Zahn: Verdienst und Geschlecht
Rand-Mittelwerte, gewichtet und ungewichtet.
dd <- read.delim("http://md.psych.bio.uni-goettingen.de/data/div/zahn-unbal-paper.txt")
# descriptives
require(Rmisc)## Loading required package: Rmisc## Loading required package: lattice## Loading required package: plyr# weighted means
mg <- Rmisc::summarySE(data = dd, "Salary", groupvars=c("Gender"))
mg## Gender N Salary sd se ci
## 1 Female 12 22.33333 4.271115 1.232965 2.713737
## 2 Male 10 22.10000 3.695342 1.168570 2.643489me <- Rmisc::summarySE(data = dd, "Salary", groupvars=c("Education"))
me## Education N Salary sd se ci
## 1 Degree 11 25.54545 1.809068 0.5454545 1.215348
## 2 No_degree 11 18.90909 2.211540 0.6668044 1.485733# mm <- Rmisc::summarySE(data = dd, "Salary", groupvars=c("Gender", "Education"))
# unweighted means
mm <- Rmisc::summarySE(data = dd, "Salary", groupvars=c("Gender", "Education"))
mm[1:2, 4]## [1] 25 17mean(mm[1:2, 4])## [1] 21mm[3:4, 4]## [1] 27 20mean(mm[3:4, 4])## [1] 23.5# compare unweighted means
# Gender: women vs men
mean(mm[1:2, 4]); mean(mm[3:4, 4])## [1] 21## [1] 23.5# Education: degree vs no degree
mean(mm[c(1,3), 4]); mean(mm[c(2,4), 4])## [1] 26## [1] 18.5# rm(mm)
# todo table| Gender | Degree | No degree | Total |
|---|---|---|---|
| female | n=8 \(\bar{x}\)=25 (sd=1.51) | n=4 \(\bar{x}\)=17 (sd=2.16) | n=12 \(\bar{x}_w\)=22.33 (sd=4.27) \(\bar{x}_u\)=21 |
| male | n=3 \(\bar{x}\)=27 (sd=2) | n=7 \(\bar{x}\)=20 (sd=1.41) | n=10 \(\bar{x}_w\)=22.1 (sd=3.7) \(\bar{x}_u\)=23.5 |
| total | n=11 \(\bar{x}_w\)=25.55 (sd=0) \(\bar{x}_u\)=25.55 |
n=11 \(\bar{x}_w\)=18.91 (sd=0) \(\bar{x}_u\)=18.91 |
n=22 \(\bar{x}\)=22.23 (sd=3.93) |
- Ungewichteter Mittelwert ist der Mittelwert der Mittelwerte
- Die Spalten-Mittelwerte haben gleiches n, daher sind gewichtetet und ungewichtete Mittelwerte gleich.
- Bei den Zeilen-Rand-Mittelwerte differieren gewichtete und ungewichtete Mittelwerte.
- Bei den Zeilen-Mittelwerten der Frauen ist die Differenz zwischen gewichtetem und ungewichtetem MW 1.33, bei den Männern 1.4
- Weighted Means entspricht dem Beachten der ungleichen Zellhäufigkeiten => Frauen verdienen besser als Männer (22.33 vs 22.1) Der Unterschied ist aber gering
- Unweighted Means entspricht dem Ignorieren der ungleichen Zellhäufigkeiten und entspricht der Interpretation der Mittelwerte der Mittelwerte => Frauen verdienen schlechter als Männer (21 vs 23.5) Der Unterschied ist deutlich größer als bei weighted Means
- Aber: Frauen in diesem Sample haben eine sehr viel höhere Wahrscheinlichkeit, einen akademischen Abschluss zu haben und deshalb besser bezahlt zu werden. Dies erklärt auch den höheren Durchschnittsverdienst bei weighted means
Modelle
SS Type I
dd <- read.delim("http://md.psych.bio.uni-goettingen.de/data/div/zahn-unbal-paper.txt")
# correlation of factors
cor(cbind(dd$Salary, as.numeric(dd$con_gender), as.numeric(dd$con_education), as.numeric(dd$con_gen_x_edu)))## [,1] [,2] [,3] [,4]
## [1,] 1.00000000 0.03028133 0.86482936 0.16764126
## [2,] 0.03028133 1.00000000 0.36514837 -0.03563483
## [3,] 0.86482936 0.36514837 1.00000000 0.09759001
## [4,] 0.16764126 -0.03563483 0.09759001 1.00000000# SS Type 1
# we define models
m.s1.g.e <- lm(Salary ~ Gender + Education, data = dd)
