Chi Squared Tests for Contingency Tables (Edexcel A Level Further Maths): Revision Note

Contingency Tables

What is a chi-squared test using contingency tables?

• A chi-squared () using contingency tables is a hypothesis test used to test whether two variables are independent of each other

  • For example whether or not favourite music genre is independent of the age group of the listener 

  • This is sometimes called a  two-way test

• This is an example of a goodness of fit test

  • We are testing whether the data fits the modelling assumption that the variables are independent

• The chi-squared () distribution is used for this test

• You will use a contingency table

  • This is a two-way table that shows the observed frequencies for the different combinations of the two variables

  • A   contingency table has  rows and  columns

    • This does not include any rows or columns used to record the row, column and grand totals

Why might I have to combine rows or columns?

• The observed values are used to calculate expected values

  • These are the expected frequencies for each combination assuming that the variables are independent

    • Your calculator may be able to calculate these for you after you input the observed frequencies

    • Or else you can calculate them using the formula

• None of the expected values used to run the test can be less than 5

  • If one of the expected values is less than 5 then you will have to combine the corresponding row or column in the table of observed values with the adjacent row or column

  • The decision between row or column will be based on which seems the most appropriate

    • For example: if the two variables are age and favourite music genre then it is more appropriate to combine age groups than types of genre

What are the degrees of freedom?

• There will be a minimum number of expected values you would need to know in order to be able to calculate all the expected values

• This minimum number is called the degrees of freedom and is often denoted by

• For a test for independence with an contingency table

  •  

  • For example: If there are 5 rows and 3 columns then you only need to know 2 of the values in 4 of the rows as the rest can be calculated using the totals

What are the steps for a chi-squared test using contingency tables?

• STEP 1: Write the hypotheses

  • : Variable X and Variable Y are independent

  • : Variable X and Variable Y are not independent

  • The hypotheses should always be stated in the context of the question

  • Make sure you clearly write what the variables are and don’t just call them 'Variable X' and 'Variable Y'

• STEP 2: Calculate the expected frequencies

  • Use the formula

•  You will need to combine rows or columns if any of the expected frequencies are less than 5

  • This process is described above

  • After combining, calculate the new expected frequencies for the modified table

  • You may also be able to enter the observed frequencies as a matrix in your calculator

    • Use the option for a 2-way test

    • Your calculator will calculate the matrix of expected frequencies

• STEP 3: Calculate the degrees of freedom for the test

  • For an contingency table (after combining)

    • Degrees of freedom is

• STEP 4: Calculate using the formula

   

  • then you will also need to determine the appropriate critical value

    • use the 'Percentage Points of the Distribution' table in the exam formula booklet

  • If you entered the observed frequencies as a matrix in your calculator

    • then your calculator's 2-way test option will give you the test statistic and the associated p-value

• STEP 5: Decide whether there is evidence to reject the null hypothesis

  • Compare the statistic with the critical value you have determined

    • If  > critical value (or ) then there is sufficient evidence to reject 

    • If  < critical value (or ) then there is insufficient evidence to reject 

• STEP 6: Write your conclusion

  • If you reject H₀

    • Variable X and variable Y are not independent

  •  If you do not reject H₀

    • Variable X and variable Y are independent

  • Be sure to state your conclusion in the context of the question