Chi-squared test | AQA A-Level Psychology Revision
- Revision Notes
- Aug 5
- 17 min read
Updated: 5 days ago
For 7182 specification, first teach in September 2025
AQA A-Level Psychology | Free Revision Notes
Estimated study time: 55 minutes
This Chi-squared test A-Level Psychology revision page explains how psychologists test for an association between two categorical variables. You will learn when Chi-squared is appropriate, distinguish observed from expected frequencies, and calculate degrees of freedom. You will also practise using these features to interpret a statistical table. This lesson builds on choosing an inferential test and levels of measurement, particularly the use of nominal data in psychological research.
Learning Objectives 🎯
By the end of this revision page, you should be able to:
Identify when the Chi-squared test should be used.
Explain why Chi-squared is a test of association.
Recognise nominal categorical data in an unfamiliar study.
Distinguish observed and expected frequencies.
Calculate expected frequencies from row, column and overall totals.
Calculate degrees of freedom.
Use degrees of freedom to locate a critical value.
Interpret a Chi-squared result using observed and critical values.
Revision Notes 📚
The Chi-squared test in A-Level Psychology
The Chi-squared test is an inferential statistical test used to investigate whether there is an association between two categorical variables.
It is appropriate when:
The researcher is testing for an association.
Both variables produce nominal data.
The results are recorded as frequencies.
A useful selection rule is:
$$\text{Association}+\text{nominal data}+\text{frequencies}=\text{Chi-squared}$$
The test is represented using the symbol:
$$\chi^2$$
This is read as Chi-squared.
What is an association?
An association occurs when the frequency found in one category appears to be related to the category of another variable.
For example, a psychologist might investigate whether there is an association between:
therapy type and whether participants improve;
attachment classification and choice of coping strategy;
participant group and preferred revision method;
sleep category and whether a memory response is correct;
type of background sound and whether participants complete a task successfully.
Both variables divide participants or observations into categories.
Example research question
A psychologist investigates:
Is there an association between whether participants receive therapy and whether their symptoms improve?
The first variable is treatment condition, with the categories:
therapy;
no therapy.
The second variable is outcome, with the categories:
improved;
did not improve.
The researcher counts how many participants fall into each combination of categories.
The study therefore produces nominal frequency data and may be analysed using Chi-squared.
Chi-squared is not a test of difference
A test of difference compares scores produced by conditions, groups or occasions.
For example:
Is there a difference in numerical memory scores between participants who revise in silence and participants who revise with background music?
This is a test of difference rather than an association between two categorical variables.
Depending on the design and level of measurement, it might require:
the sign test;
Wilcoxon;
Mann-Whitney;
a related t-test;
an unrelated t-test.
Chi-squared would not be selected simply because two groups are involved.
Chi-squared is not a test of correlation
A correlation investigates whether two measured co-variables are related.
For example:
Is the number of hours spent revising related to a numerical examination score?
Each participant provides two measured values:
$$(\text{revision time},\text{examination score})$$
This requires a correlation test rather than Chi-squared.
Depending on the level of measurement, the researcher would select:
Spearman’s rho;
Pearson’s \(r\).
These tests are covered in Spearman’s rho and Pearson’s r.
Association and correlation compared
Feature | Association | Correlation |
Type of variables | Categorical variables | Measured co-variables |
Form of results | Frequencies | Paired scores |
Level of measurement | Nominal | Ordinal or interval |
Appropriate test | Chi-squared | Spearman’s rho or Pearson’s \(r\) |
Example | Therapy type and improvement category | Sleep duration and memory score |
Recognising categorical variables
A categorical variable places participants, behaviours or responses into separate groups.
Examples include:
obeyed or did not obey;
correct or incorrect;
secure or insecure attachment;
chose Method A, Method B or Method C;
improved, stayed the same or became worse.
The categories may be represented by words, letters or numerical codes.
For example:
$$1=\text{improved}$$
$$2=\text{did not improve}$$
The use of numbers does not make the results interval data. The numbers are labels for nominal categories.
Nominal data
Nominal data consist of separate categories that do not have a meaningful numerical order.
For example:
therapy or no therapy;
indoor or outdoor setting;
correct or incorrect;
secure, insecure-avoidant or insecure-resistant.
It would not be meaningful to calculate an average category.
Chi-squared uses the number of cases falling into each category combination.
Frequency data
A frequency is the number of times a particular category or combination of categories occurs.