unique(model.matrix(m.s1.g.e))## (Intercept) GenderMale EducationNo_degree
## 1 1 0 0
## 9 1 0 1
## 13 1 1 0
## 16 1 1 1m.s1.e.g <- lm(Salary ~ Education + Gender, data = dd)
unique(model.matrix(m.s1.e.g))## (Intercept) EducationNo_degree GenderMale
## 1 1 0 0
## 9 1 1 0
## 13 1 0 1
## 16 1 1 1# Education is entered first, then Gender. Education binds all the variance it can, also covariance due to correlated factors
m.s1.eg <- lm(Salary ~ Education + Gender + Education:Gender, data = dd)
# or shorter
# m.s1.eg <- lm(Salary ~ Education*Gender, data = dd)
unique(model.matrix(m.s1.eg))## (Intercept) EducationNo_degree GenderMale EducationNo_degree:GenderMale
## 1 1 0 0 0
## 9 1 1 0 0
## 13 1 0 1 0
## 16 1 1 1 1m.s1.ge <- lm(Salary ~ Gender*Education, data = dd)
unique(model.matrix(m.s1.ge))## (Intercept) GenderMale EducationNo_degree GenderMale:EducationNo_degree
## 1 1 0 0 0
## 9 1 0 1 0
## 13 1 1 0 0
## 16 1 1 1 1# SS differ, due to sequence to enter
aov(m.s1.ge)## Call:
## aov(formula = m.s1.ge)
##
## Terms:
## Gender Education Gender:Education Residuals
## Sum of Squares 0.29697 272.39184 1.17483 50.00000
## Deg. of Freedom 1 1 1 18
##
## Residual standard error: 1.666667
## Estimated effects may be unbalancedaov(m.s1.eg)## Call:
## aov(formula = m.s1.eg)
##
## Terms:
## Education Gender Education:Gender Residuals
## Sum of Squares 242.22727 30.46154 1.17483 50.00000
## Deg. of Freedom 1 1 1 18
##
## Residual standard error: 1.666667
## Estimated effects may be unbalanced# significance of effects differ due to sequence to enter
anova(m.s1.ge)## Analysis of Variance Table
##
## Response: Salary
## Df Sum Sq Mean Sq F value Pr(>F)
## Gender 1 0.297 0.297 0.1069 0.7475
## Education 1 272.392 272.392 98.0611 1.038e-08 ***
## Gender:Education 1 1.175 1.175 0.4229 0.5237
## Residuals 18 50.000 2.778
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1anova(m.s1.eg)## Analysis of Variance Table
##
## Response: Salary
## Df Sum Sq Mean Sq F value Pr(>F)
## Education 1 242.227 242.227 87.2018 2.534e-08 ***
## Gender 1 30.462 30.462 10.9662 0.003881 **
## Education:Gender 1 1.175 1.175 0.4229 0.523690
## Residuals 18 50.000 2.778
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1SS Type II oder Type III
dd <- read.delim("http://md.psych.bio.uni-goettingen.de/data/div/zahn-unbal-paper.txt")
# Education is entered first, then Gender. Education binds all the variance it can, also covariance due to correlated factors
m.s3.eg <- aov(Salary ~ Education + Gender + Education:Gender, data = dd)
unique(model.matrix(m.s3.eg))## (Intercept) EducationNo_degree GenderMale EducationNo_degree:GenderMale
## 1 1 0 0 0
## 9 1 1 0 0
## 13 1 0 1 0
## 16 1 1 1 1m.s3.ge <- aov(Salary ~ Gender*Education, data = dd)
unique(model.matrix(m.s3.ge))## (Intercept) GenderMale EducationNo_degree GenderMale:EducationNo_degree
## 1 1 0 0 0
## 9 1 0 1 0
## 13 1 1 0 0
## 16 1 1 1 1require(car)## Loading required package: car# results using SS type I
car::Anova(m.s3.eg)## Anova Table (Type II tests)
##
## Response: Salary
## Sum Sq Df F value Pr(>F)
## Education 272.392 1 98.0611 1.038e-08 ***
## Gender 30.462 1 10.9662 0.003881 **
## Education:Gender 1.175 1 0.4229 0.523690
## Residuals 50.000 18
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1car::Anova(m.s3.ge)## Anova Table (Type II tests)
