Two Way MANOVA Fix
Ricardo
2024-10-27
Determinants of Wages from 1985 Current Population Survey
The Current Population Survey (CPS) is used to supplement census information between census years. These data consist of a random sample of 29 persons from the CPS, with information on wages and other characteristics of the workers, including sex, number of years of education, years of work experience, occupational status, region of residence and union membership. We wish to determine (i) whether wages are related to these characteristics and (ii) whether there is a gender gap in wages.
Variable Names
EDUCATION: Number of years of education.
SOUTH: Indicator variable for Southern Region (1=Person lives in South, 0=Person lives elsewhere).
SEX: Indicator variable for sex (1=Female, 0=Male).
EXPERIENCE: Number of years of work experience.
UNION: Indicator variable for union membership (1=Union member, 0=Not union member).
WAGE: Wage (dollars per hour).
AGE: Age (years).
RACE: Race (1=Other, 2=Hispanic, 3=White).
OCCUPATION: Occupational category (1=Management, 2=Sales, 3=Clerical, 4=Service, 5=Professional, 6=Other).
SECTOR: Sector (0=Other, 1=Manufacturing, 2=Construction).
MARR: Marital Status (0=Unmarried, 1=Married)library("openxlsx")
data <- read.xlsx(file.choose())
data## Education South Sex Experience Union Wage Age Sector Married
## 1 8 0 P 21 0 5.1 35 1 1
## 2 9 0 P 42 0 4.95 57 1 1
## 3 12 0 L 1 0 6.67 19 1 0
## 4 12 0 L 4 0 4 22 0 0
## 5 12 0 L 17 0 7.5 35 0 1
## 6 13 0 L 9 1 13.07 28 0 0
## 7 12 0 L 9 0 19.47 27 0 0
## 8 16 0 L 11 0 13.28 33 1 1
## 9 12 0 L 9 0 8.75 27 0 0
## 10 12 0 L 17 1 11.35 35 0 1
## 11 12 0 L 19 1 11.5 37 1 0
## 12 12 0 L 37 0 7.3 55 1 1
## 13 12 0 L 26 1 22.2 44 1 1
## 14 11 0 L 16 0 3.65 33 0 0
## 15 12 0 L 33 0 20.55 51 0 1
## 16 12 0 P 16 1 5.71 34 1 1
## 17 7 0 L 42 1 7 55 1 1
## 18 12 0 L 9 0 3.75 27 0 0
## 19 12 0 L 23 0 9.56 41 0 1
## 20 12 0 L 8 0 9.36 26 1 1
## 21 10 0 L 30 0 6.5 46 0 1
## 22 12 0 P 8 0 3.35 26 1 1
## 23 10 1 L 27 0 4.45 43 0 0
## 24 8 1 L 27 0 6.5 41 0 1
## 25 9 1 L 30 1 6.25 45 0 0
## 26 9 1 L 29 0 19.98 44 0 1
## 27 7 1 L 44 0 8 57 0 1
## 28 11 1 L 14 0 4.5 31 0 1
## 29 6 1 L 45 0 5.75 57 1 1head(data)## Education South Sex Experience Union Wage Age Sector Married
## 1 8 0 P 21 0 5.1 35 1 1
## 2 9 0 P 42 0 4.95 57 1 1
## 3 12 0 L 1 0 6.67 19 1 0
## 4 12 0 L 4 0 4 22 0 0
## 5 12 0 L 17 0 7.5 35 0 1
## 6 13 0 L 9 1 13.07 28 0 0str(data)## 'data.frame': 29 obs. of 9 variables:
## $ Education : num 8 9 12 12 12 13 12 16 12 12 ...
## $ South : num 0 0 0 0 0 0 0 0 0 0 ...
## $ Sex : chr "P" "P" "L" "L" ...
## $ Experience: num 21 42 1 4 17 9 9 11 9 17 ...
## $ Union : num 0 0 0 0 0 1 0 0 0 1 ...
## $ Wage : chr "5.1" "4.95" "6.67" "4" ...
## $ Age : num 35 57 19 22 35 28 27 33 27 35 ...
## $ Sector : num 1 1 1 0 0 0 0 1 0 0 ...
## $ Married : num 1 1 0 0 1 0 0 1 0 1 ...data[, c(1, 4, 6)] <- lapply(data[, c(1, 4, 6)], as.numeric)
x1_p <- data[,1];x1_p## [1] 8 9 12 12 12 13 12 16 12 12 12 12 12 11 12 12 7 12 12 12 10 12 10 8 9
## [26] 9 7 11 6x2_p <- data[,4];x2_p## [1] 21 42 1 4 17 9 9 11 9 17 19 37 26 16 33 16 42 9 23 8 30 8 27 27 30
## [26] 29 44 14 45x3_p <- data[,6];x3_p## [1] 5.10 4.95 6.67 4.00 7.50 13.07 19.47 13.28 8.75 11.35 11.50 7.30
## [13] 22.20 3.65 20.55 5.71 7.00 3.75 9.56 9.36 6.50 3.35 4.45 6.50
## [25] 6.25 19.98 8.00 4.50 5.75data$Sector<-as.integer(data$Sector)
data$Married<-as.integer(data$Married)
