Taller 1

#1 Con la librerıa faux de R genere tres coordenadas de color del sistema CIELab usando la distribuci´on normal multivariante (rnorm multi()), con medias para (L,a,b)=c(39.0,4.2,38.9) y desviaciones de c(1.20,0.25,1.12). Use una correlaci´on de 0.6 para todos los pares de variables. Genere 120 datos para una semilla generada a partir de su n´umero de c´edula.Prueba la hip´otesis de investigaci´on de que la media del ´ındice de color(IC) es inferior a 2.8. Si si la media muestral de IC es superior a este valor, cambie el 2.8 por un 2 % menos que la media muestral generada de sus datos y use este valor como el que probar´a en la hipotesis de investigaci´on. Use el c´alculo del IC como aparece en la referencia (1).

#ver otro archivo 

#2. En cieto a˜no de comparo la medida de NDVI(normalized difference vegetation index) obtenido por Landsat y por MODIS(Moderate Resolution Imaging Spectroradiometer) para un lote determinado. Se obtuvo la media en cada caso, la cual en el problema actual se generaron a partir de la distribuci´on normal con medias respecticas de c(0.16,0.17) y desviaciones de c(0.021,0.025). Genere datos correspondientes a 120 celdas de una grilla regular. Cada valor en celda se promedia para cada dispositivo y con esto se obtienen las estad´ısticas muestrales de cada dispositivo. T´ıpicamente, los valores positivos de NDVI est´an asociados con zonas con vegetaci´on, mientras que los valores cero y negativo corresponden a suelos desnudos y masas de agua. Con los datos generados usando su propio set.seed() comprueba la hip´otesis de que existen diferencias en las medias del NDVI atribuible a los dispositivos. Considere los siguientes escenarios: 1. Asuma varianzas iguales 2. Realice alguna prueba de igualdad de varianzas y si resultan diferentes haga los ajustes necesarios

set.seed(355)
L <- rnorm(120, 0.16, 0.021); L

##   [1] 0.1395525 0.2073858 0.1236562 0.1542815 0.1680163 0.1436127 0.1777218
##   [8] 0.1503787 0.1519672 0.1436837 0.1322546 0.1051064 0.1518875 0.1264367
##  [15] 0.1298263 0.1501895 0.1546659 0.1641859 0.1554129 0.1496153 0.1631112
##  [22] 0.1688879 0.1879625 0.1729636 0.1462325 0.1474203 0.1900412 0.1809225
##  [29] 0.1456024 0.2301891 0.1693465 0.1150605 0.1807933 0.1742627 0.1737618
##  [36] 0.1414516 0.1380789 0.1355805 0.1767371 0.1478317 0.1344604 0.1847919
##  [43] 0.1436311 0.1898646 0.1761619 0.1469070 0.1976736 0.1396557 0.1684579
##  [50] 0.1924283 0.1326426 0.1363249 0.1814244 0.1218105 0.1957081 0.1356892
##  [57] 0.1834888 0.1314879 0.1561569 0.1629119 0.1281241 0.1500706 0.1585942
##  [64] 0.1677709 0.1885849 0.1617986 0.1917852 0.1659665 0.1374260 0.1528627
##  [71] 0.1624183 0.1692320 0.1252271 0.1522780 0.1693803 0.1540560 0.1840253
##  [78] 0.1598380 0.1382301 0.1818903 0.2106029 0.1198400 0.1830580 0.1664859
##  [85] 0.1770134 0.1811287 0.1800571 0.1487675 0.1340884 0.1450900 0.2015959
##  [92] 0.1373152 0.1857415 0.1514906 0.1576445 0.1992654 0.1757176 0.1576195
##  [99] 0.1630227 0.1476079 0.1427655 0.2114548 0.1513826 0.1687252 0.1804142
## [106] 0.1718692 0.1721499 0.1903743 0.1465698 0.1347582 0.1642105 0.1905710
## [113] 0.1752341 0.1677505 0.2062763 0.1178765 0.1408514 0.1844276 0.1703392
## [120] 0.1278885