For example:
Treatment condition | Improved | Did not improve |
Therapy | \(18\) | \(12\) |
No therapy | \(10\) | \(20\) |
The value:
$$18$$
means that \(18\) participants received therapy and improved.
The value:
$$20$$
means that \(20\) participants received no therapy and did not improve.
These are frequencies rather than individual psychological scores.
Contingency tables
What is a contingency table?
Chi-squared data are commonly presented in a contingency table.
A contingency table shows the frequencies found for each combination of categories from two variables.
For example:
Improved | Did not improve | Row total | |
Therapy | \(18\) | \(12\) | \(30\) |
No therapy | \(10\) | \(20\) | \(30\) |
Column total | \(28\) | \(32\) | \(60\) |
The rows represent the categories of one variable.
The columns represent the categories of the other variable.
The table contains:
four data cells;
two row totals;
two column totals;
one overall total.
Reading a contingency table
In the example:
\(18\) participants received therapy and improved;
\(12\) received therapy and did not improve;
\(10\) received no therapy and improved;
\(20\) received no therapy and did not improve.
The first row total is:
$$18+12=30$$
The second row total is:
$$10+20=30$$
The first column total is:
$$18+10=28$$
The second column total is:
$$12+20=32$$
The overall total is:
$$30+30=60$$
It can also be found using the column totals:
$$28+32=60$$
Observed frequencies
What is an observed frequency?
An observed frequency is the actual number of cases recorded in a category combination.
Observed frequencies come directly from the results of the study.
They are represented by:
$$O$$
In the therapy example, the observed frequencies are:
$$18,\ 12,\ 10,\ 20$$
These are the results that the psychologist actually obtained.
Observed frequency example
Suppose a psychologist records the preferred revision method of students from two year groups.
Year group | Flashcards | Practice questions | Mind maps |
Year 12 | \(14\) | \(21\) | \(10\) |
Year 13 | \(8\) | \(25\) | \(12\) |
The observed frequency of Year 12 students preferring flashcards is:
$$O=14$$
The observed frequency of Year 13 students preferring practice questions is:
$$O=25$$
Observed frequencies are not estimated or predicted. They are counted from the collected data.
Features of observed frequencies
Observed frequencies:
come from the study’s results;
are actual counts;
appear in the original contingency table;
show how many cases were found in each category combination;
are compared with expected frequencies.
Expected frequencies
What is an expected frequency?
An expected frequency is the number of cases that would be expected in a cell if there were no association between the two variables.
Expected frequencies are calculated using:
the row total;
the column total;
the overall total.
They are represented by:
$$E$$
The expected frequencies describe the pattern predicted by the null hypothesis.
Expected frequencies and the null hypothesis
For a Chi-squared test, the null hypothesis predicts that there is no association between the variables.
For example:
There will be no association between treatment condition and whether participants improve.
If the null hypothesis is true, the proportions improving and not improving should not systematically depend on treatment condition.
The expected frequencies show what the table would look like if the row and column variables were not associated.
Expected frequency formula
The expected frequency for a cell is calculated using:
$$E=\frac{\text{Row total}\times\text{Column total}}{\text{Overall total}}$$
The calculation must be completed separately for each cell.
Worked example: first expected frequency
Using the therapy table:
Improved | Did not improve | Row total | |
Therapy | \(18\) | \(12\) | \(30\) |
No therapy | \(10\) | \(20\) | \(30\) |
Column total | \(28\) | \(32\) | \(60\) |
Calculate the expected frequency for participants who:
received therapy;
improved.
The relevant row total is:
$$30$$
The relevant column total is:
$$28$$
The overall total is:
$$60$$
Substitute the values:
$$E=\frac{30\times28}{60}$$
$$E=\frac{840}{60}$$
$$\boxed{E=14}$$
If there were no association between therapy and improvement, \(14\) participants would be expected in this cell.
The observed frequency was:
$$O=18$$
Worked example: second expected frequency
Calculate the expected frequency for participants who:
received therapy;
did not improve.
