# Lecture 4 by yaofenji

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FOR 220 Aerial Photo
Interpretation and
Forest Measurements

Lecture 4

Sampling Designs

Avery and Burkhart,
Chapter 3
Why/Why Not Randomly Select Samples?

We will discuss several sampling designs in this lecture

• Most statistical methods assume simple random sampling
was used.

• Sampling methods, however, will vary depending on the
objectives of the survey, the nature of the population being
sampled, and prior information about the population being
sampled.

FOR 220 Aerial Photo Interpretation and Forest Measurements
Why Not Measure Everything?

Complete Enumeration (Census)

Measure every feature of interest.
The result is a highly accurate description of the population.

Drawbacks:
Only viable with small populations.
Only cost-effective with high-valued features.

FOR 220 Aerial Photo Interpretation and Forest Measurements
Sampling Frame

Sampling Frame
The list of all possible units that might be drawn in a sample.

• Developing a reliable frame may be difficult (e.g., campgrounds)

• Access

How is a sampling frame different than a population?

FOR 220 Aerial Photo Interpretation and Forest Measurements
Sampling Designs

Sampling Design

The method of selecting [non-overlapping]
sample units to be included in a sample.

FOR 220 Aerial Photo Interpretation and Forest Measurements
Sampling Designs

Sampling Designs Covered Today:

1. Simple Random Sampling

2. Systematic Sampling

3. Stratified Random Sampling

FOR 220 Aerial Photo Interpretation and Forest Measurements
Sampling Design I: Simple Random Sampling

Assumptions:
Every possible combination of sampling units has an equal
and independent chance of being selected.

The selection of a particular unit to be sampled is not
influenced by the other units that have been selected or will
be selected.

Samples are either chosen with replacement or without
replacement.

FOR 220 Aerial Photo Interpretation and Forest Measurements
Sampling Design I: Simple Random Sampling

With and without replacement?

Small populations – samples may “remove” considerable
portion. Removes “independence”

N 12
n -6
5

FOR 220 Aerial Photo Interpretation and Forest Measurements
Sampling Design I: Simple Random Sampling

Design:

FOR 220 Aerial Photo Interpretation and Forest Measurements
Sampling Design I: Simple Random Sampling
We use the familiar equations for estimating common statistics for the population

Estimate the population                                        x        
mean:                                                       x 
 n        

          

Estimate the variance of
2           
  x 2   x 2 / n
s 

individual values:                                        
        n 1            


Compute the coefficient
CV  
s
of variation:                                                         
x

FOR 220 Aerial Photo Interpretation and Forest Measurements
Sampling Design I: Simple Random Sampling

Compute the standard error:

s2                              s2  N  n 
SE                             SE            
n  N 
OR
n                                         

(with replacement, or                (without replacement and
infinite population)                 from a finite population)

Compute confidence limits:
t = t statistic from table,
95% CI  x  t SE             determined by degrees of
freedom (n-1).

In general, it is safe to use: t = 2.0 for small n (n < 30), or t = 1.96 for large n (n > 30)

FOR 220 Aerial Photo Interpretation and Forest Measurements
Sampling Design I: Simple Random Sampling

Finite population term:
 N n
N = Population size                                     
n = sample size                                     N 

s2  N  n 
SE      N 
                             Standard error (without replacement
n                               and from a finite population)

FOR 220 Aerial Photo Interpretation and Forest Measurements
Sampling Design I: Simple Random Sampling

How many samples to take?

• Sample size should be statistically and practically
efficient.

• Enough sample units should be measured to obtain
the desired level of precision (no more, no less).

FOR 220 Aerial Photo Interpretation and Forest Measurements
Sampling Design I: Simple Random Sampling

Determine sample size (with replacement, or infinite population)

Knowing the standard deviation and the desired                             2
t s 
objective (e.g., to be within xxx board feet per acre)       n       
we can calculate the necessary minimum sample size.               E 
Stein’s Formula

Example:

s = 2000 board feet / acre                                             2
 2 (2000) 
t = 2 (assuming a 95% confidence interval)
 500  64
n           
E = want to be within +/- 500 board feet / acre                       

(E must be in same units as std. deviation)

FOR 220 Aerial Photo Interpretation and Forest Measurements
Sampling Design I: Simple Random Sampling

Determine sample size (with replacement, or infinite population):

Knowing the coefficient of variation and the                            2
 t CV   
desired objective (e.g., error is expected to be        n 
 A      

within x % of the value of the mean)                               

Example:

CV = 30%
t = 2 (assuming a 95% confidence                       2
 2 (30) 
interval)                                     5  144
n         
        
A = we want the allowable error to
be within +/- 5% of the mean
(A is expressed as a percentage of the mean)

FOR 220 Aerial Photo Interpretation and Forest Measurements
Sampling Design I: Simple Random Sampling

How many samples to take?

