Taller 1.

Universidad Externado de Colombia.

Paula Guzmán - Carlos Galvis.

Realice una prueba de hipótesis para cada asignatura (matemáticas, lectura

y escritura) para verificar si existen diferencias en el rendimiento promedio

obtenido en cada asignatura entre hombres y mujeres. Realice todas las pruebas

con significancia del 5%. Explique los resultados obtenidos.

library(readr)
StudentsPerformance <- read_csv("StudentsPerformance.csv")

## Rows: 1000 Columns: 8
## ── Column specification ────────────────────────────────────────────────────────
## Delimiter: ","
## chr (5): gender, race/ethnicity, parental level of education, lunch, test pr...
## dbl (3): math score, reading score, writing score
## 
## ℹ Use `spec()` to retrieve the full column specification for this data.
## ℹ Specify the column types or set `show_col_types = FALSE` to quiet this message.

Pruebas de matemáticas. - M1

mujeres - p1

N1 = 518
FM1 = StudentsPerformance[StudentsPerformance$gender=="female","math score"]

mean1 = (sum(FM1)/518)/100

sd1 = sqrt(var(FM1))
sd1

##            math score
## math score   15.49145

##Hombres - p2

N2 = 482
HM1<-StudentsPerformance[StudentsPerformance$gender=="male","math score"]

mean2=(sum(HM1)/482)/100
mean2

## [1] 0.6872822

sd2=sqrt(var(HM1))
sd2

##            math score
## math score   14.35628

#Prueba de hipotesis H0:{M1-M2>0} HA:{M1-M2<0}

D.E = sqrt(((sd1^2)/N1)+((sd2^2/N2)))
D.E

##            math score
## math score  0.9438701

pnorm(mean1-mean2,0,D.E, lower.tail = FALSE)

## [1] 0.5215245

M.E = 1.96*D.E
M.E

##            math score
## math score   1.849985

mean1-mean2

## [1] -0.05095011

####0.52 es mayor a 0.05 por ende no rechazo. ####0.05 es menor a 1.84 por ende no rechazo.

Prueba de escritura. - E1

mujeres - p1

NE1 = 518
FE1 = StudentsPerformance[StudentsPerformance$gender=="female","writing score"]

meanE1 = (sum(FE1)/518)/100

sdE1 = sqrt(var(FE1))
sdE1

##               writing score
## writing score      14.84484

##Hombres - p2

NE2 = 482
HME<-StudentsPerformance[StudentsPerformance$gender=="male","writing score"]

meanE2=(sum(HME)/482)/100
meanE2

## [1] 0.633112

sdE2=sqrt(var(HME))
sdE2

##               writing score
## writing score      14.11383

Prueba de hipotesis H0:{M1-M2>0} HA:{M1-M2<0}

DEE = sqrt((sdE1^2)/N1)+((sdE2^2/N2))
DEE

##               writing score
## writing score      1.065523

pnorm(meanE1-meanE2,0,DEE, lower.tail = FALSE)

## [1] 0.4657613

####0.46 es mayor a 0.05 por ende no rechaza.

MEE = 0.96*DEE
MEE

##               writing score
## writing score      1.022902

meanE1-meanE2

## [1] 0.09155978

####0.09 es menor a 1.02 por ende no rechazo. # Pruebas de lectura. -L1

mujeres - p1

NE1 = 518
FR1 = StudentsPerformance[StudentsPerformance$gender=="female","reading score"]

meanR1 = (sum(FR1)/518)/100

sdR1 = sqrt(var(FR1))
sdR1

##               reading score
## reading score      14.37825

##Hombres - p2

NE2 = 482
HRE<-StudentsPerformance[StudentsPerformance$gender=="male","reading score"]

meanR2=(sum(HRE)/482)/100
meanR2

## [1] 0.6547303

sdR2=sqrt(var(HRE))
sdR2

##               reading score
## reading score      13.93183

prubea de hipotesis H0:{M1-M2>0} HA:{M1-M2<0}

DER = sqrt((sdR1^2)/N1)+((sdR2^2/N2))
DER

##               reading score
## reading score      1.034433

pnorm(meanR1-meanR2,0,DER, lower.tail = FALSE)

## [1] 0.4725044

####0.47 es mayor a 0.05 por ende no rechaza.

MER = 0.96*DER
MER

##               reading score
## reading score     0.9930552

meanR1-meanR2

## [1] 0.07135079

####0.07 es menor a a 0.99 por ende no rechazo.

2

Muestra 1

ns= 645
means = 645/1000

####Muestra 2

nn = 642
meann = 642/1000

P = 0.648
Q = 1-P
Q

## [1] 0.352

D.E2 = sqrt(P*Q*(1/ns+1/nn))
D.E2

## [1] 0.02662568