M <- rnorm(120, 0.17, 0.025); M

##   [1] 0.2004967 0.1374094 0.1889665 0.1769820 0.1433894 0.2090428 0.1597504
##   [8] 0.1803297 0.1997638 0.1412346 0.1587425 0.1817021 0.1629829 0.1633655
##  [15] 0.1198345 0.1726941 0.1478342 0.2025956 0.1540120 0.1497173 0.1679195
##  [22] 0.1804742 0.1754229 0.1866738 0.1755304 0.1787632 0.1322037 0.2073426
##  [29] 0.2158318 0.1885453 0.1556962 0.1994932 0.1874966 0.1212086 0.2144668
##  [36] 0.1783468 0.1819250 0.1597827 0.1783029 0.1773703 0.1598931 0.1783280
##  [43] 0.1602361 0.1859944 0.1555734 0.1497820 0.1672650 0.1883578 0.1678305
##  [50] 0.1645763 0.1548610 0.1296758 0.1672025 0.1427943 0.2026380 0.1583741
##  [57] 0.1692643 0.2057877 0.1571751 0.1787596 0.1847841 0.1323778 0.1617580
##  [64] 0.1594988 0.2117563 0.1207304 0.2111912 0.1555802 0.1117560 0.1298988
##  [71] 0.2509855 0.1941586 0.2256830 0.1585186 0.1565739 0.1973486 0.1710870
##  [78] 0.1745217 0.2194679 0.1582239 0.2039720 0.1891192 0.1620289 0.1215220
##  [85] 0.1501668 0.1689345 0.2174545 0.1759562 0.1906807 0.2045973 0.1521655
##  [92] 0.1569029 0.1859828 0.1268500 0.1447323 0.1995657 0.1456394 0.1863312
##  [99] 0.1888137 0.1477821 0.1804654 0.1866268 0.1809210 0.1635523 0.1910382
## [106] 0.1531073 0.1353363 0.1259587 0.2010062 0.1859543 0.1738834 0.1615574
## [113] 0.1839937 0.1746821 0.2135268 0.1823971 0.1765713 0.1698133 0.1653591
## [120] 0.1294849

# Prueba de hipotesis para dos promedios independientes

# Verificacion de igualdad de varianzas

\[H_0: Mean NDVI{Landsat}\neq Mean NDVI{modis}\] \[H_a: Mean NDVI{Landsat}= Mean NDVI{modis}\]

Vz1 <- var(L) Vz2 <- var(M)

# Las varianzas de los dos metodos estadisticamente no son diferentes 
# A continuación se realiza prueba: 

var.test(L,M, ratio = 1)

## 
##  F test to compare two variances
## 
## data:  L and M
## F = 0.83433, num df = 119, denom df = 119, p-value = 0.3245
## alternative hypothesis: true ratio of variances is not equal to 1
## 95 percent confidence interval:
##  0.581474 1.197150
## sample estimates:
## ratio of variances 
##           0.834333

#p-valor:32.4%, se encuentra por encima del 5% por tanto LAS VARIANZAS SON IGUALES 

#prueba T, 2 muestras independientes con varianza igual

t.test(L, M, mu = 0, var.equal = T, alternative = "t")

## 
##  Two Sample t-test
## 
## data:  L and M
## t = -3.318, df = 238, p-value = 0.001049
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -0.017026230 -0.004340293
## sample estimates:
## mean of x mean of y 
## 0.1609696 0.1716529

#El p-valor es 0.1%, al ser menor que 5% mis datos muestran evidencia en CONTRA de la Ho, es decir que posiblemente no existen diferencias en las medias del NDVI atribuible a los dispositivos (Landsat y MODIS).

#3 Usando los datos del mismo ejercicio anterior pero esta vez los del NDVI solamente con Landsat, use esta medida como la medici´on lograda en el mes de junio de ese a˜no en particular. Genere otra muestra de NDVI como una media de 0.14 y una desviaci´on de 0.022 y asuma que esta medici´on se obtuvo dos meses despues para el mismo lote. Use la prueba t pareada considerada en clase y compruebe estad´ısticamente si ha disminuido la cobertura vegetal que puede medirse medianteel NDVI.Para mayor entendimiento del tema  usar la referencia 2. Nota(Sin duda la variabilidad espacial juega un papel importante en estos datos, sin embargo, por ahora solo nos interesan los promedios y no como se distribuye el NDVI dentro del lote). 

