Measuring Frequency
Basics of Frequency Analysis
Frequency assessment of plant communities is most often used to compare:
- Changes in a site over years, or
- Differences between sites
I most cases we want to know if the times or sites are “significantly
different.” This comparison of frequency data is most often conducted with a Chi
Square Analysis.
Detail Description of Chi Square Analysis
The Measuring and Monitoring book that we are using for this class has a
complete and detailed description of how a chi-square test can be used to test
differences for two populations.
Data Analysis with Chi Square
For example: Assume we examined
150 plots on bunchgrass sites with
south-facing
slopes in Idaho and Oregon. On these sites we calculated a frequency of 36%
(present 54 out of 150 plots) for
Indian
Paintbrush (Castilleja linariifolia) using a 1 m2 plot. Then, we
also examined 150 plots on sites with north-facing aspects and we found that the
frequency of paintbursh was 48% (72 plots out of 150). What we really want to
know is – does a difference of 12% indicate a real, or significant,
difference in the amount of paintbrush frequency and abundance on north and south facing slopes.
A Chi Square can give us a way to judge if the sites are really “different.”
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Data are arranged in rows of what was observed and columns of
the treatment we want to examing. In this case our "treatment" is
south- versus north-faxing slope.
-
Data from field = "Observed"
either present or absent in the plots examined.
-
The number "Expected" is
based on an equation to estimate the amount you would expect if there was no
treatment effect.
Expected= Total occurrences Observed for both treatments × total plots in on
treatment ÷ Total plots examine in both sites.
Easily calculated in a table as Expected Value = (Row
Total × Column
Total) ÷ Grand Total
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South-Facing |
North- Facing |
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|
|
Observed |
Expected |
Observed |
Expected |
|
Totals |
Present |
54 |
(126*150) ÷ 300 =
63 |
72 |
(126*150)
÷ 300 =
63 |
|
54+72 = 126 |
Absent |
150-54= 96 |
(174 *150)
÷ 300 =
87 |
150-72= 78 |
(174 *150)
÷ 300 =
87 |
|
96+78 = 174 |
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|
|
|
|
|
Total |
150 |
↑
exp. value = (row tot
× column tot) ÷ Grand Total |
150 |
↑
exp. value = (row tot
× column tot) ÷ Grand Total |
|
300 |
Chi Square Formula:
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Where:
χ2 = the chi square statistic
(calculated value)
∑ = sum of the variables
O = Observed Value
E = Expected Value |
In Example Above:
(54-63)2
63 |
+ |
(72-63)2
63 |
+ |
(96-87)2
87 |
+ |
(78-87)2
87 |
= |
?.? |
1.3 |
+ |
1.3 |
+ |
.9 |
+ |
.9 |
= |
4.4 |
A calculated χ2 statistic of 4.4 can then
be compared to a critical or table χ2
statistic to determine of the values in the comparison are different. In other
words, is the χ2 statistic that we
generated large enough to be "significant."
Evaluating a χ2 Statistic:
Look in a
Chi Square Table to determine the appropriate χ2
critical value. To find a value in a chi-square table you much determine the v (or
degrees of freedom; df) for your comparison:
v = (r-1)(c-1) or v = (number of rows in comparison - 1)
(number of columns - 1)
In our example, for a 2 x 2 comparison, v = (2-1)(2-1) = 1
- If we use a P-threshold (or
α) of 0.10, the χ2
critical or table value is 2.706.
- The χ2 statistic calculated for
these data was 4.4.
- The null hypothesis is that there is no difference between or among
samples in the comparison. (In other words, the null hypothesis in this
example. is that there is no difference in
the frequency of paintbrush on north- and south-facing slopes).
- If the value calculated is greater than the table value, we reject the
null hypothesis
- In this example, 4.4 is > 2.7 so we reject the null hypothesis,
and conclude that the frequency of paintbrush is different between north-
and south-facing slopes.
**Note on Chi-Square Tables -
1) Follow this link to a common
Chi-Square Table
2) You can also enter you calculate chi-square value in a
Chi-Square Calculator
3) Excel also has a function to calculate chi-square (check the help button
in excel) |