Resource assessment exercises: estimating a proportion

This article is part of the Resource assessment exercises. See the category page for a (chronological) table of contents.

As mentioned in the introduction there are two tree species in the example population: beech trees and oak trees. Suppose we would like to estimate the proportion of beech trees by looking at a sample of \(n=50\) again.      

s.species <- sample(trees$species, size = 50)
s.species

##  [1] 2 1 2 1 2 2 2 2 2 2 2 1 2 2 2 1 2 2 2 2 2 2 2 1 1 2 1 2 2 1 2 1 1 1 1 1 1 2
## [39] 2 2 2 2 2 2 2 2 2 1 1 2

The 1s are oak trees and the 2s are beech trees. The proportion of beech trees is estimated by

$\hat{p}=\frac{y_{i\in s} }{n}\quad\text{where}\quad y_i = \left\{ \begin{array}{l l} 1 & \quad \text{if} \,i\, \text{is a beech tree}\\ 0 & \quad \text{otherwise} \end{array} \right.$

1

In R:

    

n.beech <- length(s.species[s.species == 2])
n.beech

## [1] 34
  
p <- n.beech/n
p

## [1] 0.68

Indexing using the double equal sign \texttt{==}: the double equal sign is used to access entries that exactly match a condition. Above we select all entries \texttt{s.species} for which \texttt{s.species == 2} is true. Other operators include \verb|!=| (not equal to), \verb|<| (smaller than), \verb|>| (larger than), \verb|<=| (smaller or equal to), \verb|>=| (larger or equal to).

We estimated that 68% of the trees in the population are beech trees. The estimated number of oak trees in the population is \(\hat{q}=1-\hat{p}\). The standard error of the estimated proportion \(\hat{p}\) is given by,

$s_{\hat{p} }=\sqrt{\frac{\hat{p}\times\hat{q} }{n-1} }$

2

In R:         

q <- 1 - p
sqrt((p * q)/(n - 1))

## [1] 0.06664

Confidence intervals can be constructed in the same way as for the mean estimator above.

Since we know the species of each tree in the population, we can calculate the true proportion of beech trees,

nrow(trees[trees$species == 2, ])/nrow(trees) # true proportion of beech trees

## [1] 0.6702