les noms des colonnes sont utilises dans formula

> bartlett.test(cuckoo$length ~ cuckoo$species)

Bartlett test of homogeneity of variances

data: cuckoo$length by cuckoo$speciesBartlett's K-squared = 4.4794, df = 5, p-value = 0.4826

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Plan

1 Comparaison de moyennes et de medianes

2 Correlation et independance

3 Comparaison de distributions (normalite)

4 Comparaison de variances

5 A la decouverte des tests sous R

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Pour trouver un test sous R

Premier reflexe : help.search(”Kolmogorov”)

Sinon, le web avec les mots clefs : ”R”, ”cran”

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Pour en apprendre plus sur un test

Premier reflexe ? Regarder la documentation ...

pour verifier la nature du test ;

pour connaitre les limites d’utilisation (lire l’aide au moins une fois) ;

pour connaıtre les donnees en entree ;

pour connaıtre les differents arguments (options) et les valeurs pardefaut ;

pour connaıtre l’objet retourne par la fonction ;

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