Prueba exacta de Poisson para una muestra

poisson.test:

  • Parámetro de tasa (λ)

Hipotesis 1: \[H0:λi≤10 \frac{casos}{día}\\Ha:λi>10\frac{casos}{día}\]

# El muestreo se hizo en un lapso de 15 días
# Se encontraron 162 casos nuevos en los 15 días
# Incidencia estimada 162/15 = 10.8
poisson.test(x = 162, T = 15, r = 10, alternative = 'g')
## 
##  Exact Poisson test
## 
## data:  162 time base: 15
## number of events = 162, time base = 15, p-value = 0.1734
## alternative hypothesis: true event rate is greater than 10
## 95 percent confidence interval:
##  9.443211      Inf
## sample estimates:
## event rate 
##       10.8

Hipotesis 2: \[H_{0}:λ_{Lote_{1}}=λ_{Lote_{2}}\\H_{a}:λ_{Lote_{1}}≠λ_{Lote_{2}}\]

# El muestreo en 2 LOTES se hizo en un lapso de 15 días
# Lote 1: 54 casos nuevos en los 15 días
# Lote 2: 62 casos nuevos en los 15 días
# La hipotesis es r = 1 por lambda_lote1/lambda_lot2 = 1
# ratio muestra: (54/15)/(62/15) = 0.871
poisson.test(x = c(54, 62), T = 15, r = 1, alternative = 't')
## 
##  Comparison of Poisson rates
## 
## data:  c(54, 62) time base: 15
## count1 = 54, expected count1 = 58, p-value = 0.5159
## alternative hypothesis: true rate ratio is not equal to 1
## 95 percent confidence interval:
##  0.5933221 1.2750906
## sample estimates:
## rate ratio 
##  0.8709677

Prueba de hipotesis para un promedio

Pruebas paramétricas: Media

Datos: 1 muestra

Condiciones de validez: n > 30 o normalidad (Prueba de normalidad de Shapiro-Wilk y gráfico de normalidad)

Función R: t.test(x,…)

  • Ejemplo de uso:

\(\mu\) = media poblacional

\(\bar{x}\) = media muestral

\[\bar{x} \rightarrow \mu\] Hipótesis :

\[H0:μ_{pH}≥5\\Ha:μ_{pH}<5\]

set.seed(123)
pH = runif(n = 40, min = 4, max = 5.3)
mean(pH)
## [1] 4.732894
hist(pH,border = 4,col = "white" )

Simulacion Monte-Carlo

  • Generar una muestra de gran tamaño (\(n\))
set.seed(49)

pH_sim = replicate(n = 100,
                   runif(n = 40, min = 4, max = 5.3))
dim(pH_sim)
## [1]  40 100
media_sim = colMeans(pH_sim)
hist(media_sim,border = 4,col = "white")
abline(v=mean(pH), col='blue', lty = 2)

res_tt = t.test(x = pH, mu = 5, alternative = 'l')
ifelse(res_tt$p.value<0.05, 'Rechazo Ho', 'No rechazo Ho')
## [1] "Rechazo Ho"

\[H0:μ_{NDVI_{Directa}}≥μ_{NDVI_{Indirecto}}\\Ha:μ_{NDVI_{Directa}}<μ_{NDVI_{Indirecto}}\]

set.seed(123)

ndvi_directo = runif(n = 60, min = 0.1, max = 0.35)
ndvi_indirecto = runif(n = 60, min = 0.3, max = 0.70)

mean(ndvi_directo)
## [1] 0.2263801
mean(ndvi_indirecto)
## [1] 0.5063734
boxplot(cbind(
  ndvi_directo,
  ndvi_indirecto
),
col = "white",
border = 4
)

\[H0:σ^2_{NDVI_{Directa}}=σ^2_{NDVI_{Indirecto}}\\Ha:σ^2_{NDVI_{Directa}}≠σ^2_{NDVI_{Indirecto}}\]

