n.ajstu <- 100000
mean.ajstu <- 100
sd.ajstu <- 10

set.seed(1024)
ajstu <- rnorm2(n.ajstu, mean=mean.ajstu, sd=sd.ajstu)

mean(ajstu)
sd(ajstu)
var(ajstu)

iter <- 10000 # # of sampling 

n.4 <- 4
means4 <- rep (NA, iter)
for(i in 1:iter){
  means4[i] = mean(sample(ajstu, n.4))
}

n.25 <- 25
means25 <- rep (NA, iter)
for(i in 1:iter){
  means25[i] = mean(sample(ajstu, n.25))
}

n.100 <- 100
means100 <- rep (NA, iter)
for(i in 1:iter){
  means100[i] = mean(sample(ajstu, n.100))
}

n.400 <- 400
means400 <- rep (NA, iter)
for(i in 1:iter){
  means400[i] = mean(sample(ajstu, n.400))
}

n.900 <- 900
means900 <- rep (NA, iter)
for(i in 1:iter){
  means900[i] = mean(sample(ajstu, n.900))
}

n.1600 <- 1600
means1600 <- rep (NA, iter)
for(i in 1:iter){
  means1600[i] = mean(sample(ajstu, n.1600))
}

n.2500 <- 2500
means2500 <- rep (NA, iter)
for(i in 1:iter){
  means2500[i] = mean(sample(ajstu, n.2500))
}

h4 <- hist(means4)
h25 <- hist(means25)
h100 <- hist(means100)
h400 <- hist(means400)
h900 <- hist(means900)
h1600 <- hist(means1600)
h2500 <- hist(means2500)


plot(h4, ylim=c(0,3000), col="red")
plot(h25, add = T, col="blue")
plot(h100, add = T, col="green")
plot(h400, add = T, col="grey")
plot(h900, add = T, col="yellow")


sss <- c(4,25,100,400,900,1600,2500) # sss sample sizes
ses <- rep (NA, length(sss)) # std errors
for(i in 1:length(sss)){
  ses[i] = sqrt(var(ajstu)/sss[i])
}

ses
se.1 <- ses
se.2 <- 2 * ses 

lower.s2 <- mean(ajstu)-se.2
upper.s2 <- mean(ajstu)+se.2
data.frame(cbind(sss, ses, lower.s2, upper.s2))

# n =1600 일 경우에 
# sample의 평균이 100.15보다 작을 
# 확률은 어떻게 구해야 할까?

# n = 1600 일 경우에 
# sampling distribution은 
# Xbar ~ N(100, var(ajstu)/n.1600)
# 그리고, 위에서 standard error값은 
# sqrt(var(ajstu)/n.1600)
# 이것을 standard error라고 부른다
# 따라서
se.1600 <- sqrt(var(ajstu)/n.1600)
pnorm(100.15, mean(ajstu), se.1600)

pop <- rbinom(100000, size = 1, prob = 0.5)
par(mfrow=c(2,2)) 
iter <- 10000
n <- 5
means <- rep (NA, iter)
for(i in 1:iter){
    means[i] = mean(sample(pop, n))
}
mean(means)
hist(means, xlim=c(0,1), main=n)

iter <- 10000
n <- 25
means <- rep (NA, iter)
for(i in 1:iter){
    means[i] = mean(sample(pop, n))
}
mean(means)
hist(means, xlim=c(0,1), main=n)

iter <- 10000
n <- 100
means <- rep (NA, iter)
for(i in 1:iter){
    means[i] = mean(sample(pop, n))
}
mean(means)
hist(means, xlim=c(0,1), main=n)

iter <- 10000
n <- 900
means <- rep (NA, iter)
for(i in 1:iter){
    means[i] = mean(sample(pop, n))
}
mean(means)
sd(means)
var(means)
hist(means, xlim=c(0,1), main=n)

par(mfrow=c(1,1)) 

set.seed(2020)
pop <- rbinom(100000, size = 1, prob = 0.4)
par(mfrow=c(2,2)) 
iter <- 1000
ns <- c(25, 100, 400, 900)
l.ns <- length(ns)
for (i in 1:l.ns) {
    for(k in 1:iter) {
        means[k] = mean(sample(pop, ns[i]))
    }
    mean(means)
    sd(means)
    hist(means, xlim=c(0,1), main=n)
}
par(mfrow=c(1,1)) 

0.5가 비율인 (proportion) 모집단에 대한 여론 조사를 위해서 900명의 샘플을 취하고 이를 이용하여 모집단의 위치를 추정하자.

n <- 900
samp <- sample(pop, n)
mean(samp)
p <- mean(samp)
q <- 1-p
ser <- sqrt((p*q)/n)
ser2 <- ser * 2
p - ser2 
p + ser2