• How do we know variance in population before we
sample it?!?!

FOR 220 Aerial Photo Interpretation and Forest Measurements
Sampling Design II: Systematic Sampling

Assumptions:

The initial sampling unit is randomly selected or
established on the ground. All other sample units are
spaced at uniform intervals throughout the area sampled.

Sampling units are easy to locate.

Sampling units appear to be representative of an area.

FOR 220 Aerial Photo Interpretation and Forest Measurements
Sampling Design II: Systematic Sampling

Design:
A Grid Scheme is most common

FOR 220 Aerial Photo Interpretation and Forest Measurements
Sampling Design II: Systematic Sampling

Arguments:
For:
Regular spacing of sample units may yield efficient
estimates of populations under certain conditions.

*** Against:
Accuracy of population estimates can be low if there is
periodic or cyclic variation inherent in the population.

FOR 220 Aerial Photo Interpretation and Forest Measurements
Sampling Design II: Systematic Sampling

Arguments:
For:
There is no practical alternative to assuming that
populations are distributed in a random order across
the landscape.

Against:
Simple random sampling statistical techniques can’t
logically be applied to a systematic design unless
populations are assumed to be randomly
distributed across the landscape.

FOR 220 Aerial Photo Interpretation and Forest Measurements
Sampling Design II: Systematic Sampling

Summary:
We can (and often do) use systematic sampling to obtain estimates about the
mean of populations.

When an objective, numerical statement of precision is required, however, it
should be viewed as an approximation of the precision of the sampling effort.
(i.e. 95% confidence intervals)

Use formulas presented for simple random sampling, and where appropriate,
use the “without replacement” variations of those equations (if sampling from
a small population), otherwise use the normal SRS statistical techniques.

FOR 220 Aerial Photo Interpretation and Forest Measurements
Sampling Design III: Stratified Random Sampling

Assumptions:
A population is subdivided into subpopulations of known sizes.

A simple random sample of at least two units is drawn from each
subpopulation.

Why? To obtain a more precise estimate of the population mean. If the
variation within a subpopulation is small in relation to the total
population variance, the estimate of the population mean will be
considerably more precise than a simple random sample of the same size.

Why? To obtain an estimate of the resources within the subpopulations.

FOR 220 Aerial Photo Interpretation and Forest Measurements
Sampling Design III: Stratified Random Sampling

Design:

Simple random
samples within 3
strata

FOR 220 Aerial Photo Interpretation and Forest Measurements
Sampling Design III: Stratified Random Sampling

Estimate the overall population mean:

Where:
 L        
L = number of strata                                             Nh y h 
y st   h 1     
Nh = total number of [area] units in strata h
N = total number of [area] units in all strata                       N   
          
<< Essentially a weighted average >>                                     

We often use area for strata units (acres, hectares) in natural resource applications

FOR 220 Aerial Photo Interpretation and Forest Measurements
Sampling Design III: Stratified Random Sampling

Estimate the overall population mean:

Example:            Strata      acres      mean dbh

1           10         12.2
2           12         31.6
3           7          20.1

 (10)(12.2)  (12)(31.6)  (7)(20.1) 
y st                                         22.1 inches
                 29                  

FOR 220 Aerial Photo Interpretation and Forest Measurements
Sampling Design III: Stratified Random Sampling

Estimate the overall population standard error of the mean:
1. First compute variance within each strata

   x h  x h 2 
sh                    
2
    nh  1        
                  

Where: nh = total number of [tree or plot] units sampled in strata h

2              
  x h 2   x h 2 / nh
sh  

          nh  1             
                             
FOR 220 Aerial Photo Interpretation and Forest Measurements   Avery and Burkhart formula
Sampling Design III: Stratified Random Sampling

Estimate the overall population standard error of the mean:
2. Second, compute population standard error of the mean

 1     L Nh 2 sh 2 
SE st   2                 
(with replacement)
 N  h 1  nh       
OR

1  L  Nh sh  Nh  nh  
2  2

(without replacement)                 SE st   2           
        

 N  h1  nh  Nh  
                 

Remember: N = [area] units, n = [tree or plot] units
FOR 220 Aerial Photo Interpretation and Forest Measurements
Sampling Design III: Stratified Random Sampling

Estimate the overall population standard error of the mean:
3. Compute confidence intervals if desired

Estimated confidence intervals:

y st  t SE
st

FOR 220 Aerial Photo Interpretation and Forest Measurements
Sampling Design III: Stratified Random Sampling
Stratum (h)   Stratum Size ac. (N)       Stratum Mean
Example: Mean
1                   20                    43
2                   40                    80
3                   70                    27
4                   60                    56
5                   50                    63

(20  43)  (40  80 )  (70  27 )  (60  56 )  (50  63)
yst                                                                51 .9
(20  40  70  60  50 )

2

yst   
N  yh       h
4

N
5
h
3                              1
Sampling Design III: Stratified Random Sampling

