The chi-squared test

When to use it

Use chi-squared with frequency counts in categories (not percentages or measurements): for example the number of litter items in four zones of a regeneration area, the number of plant species found in dune zones, or how many locals and visitors gave each answer.

• Data must be raw counts; expected frequencies should be at least 5 in each category.
• Null hypothesis: there is no significant difference between the observed and expected frequencies (any difference is due to chance).

Method

1. Work out expected frequencies (E). For a 'goodness of fit' test with no reason to expect a difference, E = total / number of categories.

2. For each category calculate (O - E) squared / E.

3. Add them: chi-squared = sum of (O - E) squared / E.

4. Degrees of freedom = number of categories - 1 (for a table with rows and columns, (rows - 1) x (columns - 1)).

5. Compare with the critical value. Pearson's table: df 1: 3.84 at 0.05, 6.64 at 0.01; df 2: 5.99 and 9.21; df 3: 7.82 and 11.35. If the calculated value is equal to or greater than the critical value, the difference is significant: reject the null hypothesis.

6. Explain the result geographically.

Practice data set

Practice data (not real): litter items counted in equal-sized areas in four zones of a regeneration scheme.

Worked example

Complete the chi-squared test for the litter data at the 0.05 level.

1. Total = 80; E = 80 / 4 = 20.
2. Chi-squared = 7.20 + 0.05 + 1.25 + 3.20 = 11.70.
3. df = 4 - 1 = 3; critical value at 0.05 = 7.82.
4. 11.70 > 7.82, so reject the null hypothesis.

Answer: Chi-squared = 11.70: litter is significantly unevenly spread, highest in the older zones, suggesting the regeneration has not benefited all areas equally.

Check yourself

1. Calculate chi-squared for the litter data. Give your answer to 2 decimal places.
   Show answer

   11.70. Sum of (O - E) squared / E = 11.70.

2. Species counts in three dune zones are 18, 24 and 36. What is the expected frequency for each zone if there is no difference?
   Show answer

   26. (18 + 24 + 36) / 3 = 26.

3. Calculate chi-squared for observed counts 18, 24 and 36 with expected 26 each. Give your answer to 2 decimal places.
   Show answer

   6.46. (64 + 4 + 100) / 26 = 6.46.

4. A goodness-of-fit chi-squared test uses 5 categories. How many degrees of freedom are there?
   Show answer

   4. df = 5 - 1 = 4.

5. The species test (3 zones) gives chi-squared = 6.46. With df = 2, the critical value at 0.05 is 5.99. Conclusion?
   • Not significant: accept the null hypothesis
   • Significant: reject the null hypothesis
   • The test is invalid because df = 2
   • There is a negative correlation
   Show answer

   Significant: reject the null hypothesis. 6.46 > 5.99.

6. Which data are suitable for a chi-squared test?
   • Mean wind speeds at each site
   • Percentages of plant cover
   • Beach gradients in degrees
   • The number of people giving each answer to a question
   Show answer

   The number of people giving each answer to a question. Chi-squared needs raw frequency counts.

7. The chi-squared test should not be used when
   • the data are counts
   • expected frequencies in some categories are below 5
   • there are four categories
   • a null hypothesis has been written
   Show answer

   expected frequencies in some categories are below 5. Small expected values make the test unreliable.

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Sources

• Pearson Edexcel International Advanced Subsidiary/Advanced Level in Geography specification, Issue 3 (September 2025)
• Pearson Edexcel International Advanced Subsidiary/Advanced Level in Geography specification, Issue 3 (September 2025)
• Pearson WPS03/01 question paper, January 2022 (Spearman's rank and chi-squared critical values tables)

Last updated 8 November 2026. Spotted a mistake? Tell us.