#Prueba T student para muestras pareadas (no independientes)
#CV= Cobertura Vegetal

CVjunio<- rnorm(120, 0.16, 0.021)
CVAgosto<- rnorm(120,0.14,0.022)

\[H_0: CV{agosto} < CV{junio}\] \[H_a: CV{agosto} \geq CV{junio}\]

t.test(CVjunio,CVAgosto,mu = 0, alternative = "g",paired = T)

## 
##  Paired t-test
## 
## data:  CVjunio and CVAgosto
## t = 5.9726, df = 119, p-value = 1.243e-08
## alternative hypothesis: true difference in means is greater than 0
## 95 percent confidence interval:
##  0.01216812        Inf
## sample estimates:
## mean of the differences 
##              0.01684311

#p-valor < 5%

#los datos proporcionaron evidencia en contra de la Ho por tanto la cobertura vegetal no ha disminuido en el mes de agosto respecto al mes de junio, segun los datos del NDVI obtenidos por Landsat.

library(seqinr)

## Warning: package 'seqinr' was built under R version 4.1.3

dengue<-read.fasta("C:/Users/faile/Downloads/sequence.fasta")
dengueseq <-dengue[[1]]
length(dengueseq)

## [1] 10735

table(dengueseq)

## dengueseq
##    a    c    g    t 
## 3426 2240 2770 2299

#Ho: la proporcion de G > 25% \[H_0: G > 0.25\] \[H_a: G\leq 0.25\]

#4 Para la accecion NC 001477 pruebe si la proporcion de G es superior al 25 %.

GC(dengueseq)

## [1] 0.4666977

GC1(dengueseq)

## [1] 0.4629785

GC2(dengueseq)

## [1] 0.500559

GC3(dengueseq)

## [1] 0.4365567

2770/(3426+2240+2770+2299)

## [1] 0.2580345

prop.test(x = c(2770),
          n = c(3426+2240+2770+2299),
          alternative = 't')

## 
##  1-sample proportions test with continuity correction
## 
## data:  c(2770) out of c(3426 + 2240 + 2770 + 2299), null probability 0.5
## X-squared = 2513.1, df = 1, p-value < 2.2e-16
## alternative hypothesis: true p is not equal to 0.5
## 95 percent confidence interval:
##  0.2497989 0.2664441
## sample estimates:
##         p 
## 0.2580345

library(seqinr)
#Para las acceciones NC 001474 y NC 001477 pruebe si es diferente la proporci´on del par TG.

\[H_0: TG{(NC001474)}\neq TG{(NC001477)}\]

\[H_0: TG{(NC001474)}= TG{(NC001477)}\]

denguetable<-count(dengueseq,1);denguetable[[4]];denguetable[[3]]

## [1] 2299

## [1] 2770

prop.test(x = c(2299+2770),
          n = c(3426+2240+2770+2299),
          alternative = 't')

## 
##  1-sample proportions test with continuity correction
## 
## data:  c(2299 + 2770) out of c(3426 + 2240 + 2770 + 2299), null probability 0.5
## X-squared = 33.09, df = 1, p-value = 8.801e-09
## alternative hypothesis: true p is not equal to 0.5
## 95 percent confidence interval:
##  0.4627151 0.4816924
## sample estimates:
##         p 
## 0.4721938

library(seqinr)
dengue1<-read.fasta("C:/Users/faile/Downloads/1474.fasta")
dengue1seq <-dengue1[[1]]
length(dengue1seq)

## [1] 10723

table(dengue1seq)

## dengue1seq
##    a    c    g    t 
## 3553 2200 2713 2257

denguetable<-count(dengue1seq,1);denguetable[[4]];denguetable[[3]]

## [1] 2257

## [1] 2713

prop.test(x = c(2257+2713),
          n = c(3553+2200+2713+2257),
          alternative = 't')

## 
##  1-sample proportions test with continuity correction
## 
## data:  c(2257 + 2713) out of c(3553 + 2200 + 2713 + 2257), null probability 0.5
## X-squared = 57.029, df = 1, p-value = 4.294e-14
## alternative hypothesis: true p is not equal to 0.5
## 95 percent confidence interval:
##  0.4540195 0.4729862
## sample estimates:
##         p 
## 0.4634897

#los datos proporcionaron evidencia en contra de la Ho, ya que la prOprocion del par TG de las acceciones NC 001474 y NC 001477 es igual.