The relevant row total is:
$$30$$
The relevant column total is:
$$32$$
The overall total is:
$$60$$
Therefore:
$$E=\frac{30\times32}{60}$$
$$E=\frac{960}{60}$$
$$\boxed{E=16}$$
The observed frequency was:
$$O=12$$
Worked example: remaining expected frequencies
For participants who received no therapy and improved:
$$E=\frac{30\times28}{60}$$
$$\boxed{E=14}$$
For participants who received no therapy and did not improve:
$$E=\frac{30\times32}{60}$$
$$\boxed{E=16}$$
The completed expected-frequency table is:
Improved | Did not improve | Row total | |
Therapy | \(14\) | \(16\) | \(30\) |
No therapy | \(14\) | \(16\) | \(30\) |
Column total | \(28\) | \(32\) | \(60\) |
Comparing observed and expected frequencies
The observed and expected tables are:
Observed frequencies
Improved | Did not improve | |
Therapy | \(18\) | \(12\) |
No therapy | \(10\) | \(20\) |
Expected frequencies
Improved | Did not improve | |
Therapy | \(14\) | \(16\) |
No therapy | \(14\) | \(16\) |
The differences include:
more therapy participants improved than expected;
fewer therapy participants failed to improve than expected;
fewer no-therapy participants improved than expected;
more no-therapy participants failed to improve than expected.
Chi-squared summarises the difference between the observed and expected frequencies.
Observed and expected frequencies compared
Feature | Observed frequency | Expected frequency |
Symbol | \(O\) | \(E\) |
Meaning | Actual number recorded | Number predicted if no association exists |
Source | Collected study data | Calculated from table totals |
Role | Shows the obtained pattern | Represents the null-hypothesis pattern |
Example | \(O=18\) | \(E=14\) |
Why expected frequencies are not always equal
A common mistake is to divide the overall total equally between every cell.
This only produces the correct expected frequencies when the row and column totals are evenly balanced.
Expected frequencies must reflect the actual marginal totals.
Use:
$$E=\frac{\text{Row total}\times\text{Column total}}{\text{Overall total}}$$
Do not automatically use:
$$E=\frac{\text{Overall total}}{\text{Number of cells}}$$
Worked example with unequal row totals
Suppose the observed data are:
Group | Chose A | Chose B | Row total |
Group 1 | \(24\) | \(16\) | \(40\) |
Group 2 | \(12\) | \(8\) | \(20\) |
Column total | \(36\) | \(24\) | \(60\) |
The expected frequency for Group 1 choosing A is:
$$E=\frac{40\times36}{60}$$
$$E=\frac{1440}{60}$$
$$\boxed{E=24}$$
The expected frequency for Group 2 choosing A is:
$$E=\frac{20\times36}{60}$$
$$E=\frac{720}{60}$$
$$\boxed{E=12}$$
The expected frequencies are different because the group totals are different.
Checking expected frequencies
After calculating expected frequencies:
the expected values in each row should add to the row total;
the expected values in each column should add to the column total;
all expected values should add to the overall total.
For the therapy example:
$$14+16=30$$
for each row.
For the columns:
$$14+14=28$$
and:
$$16+16=32$$
The total is:
$$14+16+14+16=60$$
This provides a useful accuracy check.
The Chi-squared formula
The Chi-squared test statistic compares every observed frequency with its corresponding expected frequency.
The formula is:
$$\chi^2=\sum\frac{(O-E)^2}{E}$$
where:
\(O\) is the observed frequency;
\(E\) is the expected frequency;
\(\sum\) means that the values for all cells are added.
The AQA specification explicitly identifies calculation of the sign test, while Chi-squared is included among the tests students must know how to select. The formula is useful for understanding what the test measures, but the main focus here is recognising the test, understanding frequencies and calculating degrees of freedom.
What the formula shows
For each cell:
Find the difference between observed and expected frequency.
Square the difference.
Divide by the expected frequency.
Add the resulting values for all cells.
Squaring:
$$(O-E)^2$$
ensures that positive and negative differences do not cancel each other out.
A larger difference between observed and expected frequencies contributes more to the final Chi-squared value.
Small and large Chi-squared values
A small Chi-squared value indicates that:
observed frequencies are close to expected frequencies;
the obtained table is reasonably consistent with the null hypothesis.
A large Chi-squared value indicates that:
observed frequencies differ more substantially from expected frequencies;
there may be evidence of an association between the variables.
The calculated value must still be compared with a critical value before statistical significance can be determined.
Degrees of freedom
What are degrees of freedom?
Degrees of freedom are used to select the correct row from a Chi-squared critical-values table.
Degrees of freedom depend on:
the number of data rows;
the number of data columns.
They do not depend directly on the number of participants.