Example: Standard Error
Stratum (h)   Sample Size (n)   Stratum Size ac. (N)   Stratum Mean   Stratum s2
1               10                  20                  43           285
2               25                  40                  80           472
3               30                  70                  27           295
4               30                  60                  56           304
5               25                  50                  63           387

1  20 2  285 40 2  472 70 2  295 60 2  304 50 2  387 
SEst      
2 
                                         
240  10             25         30         30         25    

SEst  2.86                                                N h  sh
2    2
1
2 
SEst            
( N h )       nh
Sampling Design III: Stratified Random Sampling

Example: Confidence Interval
Remember:
Mean = 51.9
95 %CI  yst  SEst                   SE = 2.86
Overall n > 30, t = 1.96

95%CI  51.9  1.96  2.86

95%CI  51.9  5.6
Sampling Design III: Stratified Random Sampling

Minimum sample size – modified Stein’s formula:

t s
2   2
N  t   Nh  s
2                2
n     2               n                           h
AE                            N  AE
2        2

Here, N = Total N (N1 + N2 + …Nn)
AE = Allowable Error
Example: Estimate Average Age of Campers:

28      41              27        27             29
Complete enumeration:
(Census)
32      35         26             29            23
45      63         38                                 Mean age = 35.95 years
38                  46
19      31         41        56                  35
25      19         19        55

Shower
48
26      25         22        47                  27
33        54                 56
26                           32
42
28        58        48                 39

FOR 220 Aerial Photo Interpretation and Forest Measurements
Example: Estimate Average Age of Campers:

Simple
28      41              27        27             29   Random
Sampling:
32      35         26             29            23    n = 10
45      63         38        38                  46
19      31         41        56                  35   Average age =
25      19         19        55                       38.4  7.6 years

Shower
48
26      25                                            30.8 to 46.0
22        47                  27
33        54                 56
26                           32
42
28        58        48                 39    True average = 35.95

FOR 220 Aerial Photo Interpretation and Forest Measurements
Example: Estimate Average Age of Campers:

Simple
28      41              27        27             29   Random
Sampling:
32      35         26             29            23    n = 10
45      63         38        38                  46
19      31         41        56                  35   Average age =
25      19         19        55                       40.2  8.5 years

Shower
48
26      25         22        47                       31.7 to 48.7
27
33        54                 56
26                           32
42
28        58        48                 39    True average = 35.95

FOR 220 Aerial Photo Interpretation and Forest Measurements
Example: Estimate Average Age of Campers:

Systematic
27
28      41              27                       29   Sampling:
32      35         26                                 n = 10
29            23
45      63         38        38                  46
19      31         41        56                  35   Average age =
25      19         19        55                       37.7  8.0 years
Shower
48
26      25         22        47                       29.7 to 45.7
27
33        54                 56
26                           32
42
28        58        48                 39
True average = 35.95

FOR 220 Aerial Photo Interpretation and Forest Measurements
Example: Estimate Average Age of Campers:

Every 4th camper

Systematic
27
28      41              27                       29   Sampling:
32      35         26                                 n = 10
29            23
45      63         38        38                  46
19      31         41        56                       Average age =
35
25      19         19                                 37.6  8.2 years
55

Shower
48   29.4 to 45.8
26      25         22        47                  27
33        54                 56
26                           32
42
28        58        48                 39    True average = 35.95

FOR 220 Aerial Photo Interpretation and Forest Measurements
Example: Estimate Average Age of Campers:

Systematic
27
28      41              27                       29   Sampling:
32      35         26                                 n = 15
29            23
45      63         38        38                  46
19      31         41        56                       Average age =
35
25      19         19                                 38.5  5.9 years
55
Shower
48   32.6 to 44.4
26      25         22        47                  27
33        54                 56
26                           32
42
28        58        48                 39
True average = 35.95

FOR 220 Aerial Photo Interpretation and Forest Measurements
Class Example: Timber Cruise

Stand ~ 32 acres

Lets use plots and a
systematic sampling scheme
typical for a timber cruise

FOR 220 Aerial Photo Interpretation and Forest Measurements
Class Example: Timber Cruise

32 plots

cruise lines                                                  #

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5 chains apart                        #
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4 chains                                      #
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between plots                                         #
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32 acres = 320 square chains
320/16plots = 2 ac. represented by each plot
2 acres = 20 square chains
Plot spacing: (5 x 4 = 20)
Randomize location of first plot
FOR 220 Aerial Photo Interpretation and Forest Measurements
Class Example: Lab 2

1/10 acre plots
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FOR 220 Aerial Photo Interpretation and Forest Measurements
What happened here?

FOR 220 Aerial Photo Interpretation and Forest Measurements
What is this? What is the red “stuff”?

FOR 220 Aerial Photo Interpretation and Forest Measurements

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