#6 Para las acceci´on NC 001477 pruebe si el cociente de T respecto al de G es superior a la unidad. Trabaje con una secci´on que empiece 400 y termine en 555

\[H_0: T/G > 1\] \[H_a: T/G\leq 1\]

dengueseq[400:555]

##   [1] "g" "a" "c" "c" "a" "t" "g" "c" "t" "c" "c" "t" "c" "a" "t" "g" "c" "t"
##  [19] "g" "c" "t" "g" "c" "c" "c" "a" "c" "a" "g" "c" "c" "c" "t" "g" "g" "c"
##  [37] "g" "t" "t" "c" "c" "a" "t" "c" "t" "g" "a" "c" "c" "a" "c" "c" "c" "g"
##  [55] "a" "g" "g" "g" "g" "g" "a" "g" "a" "g" "c" "c" "g" "c" "a" "c" "a" "t"
##  [73] "g" "a" "t" "a" "g" "t" "t" "a" "g" "c" "a" "a" "g" "c" "a" "g" "g" "a"
##  [91] "a" "a" "g" "a" "g" "g" "a" "a" "a" "a" "t" "c" "a" "c" "t" "t" "t" "t"
## [109] "g" "t" "t" "t" "a" "a" "g" "a" "c" "c" "t" "c" "t" "g" "c" "a" "g" "g"
## [127] "t" "g" "t" "c" "a" "a" "c" "a" "t" "g" "t" "g" "c" "a" "c" "c" "c" "t"
## [145] "t" "a" "t" "t" "g" "c" "a" "a" "t" "g" "g" "a"

dengue1= dengueseq[400:555]
dengue1

##   [1] "g" "a" "c" "c" "a" "t" "g" "c" "t" "c" "c" "t" "c" "a" "t" "g" "c" "t"
##  [19] "g" "c" "t" "g" "c" "c" "c" "a" "c" "a" "g" "c" "c" "c" "t" "g" "g" "c"
##  [37] "g" "t" "t" "c" "c" "a" "t" "c" "t" "g" "a" "c" "c" "a" "c" "c" "c" "g"
##  [55] "a" "g" "g" "g" "g" "g" "a" "g" "a" "g" "c" "c" "g" "c" "a" "c" "a" "t"
##  [73] "g" "a" "t" "a" "g" "t" "t" "a" "g" "c" "a" "a" "g" "c" "a" "g" "g" "a"
##  [91] "a" "a" "g" "a" "g" "g" "a" "a" "a" "a" "t" "c" "a" "c" "t" "t" "t" "t"
## [109] "g" "t" "t" "t" "a" "a" "g" "a" "c" "c" "t" "c" "t" "g" "c" "a" "g" "g"
## [127] "t" "g" "t" "c" "a" "a" "c" "a" "t" "g" "t" "g" "c" "a" "c" "c" "c" "t"
## [145] "t" "a" "t" "t" "g" "c" "a" "a" "t" "g" "g" "a"

denguetable<-count(dengue1,1);denguetable[[4]]

## [1] 34

denguetable<-count(dengue1,1);denguetable[[3]]

## [1] 39

#Cociente T/G 
34/39

## [1] 0.8717949

#Los datos proporciona evidencia en contra de la Ho, ya que el cociente de T respecto al de G es menor a la unidad.

#7 Para las acceciones NC 001474 y NC 001477 pruebe si el cociente de TA respectoal de GC es diferentes. Trabaje con una secci´on que empiece 450 y termine en 555

\[H_0: TA/GC{(NC001474)}\neq TA/GC{(NC001477)}\] \[H_0: TA/GC{(NC001474)}= TA/GC{(NC001477)}\]

dengueseq[450:555]