Degrees of freedom are commonly represented by:
$$df$$
Degrees of freedom formula
For a contingency table:
$$df=(r-1)(c-1)$$
where:
\(r\) is the number of data rows;
\(c\) is the number of data columns.
Do not count the totals row or totals column.
Worked example: a two-by-two table
Consider:
Improved | Did not improve | |
Therapy | \(18\) | \(12\) |
No therapy | \(10\) | \(20\) |
There are:
$$r=2$$
data rows and:
$$c=2$$
data columns.
Substitute these values:
$$df=(2-1)(2-1)$$
$$df=1\times1$$
$$\boxed{df=1}$$
Why the totals are excluded
A completed contingency table may also contain:
a totals row;
a totals column.
These are not additional categories of the variables.
For example:
Improved | Did not improve | Total | |
Therapy | \(18\) | \(12\) | \(30\) |
No therapy | \(10\) | \(20\) | \(30\) |
Total | \(28\) | \(32\) | \(60\) |
The table still contains only:
$$2\text{ data rows}$$
and:
$$2\text{ data columns}$$
Therefore:
$$df=(2-1)(2-1)=1$$
It is not a three-by-three data table.
Worked example: a three-by-two table
A researcher compares three attachment classifications with whether participants seek help.
Attachment classification | Sought help | Did not seek help |
Secure | \(20\) | \(10\) |
Insecure-avoidant | \(8\) | \(17\) |
Insecure-resistant | \(13\) | \(12\) |
There are:
$$r=3$$
rows and:
$$c=2$$
columns.
Therefore:
$$df=(3-1)(2-1)$$
$$df=2\times1$$
$$\boxed{df=2}$$
Worked example: a two-by-three table
A researcher investigates whether year group is associated with preferred revision method.
Year group | Flashcards | Practice questions | Mind maps |
Year 12 | \(14\) | \(21\) | \(10\) |
Year 13 | \(8\) | \(25\) | \(12\) |
There are:
$$r=2$$
rows and:
$$c=3$$
columns.
Therefore:
$$df=(2-1)(3-1)$$
$$df=1\times2$$
$$\boxed{df=2}$$
A three-by-two table and a two-by-three table both have:
$$df=2$$
Worked example: a three-by-four table
Suppose a table contains:
three participant groups;
four response categories.
Then:
$$r=3$$
and:
$$c=4$$
Therefore:
$$df=(3-1)(4-1)$$
$$df=2\times3$$
$$\boxed{df=6}$$
Degrees of freedom examples
Number of rows | Number of columns | Calculation | Degrees of freedom |
\(2\) | \(2\) | \((2-1)(2-1)\) | \(1\) |
\(2\) | \(3\) | \((2-1)(3-1)\) | \(2\) |
\(3\) | \(2\) | \((3-1)(2-1)\) | \(2\) |
\(3\) | \(3\) | \((3-1)(3-1)\) | \(4\) |
\(3\) | \(4\) | \((3-1)(4-1)\) | \(6\) |
\(4\) | \(5\) | \((4-1)(5-1)\) | \(12\) |
Using a statistical table
Finding a critical value
After calculating degrees of freedom, the researcher uses a Chi-squared statistical table.
The correct critical value depends on:
the value of \(df\);
the selected significance level.
The general process is:
Calculate degrees of freedom.
Locate the correct \(df\) row.
Locate the required significance-level column.
Read the critical value at the intersection.
Compare the observed Chi-squared value with the critical value.
The use of critical values is introduced in probability and significance.
Observed Chi-squared value
The observed value is the Chi-squared statistic calculated from the research data.
It may be written as:
$$\chi^2_{\text{observed}}$$
This value summarises the differences between the observed and expected frequencies.
Critical Chi-squared value
The critical value is obtained from the statistical table.
It may be written as:
$$\chi^2_{\text{critical}}$$
The critical value is the threshold the observed value must reach for the result to be considered statistically significant.
Chi-squared significance rule
For Chi-squared, the result is statistically significant when:
$$\chi^2_{\text{observed}}\geq\chi^2_{\text{critical}}$$
A larger observed value indicates a greater difference between the observed and expected frequencies.
Worked significant result
Suppose:
$$\chi^2_{\text{observed}}=7.20$$
and the appropriate statistical table gives:
$$\chi^2_{\text{critical}}=5.99$$
Compare the values:
$$7.20\geq5.99$$
The result is statistically significant.
The null hypothesis should be rejected.
