INTERVALO DE CONFIANZA PARA EL TEST DEL SIGNO

n<-60
intsig<-function(n,a){c(n/2-1/2-1/2*qnorm(1-a/2)*sqrt(n),n/2+1/2+1/2*qnorm(1-a/2)*sqrt(n))}
intsig(60,0.05)
## [1] 21.90909 38.09091
x<-c(43,55,59,60,67,73,73,73,73,73,73,75,75,76,78,78,78,79,79,80,80,80,80,80,80,80,82,82,82,82,83,83,83,83,83,84,84,84,85,85,86,86,87,87,88,88,88,88,88,88,88,89,89,89,89,90,92,93,94,98)
x
##  [1] 43 55 59 60 67 73 73 73 73 73 73 75 75 76 78 78 78 79 79 80 80 80 80 80 80
## [26] 80 82 82 82 82 83 83 83 83 83 84 84 84 85 85 86 86 87 87 88 88 88 88 88 88
## [51] 88 89 89 89 89 90 92 93 94 98
yord<-sort(x)
yord
##  [1] 43 55 59 60 67 73 73 73 73 73 73 75 75 76 78 78 78 79 79 80 80 80 80 80 80
## [26] 80 82 82 82 82 83 83 83 83 83 84 84 84 85 85 86 86 87 87 88 88 88 88 88 88
## [51] 88 89 89 89 89 90 92 93 94 98
c(yord[21],yord[39])
## [1] 80 85
plot(density(x))

shapiro.test(x)
## 
##  Shapiro-Wilk normality test
## 
## data:  x
## W = 0.87764, p-value = 2.212e-05

Se rechaza H_0 de normalidad por eso tiene sentdo hacer un intervalo de cofnianza del test del signo.

Prueba de hipotesis a una cola

\[ H_0 : \theta \geq15 \] \[ H_0 : \theta <15 \]

zc<-(40 - 120/2)/(sqrt(120)/2)
zc
## [1] -3.651484
qnorm(0.05)
## [1] -1.644854

\[ \sum_{i=1}^n i = \frac{n(n+1)}{2} \] \[ \bar R = \sum_{i=1}^n i / n= \frac{n(n+1)}{2n} = (n+1)/2 \]

Intervalo de confianza

82 66 90 84 75 88 80 94 110 91

x<-c(82, 66, 90, 84, 75, 88, 80, 94, 110, 91)
n<- length(x)
n
## [1] 10
intsig(n,0.1)
## [1] 1.899258 8.100742
yor<-sort(x)
yor
##  [1]  66  75  80  82  84  88  90  91  94 110
c(yor[2],yor[8])
## [1] 75 91