Question

A. Using again the black cherry tree data set trees in R, carry out
simulations to compare the following sampling strategies for estimating
total volume of wood in the population. Here the population size N is
31 and the variable of interest is tree volume.

Strategy 1: The design simple random sampling, sample size n = 15, and
the expansion estimator tau hat = N ybar, where ybar is the sample mean.

Strategy 2: Unequal probability sampling of trees, with replacement,
with n = 15 independent draws, and the probability of drawing tree i
on any draw is proportional to tree girth (size of tree near the
ground). Use the Horvitz-Thompson estimator of the population total.
Note that the details of selecting units with unequal probability for
this design are given in the computing section of Chapter 6.

Your comparison of the two strategies should include graphics such as
histograms of estimates and calculations from the simulations of bias
and mean square error.

Write a one or two sentence conclusion on your comparison.

PLZ show R code!!

Answer 1

library(datasets)

d=trees;d

y=d$Volume;x=d$Girth

N=31;n=15;p=0.15

M=mean(y);M

#Strategy 1

ybar=numeric()

for(i in 1:100)

{

s=sample(1:N,n)

ybar[i]=mean(y[s])

}

ybar

hist(ybar)

tau_hat=N*ybar;tau_hat #Expansion estimator

#Bias and MSE

mean(ybar)

mean(ybar-M)

var(ybar)

mean((ybar-M)^2)

#Strategy 2

#Sample mean with unequality probability sample.

mean(y[s])

unique(s)

m1=mean(y[s]/x[s]);m1

p=x/sum(x)

pinc=1-(1-(p))^n;pinc #inclusion probabilities

b=sum(y[unique(s)]/pinc[unique(s)])/N;b #Horvitz estimayor

t=unique(s);y[t]

#Bias and MLE

mean(m1)

mean(m1-mu)

var(m1)

mean((m1-M)^2)

#Conclusion

#Sampling distributions of the two strategies are compared by drawing the histogram of ybar

****