Association Tests
The comparative distribution of two species can be used as an indicator of the type of relationship between them
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Interspecific competition is indicated (but not proven) if one species is more successful in the absence of another
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In such circumstances, the two species do not tend to appear in the same environment (they have different realised niches)
Interspecific competition can be assessed via a number of different research methods:
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Laboratory experiments can be conducted under controlled conditions by measuring a dependent variable when the species are both present or individually isolated
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Field manipulation research may involve the selective removal of one species to determine the impact on the other within the natural environment
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Field observations can also occur, where random sample sites are assessed (using quadrats) for the presence or absence of each species
Statistical Processing
If data has been collected via quadrat sampling, a chi-squared test can be performed to determine if there is a statistically significant association between the distribution of two species
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If two species are typically found within the same habitat, they show a positive association (e.g. predator-prey dynamic)
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If two species tend not to occur within the same habitat, they show a negative association (e.g. interspecific competition)
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If two species do not interact, there will be no association between them and their distribution will be independent of one another
Performing a Chi-Squared Test
The presence or absence of two species of mollusc (limpet and whelk) was recorded in fifty quadrats (1m2) on a rocky sea shore
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The following distribution pattern was observed:
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6 quadrats = both species
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15 quadrats = limpet only
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20 quadrats = whelk only
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9 quadrats = neither species
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Step 1: Construct a contingency table of the results
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The table shows the numbers present or absent for each species (row = one species, column = other species)
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Row totals, column totals and overall totals must also be included in the table
Limpet Present
Limpet Absent
Total
Whelk Present
6
20
26
Whelk Absent
15
9
24
Total
21
29
50
Step 2: Construct a table of frequencies
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Observed values are the values collected via quadrat sampling
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Expected values = (row total × column total) ÷ overall total
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This data is then processed to work out the chi-squared value
O
E
(O – E)2 ÷ E
Limpet Only
15
10.1
2.38
Whelk only
20
15.1
1.59
Both species
6
10.9
2.20
No species
9
13.9
1.73
Step 3: Calculate a chi-squared value
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χ2 = ∑(O – E)2 ÷ E
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2.38 + 1.59 + 2.20 + 1.73 = 7.90
Step 4: Identify the p value
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The p value indicates the probability that the results are due to chance (lower p value is more significant)
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A p value of less than 5% chance (p<0.05) is considered to be statistically significant
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The degree of freedom (df) designates what range of values fall within each significance level
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For association tests between two species the degree of freedom should always be 1
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If p<0.05 the alternative hypothesis is accepted, otherwise the null hypothesis is accepted
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Alternative hypothesis: There is an association between the two species
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Null hypothesis: There is no association between the two species (distribution is random)
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p value
0.25
0.1
0.05
0.01
df = 1
1.32
2.71
3.84
6.64
Step 5: Determine statistical significance
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The chi-squared value (7.90) is greater than the critical value for significance (3.84)
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Hence the results are statistically significant (alternative hypothesis is accepted)
