Sampling Distributions

Example: Predicting an Election Result from an Exit Poll

Assume a 50-50 chance.

rbinom(1,2271,0.50)/2271 # binomial experiment with n=2271,pi=0.50

[1] 0.4962572

results<-rbinom(1000000,2271, 0.50)/2271 # million sample proportions,
mean(results);sd(results) # each having n =2271 and pi=0.50

[1] 0.5000074

[1] 0.0104936

hist(results) # histogram of the million sample proportions

These days simple random sampling of the population of voters is not feasible in taking a sample survey about an election.

Sampling Distribution

Law of Large Numbers

mean(runif(10,0,100))

[1] 49.41412

mean(runif(1000, 0, 100))

[1] 49.19663

mean(runif(10000000,0,100))

[1] 50.0055

Central Limit Theorem

A more precise statement:

Binomial CLT

CLT_binom <- function(B,n,p) {
# B: number of iterations used to approximate the distribution of Xmean
# n: sample size
# p: success probability pi
Y <- rbinom(B,n,p)
Ymean <- Y/n # vector (length B) with the p-estimates: Algorithm 1 (2)
var.mean <-p*(1-p)/n # variance of the estimator of p
p.MC <- mean(Ymean) # Monte Carlo estimate of p
varp.MC <- var(Ymean) # MC variance estimate of var.mean
h <- hist(Ymean, col = "gray", probability=TRUE, main=paste("n=",n))
xfit<-seq(0, 1,length=5000)
yfit<-dnorm(xfit,mean=p,sd=sqrt(p*(1-p)/n))
gr <- lines(xfit, yfit, col="blue",lwd=2)
list(var.mean=var.mean, p.MC=p.MC, varp.MC=varp.MC) }

par(mfrow=c(2,2)) # multiple graphs layout in a 2x2 table format
CLT_binom(100000, 10, 0.3)

$var.mean
[1] 0.021

$p.MC
[1] 0.299299

$varp.MC
[1] 0.02101982

CLT_binom(100000, 30, 0.3)

$var.mean
[1] 0.007

$p.MC
[1] 0.3001283

$varp.MC
[1] 0.007001842

CLT_binom(100000, 100, 0.3)

$var.mean
[1] 0.0021

$p.MC
[1] 0.3000896

$varp.MC
[1] 0.002099831

CLT_binom(100000, 1000, 0.3)

$var.mean
[1] 0.00021

$p.MC
[1] 0.3000255

$varp.MC
[1] 0.0002105071

Poisson CLT

pois_CLT <- function(n, mu, B) {
# n: vector of 2 sample sizes [e.g. n <- c(10, 100)]
# mu: mean parameter of Poisson distribution
# B: number of simulated random samples from the Poisson
par(mfrow = c(2, 2))
for (i in 1:2){
Y <- numeric(length=n[i]*B)
Y <- matrix(rpois(n[i]*B, mu), ncol=n[i])
Ymean <- apply(Y, 1, mean) # or, can do this with rowMeans(Y)
barplot(table(Y[1,]), main=paste("n=", n[i]), xlab="y",
col="lightsteelblue") # sample data dist. for first sample
hist(Ymean, main=paste("n=",n[i]), xlab=expression(bar(y)),
col="lightsteelblue") # histogram of B sample mean values
} }

# implement:with 100000 random sample sizes of 10 and 100, mean = 0.7
n <- c(10, 100)
pois_CLT(n, 0.7, 100000)

Demonstrations: Web Apps (artofstat.com)

Exercises

Verify Fig. 3.4 (below) this using simulations.