##
## Response: Salary
## Sum Sq Df F value Pr(>F)
## Gender 30.462 1 10.9662 0.003881 **
## Education 272.392 1 98.0611 1.038e-08 ***
## Gender:Education 1.175 1 0.4229 0.523690
## Residuals 50.000 18
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1A.m.s3.eg <- car::Anova(m.s3.eg, type="III")
A.m.s3.ge <- car::Anova(m.s3.ge, type="III")
A.m.s2.eg <- car::Anova(m.s3.eg, type="II")
A.m.s2.ge <- car::Anova(m.s3.ge, type="II")
# we define SS type I models to compare with
m.s1.eg <- lm(Salary ~ Education*Gender, data = dd)
m.s1.ge <- lm(Salary ~ Gender*Education, data = dd)
# we compare SS type III and SS type I
A.m.s3.eg$`Sum Sq`## [1] 5000.000000 170.666667 8.727273 1.174825 50.000000A.m.s3.ge$`Sum Sq`## [1] 5000.000000 8.727273 170.666667 1.174825 50.000000A.m.s2.eg$`Sum Sq`## [1] 272.391841 30.461538 1.174825 50.000000A.m.s2.ge$`Sum Sq`## [1] 30.461538 272.391841 1.174825 50.000000summary(aov(m.s1.eg))[[1]]$`Sum Sq`## [1] 242.227273 30.461538 1.174825 50.000000summary(aov(m.s1.ge))[[1]]$`Sum Sq`## [1] 0.2969697 272.3918415 1.1748252 50.0000000A.m.s3.eg## Anova Table (Type III tests)
##
## Response: Salary
## Sum Sq Df F value Pr(>F)
## (Intercept) 5000.0 1 1800.0000 < 2.2e-16 ***
## Education 170.7 1 61.4400 3.273e-07 ***
## Gender 8.7 1 3.1418 0.09323 .
## Education:Gender 1.2 1 0.4229 0.52369
## Residuals 50.0 18
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1A.m.s2.eg## Anova Table (Type II tests)
##
## Response: Salary
## Sum Sq Df F value Pr(>F)
## Education 272.392 1 98.0611 1.038e-08 ***
## Gender 30.462 1 10.9662 0.003881 **
## Education:Gender 1.175 1 0.4229 0.523690
## Residuals 50.000 18
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1A.m.s2.ge## Anova Table (Type II tests)
##
## Response: Salary
## Sum Sq Df F value Pr(>F)
## Gender 30.462 1 10.9662 0.003881 **
## Education 272.392 1 98.0611 1.038e-08 ***
## Gender:Education 1.175 1 0.4229 0.523690
## Residuals 50.000 18
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1aov(m.s1.eg)## Call:
## aov(formula = m.s1.eg)
##
## Terms:
## Education Gender Education:Gender Residuals
## Sum of Squares 242.22727 30.46154 1.17483 50.00000
## Deg. of Freedom 1 1 1 18
##
## Residual standard error: 1.666667
## Estimated effects may be unbalancedaov(m.s1.ge)## Call:
## aov(formula = m.s1.ge)
##
## Terms:
## Gender Education Gender:Education Residuals
## Sum of Squares 0.29697 272.39184 1.17483 50.00000
## Deg. of Freedom 1 1 1 18
##
## Residual standard error: 1.666667
## Estimated effects may be unbalancedcar::Anova(aov(Salary ~ Education + Gender + Education:Gender, data = dd))## Anova Table (Type II tests)
##
## Response: Salary
## Sum Sq Df F value Pr(>F)
## Education 272.392 1 98.0611 1.038e-08 ***
## Gender 30.462 1 10.9662 0.003881 **
## Education:Gender 1.175 1 0.4229 0.523690
## Residuals 50.000 18
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Je nach Art der Berechnung, also in Abhängigkeit vom verwendeten Quadratsummentyp bekommen wir unterschiedliche Bewertungen der Effekte und beantworten unterschiedliche Hypothesen.
Je nach Berechnungsart würden wir also schlussfolgern, dass es einen geschlechtsabhängigen Unterschied in der Bezahlung gibt, oder nicht.
