TALLER 421
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Problema 1
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x_bar <- 3250
sigma2 <- 1000
sigma <- sqrt(sigma2)
n <- 12
# (a)(b). Intervalos de confianza
nivel <- 0.95 # Cambiar el valor: 0.90,0.95 y 0.99
alpha <- 1 - nivel
z <- qnorm(1 - alpha/2)
error <- z * sigma / sqrt(n)
LI <- x_bar - error
LS <- x_bar + error
IC <- c(LI, LS)
IC
## [1] 3232.108 3267.892
# Ancho del intervalo
LS - LI
## [1] 35.78388
2*error
## [1] 35.78388
# (c). Tamaño de muestra requerido
E <- 15 # error máximo permitido
conf <- 0.99 # nivel de confianza
alpha <- 1 - conf
z <- qnorm(1 - alpha/2)
n_R <- (z * sigma / E)^2
n_R
## [1] 29.48843
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Problema 2
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x <- c(16.8,17.2,17.4,16.9,16.5,17.1)
shapiro.test(x) # (a) prueba de la normalidad
##
## Shapiro-Wilk normality test
##
## data: x
## W = 0.98779, p-value = 0.9831
# (b) Estadísticos muestrales
n <- length(x)
media <- mean(x)
s <- sqrt(var(x))
conf <- 0.99 # nivel de confianza
alpha <- 1 - conf # Nivel de significancia
t <- qt(1 - alpha/2, df = n - 1) # Valor critico
error <- t * s / sqrt(n)
LI <- media - error
LS <- media + error
IC <- c(LI, LS)
IC
## [1] 16.45847 17.50820
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Problema 3
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n <- 51 # Tamaño de muestra
s <- 0.37 # desviacion estandar
conf <- 0.95 # confianza
alpha <- 1 - conf # Significancia
gl <- n - 1 # grados de libertad
# (a) Pregunta - Supuesto: Normalidad de la variable
Chi.Der <- qchisq(alpha/2 , df = gl)
Chi.Izq <- qchisq(1-alpha/2, df = gl)
Li <- (n-1)*s^2/Chi.Izq
Ls <- (n-1)*s^2/Chi.Der
c(Li,Ls)
## [1] 0.09584124 0.21154381
c(sqrt(Li),sqrt(Ls))
## [1] 0.3095824 0.4599389
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Problema 4
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n <- 1000
x <- 853
conf <- 0.95
alpha <- 1 - conf
p <- x/n
# Supuestos: np>5 o n(1−p)>5
n*p > 5
## [1] TRUE
n*(1-p) > 5
## [1] TRUE
Media <- p
Var <- p*(1-p)/n
z <- qnorm(1 - alpha/2)
Li <- p - z * sqrt(p*(1-p)/n)
Ls <- p + z * sqrt(p*(1-p)/n)
c(Li,Ls)
## [1] 0.8310527 0.8749473
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Problema 5
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Datos <- c(1.542, 1.622, 1.440, 1.459, 1.598, 1.585, 1.466, 1.608, 1.533, 1.498,
1.532, 1.546, 1.520, 1.532, 1.600, 1.466, 1.494, 78.000, 1.523, 1.504,
1.499, 1.548, 1.542, 1.397, 1.545, 1.611, 1.626, 1.511, 1.487, 1.558)
# (a) Normalidad de los datos: Ho. Normalidad
# ---------------------------------------------------------------------------- #
XSin.atipicos <- Datos[Datos < 10]
shapiro.test(Datos)
##
## Shapiro-Wilk normality test
##
## data: Datos
## W = 0.18266, p-value = 8.189e-12
shapiro.test(XSin.atipicos) # Sin valores atipicos
##
## Shapiro-Wilk normality test
##
## data: XSin.atipicos
## W = 0.97294, p-value = 0.6417
X <- XSin.atipicos
# (b) Intervalo de confianza del 99% para la media del contenido de alquitra
# ---------------------------------------------------------------------------- #
n <- length(X)
x_bar <- mean(X)
sd <- sd(X)
alpha <- 0.01
conf <- 1 - alpha
# (+) IC: Para la media - Varianza Poblacional desconcida
z <- qnorm(1 - alpha/2)
error <- z * sd / sqrt(n)
LI <- x_bar - error
LS <- x_bar + error
IC <- c(LI, LS)
IC
## [1] 1.503627 1.557890
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Problema 6
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X <- c(3, 4, 2, 5, 4, 8, 2, 9, 3, 6, 2, 8, 3, 3, 5, 6, 3,
7, 2, 8, 4, 4, 4, 0, 5, 2, 3, 0, 4, 8)
shapiro.test(X) # Ho. Normalidad
##
## Shapiro-Wilk normality test
##
## data: X
## W = 0.93955, p-value = 0.08848
# (a) IC para la media - Confianza del 95%
# ---------------------------------------------------------------------------- #
n <- length(X) # Tamaño de muestra
x_bar <- mean(X)
sd <- sd(X) # desviacion estandar
conf <- 0.95 # confianza
alpha <- 1 - conf # Significancia
z <- qnorm(1 - alpha/2)
error <- z * sd / sqrt(n)
LI <- x_bar - error
LS <- x_bar + error
ICmedia <- c(LI, LS)
ICmedia
## [1] 3.378833 5.087834
# (b) IC para varianza - Confianza del 98%
# ---------------------------------------------------------------------------- #
gl <- n - 1 # grados de libertad
Chi.Der <- qchisq(alpha/2 , df = gl) # Chi (α/2)
Chi.Izq <- qchisq(1-alpha/2, df = gl) # Chi (1-α/2)
Li <- (n-1)*sd^2/Chi.Izq
Ls <- (n-1)*sd^2/Chi.Der
ICvar <- c(Li,Ls)
ICvar
## [1] 3.616763 10.305099
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Problema 7
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# Tamaño de muestra (n)
sigma <- 40
E <- 15
alpha <- 0.05
z <- qnorm(1 - alpha/2)
n <- (z*sigma/E)^2
n <- ceiling(n)
n
## [1] 28
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Problema 8
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CE <- c(428, 419, 458, 439, 441, 456, 463, 429, 438, 445, 441, 463)
CD <- c(462, 448, 435, 465, 429, 472, 453, 459, 427, 468, 452, 447)