A suitable conclusion would be:
The observed Chi-squared value was greater than the critical value, so the result was statistically significant at the selected level. The null hypothesis was rejected. There was a statistically significant association between the two categorical variables.
Worked non-significant result
Suppose:
$$\chi^2_{\text{observed}}=4.35$$
and:
$$\chi^2_{\text{critical}}=5.99$$
Compare the values:
$$4.35<5.99$$
The result is not statistically significant.
The null hypothesis should be retained.
Equality reaches the threshold
Suppose:
$$\chi^2_{\text{observed}}=5.99$$
and:
$$\chi^2_{\text{critical}}=5.99$$
Because:
$$5.99\geq5.99$$
the result is statistically significant.
The equality symbol is important.
Writing a complete statistical conclusion
A complete Chi-squared conclusion should include:
The observed value.
The critical value.
The comparison between them.
The significance level.
Whether the result is statistically significant.
Whether the null hypothesis is rejected or retained.
A contextual statement about the association.
For example:
The observed Chi-squared value of \(7.20\) was greater than the critical value of \(5.99\). The result was statistically significant at \(p\leq0.05\), so the null hypothesis was rejected. There was a statistically significant association between treatment condition and whether participants improved.
Chi-squared compared with other tests
Chi-squared and the sign test
Chi-squared and the sign test both involve nominal data, but they have different purposes.
Feature | Chi-squared | Sign test |
Purpose | Association | Difference |
Data | Category frequencies | Positive and negative signs |
Relationship between results | Frequencies in a contingency table | Related paired differences |
Example | Treatment type and improvement | Anxiety increased or decreased after treatment |
The calculation of the sign test is covered in the sign test.
Chi-squared and Spearman’s rho
Both may investigate whether variables are linked, but the data are different.
Feature | Chi-squared | Spearman’s rho |
Purpose | Association | Correlation |
Data | Nominal category frequencies | Ordinal paired scores |
Display | Contingency table | Paired ranks or scattergram |
Example | Therapy type and improvement category | Stress rank and sleep-quality rank |
Chi-squared and Pearson’s \(r\)
Feature | Chi-squared | Pearson’s \(r\) |
Purpose | Association | Correlation |
Level of measurement | Nominal | Interval |
Form of data | Frequencies | Paired numerical measurements |
Example | Group and response category | Sleep duration and memory score |
Complete test-selection comparison
Statistical test | Purpose | Data relationship | Level of measurement |
Sign test | Difference | Related | Nominal |
Wilcoxon | Difference | Related | Ordinal |
Mann-Whitney | Difference | Unrelated | Ordinal |
Related t-test | Difference | Related | Interval |
Unrelated t-test | Difference | Unrelated | Interval |
Spearman’s rho | Correlation | Paired scores | Ordinal |
Pearson’s \(r\) | Correlation | Paired scores | Interval |
Chi-squared | Association | Categorical frequencies | Nominal |
Selecting Chi-squared from a scenario
The three-question method
Ask the following questions.
Question 1: Is the researcher investigating an association?
Look for wording such as:
association;
linked to;
category;
whether membership of one category is associated with another.
Question 2: Are both variables categorical?
Examples include:
therapy or no therapy;
improved or not improved;
secure, avoidant or resistant;
correct or incorrect.
Question 3: Are the results frequencies?
The data should record the number of cases in each category combination.
If the answers to all three questions are yes, Chi-squared is likely to be appropriate.
Worked selection example 1
A psychologist investigates whether attachment classification is associated with whether adults seek social support.
Purpose
$$\text{Association}$$
Level of measurement
Both variables consist of categories:
$$\text{Nominal}$$
Form of data
The psychologist counts the number of participants in each category combination:
$$\text{Frequencies}$$
Test
$$\boxed{\text{Chi-squared}}$$
Worked selection example 2
A researcher records:
whether participants revise alone or with others;
whether they pass or fail a test.
The researcher investigates whether revision setting is associated with test outcome.
The variables are categorical and the results are frequencies.
The appropriate test is:
$$\boxed{\text{Chi-squared}}$$
Worked selection example 3
A psychologist investigates whether anxiety ratings differ before and after treatment.
The same participants provide ratings on an ordered scale.
This study investigates:
$$\text{A difference}$$
using:
$$\text{Related ordinal data}$$
The appropriate test is:
$$\boxed{\text{Wilcoxon}}$$
Chi-squared is not appropriate because the researcher is comparing ordered scores rather than examining an association between two categorical variables.