Berechnungen und SS mit ezANOVA()
dd <- read.delim("http://md.psych.bio.uni-goettingen.de/data/div/zahn-unbal-paper.txt")
dd$subj <- factor(dd$nr)
require(ez)## Loading required package: ez# fit model using ezANOVA
m.s1g <- ezANOVA(data=dd, dv=.(Salary), wid = .(subj), between=.(Gender, Education), detailed=TRUE, type=1, return_aov=T)## Warning: Data is unbalanced (unequal N per group). Make sure you specified
## a well-considered value for the type argument to ezANOVA().## Warning: Using "type==1" is highly questionable when data are unbalanced
## and there is more than one variable. Hopefully you are doing this for
## demonstration purposes only!ezANOVA(data=dd, dv=.(Salary), wid = .(subj), between=.(Gender, Education), detailed=TRUE, type=1)## Warning: Data is unbalanced (unequal N per group). Make sure you specified
## a well-considered value for the type argument to ezANOVA().
## Warning: Using "type==1" is highly questionable when data are unbalanced
## and there is more than one variable. Hopefully you are doing this for
## demonstration purposes only!## $ANOVA
## Effect DFn DFd SSn SSd F p p<.05
## 1 Gender 1 18 0.2969697 50 0.1069091 7.474625e-01
## 2 Education 1 18 272.3918415 50 98.0610629 1.038093e-08 *
## 3 Gender:Education 1 18 1.1748252 50 0.4229371 5.236898e-01
## ges
## 1 0.005904326
## 2 0.844909227
## 3 0.022957092m.s1g## $ANOVA
## Effect DFn DFd SSn SSd F p p<.05
## 1 Gender 1 18 0.2969697 50 0.1069091 7.474625e-01
## 2 Education 1 18 272.3918415 50 98.0610629 1.038093e-08 *
## 3 Gender:Education 1 18 1.1748252 50 0.4229371 5.236898e-01
## ges
## 1 0.005904326
## 2 0.844909227
## 3 0.022957092
##
## $aov
## Call:
## aov(formula = formula(aov_formula), data = data)
##
## Terms:
## Gender Education Gender:Education Residuals
## Sum of Squares 0.29697 272.39184 1.17483 50.00000
## Deg. of Freedom 1 1 1 18
##
## Residual standard error: 1.666667
## Estimated effects may be unbalancedsummary(m.s1g$aov)## Df Sum Sq Mean Sq F value Pr(>F)
## Gender 1 0.30 0.30 0.107 0.747
## Education 1 272.39 272.39 98.061 1.04e-08 ***
## Gender:Education 1 1.17 1.17 0.423 0.524
## Residuals 18 50.00 2.78
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1# we compare it to aov which uses SS type 1
#aov(Salary ~ Education + Gender + Education:Gender, data = dd)
summary(aov(Salary ~ Gender*Education, data = dd))## Df Sum Sq Mean Sq F value Pr(>F)
## Gender 1 0.30 0.30 0.107 0.747
## Education 1 272.39 272.39 98.061 1.04e-08 ***
## Gender:Education 1 1.17 1.17 0.423 0.524
## Residuals 18 50.00 2.78
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1# and now we change sequence to enter
ezANOVA(data=dd, dv=.(Salary), wid = .(subj), between=.(Education, Gender), detailed=TRUE, type=1)## Warning: Data is unbalanced (unequal N per group). Make sure you specified
## a well-considered value for the type argument to ezANOVA().
## Warning: Using "type==1" is highly questionable when data are unbalanced
## and there is more than one variable. Hopefully you are doing this for
## demonstration purposes only!## $ANOVA
## Effect DFn DFd SSn SSd F p p<.05
## 1 Education 1 18 242.227273 50 87.2018182 2.534451e-08 *
## 2 Gender 1 18 30.461538 50 10.9661538 3.881337e-03 *
## 3 Education:Gender 1 18 1.174825 50 0.4229371 5.236898e-01
## ges
## 1 0.82890030
## 2 0.37858509
## 3 0.02295709m.s1e <- ezANOVA(data=dd, dv=.(Salary), wid = .(subj), between=.(Education, Gender), detailed=TRUE, type=1, return_aov=T)## Warning: Data is unbalanced (unequal N per group). Make sure you specified
