Chapter 12 Sample surveys

  • Simulate a survey of 25 people taken from a population of size 200 with 2 types of people in the proportion 1:3.

  • This is equivalent to taking 25 draws from a box containing 50 1’s and 150 0’s.

  • Note: A survey is technically without replacement, but here we will approximate by the box model which assumes replacement.


12.1 Box model (assuming replacement)

  • Draw the model (Ext)
library("DiagrammeR")

grViz(" 
  digraph CFA {

  # All
  node [fontname = Helvetica, fontcolor = White]

    # Box
    node [shape = box, style=filled, color=SteelBlue4, width=2 label='50x1s  150x0s'][fillcolor = SteelBlue4]
    a ; 

    # Sample
    node [shape = circle, style=filled, color=IndianRed, width=0.5, label='-'][fillcolor = IndianRed]
    b ; 

    # Draws
    a -> b [fontname = Helvetica,label = '25 draws',fontsize=8]
    b -> a  [fontname = Helvetica,color=grey,arrowsize = 0.5]
  }

")
  • Define the box (mean, SD)
box=c(rep(1,50),rep(0,150))  # Define the box/population
n = 25 # Number of draws (sample size)

# mean & SD of box
meanbox = mean(box)
sdbox = popsd(box)
  • The Sample Sum = number of 1’s in the sample.
n*meanbox  # EV
sqrt(n)*sdbox  # SE
  • The Sample Mean = proportion of 1’s in the sample (EV & SE)
meanbox  # EV
sdbox/sqrt(n)  # SE


12.2 Bootstrapping a box model

  • A survey of 10 people is taken from a population of size 200 with 2 types of people, resulting in 4 0’s and 6 1’s.
  • In practise, we don’t normally know the proportions within the population, and so we can’t define the “box”.
  • So one approach is to “bootstrap”: we substitute the proportions of 0’s and 1’s in the sample for the 0’s and 1’s in the population (box).
  • This estimate is good when the sample is reasonably large.
# Bootstrap box model
boxest0 = 200*4/10
boxest1 = 200*6/10
box = c(rep(1, boxest1), rep(0, boxest0))
n = 10 # Size of sample (= number of draws)

# Mean & SD of box
meanbox = mean(box)  # population proportion of 1's
sdbox = popsd(box)

# EV & SE of Sample Mean (= sample proportion of 1's)
meanbox
sdbox/sqrt(n)

  • Note the expected value of sample proportion of 1’s (0.6), is equal to the population proportion of 1s (0.6), as it should be.

  • Hence, we expect 60% 1s in the sample with a SE of around 15 percentage points.


12.3 Confidence Intervals for population proportion

  • A survey of 30 people is taken from a population of size 600 with 2 types of people in the proportion 1:2.

  • This is equivalent to taking 30 draws from a box containing 200 1’s and 400 0’s (assuming replacement).

  • Define the box (mean, SD)

box=c(rep(1,200),rep(0,400))  # Define the box/population
n = 30 # Number of draws (sample size)

# mean & SD of box
meanbox = mean(box)
sdbox = popsd(box)
  • An approximate 68% confidence interval (CI) for the population proportion is \[\mbox{sample proportion} \pm 1 \times \mbox{SE for sample proportion} \]
# EV & SE of sample proportion of 1's
ev = meanbox
se = sdbox/sqrt(n)

# 68% CI
c(ev - 1*se,ev + 1*se)
  • An approximate 95% confidence interval (CI) for the population percentage is \[\mbox{sample proportion} \pm 2 \times \mbox{SE for sample proportion} \]
# 95% CI
c(ev - 2*se,ev + 2*se)
  • An approximate 99.7% confidence interval (CI) for the population percentage is \[\mbox{sample proportion} \pm 3 \times \mbox{SE for sample proportion} \]
# 99.7% CI
c(ev - 3*se,ev + 3*se)