# (a) Obtener un intervalo de confianza del 95% para la diferencia de medias.
# ---------------------------------------------------------------------------- #
di <- CE - CD
n <- length(di)
Xm <- mean(di)
sd <- sd(di)
alpha <- 0.05
t <- qt(1 - alpha/2, df = n - 1)
error <- t*sd/sqrt(n)
li <- Xm - error
ls <- Xm + error
c(li,ls)
## [1] -21.417728 5.251061
t.test(CE,CD, paired = TRUE, # pareadas
conf.level = 0.95)$conf.int # nivel de confianza
## [1] -21.417728 5.251061
## attr(,"conf.level")
## [1] 0.95
# (b) Determinar si existe una diferencia real entre las dos campañas. ¿Cuál es más conveniente?
# ---------------------------------------------------------------------------- #
n1 <- length(CD) ; n2 <- length(CD)
x1m <- mean(CE) ; x2m <- mean(CD)
s1 <- sd(CE) ; s2 <- sd(CD)
# Condicion 1. Normalidad (Ambas muestras)
shapiro.test(CE)
##
## Shapiro-Wilk normality test
##
## data: CE
## W = 0.94118, p-value = 0.5135
shapiro.test(CD)
##
## Shapiro-Wilk normality test
##
## data: CD
## W = 0.94131, p-value = 0.5152
# Condicion 2. Igualdad de varianzas o no
var.test(CE,CD) # No se rechaza Ho
##
## F test to compare two variances
##
## data: CE and CD
## F = 0.91354, num df = 11, denom df = 11, p-value = 0.8835
## alternative hypothesis: true ratio of variances is not equal to 1
## 95 percent confidence interval:
## 0.2629879 3.1733660
## sample estimates:
## ratio of variances
## 0.9135409
# [1]. IC. Varianzas iguales
gl <- n1+n2-2
sp.cua <- sqrt(((n1-1)*s1^2 + (n2-1)*s2^2)/gl) # Desviación estándar combinada (o agrupada)
To <- (x1m-x2m)/ sqrt(sp.cua)* sqrt(1/n1 + 1/n2) # Estadistico de prueba
t.Crit <- qt(1 - alpha/2, df = gl)
To > t.Crit # Rechaza Ho si TRUE
## [1] FALSE
# (+) automatico
t.test(CE, CD, conf.level = 0.95, var.equal = TRUE)
##
## Two Sample t-test
##
## data: CE and CD
## t = -1.355, df = 22, p-value = 0.1892
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -20.455121 4.288454
## sample estimates:
## mean of x mean of y
## 443.3333 451.4167
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Problema 9
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n <- 87 # Estaciones
x <- 13 # Estaciones con fuga
# (a) IC para la proporcion al 95%
# ---------------------------------------------------------------------------- #
conf <- 0.95
alpha <- 1 - conf
p <- x/n # Proporcion muestral
# Supuestos: np>5 o n(1−p)>5
n*p > 5
## [1] TRUE
n*(1-p) > 5
## [1] TRUE
Media <- p
Var <- p*(1-p)/n
z <- qnorm(1 - alpha/2)
Li <- p - z * sqrt(p*(1-p)/n)
Ls <- p + z * sqrt(p*(1-p)/n)
c(Li,Ls)
## [1] 0.07451236 0.22433821
prop.test(x,n, conf.level = 0.95)
##
## 1-sample proportions test with continuity correction
##
## data: x out of n, null probability 0.5
## X-squared = 41.379, df = 1, p-value = 1.254e-10
## alternative hypothesis: true p is not equal to 0.5
## 95 percent confidence interval:
## 0.08505843 0.24562438
## sample estimates:
## p
## 0.1494253
# (b) Tamaño de muestra y ajuste para poblacion finita
# ---------------------------------------------------------------------------- #
e <- 0.03
nCal <- (z^2)*p*(1-p)/e^2
nCal
## [1] 542.4881