## a well-considered value for the type argument to ezANOVA().
## Warning: Using "type==1" is highly questionable when data are unbalanced
## and there is more than one variable. Hopefully you are doing this for
## demonstration purposes only!m.s1e## $ANOVA
## Effect DFn DFd SSn SSd F p p<.05
## 1 Education 1 18 242.227273 50 87.2018182 2.534451e-08 *
## 2 Gender 1 18 30.461538 50 10.9661538 3.881337e-03 *
## 3 Education:Gender 1 18 1.174825 50 0.4229371 5.236898e-01
## ges
## 1 0.82890030
## 2 0.37858509
## 3 0.02295709
##
## $aov
## Call:
## aov(formula = formula(aov_formula), data = data)
##
## Terms:
## Education Gender Education:Gender Residuals
## Sum of Squares 242.22727 30.46154 1.17483 50.00000
## Deg. of Freedom 1 1 1 18
##
## Residual standard error: 1.666667
## Estimated effects may be unbalancedsummary(m.s1e$aov)## Df Sum Sq Mean Sq F value Pr(>F)
## Education 1 242.23 242.23 87.202 2.53e-08 ***
## Gender 1 30.46 30.46 10.966 0.00388 **
## Education:Gender 1 1.17 1.17 0.423 0.52369
## Residuals 18 50.00 2.78
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1# we now run an ANOVA SS type 2
ezANOVA(data=dd, dv=.(Salary), wid = .(subj), between=.(Gender, Education), detailed=TRUE, type=2)## Warning: Data is unbalanced (unequal N per group). Make sure you specified
## a well-considered value for the type argument to ezANOVA().## $ANOVA
## Effect DFn DFd SSn SSd F p p<.05
## 1 Gender 1 18 30.461538 50 10.9661538 3.881337e-03 *
## 2 Education 1 18 272.391841 50 98.0610629 1.038093e-08 *
## 3 Gender:Education 1 18 1.174825 50 0.4229371 5.236898e-01
## ges
## 1 0.37858509
## 2 0.84490923
## 3 0.02295709
##
## $`Levene's Test for Homogeneity of Variance`
## DFn DFd SSn SSd F p p<.05
## 1 3 18 0.3398268 17.02381 0.1197711 0.9472899m.s2 <- ezANOVA(data=dd, dv=.(Salary), wid = .(subj), between=.(Gender, Education), detailed=TRUE, type=2, return_aov=T)## Warning: Data is unbalanced (unequal N per group). Make sure you specified
## a well-considered value for the type argument to ezANOVA().m.s2## $ANOVA
## Effect DFn DFd SSn SSd F p p<.05
## 1 Gender 1 18 30.461538 50 10.9661538 3.881337e-03 *
## 2 Education 1 18 272.391841 50 98.0610629 1.038093e-08 *
## 3 Gender:Education 1 18 1.174825 50 0.4229371 5.236898e-01
## ges
## 1 0.37858509
## 2 0.84490923
## 3 0.02295709
##
## $`Levene's Test for Homogeneity of Variance`
## DFn DFd SSn SSd F p p<.05
## 1 3 18 0.3398268 17.02381 0.1197711 0.9472899
##
## $aov
## Call:
## aov(formula = formula(aov_formula), data = data)
##
## Terms:
## Gender Education Gender:Education Residuals
## Sum of Squares 0.29697 272.39184 1.17483 50.00000
## Deg. of Freedom 1 1 1 18
##
## Residual standard error: 1.666667
## Estimated effects may be unbalancedsummary(m.s2$aov)## Df Sum Sq Mean Sq F value Pr(>F)
## Gender 1 0.30 0.30 0.107 0.747
## Education 1 272.39 272.39 98.061 1.04e-08 ***
## Gender:Education 1 1.17 1.17 0.423 0.524
## Residuals 18 50.00 2.78
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1car::Anova(aov(Salary ~ Education * Gender, data = dd), type=2)## Anova Table (Type II tests)
##
## Response: Salary
## Sum Sq Df F value Pr(>F)
## Education 272.392 1 98.0611 1.038e-08 ***
## Gender 30.462 1 10.9662 0.003881 **
## Education:Gender 1.175 1 0.4229 0.523690
## Residuals 50.000 18
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1car::Anova(aov(Salary ~ Gender * Education, data = dd), type=2)## Anova Table (Type II tests)
##
## Response: Salary
## Sum Sq Df F value Pr(>F)
## Gender 30.462 1 10.9662 0.003881 **
## Education 272.392 1 98.0611 1.038e-08 ***
## Gender:Education 1.175 1 0.4229 0.523690
## Residuals 50.000 18
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1# ... sequence to enter isn't important any more