dataf<-data[1:29,c(1,4,6)];dataf## Education Experience Wage
## 1 8 21 5.10
## 2 9 42 4.95
## 3 12 1 6.67
## 4 12 4 4.00
## 5 12 17 7.50
## 6 13 9 13.07
## 7 12 9 19.47
## 8 16 11 13.28
## 9 12 9 8.75
## 10 12 17 11.35
## 11 12 19 11.50
## 12 12 37 7.30
## 13 12 26 22.20
## 14 11 16 3.65
## 15 12 33 20.55
## 16 12 16 5.71
## 17 7 42 7.00
## 18 12 9 3.75
## 19 12 23 9.56
## 20 12 8 9.36
## 21 10 30 6.50
## 22 12 8 3.35
## 23 10 27 4.45
## 24 8 27 6.50
## 25 9 30 6.25
## 26 9 29 19.98
## 27 7 44 8.00
## 28 11 14 4.50
## 29 6 45 5.75str(data)## 'data.frame': 29 obs. of 9 variables:
## $ Education : num 8 9 12 12 12 13 12 16 12 12 ...
## $ South : num 0 0 0 0 0 0 0 0 0 0 ...
## $ Sex : chr "P" "P" "L" "L" ...
## $ Experience: num 21 42 1 4 17 9 9 11 9 17 ...
## $ Union : num 0 0 0 0 0 1 0 0 0 1 ...
## $ Wage : num 5.1 4.95 6.67 4 7.5 ...
## $ Age : num 35 57 19 22 35 28 27 33 27 35 ...
## $ Sector : int 1 1 1 0 0 0 0 1 0 0 ...
## $ Married : int 1 1 0 0 1 0 0 1 0 1 ...Uji Normalitas Multivariat
Uji Hipotesis
H0 : Data berdistribusi normal multivariat
H1 : Data tidak berdistribusi normal multivariat
Taraf Signifikansi
alpha = 5% = 0,05
library(MVN)## Warning: package 'MVN' was built under R version 4.4.1test = mvn(data[1:29,c(1,4,6)], mvnTest = "mardia", univariateTest = "SW", multivariatePlot = "qq")
test## $multivariateNormality
## Test Statistic p value Result
## 1 Mardia Skewness 13.2822084762488 0.208318454443323 YES
## 2 Mardia Kurtosis -0.214329504809508 0.830290111083336 YES
## 3 MVN <NA> <NA> YES
##
## $univariateNormality
## Test Variable Statistic p value Normality
## 1 Shapiro-Wilk Education 0.8590 0.0012 NO
## 2 Shapiro-Wilk Experience 0.9451 0.1365 YES
## 3 Shapiro-Wilk Wage 0.8277 0.0003 NO
##
## $Descriptives
## n Mean Std.Dev Median Min Max 25th 75th Skew
## Education 29 10.827586 2.172375 12 6.00 16.0 9.0 12.00 -0.3918675
## Experience 29 21.482759 12.740881 19 1.00 45.0 9.0 30.00 0.3390871
## Wage 29 8.965517 5.436784 7 3.35 22.2 5.1 11.35 1.1815078
## Kurtosis
## Education -0.0565370
## Experience -1.0901924
## Wage 0.2018184P-Value Uji Normalitas Multivariat
Mardia Skewness : 0.20831845
Mardia Kurtosis : 0.830290111
Kriteria Uji
Pada Mardia’s Test Skewness, tolak H0 jika p-value < alpha
Pada Mardia’s Test Kurtosis, tolak H0 hika p-value < alpha
Kesimpulan
Berdasarkan hasil perhitungan diatas diketahui bahwa data berdistribusi normal multivariat dikarenakan p value dari kurtosis dan skewness > apha