Worked selection example 4
A researcher investigates whether hours of revision are related to examination scores.
Both variables are measured numerically using equal units.
This study investigates:
$$\text{A correlation between interval variables}$$
The appropriate test is:
$$\boxed{\text{Pearson's }r}$$
Chi-squared is not appropriate because the results are paired numerical scores rather than category frequencies.
Writing a strong test justification
A complete justification should identify:
The study investigates an association.
Both variables produce nominal categories.
The results are frequency data.
For example:
The Chi-squared test is appropriate because the researcher is investigating an association between treatment condition and improvement, both variables are nominal categories, and the data are frequencies showing how many participants fall into each category combination.
An answer such as:
Chi-squared is appropriate because the data are nominal.
is incomplete because it does not identify:
the research purpose;
the use of frequencies.
Key Words 🔑
Key word | Student-friendly definition | How it may be used in an exam |
Chi-squared test | An inferential test used to investigate an association between two nominal categorical variables using frequency data. | You may identify the test, justify its use or interpret its result. |
Association | A relationship between two categorical variables. | Chi-squared tests whether category membership in one variable is associated with another. |
Categorical variable | A variable that places cases into separate groups or categories. | Both variables in a Chi-squared test are categorical. |
Nominal data | Data consisting of separate categories without a meaningful numerical order. | Chi-squared requires nominal category data. |
Frequency | The number of cases in a category or combination of categories. | Chi-squared is calculated using frequencies. |
Contingency table | A table showing frequencies for combinations of categories from two variables. | You may use it to identify observed frequencies and calculate degrees of freedom. |
Observed frequency | The actual number of cases recorded in a cell. | It is represented by \(O\). |
Expected frequency | The number predicted in a cell if there is no association between the variables. | It is represented by \(E\) and calculated from table totals. |
Row total | The total frequency across one row of a contingency table. | It is used to calculate expected frequencies. |
Column total | The total frequency down one column of a contingency table. | It is also used in the expected-frequency formula. |
Overall total | The total number of cases across the whole table. | It forms the denominator in the expected-frequency formula. |
Degrees of freedom | A value calculated from the number of rows and columns and used to select a critical value. | For Chi-squared, use \((r-1)(c-1)\). |
Observed value | The calculated Chi-squared statistic from the study’s results. | It is compared with a critical value. |
Critical value | The threshold found in a statistical table. | The observed Chi-squared value must meet or exceed it. |
Statistical significance | A judgement that a result meets the selected probability criterion. | A significant Chi-squared result suggests an association. |
Null hypothesis | A prediction that there is no association between the variables. | It is rejected when the result is statistically significant. |
Hints from the Examiner Reports 💡
No lesson-specific examiner guidance was identified in the provided reports.
Common Mistakes ⚠️
Mistake: Selecting Chi-squared for every study containing nominal data.
Why this is incorrect:
Nominal data may also be used with the sign test.
Chi-squared specifically tests an association between categorical variables using frequencies.
How to improve:
Identify the research purpose before selecting the test.
Mistake: Selecting Chi-squared for a difference between two sets of scores.
Why this is incorrect:
Chi-squared tests an association between categories rather than a difference between measured scores.
How to improve:
Look for frequencies in a contingency table rather than two sets of performance scores.
Mistake: Selecting Chi-squared for a correlation between measured co-variables.
Why this is incorrect:
A correlation uses paired ordinal or interval scores.
Chi-squared uses nominal category frequencies.
How to improve:
Distinguish paired measurements from counts within categories.
Mistake: Confusing observed and expected frequencies.
Why this is incorrect:
Observed frequencies come from the collected results. Expected frequencies are calculated from the totals under the null hypothesis.
How to improve:
Remember:
Observed means obtained. Expected means calculated if no association exists.
Mistake: Dividing the overall total equally between every cell.
Why this is incorrect:
Expected frequencies must reflect both the row and column totals.
How to improve:
Use:
$$E=\frac{\text{Row total}\times\text{Column total}}{\text{Overall total}}$$
Mistake: Using the wrong row or column total.
Why this is incorrect:
Each cell must use the total from its own row and its own column.
How to improve:
Trace from the cell horizontally to its row total and vertically to its column total before substituting values.
Mistake: Treating the expected frequency as the predicted research result.