var(ndvi_indirecto)/var(ndvi_directo)
## [1] 2.275008
var.test(x = ndvi_directo, y = ndvi_indirecto, ratio = 1,
         alternative = 't')
## 
##  F test to compare two variances
## 
## data:  ndvi_directo and ndvi_indirecto
## F = 0.43956, num df = 59, denom df = 59, p-value = 0.001928
## alternative hypothesis: true ratio of variances is not equal to 1
## 95 percent confidence interval:
##  0.2625593 0.7358798
## sample estimates:
## ratio of variances 
##           0.439559
t.test(x = ndvi_directo, y = ndvi_indirecto,
       mu = 0, paired = FALSE, var.equal = FALSE,
       alternative = 'l')
## 
##  Welch Two Sample t-test
## 
## data:  ndvi_directo and ndvi_indirecto
## t = -16.196, df = 102.47, p-value < 2.2e-16
## alternative hypothesis: true difference in means is less than 0
## 95 percent confidence interval:
##        -Inf -0.2512988
## sample estimates:
## mean of x mean of y 
## 0.2263801 0.5063734
ndvi_directo_2 = sort.int(ndvi_directo, 30)
ndvi_indirecto_2 = sort.int(ndvi_indirecto, 20)
plot(ndvi_indirecto_2, type = 'l', ylim = c(0,0.7))
points(ndvi_directo_2, type = 'l', col = 4)

abline(h = mean(ndvi_indirecto_2))
abline(h = mean(ndvi_directo_2), col = 4)

d_ndvi_d_2 = diff(ndvi_directo_2)
d_ndvi_i_2 = diff(ndvi_indirecto_2)

plot(d_ndvi_d_2, type = 'l', ylim = c(-0.2,0.2))
points(d_ndvi_i_2, type = 'l', col = 4)

abline(h = mean(d_ndvi_d_2))
abline(h = mean(d_ndvi_i_2), col=4)

var.test(d_ndvi_d_2, d_ndvi_i_2,
         ratio = 1, alternative = 't')
## 
##  F test to compare two variances
## 
## data:  d_ndvi_d_2 and d_ndvi_i_2
## F = 0.19702, num df = 58, denom df = 58, p-value = 4.507e-09
## alternative hypothesis: true ratio of variances is not equal to 1
## 95 percent confidence interval:
##  0.1171544 0.3313260
## sample estimates:
## ratio of variances 
##          0.1970185
t.test(d_ndvi_d_2, d_ndvi_i_2,
       mu = 0, alternative = 't', var.equal = FALSE)
## 
##  Welch Two Sample t-test
## 
## data:  d_ndvi_d_2 and d_ndvi_i_2
## t = -0.1834, df = 80, p-value = 0.8549
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -0.01892964  0.01573499
## sample estimates:
##   mean of x   mean of y 
## 0.003513990 0.005111318

Prueba de hipotesis para dos muestras pareadas

Pruebas paramétricas: Media

Datos: 2 muestras

Condiciones de validez: n > 30 o normalidad (Prueba de normalidad de Shapiro-Wilk y gráfico de normalidad)

Función R: t.test(x,y,paired=T)

Hipotesis 1: \[H0:μ_{AE_{corte 1}}=μ_{AE_{corte 2}}\\Ha:μ_{AE_{corte 1}}≠_{AE_{corte 2}}\]

set.seed(123)

AE_c1 = rnorm(30, 7, 0.5)
AE_c2 = rnorm(30, 6.5, 0.45)

t.test(x = AE_c1, y = AE_c2,
       mu = 0, paired = TRUE,
       alternative = 't')
## 
##  Paired t-test
## 
## data:  AE_c1 and AE_c2
## t = 3.2719, df = 29, p-value = 0.00276
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  0.1485385 0.6438532
## sample estimates:
## mean of the differences 
##               0.3961959
boxplot(cbind(AE_c1, AE_c2),
        col = "white",
        border = 4)
points(1:2, c(mean(AE_c1), mean(AE_c2)), pch=16, col='blue')