1. Chi Square Test X2

2. Chi Square is a test used to see if two pieces of data are significantly different or due to chance
In Biology, we use this test a lot to see if our data is significant. In a population lab, we would see if 2 species found are associated with each other

3. Quadrant Sampling
This method is only suitable for plants and other organisms that are not motile.
Choose a random number to determine the length and width of your area
If the absence or presence of more than one species is recorded in every quadrat during sampling of a habitat, it is possible to test for an association between species.

6. A quadrat is a wire shaped into a square of a known size , such as 10x10 meters or 100m2.
If you want to know the population size of two plant species, take random samples of this area by throwing down the quadrat and recording the population numbers in each subunit of the quadrat.

7. Setting up quadrats Setting up quadrats
This is a grid of 100 quadrats, each 10 m on a side. Dr.Rodrigue
Environmental Systems

8. Count how many individuals there are inside the quadrat of the plant population being studied. Repeat steps 2 and 3 as many times as possible.
Measure the total size of the area occupied by the population, in square meters.
Survey of native and exotic species

9. Calculate the mean number of plants per quadrat
Calculate the mean number of plants per quadrat. Then calculate the estimated population size using the following equation:
Population size = mean number per quadrat X total area
area of each quadrat

10. Populations are often unevenly distributed because some parts of the habitat are more suitable for a species than others.
If two species occur in the same parts of a habitat, they will tend to be found in the same quadrats.
This is known as a Positive association

11. H0 -Two species are distributed independently
There are 2 hypotheses:
H0 -Two species are distributed independently
The Null Hypothesis
H1 – Two species are associated (either positively so they tend to occur together or negatively so they tend to occur apart)
We can test these hypotheses using a statistical procedure – the chi square test

12. Method for Chi Square
Draw up a contingency table of observed frequencies.
Species A present
Species A absent
Row totals
Species B present
Species B absent
Column totals

13. Calculate the row and column totals.
Adding the row and column totals should give the same grand total in the lower right cell.

14. Calculate the expected frequencies, assuming the independent distribution for each of the four species combinations.
Each expected frequency is calculated from values on the contingency table using this equation.
Expected frequency = row total x column total
grand total

15. Calculate the degree’s of freedom using this equation: DF = (m-1)(n-1)
Where m and n are the number of rows and columns in the contingency table

16. Find the critical region for chi-squared from a table of chi-square values, using the degrees of freedom that you calculated. It should have a significance level (p) of 0.05 (5%)

17. X2 = Σ 𝑜−𝑒 2 e O – observed e- expected Σ- the sum of
Calculate the chi-squared using this equation:
X2 = Σ 𝑜−𝑒 2 e
O – observed
e- expected
Σ- the sum of

18. What is statistically significant
H0 - the null hypothesis with the belief that there is no relationship between the two
H1 – There is a relationship
The usual procedure is to test the null hypothesis with the expectation of showing that it is false.
If you say that the results were statistically significant, it means that if the null hypothesis was true, the probability of getting results as extreme as the oberved results would be very small.

20. example
In a certain town, there are about one million eligible voters. A simple random sample of 10,000 eligible voters were chosen to study the relationship between sex and participation in the last election. The results are summarized in the following 2x2 contingency table:
Men
Women
Voted
2792
3591
Didn’t vote
1486
2131

21. We want to check whether being a man or a woman (columns) is independent of having voted in the last election (rows). In other words is ‘sex and voting independent’?
Null – sex is independent of voting
Alternative – sex and voting are dependent

22. We now need to complete our contingency table.
Men
Women
Total
voted
2792
3591
6383
Didn’t vote
1486
5722
3617
4278
10000

23. Expected Table Men Women Total Voted 2731 3652 6383 Didn’t vote 1547
2070
3617
Totals
4278
5722
10000
Remember: expected frequencies = row totals x column totals
grand total
So: 6383 x 4278 / = 2731
6383 x 5722 / = 3652

24. Now we have the observed table and the expected table under the null hypothesis of independence. Now we need to compute X2
(O – e)2 e

25. So…..
(2792 – 2731)2 = 1.36
2731
(3591 – 3652) = 1.0
3652
Etc . X2 = = 6.6
Degrees of freedom 2 – 1 = 1

27. Since X2 is 6.6 which has a p value of 1%, we have to reject the NULL hypothesis. The data supports the hypothesis that sex and voting are dependent in this town.

28. http://www. youtube. com/watch