Uji Homogenitas
Dikarenakan ada 2 grup yaitu sector (0=other,1=manufacturing) dan married(0=unmarried,1=married) maka dilakukan 2 uji homogenitas dengan ghipotesis,taraf signifikansi, dan uji yang sama yakni enggunakan Box’s M dengan taraf signifikansi 5%
Hipotesis
H0 = Matriks Kovarians Grup sama
H1 = Matriks Kovarians Grup tidak sama
Taraf Signifikansi
alpha = 5% = 0,05
# Sector
library(biotools)## Warning: package 'biotools' was built under R version 4.4.1## Loading required package: MASS## ---
## biotools version 4.2head(data[,8])## [1] 1 1 1 0 0 0boxM(data = dataf, grouping = data[,8])##
## Box's M-test for Homogeneity of Covariance Matrices
##
## data: dataf
## Chi-Sq (approx.) = 6.7137, df = 6, p-value = 0.3481# Married
head(data[,9])## [1] 1 1 0 0 1 0boxM(data = dataf, grouping = data[,9])##
## Box's M-test for Homogeneity of Covariance Matrices
##
## data: dataf
## Chi-Sq (approx.) = 10.367, df = 6, p-value = 0.11Kriteria Uji
Tolak H0 jika p-value < alpha, terima dalam hal lainnya
Keputusan
Pada faktor sektor, P-value (0,3481) > alpha (0,05)
Pada faktor married, P-value (0,11) > alpha (0,05)
Kesimpulan
Pada faktor sektor, dengan taraf signifikansi 5% dapat disimpulkan bahwa data memiliki kovarians yang sama dan memenuhi asumsi homogenitas
Pada faktor married, dengan taraf signifikansi 5% dapat disimpulkan bahwa data memiliki kovarians yang sama dan memenuhi asumsi homohhenitas
Two Way MANOVA
H0 : Faktor sektor tidak berpengaruh terhadap gaji seseorang
H1 : Setidaknya ada satu yang tidak berpengaruh terhadap gaji seseorang
H0 : Faktor married berpengaruh terhadap gaji seseorang
H1 : Setidaknya ada satu faktor yang tidak berpengaruh terhadap gaji
sector <- as.factor(data$Sector[1:29])
married <- as.factor(data$Married[1:29])
manova <- manova(cbind(x1_p, x2_p, x3_p) ~ sector * married, data = data)
# Menampilkan hasil
summary(manova)## Df Pillai approx F num Df den Df Pr(>F)
## sector 1 0.044373 0.35599 3 23 0.78526
## married 1 0.243258 2.46448 3 23 0.08792 .
## sector:married 1 0.026406 0.20794 3 23 0.88985
## Residuals 25
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Kesimpulan
- Pada faktor sector diketahui p-value (0,78526) > alpha (0,05) maka terima H0, sehingga faktor sektor tidak berpengaruh terhadap gaji seseorang
- Pada faktor married diketahui p-value (0,08792) > alpha (0,05) maka terima H0, sehingga faktor married tidak berpengaruh terhadap gaji seseorang