Topic 4.1: CHI-SquARED TesT
Worked Example
A chi-squared test can be applied to quadrat sampling data to determine if there
is a statistically significant association between the distribution of two species
The following distribution was found:
• 6 quadrats = both species
• 20 quadrats = queen scallop only
• 15 quadrats = king scallop only
• 9 quadrats = neither species
Present Absent Total
KING
Case Study: Scallops on a Rocky Shore
The presence / absence of two scallop species is recorded in 50 quadrats (1m2)
QUEEN
Present
6
15
21
Absent
20
9
29
Total
26
24
50
Step 1: Identify Expected Frequencies
There are two distinct possibilities regarding associations between the two species:
• Null Hypothesis (H0) – There is no association (i.e. distribution is random)
• Alternative Hypothesis (H1) – There is an association (positive or negative)
The expected frequencies can be calculated using the following formula:
• Expected frequency = (Row total × Column total) ÷ Grand total
Present Absent Total
21 × 26 21 × 24
50
50
29 × 26 29 × 24
Absent
50
50
Present
KING
A table must be constructed that identifies expected frequencies of distribution
• This data will be compared against the observed values previously identified
QUEEN
Total
26
24
21
29
50
Step 2: Apply the Chi-Squared Formula
The chi-squared (𝟀2) formula calculates a value based on a comparison of the
observed frequencies (O) and the expected frequencies (E)
Present Absent
(O – E)2
E
Based on the worked example, the value calculated by the chi-squared test is:
• 𝟀2 = 2.20 + 2.38 + 1.59 + 1.73 = 7.90
A degree of freedom (df) will also be required to determine statistical significance
• df = (number of rows – 1) × (number of columns – 1)
Raw data table had two rows and two columns, so degree of freedom equals one
Present
∑
O
E
(O – E)2
E
6
10.9
15
10.1
2.20
2.38
Absent
=
KING
𝟀2
QUEEN
O
E
(O – E)2
E
20
15.1
9
13.9
1.59
1.73
Step 3: Determine Significance
The chi-squared value is used to determine statistical significance (p value)
• p<0.05 is considered significant (less than 5% likelihood results due to chance)
The null hypothesis can be rejected and the alternative hypothesis accepted
Because the species do not tend to co-exist, we might infer a negative association
1
p value
Based on the worked example, a value of 7.90 lies above a p value of 0.01
• This means results are significant (less than 1% probability it is due to chance)
df
0.01
6.635
0.05
3.841
0.1
2.706
Values that
are greater
than this are
statistically
significant