Why this is incorrect:
The expected frequency represents the pattern expected if the null hypothesis is true and no association exists.
How to improve:
Link expected frequencies explicitly to the null hypothesis.
Mistake: Counting the totals row and totals column when calculating degrees of freedom.
Why this is incorrect:
Totals are not categories of the variables.
How to improve:
Count only the rows and columns containing observed data.
Mistake: Adding the numbers of rows and columns.
Why this is incorrect:
Degrees of freedom are not calculated using:
$$r+c$$
How to improve:
Use:
$$df=(r-1)(c-1)$$
Mistake: Subtracting one only once.
Why this is incorrect:
One must be subtracted from both the number of rows and the number of columns.
How to improve:
Write both brackets before calculating:
$$(r-1)(c-1)$$
Mistake: Assuming degrees of freedom equal the number of participants.
Why this is incorrect:
For Chi-squared, degrees of freedom depend on the dimensions of the contingency table.
How to improve:
Count the data rows and data columns.
Mistake: Applying the smaller-observed-value significance rule.
Why this is incorrect:
For Chi-squared, a larger observed value represents a greater difference between observed and expected frequencies.
How to improve:
Remember:
$$\chi^2_{\text{observed}}\geq\chi^2_{\text{critical}}$$
means the result is significant.
Mistake: Saying that a significant result proves a causal relationship.
Why this is incorrect:
Chi-squared identifies a statistically significant association between categorical variables. It does not automatically establish causation.
How to improve:
Use the term association in the conclusion.
Exam-Style Questions ✍️
Question 1
State when the Chi-squared test should be used.[3 marks]
Question 2
Explain one difference between an observed frequency and an expected frequency.[2 marks]
Question 3
A psychologist records:
whether participants received therapy or no therapy;
whether their symptoms improved or did not improve.
The psychologist wants to investigate whether treatment condition is associated with improvement.
Identify an appropriate inferential statistical test. Explain your answer.[4 marks]
Question 4
The table shows the number of participants choosing two revision methods.
Group | Flashcards | Practice questions | Row total |
Group A | \(18\) | \(12\) | \(30\) |
Group B | \(10\) | \(20\) | \(30\) |
Column total | \(28\) | \(32\) | \(60\) |
Calculate the expected frequency for:
a) Group A choosing flashcards;[2 marks]
b) Group A choosing practice questions;[2 marks]
c) Group B choosing flashcards;[2 marks]
d) Group B choosing practice questions.[2 marks]
Question 5
A contingency table contains:
two data rows;
three data columns.
Calculate the degrees of freedom. Show your working.[2 marks]
Question 6
A researcher uses a contingency table containing:
four categories for Variable A;
three categories for Variable B.
Calculate the degrees of freedom. Show your working.[2 marks]
Question 7
The following table includes totals.
Response A | Response B | Total | |
Condition 1 | \(15\) | \(10\) | \(25\) |
Condition 2 | \(9\) | \(16\) | \(25\) |
Total | \(24\) | \(26\) | \(50\) |
A student claims that there are three rows and three columns, so:
$$df=(3-1)(3-1)=4$$
Explain why the student is incorrect and calculate the correct degrees of freedom.[3 marks]
Question 8
A Chi-squared test produces:
$$\chi^2_{\text{observed}}=8.14$$
The critical value is:
$$\chi^2_{\text{critical}}=5.99$$
a) Determine whether the result is statistically significant.[1 mark]
b) Explain your answer using the observed and critical values.[2 marks]
c) State what should happen to the null hypothesis.[1 mark]
Question 9
A researcher investigates whether attachment classification is associated with preferred coping strategy.
The result is not statistically significant.
Write an appropriate conclusion referring to:
the association;
the null hypothesis;
the appropriate probability symbol at the \(0.05\) significance level.
[3 marks]
Question 10
A psychologist investigates whether year group is associated with preferred revision method.
The contingency table has:
two year-group rows;
three revision-method columns.
The calculated Chi-squared value is:
$$\chi^2_{\text{observed}}=6.40$$
The critical value at:
$$p\leq0.05$$
is:
$$\chi^2_{\text{critical}}=5.99$$
Write a complete statistical conclusion. Your answer should:
calculate and state the degrees of freedom;
compare the observed and critical values;
state whether the result is statistically significant;
state whether the null hypothesis should be rejected or retained;
refer to the association between year group and preferred revision method.
[5 marks]
Comments