##   [1] "c" "c" "c" "g" "a" "g" "g" "g" "g" "g" "a" "g" "a" "g" "c" "c" "g" "c"
##  [19] "a" "c" "a" "t" "g" "a" "t" "a" "g" "t" "t" "a" "g" "c" "a" "a" "g" "c"
##  [37] "a" "g" "g" "a" "a" "a" "g" "a" "g" "g" "a" "a" "a" "a" "t" "c" "a" "c"
##  [55] "t" "t" "t" "t" "g" "t" "t" "t" "a" "a" "g" "a" "c" "c" "t" "c" "t" "g"
##  [73] "c" "a" "g" "g" "t" "g" "t" "c" "a" "a" "c" "a" "t" "g" "t" "g" "c" "a"
##  [91] "c" "c" "c" "t" "t" "a" "t" "t" "g" "c" "a" "a" "t" "g" "g" "a"

dengue1= dengueseq[450:555]
dengue1

##   [1] "c" "c" "c" "g" "a" "g" "g" "g" "g" "g" "a" "g" "a" "g" "c" "c" "g" "c"
##  [19] "a" "c" "a" "t" "g" "a" "t" "a" "g" "t" "t" "a" "g" "c" "a" "a" "g" "c"
##  [37] "a" "g" "g" "a" "a" "a" "g" "a" "g" "g" "a" "a" "a" "a" "t" "c" "a" "c"
##  [55] "t" "t" "t" "t" "g" "t" "t" "t" "a" "a" "g" "a" "c" "c" "t" "c" "t" "g"
##  [73] "c" "a" "g" "g" "t" "g" "t" "c" "a" "a" "c" "a" "t" "g" "t" "g" "c" "a"
##  [91] "c" "c" "c" "t" "t" "a" "t" "t" "g" "c" "a" "a" "t" "g" "g" "a"

denguetable<-count(dengue1,1);denguetable[[4]];denguetable[[1]]

## [1] 23

## [1] 32

#Cociente TA
23/32

## [1] 0.71875

denguetable<-count(dengue1,1);denguetable[[3]];denguetable[[2]]

## [1] 29

## [1] 22

#Cociente GC
29/22

## [1] 1.318182

#Cociente TA/GC
0.71875/1.318182

## [1] 0.5452585

#0.5452585

dengue1seq[450:555]

##   [1] "c" "a" "c" "a" "c" "g" "t" "a" "a" "c" "g" "g" "a" "g" "a" "a" "c" "c"
##  [19] "a" "c" "a" "c" "a" "t" "g" "a" "t" "c" "g" "t" "c" "a" "g" "c" "a" "g"
##  [37] "a" "c" "a" "a" "g" "a" "g" "a" "a" "a" "g" "g" "g" "a" "a" "a" "a" "g"
##  [55] "t" "c" "t" "t" "c" "t" "g" "t" "t" "t" "a" "a" "a" "a" "c" "a" "g" "a"
##  [73] "g" "g" "a" "t" "g" "g" "c" "g" "t" "g" "a" "a" "c" "a" "t" "g" "t" "g"
##  [91] "t" "a" "c" "c" "c" "t" "c" "a" "t" "g" "g" "c" "c" "a" "t" "g"

dengue2= dengue1seq[450:555]
dengue2

##   [1] "c" "a" "c" "a" "c" "g" "t" "a" "a" "c" "g" "g" "a" "g" "a" "a" "c" "c"
##  [19] "a" "c" "a" "c" "a" "t" "g" "a" "t" "c" "g" "t" "c" "a" "g" "c" "a" "g"
##  [37] "a" "c" "a" "a" "g" "a" "g" "a" "a" "a" "g" "g" "g" "a" "a" "a" "a" "g"
##  [55] "t" "c" "t" "t" "c" "t" "g" "t" "t" "t" "a" "a" "a" "a" "c" "a" "g" "a"
##  [73] "g" "g" "a" "t" "g" "g" "c" "g" "t" "g" "a" "a" "c" "a" "t" "g" "t" "g"
##  [91] "t" "a" "c" "c" "c" "t" "c" "a" "t" "g" "g" "c" "c" "a" "t" "g"

denguetable<-count(dengue2,1);denguetable[[4]];denguetable[[1]]

## [1] 19

## [1] 37

#Cociente TA
19/37

## [1] 0.5135135

denguetable<-count(dengue2,1);denguetable[[3]];denguetable[[2]]

## [1] 27

## [1] 23

#Cociente GC
27/23

## [1] 1.173913

#Cociente TA/GC
0.5135135/1.713913

## [1] 0.2996147

#0.2996147

#Los datos proporciona evidencia a favor de la Ho, ya que el cociente de TA respecto al de GC de las acceciones NC 001474 y NC 001477 es diferente.