Resource assessment exercises: finite population correction

From AWF-Wiki
Revision as of 10:42, 23 June 2014 by Lburgr (Talk | contribs)

Jump to: navigation, search
Construction.png sorry: 

This section is still under construction! This article was last modified on 06/23/2014. If you have comments please use the Discussion page or contribute to the article!


Finite population correction

If we observe all values \(y_{i\in U}\) we talk about a census. The mean is calculated, not estimated, i.e.,

    

## [1] 21.05
## [1] 21.05

As noted above, SRSwoR stands for simple random sampling without replacement (woR). However, if we take a sample with replacement (SRS; set replace = TRUE) we get a slightly different value. This would be an estimate.

       

## [1] 20.96

If we are interested in a population parameter and take an SRSwoR of size \(n=N\), we get the true population value. There is no doubt about its value (assuming that measurement errors are absent). However, if we estimate the standard error for the \(n=N\) sample we get a positive value instead of a zero \(s_{\bar{y}}\).

## [1] 0.07416

This cannot be! The estimator of the standard error holds for sampling with replacement. For sampling without replacement we have to correct for the fact that we took a relatively large sample. The finite population correction (fpc) for a relatively large sample is defined as,

\(\text{fpc}=1-\frac{n}{N}. \tag{1}\)

Obviously, if \(n=N\) the fpc becomes zero. Suppose we take a sample of size \(n=25,000\) from trees, then

     
  

## [1] 0.08099

    
  

## [1] 0.03307

  

## [1] 0.03307

For the parametric standard error the fpc becomes,

\(\text{fpc}=\frac{N-n}{N-1}. \tag{2}\)

As a rule of thumb, we apply the fpc when the sampling fraction

\(f=\frac{n}{N} \tag{3},\)

exceeds 0.05, i.e., 5 percent.

Required sample size determination

Above we took a sample S of size \(n=50\).

##  [1] 22  8 18 43 21 44 17 25 32 10 11 17  9 10 56 14 14 10 20  8 37 14 55 29 33
## [26] 17 10 15 29  8 21  9  9 24 21 28 19 58 16 16 15 20  5  9 14 30 11  9 12 27

The width of the confidence interval was,

      

## [1] 7.364

Suppose a confidence interval of \(A=3\) cm is desired. How large should the sample size, \(n\), be? This can be estimated,

\(A=t_{\alpha,n-1}\frac{s}{\sqrt{n}}\rightarrow n=\frac{t_{\alpha, n-1}^2s^2}{A^2} \tag{4}\)

We will use the sample S (\(n=50\)) to estimate how many observations we need in our sample. In :

  
      

## [1] 301.2

We always need to round up!

  

We estimate the width of the confidence interval using the new sample size of \(n=302\).

  
      

## [1] 2.759
Personal tools
Namespaces

Variants
Actions
Navigation
Development
Toolbox
Print/export