1. ECOSYSTEMS & ENERGY FLOW
4.1 & 4.2

2. Field work – associations between species (p207)
Testing for association between two species using the chi-squared test with data obtained by quadrat sampling

3. Quadrats sampling
Quadrats are square sample areas, often marked by a quadrat frame.
Quadrat sampling involves repeatedly placing a quadrat frame at random positions in a habitat and recording numbers of organisms present.
Goal is to obtain reliable estimates of population sizes.
Not useful for motile organisms.

5. If the presence or absence of more than one species is recorded in every quadrat during sampling of habitat, it is possible to test for an association between species.
Positive Association: if two species occur in the same parts of a habitat, they will tend to be found in the same quadrat.
Negative Association: distribution of two species is independent

6. Two Possible Hypothesis
H0: two species are distributed independently (null hypothesis)
H1: two species are associated (either positively so they tend to occur together or negatively so they tend to occur apart) (alternative hypothesis)
We can test these hypothesis using a statistical procedure called the chi-squared test.

7. Chi-squared testing
Chi-squared is a statistical test commonly used to compare observed data with data we would expect to obtain according to a specific hypothesis.
Example: If according to Mendel’s laws, you expect 10 of 20 offspring to be male, and the actual observed number is 8 males. Then you might want to know whether the deviation (difference between observed and expected) is the result of chance or were they due to other factors.

8. Chi-squared testing
The usual procedure is to test the H0, with the expectation of showing that it is false.
A statistic is calculated using the results of the research and is compared with a range of possible values called the critical region.
If the calculated statistic exceeds the critical region, the H0 is considered to be false and is therefore rejected.
H0 is accepted
H0 is rejected

9. Statistical Significance
For the results to be “statistically significant” means that if the H0 was true, the probability of getting results as extreme as the observed results would be very small. This is known as the significance level.
It is the cut-off point for the probability of rejecting the null hypothesis when in fact it was true. A level of 5% is often chosen.

10. Chi-squared testing
Draw a contingency table of observed frequencies. These are the # of quadrats containing or not containing the two species.
Species A present
Species A absent
Row Totals
Species B present
Species B absent
Column Totals

11. Chi-squared testing
Draw a contingency table of observed frequencies. These are the # of quadrats containing or not containing the two species.
Species A present
Species A absent
Row Totals
Species B present
Species B absent
Column Totals
207 67
274
1795
811
984
Grand Total
1018
1051
2069

12. Chi-squared testing
Calculate the expected frequencies, assuming independent distribution, for each of the four species combinations.
Calculate the number of degrees of freedom using this equation:
Expected frequency
row total x column total
grand total
The concept of degrees of freedom is central to the principle of estimating statistics of populations from samples of them. "Degrees of freedom" is commonly abbreviated to df. Think of df as a mathematical restriction that needs to be put in place when estimating one statistic from an estimate of another.
Where m and n are the number of rows and number of columns
Degrees of freedom
(m-1)(n-1)

13. Chi-squared testing
Find the critical region for chi- squared from a table of chi- squared values, using the degrees of freedom that you have calculated and a significance level (p) of or 5%. The critical region is any value of chi-squared larger than the value in the table.

14. Significance level (P Value)
The relative standard commonly used in biological research is p > 0.05
The p value is the probability that the deviation of the observed from that expected is due to chance alone (no other forces acting).
A small p-value (like indicates strong evidence against the null hypothesis.

15. Chi-squared testing Calculate the chi-squared using this equation.X2
(f0 - fe)2 fe
Where f0 is the observed frequency
fe is the expected frequency
is the sum of.

16. Chi-squared testing
Compare the calculated value of chi-squared with the critical region.
If the X2 ≤ CV, then ACCEPT the Null Hypothesis
i.e. there is no evidence at the 5% level for an association between species
If the X2 > CV, then REJECT the Null Hypothesis
i.e. there is evidence at the 5% level for an association between the two species

17. Example p209
Figure shows an area of the summit of Caer Caradoc, a hill in Shropshire, England.
The area is grazed by sheep in summer and hill walkers cross it on grassy paths. There are raised hummocks with heather (Calluna vulgaris) growing in them. A visual survey of this site suggested that Rhytidiadelphus squarrosus, a species of moss growing in this area, was associated with these heather hummocks. The presence or absence of the heather and the moss was recorded in a sample of quadrats, positioned randomly.

