Transcript Chi Square Test X2 - Anoka-Hennepin School District 11

Chi Square Test
2
X
• 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
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.
• 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.
Setting up quadrats
• Setting up quadrats
This is a grid of 100 quadrats, each 10 m
on a side.
http://www.csulb.edu/~rodrigue/geog20
0/lab4.html; Dr.Rodrigue
http://home.iae.nl/users/isse/Files/environmental_systems.htm;
Environmental Systems
3.
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.
4.
Measure the total size of the area
occupied by the population, in square
meters.
http://www.waite.adelaide.edu.au/school/Habitat/survey.html;
Survey of native and exotic species
5. 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
• 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
• 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
Method for Chi Square
• Draw up a contingency table of observed
frequencies.
Species A
present
Species B present
Species B absent
Column totals
Species A absent
Row totals
• Calculate the row and column totals.
• Adding the row and column totals should give
the same grand total in the lower right cell.
• 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
• 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
• 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%)
• Calculate the chi-squared using this equation:
X2 = Σ 𝑜 − 𝑒 2
e
O – observed
e- expected
Σ- the sum of
• 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.
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
• 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
• We now need to complete our contingency
table.
Men
Women
Total
voted
2792
3591
6383
Didn’t vote
1486
5722
3617
Total
4278
5722
10000
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 / 10000 = 2731
6383 x 5722 / 10000 = 3652
• 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
• So…..
• (2792 – 2731)2 = 1.36
2731
(3591 – 3652)2
= 1.0
3652
Etc . X2 = 1.4+1.0+2.4+1.8 = 6.6
Degrees of freedom 2 – 1 = 1
• 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.
http://www.youtube.com/watch?v=WXPBoFDqNVk&safety_mode=true&persist_safety
_mode=1