Chi-square tests (Animal Science)
Chi-Squared Tests
For quantitative data (numerical measurements) we use t tests but they cannot be used for qualitative data (non numerical). For qualitative data we use Chi-Squared ($\chi^2$) Tests. Chi-Squared tests provide an objective method of investigating the probabilities of individuals being in specific categories/groups, hence why it is used for qualitative data.
The Chi-Squared test is used to:
- Test the goodness of fit of data to a hypothetical pattern, for example, does the data follow a Mendelian ratio?
- Test the association between attributes (using Contingency Tables), for example, is there an association between eye colour and coat colour?
There are rules to follow when using a Chi-Square test:
- No category has a mean of less than 5.
- Division between categories should be as natural as possible.
- Keep to the raw data.
- Avoid categories that vastly exceed others.
The Method
Both R and Minitab perform Chi-Squared tests. However, it is useful to learn how to carry out the test by hand. The steps are as follows:
- After collecting data we calculate the Chi-Square statistic/value:
\begin{equation} \chi^2 = \sum {\frac{(O-E)^2}{E}} \end{equation}
where:
\begin{align} O &= \text{Observed value} \\ E &= \text{Expected value} \\ \sum{} &= \text{Sum of} \end{align}
- We then compare this statistic to a $\chi^2$ table on the appropriate degrees of freedom depending on the type of Chi-Squared test (see Worked Examples for more details).
- If the Chi-Square value does not exceed the critical value then the current hypothesis will be accepted, i.e. the data does follow the hypothetical pattern.
- If it does exceed the critical value then an alternative hypothesis must be devised.
Worked Example 1
Worked Example
We shall now look at an example where we examine the numbers of dogs expressing the different genotypes found in a generation from a Mendelian breeding trial. The Merle (M) gene in dogs gives all white coat colour when present as a homozygote (MM). For the heterozygote (Mm) there will be a dappled coat of normal and dilute pigmentation and white markings on the shoulders and head. The mm genotype is normal coat colour. Theory says that the animal types should be found in a ratio of $1$ (MM) : $2$ (Mm) : $1$ (mm) if two Mm are mated.
After carrying out a series of matings the results are pooled (see table below) and the Mendelian pattern is examined.
|
Coat Type |
Observed $(O)$ |
|---|---|
|
White |
23 |
|
Merle |
58 |
|
Normal |
39 |
|
Total |
120 |
- State the hypothesis.
- Calculate the Chi-Squared statistic and report the conclusions.
Worked Example 2
Worked Example
The hair length and eye colour of all the cats visiting a veterinary practice over five days were recorded.
The corresponding contingency table is given below:
|
Blue Eyes |
Yellow Eyes |
Green Eyes |
Total |
|
|---|---|---|---|---|
|
'Short Hair' |
25 |
48 |
13 |
86 |
|
'Long Hair' |
8 |
32 |
24 |
64 |
|
'Total' |
33 |
80 |
37 |
150 |
- Determine whether or not there is an association between eye colour and hair length in this group of cats.
Solution
The hypothesis we are testing is: There is no association between eye colour and hair length in this particular group of cats. If this is rejected then there must be an association.
We know we need to use a Chi-Squared test because we are dealing with qualitative data, eye colour, short hair and long hair are not numerical.
Like in the above example, we need to calculate the expected values (1 d.p.) for the contingency table.
- Short hair & blue eyes: $E =$ $\dfrac {\text{(Total for Short Hair)} \times \text{(Total for Blue Eyes)}~}{\text{Total Number of Cats}~} = \dfrac{86 \times 33}{150} = 18.9$
- Short hair & yellow eyes: $E =$ $\dfrac {\text{(Total for Short Hair)} \times \text{(Total for Yellow Eyes)}~}{\text{Total Number of Cats}~} = \dfrac{86 \times 80}{150} = 45.9$
- Short hair & green eyes: $E =$ $\dfrac {\text{(Total for Short Hair)} \times \text{(Total for Green Eyes)}~}{\text{Total Number of Cats}~} = \dfrac{86 \times 37}{150} = 21.2$
- Long hair & blue eyes: $E =$ $\dfrac {\text{(Total for Long Hair)} \times \text{(Total for Blue Eyes)}~}{\text{Total Number of Cats}~} = \dfrac{64 \times 33}{150} = 14.1$
- Long hair & yellow eyes: $E =$ $\dfrac {\text{(Total for Long Hair)} \times \text{(Total for Yellow Eyes)}~}{\text{Total Number of Cats}~} = \dfrac{64 \times 80}{150} = 34.1$
- Long hair & green eyes: $E =$ $\dfrac {\text{(Total for Long Hair)} \times \text{(Total for Green Eyes)}~}{\text{Total Number of Cats}~} = \dfrac{64 \times 37}{150} = 15.8$
Here is the contingency table for the expected $(E)$ numbers:
|
Blue Eyes |
Yellow Eyes |
Green Eyes |
Total |
|
|---|---|---|---|---|
|
'Short Hair' |
18.9 |
45.9 |
21.2 |
86 |
|
'Long Hair' |
14.1 |
34.1 |
15.8 |
64 |
|
'Total' |
33 |
80 |
37 |
150 |
We now need to calculate the Chi-Squared values (1 d.p.) using the formula: $\chi^2 = \frac{(O-E)^2}{E}$
- Short hair & blue eyes: $\chi^2 =$ $\frac{(25-18.9)^2}{18.9} = 2.0$
- Short hair & yellow eyes: $\chi^2 =$ $\frac{(48-45.9)^2}{45.9} = 0.1$
- Short hair & green eyes: $\chi^2 =$ $\frac{(13-21.2)^2}{21.2} = 3.2$
- Long hair & blue eyes: $\chi^2 =$ $\frac{(8-14.1)^2}{14.1} = 2.6$
- Long hair & yellow eyes: $\chi^2 =$ $\frac{(32-34.1)^2}{34.1} = 0.1$
- Long hair & green eyes: $\chi^2 =$ $\frac{(24-15.8)^2}{15.8} = 4.3$
Here is the table for the $\chi^2$ values.
|
Blue Eyes |
Yellow Eyes |
Green Eyes |
|
|---|---|---|---|
|
'Short Hair' |
2.0 |
0.1 |
3.2 |
|
'Long Hair' |
2.6 |
0.1 |
4.3 |
Our Chi-Squared statistic is $2.0 + 0.1 + 3.2 + 2.6 + 0.1 + 4.3 = 12.3$.
We compare this to a Chi-Squared table on $(\text{Number of Rows} - 1) \times (\text{Number of Columns} -1) = (1 \times 2) = 2$ degrees of freedom. The corresponding $\chi^2$ value is $5.991$ at $P = 0.05$ level.
$12.3$ exceeds $5.991$ so we must reject that there is no association between hair length and eye colour in this group of cats and conclude that an association must therefore exist. Note: Minitab and R both yield $P = 0.002$, which is very significant (i.e. strong evidence that there is an association).
Here are video tutorials for this example in R Studio and Minitab (ver. 16):
Test Yourself
Try our Numbas test on hypothesis testing: Practising confidence intervals and hypothesis tests.
See Also
To develop these ideas further see the other sections of Hypothesis Tests (Animal Science).
For additional information on topics covered in this section see the main site's page on hypothesis testing.