# we now run an ANOVA SS type 3
ezANOVA(data=dd, dv=.(Salary), wid = .(subj), between=.(Gender, Education), detailed=TRUE, type=3)## Warning: Data is unbalanced (unequal N per group). Make sure you specified
## a well-considered value for the type argument to ezANOVA().## $ANOVA
## Effect DFn DFd SSn SSd F p p<.05
## 1 (Intercept) 1 18 9305.790210 50 3350.0844755 6.611923e-22 *
## 2 Gender 1 18 29.370629 50 10.5734266 4.428981e-03 *
## 3 Education 1 18 264.335664 50 95.1608392 1.306343e-08 *
## 4 Gender:Education 1 18 1.174825 50 0.4229371 5.236898e-01
## ges
## 1 0.99465572
## 2 0.37004405
## 3 0.84093437
## 4 0.02295709
##
## $`Levene's Test for Homogeneity of Variance`
## DFn DFd SSn SSd F p p<.05
## 1 3 18 0.3398268 17.02381 0.1197711 0.9472899m.s3 <- ezANOVA(data=dd, dv=.(Salary), wid = .(subj), between=.(Gender, Education), detailed=TRUE, type=3, return_aov=T)## Warning: Data is unbalanced (unequal N per group). Make sure you specified
## a well-considered value for the type argument to ezANOVA().m.s3## $ANOVA
## Effect DFn DFd SSn SSd F p p<.05
## 1 (Intercept) 1 18 9305.790210 50 3350.0844755 6.611923e-22 *
## 2 Gender 1 18 29.370629 50 10.5734266 4.428981e-03 *
## 3 Education 1 18 264.335664 50 95.1608392 1.306343e-08 *
## 4 Gender:Education 1 18 1.174825 50 0.4229371 5.236898e-01
## ges
## 1 0.99465572
## 2 0.37004405
## 3 0.84093437
## 4 0.02295709
##
## $`Levene's Test for Homogeneity of Variance`
## DFn DFd SSn SSd F p p<.05
## 1 3 18 0.3398268 17.02381 0.1197711 0.9472899
##
## $aov
## Call:
## aov(formula = formula(aov_formula), data = data)
##
## Terms:
## Gender Education Gender:Education Residuals
## Sum of Squares 0.29697 272.39184 1.17483 50.00000
## Deg. of Freedom 1 1 1 18
##
## Residual standard error: 1.666667
## Estimated effects may be unbalancedsummary(m.s3$aov)## Df Sum Sq Mean Sq F value Pr(>F)
## Gender 1 0.30 0.30 0.107 0.747
## Education 1 272.39 272.39 98.061 1.04e-08 ***
## Gender:Education 1 1.17 1.17 0.423 0.524
## Residuals 18 50.00 2.78
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1car::Anova(aov(Salary ~ Education * Gender, data = dd), type=3)## Anova Table (Type III tests)
##
## Response: Salary
## Sum Sq Df F value Pr(>F)
## (Intercept) 5000.0 1 1800.0000 < 2.2e-16 ***
## Education 170.7 1 61.4400 3.273e-07 ***
## Gender 8.7 1 3.1418 0.09323 .
## Education:Gender 1.2 1 0.4229 0.52369
## Residuals 50.0 18
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1car::Anova(aov(Salary ~ Gender * Education, data = dd), type=3)## Anova Table (Type III tests)
##
## Response: Salary
## Sum Sq Df F value Pr(>F)
## (Intercept) 5000.0 1 1800.0000 < 2.2e-16 ***
## Gender 8.7 1 3.1418 0.09323 .
## Education 170.7 1 61.4400 3.273e-07 ***
## Gender:Education 1.2 1 0.4229 0.52369
## Residuals 50.0 18
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1# ... sequence to enter isn't important any more
# a final plot via ezPlot()
#ezPlot(data=dd, dv=.(Salary), wid = .(subj), between=.(Gender, Education), x=Education)
ezPlot(data=dd, dv=.(Salary), wid = .(subj), between=.(Gender, Education), x=Education, split=.(Gender))## Warning: Data is unbalanced (unequal N per group). Make sure you specified
## a well-considered value for the type argument to ezANOVA().## Warning in ezStats(data = data, dv = dv, wid = wid, within = within,
## within_full = within_full, : Unbalanced groups. Mean N will be used in
## computation of FLSD
ezPlot(data=dd, dv=.(Salary), wid = .(subj), between=.(Gender, Education), x=Gender, split=.(Education))## Warning: Data is unbalanced (unequal N per group). Make sure you specified
## a well-considered value for the type argument to ezANOVA().
## Warning: Unbalanced groups. Mean N will be used in computation of FLSD