18. The presence or absence of the heather and the moss was recorded in a sample of 100 quadrats, positioned randomly.
Results:
Species
Frequency
Heather only 9
Moss only 7
Both species 57
Neither species 27
Is there an association (positive or negative) between these two species?

19. Two possible hypothesis
H0: heather and moss are distributed independently (null hypothesis)
H1: heather and moss are associated (alternative hypothesis)
Test the hypothesis using the chi-squared test.

20. Question 1
Construct a contingency table with your observed values (f0).
Heather present
Heather absent
Row Totals
Moss present
Moss absent
Column Totals 57 7 64 9 27 36 66 34
100

21. Question 2
Calculate expected values (fe), assuming no association between the species.
Based on the row totals, moss should be present 64% of the time and absent 36% of the time; this should hold in all four cells.
Based on the column totals, heather should be present 66% of the time and absent 34% of the time

22. Question 3
Calculate the degrees of freedom: Degrees of freedom = (m-1)(n-1) = (2-1)(2-1) = 1

23. Question 4
Find the critical region for chi-squared at a significance level of 5%
The critical region (obtained from a table of chi-squared values) is 3.841or larger.

24. Question 5 Calculate chi-squared:
X2 = ( )2/ (7-21.8) 2/ (9-23.8) 2/ ( ) 2/12.2 = =

25. Question 6
Compare the calculated value of chi-squared with the critical region. The calculated value of chi-squared ( ) is in the critical region (3.841or larger) so there is evidence at the 5% level for an association between the two species; we can reject the null hypothesis. χ2 > critical value, therefore there is a positive association between the 2 species
A statistic is calculated using the results of the research and is compared with a range of possible values called the critical region.

26. Answer to Practice Problem:
Douglas Fir and Dwarf Mistletoe

27. X2 = 7.70 > critical value
Since our calculated value is 7.70 then we can reject the null hypothesis and the alternative hypothesis is accepted. Basically, the means that there is astatistically significant association between Douglas Fir and Dwarf Mistletoe, and the distributions of the two species are not independent of each other.

28. Answers to practice q
Sage and Bramble

29. Contingency Table Observed values Sage Present Sage Absent Total
Sage Present
Sage Absent
Total
Bramble present 51 40 91
Bramble absent 15 66
106

30. Expected Values Sage Present Sage Absent Total Bramble present
Sage Present
Sage Absent
Total
Bramble present
91 x 66 /106 =
55.66
91 x 40 / 106 =
34.34 91
Bramble absent
15 x 66 / 106 =
9.34
15 x 40 / 106 =
5.66 15 66 40
106

31. Chi-squared test X2 = [(51-55.66)2/55.66 ] + [ (40-34.34)2 /34.34 ]
+ [ ( )2/9.34 ] + [ (0-5.66)2/5.66 ]
X2 =
X2 = 10.41

32. Conclusion X2 > critical value (for df = 1) 10.41 > 3. 841
>
Therefore, the null hypothesis is rejected. There is a positive correlation between the sage and bramble

33. Mesocosms
Small, closed-off experimental systems set up as ecological experiments
Fenced-off enclosures in grassland or forest could be used as terrestrial mesocosms
Tanks set up in the lab could be used as aquatic mesocosms
Can be used to test effects of varying certain conditions on ecosystem stability as well as the sustainability of ecosystems

34. Mesocosms
For example, tanks could be set up with and without fish to investigate the effects of fish on aquatic ecosystems.

35. Practical 5
Setting up sealed mesocosms to try to establish sustainability.
We didn’t get to this, but lets try this simulation….
/virtual_ecosphere.swf

36. Mesocosms What is required for successful closed ecosystem?
Energy source
sun
Producers/autotrophs
Soil, CO2, water
Consumers O2
Decomposers
Soil, organic material, O2