##############################
### RESENJA NEKIH ZADATAKA ###
##############################
# 1.
trees
N <- length(trees$Girth)
s <- sample(1:N,10)
mean(trees[s,]$Volume)
# verovatnoce izbora
(p <- trees$Girth/sum(trees$Girth))
sum(p)
# izbor uzorka sa ponavljanjem sa verovatnocom izbora proporcionalnoj girth
n <- 10
(s <- sample(1:N, n, repl=T, prob=p))
(uzorak <- trees$Volume[s])
(m <- mean(uzorak))
# 2.
baza1 <- read.table("D:\...\baza1.txt")
baza1
s<-sample(length(baza1$V1), 8)
uzorak<-baza1[s,]
mean(uzorak$V2)
# 3.
baza2 <- read.table("D:\...\baza2.txt")
baza2
p_i <- baza2$V1/sum(baza2$V1)
s <- sample(1:10,3,prob = p_i,replace = T)
(uzorak <- baza2[s,])
(ocena_proiz <- sum(uzorak$V2/p_i[s])/3)
# 4. 
baza3 <- read.table("D:/.../baza3.txt")
attach(baza3)
n <- 30
(N <- length(V1))
(N1 <- length(baza3[V1==1,]$V1))
(N2 <- length(baza3[V1==2,]$V1))
(N3 <- length(baza3[V1==3,]$V1))
(n1 <- round(n/N*N1))
(n2 <- round(n/N*N2))
(n3 <- round(n/N*N3))
(uzorak_str1 <- baza3[V1==1,][sample(1:N1,n1),])
(uzorak_str2 <- baza3[V1==2,][sample(1:N2,n2),])
(uzorak_str3 <- baza3[V1==3,][sample(1:N3,n3),])
x_h <- c(mean(uzorak_str1$V2),mean(uzorak_str2$V2),mean(uzorak_str3$V2))
N_h <- c(N1,N2,N3)
(broj_poseta <- sum(x_h*N_h)/N)
n_h <- c(n1,n2,n3)
S2_h <- c(var(baza3[V1==1,]$V2),var(baza3[V1==2,]$V2),var(baza3[V1==3,]$V2))
(disp_ocene <- sum(N_h*S2_h)*(N-n)/(N*N*n))
s2_h <- c(var(uzorak_str1$V2),var(uzorak_str2$V2),var(uzorak_str3$V2))
(ocena_disp <- sum(N_h*s2_h)*(N-n)/(N*N*n))
# 5.
baza4 <- read.delim("D:/.../baza4.txt", header=FALSE)
(y_srednje <- 5475/75)
(b <- cov(baza4$Y,baza4$X)/var(baza4$Y))
(x_lr <- mean(baza4$X)+b*(y_srednje-mean(baza4$Y)))
(t_lr <- 75*x_lr)
# 6. 
M <- 900
k <- round(900/120)
start <- sample(1:k, 1)
(s <- seq(start, M, k))
(uzorak <-c(s,cumsum(c(max(s)+k-M,rep(k,120-round(900/k)-1)))[1:(120-length(s))]))