4.1 Chi-squared test
IB Biology Skill:
• Skill: Testing for association between two species using the chi-squared test with data obtained by
quadrat sampling.
• Skill: Recognizing and interpreting statistical significance.
Background Information:
There are numerous uses for this test. Our example shows its use in the simplest way possible.
All chi-squared tests are concerned with counts of things (frequencies) that you can put into
categories. For example, you might be investigating flower colour and have frequencies of red
flowers and white flowers. Or you might be investigating human health and have frequencies of
smokers and non-smokers.
The test looks at the frequencies you obtained and compares them with the frequencies you
might expect given your null hypothesis. The null hypothesis is this:
There is no significant difference between the observed and expected frequencies
The only difficult part of a chi-squared test is working out what the expected values are. In our
example this is very simple. It can get more complicated in different circumstances. If you use
this test, each measurement has to be independent of the others and all your expected
frequencies should be bigger than 5.
Problem :
Let us surmise that you have been wandering about on a seashore and you have noticed that a
small snail (the flat periwinkle) seems to live only on seaweeds of various kinds. You decide to
investigate whether the animals prefer certain kinds of seaweed by counting numbers of
animals on different species. You end up with the following data:
1. Null Hypothesis: Our null hypothesis is stated above. Put more specifically for this
example what we are saying is this: "There is no difference in frequencies of flat
periwinkles over the 5 kinds of seaweed".
alternative hypothesis (HA): There is a real difference between numbers of periwinkles
associated with different species
1. Calculated EXPECTED Frequencies if our Null Hypothesis:
In our example this is a simple task. Our null hypothesis is that there is no difference
between observed and expected frequencies. If this were exactly the case there would be
no differences in the frequencies over all of our categories. The best estimate we could
make therefore would be to add up all our observed frequencies and divide by the number
of categories. (Calculate the mean). So our expected frequency for each category would be:
45 + 38 + 10 + 5 + 2 = 100
100/5 = 20 = expected frequency
2. Calculate the value of our test statistic that is specific to our data. The formula for Chisquared is this:
An easy way to do this is to tabulate the calculations:
Seaweed
serrated
wrack
bladder
wrack
egg wrack
spiral wrack
other algae
TOTALS
Observed
frequencies
Expected
frequencies
O–E
O - E squared
O-E squared/E
45
20
25
625
31.3
38
20
18
324
16.2
10
5
2
100
20
20
20
100
-10
-15
-18
100
225
324
5
11.3
16.2
79.9
3. Identify the Chi Square Value in the above table (HINT:The total of our final column
represents the value of the test statistic that is specific to our data) = 79.9
.
4. Calculate the degrees of Freedom: Degrees of freedom is one less than the number of
categories/ types of seaweed:
(5 - 1 = 4).
5. Identify the Significance Level: There is no rule about this but with fieldwork data most
people reckon that 5% significance is an adequate level of acceptance or rejectance of
the null hypothesis. Remember this means that you would expect to be correct in
accepting or rejecting your null hypothesis 95% of the time, 5% of the time you might
get a different result due to chance.
As with all tests of this kind, we next compare our calculated value with the critical value
(obtained from a table of critical values of chi-squared).

Our calculated Value:
79.9 .


Critical value of Chi-squared at 5% significance and 4 degrees of freedom is 9.48.
Do you accept of Reject the NULL HYPOTHESIS? REJECT.
o If our calculate value is bigger that the critical value then we must reject the null
hypothesis.
o If our calculate value is less than the critical value then we accept the null
hypotheis
6. Discuss what it means to accept or reject the null hypothesis




In rejecting the null hypothesis, we are saying that the snails are not homogeneously
scattered about the various sorts of seaweed but seem to prefer living on certain
species.
A cursory glance at the original data will of course inform of this in our simple
example. It is extremely clear that the snails like serrated wrack and bladder wrack a
lot and don't much like any of the others. The point about all of this is that it allows us
to state exactly how certain we are about our null hypotheses.
It is worth pointing out that statistics of this kind tell you nothing about the biology of
the situation. All we are saying is that our observed frequencies are different to our
expected ones. You could criticise our approach by pointing out that it might be that
there are not equal amounts of each kind of seaweed on the shore for the animals to
live on. We have implied that there are because we have taken all our expected
frequencies to be the same value. We could allow for this by estimating how much of
each seaweed was available to the animals and weighting our expected frequencies
accordingly.