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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
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
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
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
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
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 6 / 76
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.
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 7 / 76
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
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 9 / 76
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 10 / 76
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 11 / 76
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 12 / 76
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 13 / 76
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
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 15 / 76
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)
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 16 / 76
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 17 / 76
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 18 / 76
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)
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 19 / 76
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 20 / 76
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 21 / 76
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)
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 22 / 76
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 23 / 76
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
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 25 / 76
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 26 / 76
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 27 / 76
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 28 / 76
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)
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 29 / 76
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 30 / 76
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 31 / 76
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
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 33 / 76
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 34 / 76
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 35 / 76
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
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 37 / 76
Etude de correlation
variablesquantitatives
variablesqualitatives
test du χ2
chisq.test()
test dePearson
cor.test()
Regressionlineaire
lm()
test deSpearman
cor.test()
normalite
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 38 / 76
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 39 / 76
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 40 / 76
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")●
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70 80 90 100 110 120 130
6080
100
120
140
smoke$Smoking
smok
e$M
orta
lity
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 41 / 76
Test de correlation de Pearson
Utilisation de la fonction cor.test()
x,y les valeurs des deux variables
method ”pearson”
> cor.test(smoke$Mortality, smoke$Smoking, method = "pearson")
Pearson's product-moment correlation
data: smoke$Mortality and smoke$Smokingt = 4.9222, df = 23, p-value = 5.658e-05alternative hypothesis: true correlation is not equal to 095 percent confidence interval:0.4478559 0.8662224sample estimates:
cor0.7162398
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 42 / 76
Regression lineaire
Rappels de stats
1 Construire un modele lineaire de la forme Y = aX + b et test ducoefficient de determination
2 Conditions d’application : normalite des variables, lien lineaire entreles variables
3 Hypotheses :
HO : R2 == 0H1 : R2 6= 0
4 Statistique : sous H0, F = R2(n−2)1−r2 suit une loi de Fischer-Snedecors a
1 et n-2 ddl
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 43 / 76
Regression lineaire
Utilisation de la fonction lm()
formula la formule du modele lineaire
data le tableau de donnee (si les noms de colonnes sont utilisesdans formula
> model = lm(smoke$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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 44 / 76
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 45 / 76
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 46 / 76
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)
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 47 / 76
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
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Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 48 / 76
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 49 / 76
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 50 / 76
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 51 / 76
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")))
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 52 / 76
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 53 / 76
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 54 / 76
Comparaison de distributions
2 distributionsquelconques
test de normalite
test deShapiro
shapiro.test()
test deKolmogorov-
Smirnov
ks.test()
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 55 / 76
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
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 56 / 76
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.
Loıc PONGER ([email protected]) Les test statistiques elementaires avec R 19 fevrier 2010 57 / 76
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
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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)
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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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